Kerala Syllabus 6th Standard Maths Solutions Guide

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Kerala State Syllabus 6th Standard Maths Textbooks Solutions

Kerala Syllabus 6th Standard Maths Guide

Kerala State Syllabus 6th Standard Maths Textbooks Solutions Part 1

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Kerala Syllabus 6th Standard Maths Solutions Chapter 4 Volume

You can Download Volume Questions and Answers, Activity, Notes, Kerala Syllabus 6th Standard Maths Solutions Chapter 4 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 6th Standard Maths Solutions Chapter 4 Volume

Volume Text Book Questions and Answers

Large and small Textbook Page No. 57

Athira has collected many things and has arranged them into different lots.
Kerala Syllabus 6th Standard Maths Solutions Chapter 4 Volume 1
Look at two things from the first lot.
Kerala Syllabus 6th Standard Maths Solutions Chapter 4 Volume 2
Which is bigger?
How did you find out?
Now look at two things from the second lot:
Kerala Syllabus 6th Standard Maths Solutions Chapter 4 Volume 3
How do we find out which is bigger?
To find out the bigger of two sticks, we need only measure their lengths.
What about two rectangles?
We have to calculate their area, right?
Answer:
Yes,

Explanation:
Yes, by calculating the area we can find which of the two envelopes is bigger.

Kerala Syllabus 6th Standard Maths Solutions Chapter 4 Volume

Rectangle blocks
Look at two wooden blocks from Athira’s collection. Which is larger?
Kerala Syllabus 6th Standard Maths Solutions Chapter 4 Volume 4
How did you decide?
Now look at these two.
Which is larger?
Kerala Syllabus 6th Standard Maths Solutions Chapter 4 Volume 5
Let’s see how we can decide.
Answer:
The block with more volume is larger,

Explanation:
We can decide about a larger wooden block by comparing their volumes. Volume = length X breadth X height.

Size of a rectangle block Textbook Page No. 59

Look at these rectangular blocks:
Kerala Syllabus 6th Standard Maths Solutions Chapter 4 Volume 6

They are all made by stacking smaller blocks of the same size.
Which of them is the largest?
We need only count the little blocks in each, right?
Can you find how many little blocks make up each of large blocks below?
Is there a quick way to find the number of little blocks in each, without actually counting all?
Kerala Syllabus 6th Standard Maths Solutions Chapter 4 Volume 7
Answer:
Yes, we can find the number of little blocks which make up the large block. Yes, there is quick way to find the number of little blocks ; Count the number of blocks in length , breadth and height wise.
By the product of these values we can get the total number of small blocks.

Kerala Syllabus 6th Standard Maths Solutions Chapter 4 Volume 8
This rectangular block contains 64 smaller blocks. If one small block is removed from each corner of the large block
above, how many would be left?
Answer:
56 small blocks are left,

Explanation:
Total number of small blocks in rectangular block = 64, Number of corners in a rectangular block = 8, Number of small blocks left in the rectangular block = 64 – 8 = 56.

Kerala Syllabus 6th Standard Maths Solutions Chapter 4 Volume 9
Which of these is the largest?
And the smallest?
Answer:
Second block is larger than the first,

Explanation:
Volume of the first rectangular block = length X breadth X height = 3 X 3 X 3 = 27 cubic units,
Volume of the second rectangular block = length X breadth X height = 5 X 3 X 4 = 60 cubic units,
Therefore, second block is larger than the first.

Look at these blocks:
Kerala Syllabus 6th Standard Maths Solutions Chapter 4 Volume 10
How many small blocks are there in each?
Do they have the same size?
To compare sizes by just counting, what kind of little blocks should be used in both?
Answer:
Same shape and size,

Explanation:
Each block consists of 12 small blocks. Yes, the two blocks are of same size. The little blocks used should be of same shape and size.

Kerala Syllabus 6th Standard Maths Solutions Chapter 4 Volume

Size as number Textbook Page No. 60

Look at this picture:
Kerala Syllabus 6th Standard Maths Solutions Chapter 4 Volume 11
What is the area of the rectangle?
How many small squares of side 1 centimetre are in it?
4 X 3 = 12
The area of a square of side 1 centimetre is 1 square centimetre; the area of the whole rectanlge is 12 square centimetres.
Now look at the rectangular block:
Kerala Syllabus 6th Standard Maths Solutions Chapter 4 Volume 12
It is made by stacking cubes of side 1 centimetre.
Kerala Syllabus 6th Standard Maths Solutions Chapter 4 Volume 13
How many?
So, the size of this block is equal to 24 such cubes.
Size measured like this is called volume in mathematics.
We say that a cube of length, breadth and height 1 centimetre has a volume of 1 cubic centimetre.
24 such cubes make up the large block in the picture.
Its volume is 24 cubic centimetre.

Kerala Syllabus 6th Standard Maths Solutions Chapter 4 Volume 14
All sides of the large cube shown above are painted. How many small cubes would have no paint at all?
Answer:
One cube is left,

Explanation:
Only one cube is left with no paint on it which is in the middle of the large cube.

Kerala Syllabus 6th Standard Maths Solutions Chapter 4 Volume

All blocks shown below are made up of cubes of side 1 centimetre. Calculate the volume of each:
Kerala Syllabus 6th Standard Maths Solutions Chapter 4 Volume 15
Answer:
64 cubic centimetres,12 cubic centimetres,64 cubic centimetres,36 cubic centimetres,63 cubic centimetres,12 cubic centimetres,

Explanation:
Kerala-Syllabus-6th-Standard-Maths-Solutions-Chapter-4-Volume-15
Total number of cubes = length X breadth X height = 4 X 4 X 4 = 64, if side of each small cube is 1 centimetre then Volume of each small cube = length X breadth X height = 1 X 1 X 1 = 1 cubic centimetres, Therefore volume of the figure = total number of cubes X volume of each small cube = 64 cubic centimetres.
Kerala-Syllabus-6th-Standard-Maths-Solutions-Chapter-4-Volume-16
Total number of cubes = length X breadth X height = 2 X 2 X 3 = 12, if side of each small cube is 1 centimetre then
Volume of each small cube = length X breadth X height = 1 X 1 X 1 = 1 cubic centimetres, Therefore volume of the figure = total number of cubes X volume of each small cube = 12 cubic centimetres.
Kerala-Syllabus-6th-Standard-Maths-Solutions-Chapter-4-Volume-17
Total number of cubes = length X breadth X height = 4 X 4 X 4 = 64, if side of each small cube is 1 centimetre then
Volume of each small cube = length X breadth X height = 1 X 1 X 1 = 1 cubic centimetres, Therefore volume of the figure = total number of cubes X volume of each small cube = 64 cubic centimetres.
Kerala-Syllabus-6th-Standard-Maths-Solutions-Chapter-4-Volume-18
Total number of cubes = length X breadth X height = 4 X 3 X 3 = 36, if side of each small cube is 1 centimetre then
Volume of each small cube = length X breadth X height = 1 X 1 X 1 = 1 cubic centimetres, Therefore volume of the figure = total number of cubes X volume of each small cube = 36 cubic centimetres.
Kerala-Syllabus-6th-Standard-Maths-Solutions-Chapter-4-Volume-19
Total number of cubes = length X breadth X height = 7  X 3 X 3 = 63, if side of each small cube is 1 centimetre then
Volume of each small cube = length X breadth X height = 1 X 1 X 1 = 1 cubic centimetres, Therefore volume of the figure = total number of cubes X volume of each small cube = 63 cubic centimetres.

Kerala-Syllabus-6th-Standard-Maths-Solutions-Chapter-4-Volume-20
Total number of cubes = length X breadth X height = 2 X 3 X 2 = 12, if side of each small cube is 1 centimetre then
Volume of each small cube = length X breadth X height = 1 X 1 X 1 = 1 cubic centimetres, Therefore volume of the figure = total number of cubes X volume of each small cube = 12 cubic centimetres.

Volume calculation Textbook Page No. 62

See this rectangular block:
Kerala Syllabus 6th Standard Maths Solutions Chapter 4 Volume 16
How do we calculate its volume?
Answer:
Volume is 15 cubic centimetre,

Explanation:
Volume of the rectangular block = length X breadth X height = 5 X 3 X 1 = 15 cubic centimetre.
Therefore, it’s volume is 15 cubic centimetre.

For that, we must find out how many cubes of side 1 centimetre we need to make it.
Kerala Syllabus 6th Standard Maths Solutions Chapter 4 Volume 17
So, its volume is 15 cubic centimetres.
What about this block?
Kerala Syllabus 6th Standard Maths Solutions Chapter 4 Volume 18
This can be made by stacking one over another, two blocks seen first:
Kerala Syllabus 6th Standard Maths Solutions Chapter 4 Volume 19
So, to make it, how many cubes of side 1 centimetre do we need?

Kerala Syllabus 6th Standard Maths Solutions Chapter 4 Volume 20
Thus the volume of this block is 30 cubic centimetres.

Like this, calculate the volume of each of the rectangular blocks shown below and write it beside each:

Kerala Syllabus 6th Standard Maths Solutions Chapter 4 Volume 29
Kerala Syllabus 6th Standard Maths Solutions Chapter 4 Volume 22
Answer:
1) 56 cubic centimetres,
2) 54 cubic centimetres,
3) 125 cubic centimetres,
4) 100 cubic centimetres,

Explanation:
Kerala-Syllabus-6th-Standard-Maths-Solutions-Chapter-4-Volume-29
1) Length = 7 centimetres, Breadth = 4 centimetres, Height = 2 centimetres,
Volume of the rectangular block = length X breadth X height = 7 X 4 X 2 = 56 cubic centimetres.
Kerala-Syllabus-6th-Standard-Maths-Solutions-Chapter-4-Volume-31

2) Length = 6 centimetres, Breadth = 3 centimetres, Height = 3 centimetres
Volume of the rectangular block = length X breadth X height = 6 X 3 X 3 = 54 cubic centimetres.
Kerala-Syllabus-6th-Standard-Maths-Solutions-Chapter-4-Volume-24
3) Length = 5 centimetres, Breadth = 5 centimetres, Height = 5 centimetres,
Volume of the rectangular block = length X breadth X height = 5 X 5 X 5 = 125 cubic centimetres.
Kerala-Syllabus-6th-Standard-Maths-Solutions-Chapter-4-Volume-25
4) Length = 5 centimetres, Breadth = 4 centimetres, Height = 5 centimetres
Volume of the rectangular block = length X breadth X height = 5 X 4 X 5 = 100 cubic centimetres.

So, now, you know how to calculate the volume of a rectangular block, dont’ you?
The volume of a rectangular block is the product of its length, breadth and height.

Question 1.
The length, breadth and height of a brick are 21 centimetres, 15 centimetres and 7 centimetres. What is its volume?
Answer:
Volume is 2205 cubic centimetres,

Explanation:
Volume of the rectangular block = length X breadth X height = 21 X 15 X 7 = 2,205 cubic centimetres.

Question 2.
A rectangular cube of iron is of side 8 centimetres. What is its volume? 1 cubic centimetre of iron weighs 8 grams. What is the weight of the large cube?
Answer:
Weight of large cube is 4096 grams,

Explanation:
Volume of a rectangular cube = side X side X side = 8 X 8 X 8 = 512 cubic centimetres, Weight of 1 cubic centimetre of iron = 8 grams, Weight of the large cube = 512 cubic centimetres X 8 grams = 4,096 grams, Therefore, weight of the large cube is 4,096 grams or 4.096 kilo grams.

Kerala Syllabus 6th Standard Maths Solutions Chapter 4 Volume

Volume and length Textbook Page No. 65

A wooden block of length 8 centimetres and breadth 4 centimetres has a volume of 180 cubic centimetres. What is its height?
Volume is the product of length, breadth and height.
So in this problem, the product of 9 and 4 multiplied by the height is 180.
That is, 36 multiplied by the height gives 180.
So to find out the height, we need only divide 180 by 36.
The table shows measurement of some rectangular blocks. Calculate the missing measures.
Kerala Syllabus 6th Standard Maths Solutions Chapter 4 Volume 23
Answer:
Kerala-Syllabus-6th-Standard-Maths-Solutions-Chapter-4-Volume-32

Explanation:
1) Length = 3 centimetres, Breadth = 8 centimetres, Height = 7 centimetres,
Volume of the rectangular block = length X breadth X height = 3 X 8 X 7 = 168 cubic centimetres.
2)Length = 6 centimetres, Breadth = 4 centimetres, Height = 5 centimetres,
Volume of the rectangular block = length X breadth X height = 6 X 4 X 5 = 120 cubic centimetres.
3)Length = 6 centimetres, Breadth = 4 centimetres, Volume of the rectangular block = length X breadth X height = 6 X 4 X height = 48 cubic centimetres , Height = \(\frac{48}{24}\) = 2 centimetres.
4)Length = 8 centimetres, Height = 2 centimetres, Volume of the rectangular block = length X breadth X height = 8 X breadth X 2 = 48 cubic centimetres, Breadth = \(\frac{48}{16}\) = 3 centimetres.
5) Breadth = 2 centimetres, Height = 2 centimetres, Volume of the rectangular block = length X breadth X height = length X 2 X 2 = 48 cubic centimetres, Length = \(\frac{48}{4}\) = 12 centimetres.
6) Breadth = 2 centimetres, Height = 4 centimetres, Volume of the rectangular block = length X breadth X height = length X 2 X 4 = 80 cubic centimetres, Length = \(\frac{80}{8}\) = 10 centimetres.
7)Length = 14 centimetres, Height = 5 centimetres, Volume of the rectangular block = length X breadth X height = 14 X breadth X 5 = 210 cubic centimetres, Breadth = \(\frac{210}{70}\) = 3 centimetres.

Area and volume

What is the area of a rectangle of length 8 centimetres and breadth 2 centimetres?
What about the volume of a rectangular block of length 8 centimetres, breadth 2 centimetres and height 1 centimetre?
Answer:
Area is 16 square centimetres, Volume is 16 cubic centimetres,

Explanation:
Length= 8 centimetres, Breadth= 2 centimetres, Height= 1 centimetre. Area of rectangle = length X breadth = 8 X 2 = 16 square centimetres, Volume of rectangle = length X breadth X height = 8 X 2 X 1 = 16 cubic centimetres.

New shapes Textbook Page No. 66

We can make shapes other than rectangular block, by stacking cubes. For example, see this:
Kerala Syllabus 6th Standard Maths Solutions Chapter 4 Volume 24

It is made by stacking cubes of side I centimetre. Can you calculate its volume?
How many cubes are there at the very bottom?
And in the step just above it?
Thus we can count the number of cubes in each step.
How many cubes in all?
What is the volume of the stairs?
Answer:
Volume of the stairs is 216 cubic centimetres,

Explanation:
Yes, we can calculate the volume of the given figure. There are total of 81 cubes in the bottom line because there are 9 cubes vertically and 9 cubes horizontally arranged product of these gives the number of cubes. There are total of 63 cubes in the bottom line because there are 7 cubes vertically and 9 cubes horizontally arranged product of these gives the number of cubes.
There are total of 216 cubes in the given figure obtained by adding the cubes of each line.
Volume of the stairs = total number of cubes ( since 1 cube is of 1 cubic centimetre in volume ) = 81 + 63 + 45 + 27 = 216 cubic centimetre.

Now look at this figure:
Kerala Syllabus 6th Standard Maths Solutions Chapter 4 Volume 25
It is made by stacking square blocks. The bottom block is of side 9 centimetres. As we move up, the sides decrease by 2 centimetres at each step.

What is the volume of a rectangular block of length 4 centimetre, breadth 3 centimetre and
height 1 centimetre? If the length, breadth and height are doubled, what happens to the volume?
Answer:
Volume of rectangular block is 12 cubic centimetres. If the length, breadth and height of a rectangular block are doubled then the volume increases by 8 times.

Explanation:
Length = 4 centimetre, Breadth = 3 centimetre, Height = 1 centimetre,
Volume of rectangular block = length X breadth X height = 4 X 3 X 1 = 12 cubic centimetres.

If the length, breadth and height are doubled then Length = 8 centimetre, Breadth = 6 centimetre
Height = 2 centimetre,Volume of rectangular block = length X breadth X height = 8 X 6 X 2 = 96 cubic centimetres.

Therefore, if the length, breadth and height of a rectangular block are doubled then the volume increases by 8 times.

All blocks are of height 1 centimetre. What is the volume of this tower?
Just calculate the volume of each square block and add. Try it!
Answer:
Volume of the tower is 165 cubic centimetres,

Explanation:
Length = 9 centimetre, Breadth = 9 centimetre, Height = 1 centimetre, Volume of first square block = length X breadth X height = 9 X 9 X 1 = 81 cubic centimetres.
Length = 7 centimetre, Breadth = 7 centimetre , Height = 1 centimetre, Volume of second square block = length X breadth X height = 7 X 7 X 1 = 49 cubic centimetres.
Length = 5 centimetre, Breadth = 5 centimetre, Height = 1 centimetre, Volume of third square block = length X breadth X height = 5 X 5 X 1 = 25 cubic centimetres.
Length = 3 centimetre, Breadth = 3 centimetre, Height = 1 centimetre, Volume of fourth square block = length X breadth X height = 3 X 3 X 1 = 9 cubic centimetres.
Length = 1 centimetre, Breadth = 1 centimetre,  Height = 1 centimetre, Volume of fifth square block = length X breadth X height = 1 X 1 X 1 = 1 cubic centimetres. Volume of the tower = 81 + 49 + 25 + 9 + 1 = 165 cubic centimetres.

Calculate the volumes of the figures shown below. All lengths are in centimetres.
Kerala Syllabus 6th Standard Maths Solutions Chapter 4 Volume 26
Answer:
1) 416 cubic centimetres, 2) 448 cubic centimetres, 3) 324 cubic centimetres,

Explanation:
Kerala-Syllabus-6th-Standard-Maths-Solutions-Chapter-4-Volume-28
1) Volume of the given figure = Volume of vertical cuboid + Volume of horizontal cuboid + Volume of horizontal cuboid + Volume of cube, Length = 20 centimetre, Breadth = 4 centimetre, Height = 2 centimetre, Volume of vertical cuboid = length X breadth X height = 20 X 4 X 2 = 160 cubic centimetres, Length = 12 centimetre, Breadth = 4 centimetre, Height = 2 centimetre, Volume of horizontal cuboid = length X breadth X height = 12 X 4 X 2 = 96 cubic centimetres, Length of side of cube = 4 centimetres, Volume of cube = side X side X side = 4 X 4 X 4 = 64 cubic centimetres, Volume of the given figure = 160 cubic centimetres + 96 cubic centimetres + 96 cubic centimetres + 64 cubic centimetres = 416 cubic centimetres.
Kerala-Syllabus-6th-Standard-Maths-Solutions-Chapter-4-Volume-27
2) Volume of the given figure = Volume of vertical cuboid + Volume of vertical cuboid + Volume of cube, Length = 16 centimetre, Breadth = 4 centimetre, Height = 3 centimetre, Volume of vertical cuboid = length X breadth X height = 16 X 4 X 3 = 192 cubic centimetres, Length of side of cube = 4 centimetres, Volume of cube = side X side X side = 4 X 4 X 4 = 64 cubic centimetres, Volume of the given figure = 192 cubic centimetres + 192 cubic centimetres + 64 cubic centimetres = 448 cubic centimetres.
Kerala-Syllabus-6th-Standard-Maths-Solutions-Chapter-4-Volume-33
3) Volume of the given figure = Volume of vertical cuboid + Volume of horizontal cuboid, Length = 11 centimetre, Breadth = 4 centimetre, Height = 3 centimetre, Volume of vertical cuboid = length X breadth X height = 11 X 4 X 3 = 132 cubic centimetres, Length = 16 centimetre, Breadth = 4 centimetre, Height =3 centimetre, Volume of horizontal cuboid = length X breadth X height = 16 X 4 X 3 = 192 cubic centimetres, Volume of the given figure = 132 cubic centimetres + 192 cubic centimetres = 324 cubic centimetres.

Kerala Syllabus 6th Standard Maths Solutions Chapter 4 Volume

Large measures Textbook Page No. 67

What is the volume of a cube of side 1 metre?
I metre means 100 centimetres?
So, we must calculate the volume of a cube of side 100 centimetres. How much is it?
We say that the volume of cube of 1 metre is 1 cubic metre.
So,
1 cubic metre = 1000000 cubic centimetre.
Volume of large objects are often said as cubic metres.

Question 1.
A truck is loaded with sand, 4 metre long, 2 metre wide and 1 metre high. The price of 1 cubic metre of sand is 1000
rupees. What is the price of this truck load?
Answer:
Price of the truck loaded with sand is 8000 rupees,

Explanation:
Volume of the truck loaded with sand =length X breadth X height = 4 X 2 X 1 = 8 cubic metres.
Price of 1 cubic metre of sand = 1000 rupees, Price of the truck loaded with sand = 8 cubic metres X 1000 rupees = 8000 rupees, Therefore, price of the truck loaded with sand is 8000 rupees.

Question 2.
What is the volume in cubic metres of a platform 6 metre long, 1 metre wide and 50 centimetre high?
Answer:
Volume of the rectangular block is 3 cubic metres,

Explanation:
Length = 6 metre, Breadth = 1 metre, Height = 50 cm = \(\frac{1}{2}\) metre, Volume of the rectangular block = length X breadth X height = 6 X 1 X \(\frac{1}{2}\) = 3 cubic metres.

Question 3.
What is the volume of a piece of wood which is 4 metres long, \(\frac{1}{2}\) metre wide and 25 centimetre high? The price of 1 cubic metre of wood is 60000 rupees. What is the price of this piece of wood?
Answer:
Price of the piece of wood is 30,000 rupees,

Explanation:
Length = 4 metre, Breadth = \(\frac{1}{2}\) metre, Height = 25 cm = \(\frac{1}{4}\) metre,
Volume of a piece of wood = length X breadth X height = 4 X \(\frac{1}{2}\) X \(\frac{1}{4}\) = \(\frac{1}{2}\) cubic metre, Price of 1 cubic metre of wood = 60000 rupees, Price of the piece of wood = \(\frac{1}{2}\) cubic metre X 60000 rupees = 30000 rupees, Therefore, price of the piece of wood is 30,000 rupees.

Capacity

Look at this hollow box:
Kerala Syllabus 6th Standard Maths Solutions Chapter 4 Volume 27
It is made with thick wooden planks. Because of the thickness, its inner length, breadth and height are less than the outer measurements.

The inner length, breadth and height are 40 centimetres, 20 centimetres and 10 centimetres.
So, a rectangular block of these measurement can exactly fit into the space within this box.
The volume of this rectangular block is the volume whithin the box.
This volume is called the capacity of the box.
Thus the capacity of this box is;
40 X 20 X 10 = 8000 cc
So, what is the capacity of a box whose inner length, breadth and height are 50 centimetres, 25 centimetres and 20 centimetres?
Answer:
Capacity of the box is 25,000 cubic centimetres,

Explanation:
Length = 50 centimetres, Breadth = 25 centimetres, Height = 20 centimetres, Thus the capacity of the box = 50 X 25 X 20 = 25,000 cubic centimetres.

Litre and cubic metre

1 litre is 1000 cubic centimetres and 1 cubic metres is 1000000 cubic centimetres. So, 1 cubic metre is 1000 litres.

Liquid measures

What is the capacity of a cubical vessel of inner side 10 centimetres?
10 × 10 × 10 = 1000 cubic centimetres
1 litre is the amount of water that fills this vessel.
That is
1 litre = 1000 cubic centimetres
We can look at this in another way. Ifa cube of side 10 centimetres in completely immersed in a vessel, filled with water then the amount of water that overflows would be 1 litre.

So, how many litres of water does if a vessel of length 2 centimetres, breadth 15 centimetres and height 10 centimetres contain?

Let’s look at another problem:

A rectangular tank of length 4 metres and height 2\(\frac{1}{2}\) metres can contain 15000 litres of water. What is the breadth of the tank?

If we find the product of length, breadth and height in metres. we get the volume in cubic metres.
Here the volume is given to be 15000 litres.
That is, 15 cubic metres.

The product of length and height is
4 X 2\(\frac{1}{2}\) = 10
So, breadth multiplied by 10 is 15.
From this, we can calculate the width as \(\frac{15}{10}\) = 1\(\frac{1}{2}\) metre.

Now suppose this tank contains 6000 litres of water. What is the height of the water?
The amount of water is 6 cubic metres. So, the product of the length and breadth of the tank and the height of the water, all in metres is 6.
Product of length and breadth is; 4 X 1\(\frac{1}{2}\) = 6
So, height is 6 ÷ 6 = 1 metre.

Kerala Syllabus 6th Standard Maths Solutions Chapter 4 Volume

In the water Textbook Page No. 69

A vessel is filled with water. If a cube of side 1 centimetre is immersed into it, how many cubic centimetre of water would overflow? What if 20 such cubes are immersed?
Kerala Syllabus 6th Standard Maths Solutions Chapter 4 Volume 28
Answer:
If 1 cube is immersed then 1 cubic centimetre of water would overflow, If 20 cubes are immersed then 20 cubic centimetre of water would overflow,

Explanation:
Volume of the object added in the vessel = Volume of water displaced or overflowed, Volume of 1 small cube = side X side X side = 1 X 1 X 1 = 1 cubic centimetre. Therefore, 1 cubic centimetre of water were overflowed. Volume of 20 small cube = 20 (Volume of 1 small cube ) = 20 (side X side X side) = 20 (1 X 1 X 1) = 20 cubic centimetres. Therefore, 20 cubic centimetres of water were overflowed.

Raising water

A swimming pool is 25 metres long, 10 metres wide and 2 metre deep. It is half filled. How many litres of water does it contain now?
25 × 10 × 1 = 250 cubic metres
= 250000 litre
Suppose the water level is increased by 1 centimetre. How many more litres of water does it contain now?
Answer:
25,2500 litres of water will be increased in the swimming pool,

Water level is increased by 1 centimetre, Length = 25 metres, Breadth = 10 metres, Height = 2.01 metres, Volume of swimming pool = length X breadth X height = 25 X 10 X 2.01 = 502.5 cubic metres = 502500 litres, Number of more litres of water to be added = Volume of swimming pool after increasing 1 centimetres – Volume of swimming pool = 502500 litres – 250000 litres = 25,2500 litres.

Textbook Page No. 70

Question 1.
The inner sides of a cubical box are of length 4 centimetres. What is its capacity? How many cubes of side 2 centimetres can be stacked inside it?
Answer:
8 cubes can be stacked inside the cubical box,

Explanation:
Length = 4 centimetres, Volume of inner sides of a cubical box = length X length X length = 4 X 4 X 4 = 64 cubic centimetres. If side = 2 centimetres, Volume of the cube = side X side X side = 2 X 2 X 2 = 8 cubic centimetres, Cubes stacked inside the cubical box = \(\frac{Volume of inner sides of a cubical box}{Volume of the cube}\) = \(\frac{64}{8}\) = 8 cubes.

Question 2.
The inner side of a rectangular tank are 70 centimetres, 80 centimetres, 90 centimetres. How many litres of
water can it contain?
Answer:
504 litres of water are in the rectangular tank,

Explanation:
Length = 70 centimetres, Breadth = 80 centimetres, Height = 90 centimetres, Capacity of the rectangular tank = 70 X 80 X 90 = 504000 cubic centimetres = 0.504 cubic metres = 504 litres ( since 1 cubic metre = 1000 litres).

Question 3.
The length and breadth of a rectangular box are 90 centimetres and 40 centimetres. It contains 180 litres of water. How high is the water level?
Answer:
Height of the rectangular box is 50 centimetres,

Explanation:
Length = 90 centimetres, Breadth = 40 centimetres, Capacity of the rectangular box = 180 litres = 180000 cubic centimetres ( since 1 cubic centimetre= 0.001 litre), Volume of the rectangular box = Capacity of rectangular box = length X breadth X height = 90 X 40 X height = 180000 cubic centimetres, Product of length and breadth = 90 X 40 = 3600 square centimetres, Height of the rectangular box = \(\frac{180000}{3600}\) = 50 centimetres.

Question 4.
The inner length, breadth and height of a tank are 80 centimetres, 60 centimetres and 15 centimetres, and it contains water 15 centimetre high. How much more water is needed to fill it?
Answer:
The tank is already filled with 72 litres because the tank is filled 15 centimetres high,

Explanation:
Length = 80 centimetres, Breadth = 60 centimetres, Height = 15 centimetres, Capacity of the rectangular tank = 80 X 60 X 15 = 72000 cubic centimetres = 0.072 cubic metres = 72 litres ( since 1 cubic metre = 1000 litres).

Question 5.
The panchayat decided to make a rectangular pond. The length, breadth and depth were decided to be 20 metres, 15 metres and 2 metres. The soil dug out was removed in a truck which can cariy a load of length 3 metres, breadth 2 metres and height 1 metre. How many truck loads of soil have to be moved?
Answer:
100 loads of soil have to be moved in the truck to form a rectangular pond,

Explanation:
Length = 20 metres, Breadth = 15 metres, Height = 2 metres, Volume of the rectangular pond = length X breadth X height = 20 X 15 X 2 = 600 cubic metres. If dimensions of truck are : Length = 3 metres, Breadth = 2 metres, Height = 1 metres, Volume of sand loaded in truck = length X breadth X height = 3 X 2 X 1 = 6 cubic metres. Number of truck loads of soil = \(\frac{Volume of the rectangular pond }{Volume of sand loaded in truck}\) = \(\frac{600}{6}\) = 100 loads.

Question 6.
The inner length and breadth of an aquarium are 60 centimetres and 30 centimetres. It is half filled with water. When a stone is immersed in it, the water level rose by 10 centimetres. What is the volume of the stone?
Answer:
Volume of the stone is 18,000 cubic centimetres,

Explanation:
Length = 60 centimetres, Breadth = 30 centimetres, Height = h centimetres, Volume of the aquarium = length X breadth X height = 60 X 30 X h = 1800 h cubic centimetres. After immersing a stone then Length = 60 centimetres, Breadth = 30 centimetres, Height = (h+10) centimetres, Volume of the aquarium after adding a stone = length X breadth X height = 60 X 30 X (h+10) = 1800 (h+10) cubic centimetres, Volume of the stone = Volume of the aquarium after adding a stone – Volume of the aquarium = 1800 (h+10) – 1800 = 1800 h + 18000 – 1800 h = 18000 cubic centimetres.

Kerala Syllabus 6th Standard Maths Solutions Chapter 4 Volume

Question 7.
A rectangular iron block has height 20 centimetres, breadth 10 centimetres and height 5 centimetres. It is melt and recast into a cube. What is the length of a side of this cube?
Answer:
Length of side of the cube is 10 centimetres,

Explanation:
Length = 20 centimetres, Breadth = 10 centimetres, Height = 5 centimetres,
Volume of the rectangular iron block = length X breadth X height = 20 X 10 X 5 = 1000 cubic centimetres, The rectangular iron box is recast into a cube. Volume of the cube = Volume of the rectangular box = 1000 cubic centimetres, side X side X side = 1000 cubic centimetres, side of cube = 10 centimetres.

Question 8.
A tank 2\(\frac{1}{2}\) metre long and 1 metre wide is to contain 10000 litres. How high must be the tank?
Answer:
Height of the tank is 4 metres,

Explanation:
Length = 2.5 metres, Breadth = 1 metre, Capacity of the tank = 10000 litres =10 cubic metres ( since 1 cubic metre= 1000 litre), Volume of the tank = Capacity of the tank = length X breadth X height = 2.5 X 1 X height = 10 cubic metres, Product of length and breadth = 2.5 X 1 = 2.5 square metres, Height of the tank = 10 / 2.5 = 4 metres.

Question 9.
From the four corners of a square piece of paper of side 12 centimetres, small squares of side 1 centimetre are cut
off. The edges of this are bent up and joined to form a container of height 1 centimetre. What is the capacity of the container? If squares of side 2 centimetres are cut off, what would be the capacity?
Answer:
Capacity of the container is 100 cubic centimetres, The capacity of square piece after cut off of squares is 84 cubic centimetres,

Explanation:
Dimensions of container are: Length = 12 – 1 – 1 = 10 centimetres, Breadth = 12 – 1 – 1 = 10 centimetres, Height = 1 centimetre, Capacity of the container = length X breadth X height = 10 X 10 X 1 = 100 cubic centimetres, If side of square = 2 centimetres, Capacity of the square = length X breadth X height = 2 X 2 X 1 = 4 cubic centimetres. Capacity of the square piece after cut  off of squares =  Capacity of the container – 4 (Capacity of the square) = 100 – 4(4) = 84 cubic centimetres.

Kerala Syllabus 6th Standard Maths Solutions Chapter 5 Decimal Forms

You can Download Decimal Forms Questions and Answers, Activity, Notes, Kerala Syllabus 6th Standard Maths Solutions Chapter 5 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 6th Standard Maths Solutions Chapter 5 Decimal Forms

Decimal Forms Text Book Questions and Answers

Measuring length Textbook Page No. 73

Kerala Syllabus 6th Standard Maths Solutions Chapter 5 Decimal Forms 1
What is length of this pencil?
6 centimetres and 7 millilitres.
How about putting it in millimetres only? 67 millimetres.
Can you say it in centimetres only?
One centimetre means 10 millimetres.
Putting it the other way round, one millimetre is a tenth of a centimetre.
That is, \(\frac{1}{10}\) centimetre
1 millilitre = \(\frac{1}{10}\) centimetre
So, 7 millimetres is \(\frac{7}{10}\) centimetres.
Now can’t you say the length of the pencil in just centimetres?
6 centimetres, 7 millilitres = 6\(\frac{7}{10}\) centimetres.

We also write this as 6.7 centimetre. To be read 6 point 7 centimetre. It is called the decimal form of 6\(\frac{7}{10}\) centimetres.

Like this, 7 centimetre, 9 millimetre is \(\frac{9}{10}\) centimetre. And we write it as 7.9 centimetre in decimal form.

Now measure the length of your pencil and write it in decimal form.

My pencil is exactly 8 centimetres. How do I write it in decimal form?
Kerala Syllabus 6th Standard Maths Solutions Chapter 5 Decimal Forms 2

Just write 8.0
Kerala Syllabus 6th Standard Maths Solutions Chapter 5 Decimal Forms 3

Since in 8 centimetres, there is no millimetre left over, we may write it as 8.0 centimetres also.
Lengths less than one centimetre is put as only millimetres. How do we write such lengths as centimetres?
Answer:
8.0 centimetres,

Explanation:
We write 8 centimetres in decimal form as 8.0 centimetres.

For example, 6 millimetres means \(\frac{6}{10}\) centimetres and so we write it as 0.6 centimetres (read 0 point 6 centimetres)
Like this, 4 millimetre = \(\frac{4}{10}\) centimetre = 0.4 centimetre.

Kerala Syllabus 6th Standard Maths Solutions Chapter 5 Decimal Forms

Different measures

Lengths greater than one centimetre are usually said in metres. How many centimetres make a metre?
In reverse, what fraction of a metre is a centimetre?
1 centimetre = \(\frac{1}{100}\) metre.
Kerala Syllabus 6th Standard Maths Solutions Chapter 5 Decimal Forms 4

Sajin measured the length of a table as 1 metre and 13 centimetres. How do we say it in metres only?

13 centimetres means \(\frac{13}{100}\) of a metre.
That is, \(\frac{13}{100}\) metre.
1 metre and 13 centimetre means 1\(\frac{13}{100}\) metre. We can write this us 1.13 metres in decimal form.
Kerala Syllabus 6th Standard Maths Solutions Chapter 5 Decimal Forms 5
Like this,
3 metres, 45 centimetres = 3\(\frac{45}{100}\) metre = 3.45 metres.
Now how do we write 34 centimetres in terms ola metre?
34 centimetre = \(\frac{34}{100}\) metre = 0.34 metre.

Vinu measured the length of a table as 1 metre, 12 centimetres, 4 millimetres.
How do we say it in terms of a metre?
12 centimetres means 120 millimetres.
With 4 millimetres more, it is 124 millimetres.
1 millimetre is \(\frac{1}{100}\) of a metre.
So, 124 millimetres = \(\frac{124}{100}\) metre.
1 metre and 124 millimetre together is 1\(\frac{124}{100}\) metre.
Its decimal form is 1.124 metre.
Thus 5 metre, 32 centimetres, 4 millimetres in decimal form is,
5 metre, 324 millilitre = 5\(\frac{324}{1000}\) = 5.324 metre.

Millimetre and metre
1 m = 100 cm
1 cm = 10 mm
1 m = 1000 mm
So,
1 cm = \(\frac{1}{100}\) m
1 mm = \(\frac{1}{10}\) cm
1 mm = \(\frac{1}{1000}\) m

We can write other measurements also in the decimal form.
One gram is \(\frac{1}{1000}\) of a kilogram.
So, 5 kilograms and 315 grams we can write as 5\(\frac{315}{1000}\) kilograms.
Its, decimal form is 5.3 15 kilograms.

Like this,
4 grams 250 milligrams = 4\(\frac{250}{100}\) gram = 4.250 grams.
A millilitre is \(\frac{1}{1000}\) litre.
So,
725 millilitre = \(\frac{725}{1000}\) litre = 0.725 litre.

Kerala Syllabus 6th Standard Maths Solutions Chapter 5 Decimal Forms

Write the following measurements in fractional and in decimal form.
Kerala Syllabus 6th Standard Maths Solutions Chapter 5 Decimal Forms 6
Answer:
Kerala State Syllabus 6th Standard Maths Solutions Chapter 5 Decimal Forms-1

Explanation:
Wrote the given measurements in fractional and in decimal form above as 1. 4 cm 3 mm in fractional form is as 1 cm = 10 mm so 4 X 10 mm + 3 mm = 40 + 3 = 43 mm or \(\frac{43}{1}\) mm and in decimal form is 4.3 cm, 2. 5 mm in fractional form is as 1 mm = \(\frac{1}{10}\) cm so it is \(\frac{5}{10}\) cm and in decimal form it is 0.5 cm, 3. 10 m 25 cm in fractional form is as 1 cm = \(\frac{1}{100}\) m so it is \(\frac{1025}{100}\) m and in decimal form it is 10.25 m, 4. 2 kg 125 g in fractional form as 1 g = \(\frac{1}{1000}\) kg is \(\frac{2125}{1000}\) kg and in decimal form it is 2.125 kg, 5. 16 l 275 ml in fractional form as 1 ml = \(\frac{1}{1000}\) l so it is \(\frac{16275}{1000}\) l in decimal form it is 16.275 l, 6. 13l 225 ml in fractional form as 1 ml = \(\frac{1}{1000}\) l so it is \(\frac{13225}{1000}\) l in decimal form it is 13.225 l 7. 325 ml in fractional form is as 1 ml = \(\frac{1}{1000}\) l so it is \(\frac{325}{1000}\) l in decimal form it is 0.325 l.

In reverse Textbook Page No. 77

Kerala Syllabus 6th Standard Maths Solutions Chapter 5 Decimal Forms 7

1.45 metre as a fraction is 1\(\frac{45}{1000}\) metre.
How much in metre and centimetre?
1 metre 45 centimetre.
That is 145 centimetres.
So, 1.45 metre means 145 centimetres.
Like this, how about writing 0.95 metre in centimetre?
How much centimetre is this?
Answer:
145 centimetre shirt, 95 centimetre pants,

Explanation:
Given 1.45 metre for a shirt, 0.95 metre for pants, So in centimetres 1 m is equal to 100 centimetre so 1.45 metre is 100 + 45 = 145 centimetre for shirt. Now 0.95 metre = 0.95 X 100 centimetre = 95 centimetre pants.

Next try converting 0.425 kilograms into grams?
Kerala Syllabus 6th Standard Maths Solutions Chapter 5 Decimal Forms 8
0.425 kilograms = \(\frac{425}{1000}\) kilograms = 425 gram.

Fill up the table.

Kerala Syllabus 6th Standard Maths Solutions Chapter 5 Decimal Forms 9
Answer:
Kerala State Syllabus 6th Standard Maths Solutions Chapter 5 Decimal Forms-2

Explanation:
Filled the given table as shown above,1) 3.2 cm in expanded form is as 1 cm = 10 mm so 3 X 10 mm + 2 mm = 32 mm and 1 mm = \(\frac{1}{10}\) cm, the fraction form is 3\(\frac{2}{10}\) cm, 2) 1 mm = \(\frac{1}{10}\) cm so 7 mm = \(\frac{7}{10}\) cm and in decimal form it is 0.7 cm, 3) 3.41 m in expanded form is as 1 m = 100 cm so 3 X 100 cm + 41 cm = 341 cm and 1 cm = \(\frac{1}{100}\) m, the fraction form is 3\(\frac{41}{100}\) cm, 4) \(\frac{62}{10}\) m = 6.2 m and 1m = 100 cm so 6.2 m = 6.2 X 100 = 620 cm, 5) 5.346 kg in expanded form is as 1 kg = 1000 g so 5 X 1000 g + 346 g = 5346 g and 1 kg =1000 g, the fraction form is \(\frac{5346}{1000}\) kg, 6) 1 kg = 1000 g and 1 g = \(\frac{1}{1000}\) kg so 425 g in decimal form is 0.425 kg and in fractional form is \(\frac{425}{1000}\) kg, 7) 2.375 l in expanded form is as 1 l = 1000 ml so 2 x 1000 ml + 375 ml = 2375 ml and 1 ml = \(\frac{1}{1000}\) l, the fraction form is \(\frac{2375}{1000}\) l, 8) 1.350 l in expanded form is as 1 l = 1000 ml so 1 X 1000 ml + 350 ml = 1350 ml and 1 ml = \(\frac{1}{1000}\) l, the fraction form is \(\frac{1350}{1000}\) l, 9) 1 l = 1000 ml and 1 ml = \(\frac{1}{1000}\) l then \(\frac{625}{1000}\) l in decimal form is 0.625 l and in expanded form 0.625 l X 1000 ml = 625 ml.

Kerala Syllabus 6th Standard Maths Solutions Chapter 5 Decimal Forms

One fractions, many from

The heights of the children in a class are recorded. Ravi is 1 metre, 34 centimetre tall. This was written 1.34 metres. Naufal is 1 metre, 30 centimetres tall and this was written 1.30 metres.
Lissi had a doubt!
30 centimetres means \(\frac{30}{1000}\) metre. This can be written \(\frac{3}{10}\) metre.
So, why not write Ravi’s height as 1.3 metres?

“Both are right,” the teacher said.
Since, \(\frac{3}{10}\) = \(\frac{30}{100}\), we can write the decimal form of \(\frac{3}{10}\) as 0.3 or 0.30.
Then Ravi had a doubt: Since \(\frac{3}{10}\) = \(\frac{300}{1000}\), we can write, 30 centimetres as 0.300 metres.
“It is also right,” the teacher continued. How we write decimals is a matter of convenience.

For example, look at some lengths measured in metre and centimetre.
1 metre 25 centimetres
1 metre 30 centimetres
1 metre 32 centimetres
It is convenient to write these like this:
1.25 metre
1.30 metre
1.32 metre

If we measure millimetres also like this:
1 metre 25 centimetres 4 millimetres
1 metre 30 centimetres
1 metre 32 centimetres
It is better to write them as:
1.254 metre
1.300 metre
1.320 metre
Like this how can we write the decimal form of 2 kilogram, 400 gram?
What about 3 litres. 500 millilitres?
Answer:
2.400 kilograms, 3.500 litres,

Explanation:
1 kilogram = 1000 grams so 400 grams = \(\frac{400}{1000}\) kilogram then 2 kilogram + 0.400 kilograms = 2.400 kilograms, 1 litre = 1000 millilitres so 500 millilitres = \(\frac{500}{1000}\) litre then 2 litres + 0.400 litres = 2.400 litres and 3 litres.500 millilitres = 3.500 litres.

Place value Textbook Page No. 80

We have seen how we can write various measurements as fractions and in decimal forms.
If we look at just the numbers denoting these measurements, we see that they are fractions with 10, 100, 1000 so on as denominators.

For example, just as we wrote 2 centimetres, 3 millimetres as 2\(\frac{3}{10}\) and then as 2.3, we can write 2\(\frac{3}{10}\) as 2.3, whatever, be the measurement.
That is, 2.3 is the decimal form of 2\(\frac{3}{10}\).
Similarly, 4.37 is the decimal from of 4\(\frac{37}{100}\)
We can write
2\(\frac{3}{10}\) = 2.3
4\(\frac{37}{100}\) = 4.37
and so on.

On the otherhand, numbers in decimal form can be written as fractions:
247.3 = 247\(\frac{3}{10}\) = 247 + \(\frac{3}{10}\)
The number 247 in this can be split into hundreds, tens and ones:
247 = (2 × 100) + (4 × 10) + (7 × 1)
So, we can write 247.3 as
247.3 = (2 × 100) + (4 × 10) + (7 × 1) + (3 × \(\frac{1}{10}\))
How about 247.39?

First we write
247.39 = 247\(\frac{39}{100}\) = 247 + \(\frac{39}{100}\)
Then split \(\frac{39}{100}\) like this:
\(\frac{39}{100}\) = \(\frac{30+9}{100}\) = \(\frac{30}{100}\) + \(\frac{9}{100}\) = \(\frac{3}{10}\) + \(\frac{9}{100}\) = (3 × \(\frac{1}{100}\)) + (9 × \(\frac{1}{100}\))
So, we can write 247.39 like this
247.39 = (2 × 100) + (4 × 10) + (7 × 1) + (3 × \(\frac{1}{10}\)) + (9 × \(\frac{1}{100}\))
In general, we can say this:

In a decimal form, we put the dot to separate the whole number part and the fraction part. Digits to the left of the dot denote multiples of one. ten, hunded and so on; digits on the right denote multiples of tenth, hundredth, thousandth and so on.

For example. 247.39 can be split like this:
Kerala Syllabus 6th Standard Maths Solutions Chapter 5 Decimal Forms 10
Can you split the numbers below like this.
1.42 16.8 126.360 1.064 3.002 0.007
Answer:
Yes we can split the numbers,

Explanation:
Kerala-Syllabus-6th-Standard-Maths-Solutions-Chapter-5-Decimal-Forms-10
1.42 = (1 X 1) + (4 X \(\frac{1}{10}\)) + (2 X \(\frac{1}{100}\)),
Kerala-Syllabus-6th-Standard-Maths-Solutions-Chapter-5-Decimal-Forms-12
16.8 = (1 X 10) + (6 X 1) + (8 X \(\frac{1}{10}\)),
Kerala-Syllabus-6th-Standard-Maths-Solutions-Chapter-5-Decimal-Forms-13
126.360 = (1 X 100) + (2 X 10) + (6 X 1) + (3 X \(\frac{1}{10}\)) + (6 X \(\frac{1}{100}\)),Kerala-Syllabus-6th-Standard-Maths-Solutions-Chapter-5-Decimal-Forms-15
1.064 = (1 X 1) + (0 X \(\frac{1}{10}\)) + (6 X \(\frac{1}{100}\)) + (4 X  \(\frac{1}{1000}\)),
Kerala-Syllabus-6th-Standard-Maths-Solutions-Chapter-5-Decimal-Forms-16
3.002 = (3 X 1) + (0 X \(\frac{1}{10}\)) + (0 X \(\frac{1}{100}\)) + (2 X \(\frac{1}{1000}\)),
Kerala-Syllabus-6th-Standard-Maths-Solutions-Chapter-5-Decimal-Forms-17
0.007 = (0 X 1) + (0 X \(\frac{1}{10}\)) + (0 X \(\frac{1}{100}\)) + (7 X \(\frac{1}{1000}\)).

Fraction and decimal Textbook Page No. 81

\(\frac{1}{2}\) centimetres means, 5 millimetres. Its decimal form is 0.5 centimetre. So the decimal form of the fmction \(\frac{1}{2}\) is 0.5
\(\frac{1}{2}\) = \(\frac{5}{10}\) right?
Similarly, what is the decimal form of \(\frac{1}{5}\)?

Kerala Syllabus 6th Standard Maths Solutions Chapter 5 Decimal Forms

Measurements again

Let’s look at the decimal form of some measurements again. For example, what is the decimal form of 23 metre, 40 centimetre. As seen earlier.
23 metre 40 centimetre = 23\(\frac{40}{100}\) metre = 23.40 metre

Looking at just the numbers;
\(\frac{40}{100}\) = \(\frac{4}{100}\)
23\(\frac{40}{100}\) = 23\(\frac{4}{10}\) = (2 X 10) + (3 X 1) + (4 X \(\frac{4}{10}\)) = 23.4
So, we can write 23 metre, 40 centimetre either as 23.40 metre or as 23.4 metre.
What about 23 metre, 4 centimetre?
23 metre 4 centimetres = 23\(\frac{4}{100}\) metre
Writing just the numbers,
23\(\frac{4}{100}\) = (2 X 10) + (3 X 1) + (4 X \(\frac{1}{100}\))
= (2 X 10) + (3 X 1) + (0 X \(\frac{1}{10}\)) + (4 X \(\frac{1}{100}\))
= 23.04
Here the 0 just after the dot shows that the fractional part of the number has no tenths (The 0 in 307 shows that, after 3 hundreds, this number has no tens, right?)

Thus we write 23 metres, 4 centimetres as 23.04 metres. How about 23 metres and 4 millimetres? 23 metres 4 millimetres
= 23\(\frac{4}{1000}\) metres
Writing only the numbers,
23\(\frac{4}{1000}\) = (2 X 10) + (3 X 1) + (4 X \(\frac{1}{1000}\))
= (2 X 10) + (3 X 1) + (0 X \(\frac{1}{10}\)) + (0 X \(\frac{1}{100}\)) + (4 X \(\frac{1}{1000}\))
= 23.004
Thus
23 metre 4 millimetres = 23.004 metre

Kerala Syllabus 6th Standard Maths Solutions Chapter 5 Decimal Forms

Some other fractions 

We cannot write \(\frac{1}{4}\) as a fraction with denominator 10. But we have \(\frac{1}{4}\) = \(\frac{25}{100}\). So the decimal form of \(\frac{1}{4}\) is 0.25. What is the decimal form of \(\frac{3}{4}\)? And \(\frac{3}{8}\)?
Answer:
Decimal form of \(\frac{3}{4}\) is 0.75 and decimal form of \(\frac{3}{8}\) is 0.375,

Explanation:
Writing, \(\frac{3}{4}\) = \(\frac{3}{4}\) X \(\frac{25}{25}\) = \(\frac{75}{100}\) = 0.75 is the decimal form, \(\frac{3}{8}\) = \(\frac{3}{8}\) X \(\frac{125}{125}\) = \(\frac{375}{1000}\) = 0.375 is the decimal form.

Fill up this table.

Kerala Syllabus 6th Standard Maths Solutions Chapter 5 Decimal Forms 11
Answer:
Kerala-Syllabus-6th-Standard-Maths-Solutions-Chapter-5-Decimal-Forms-19

Explanation:
1) 45 cm in fraction form is \(\frac{45}{100}\) m because 1 m = 100 cm and in decimal form it is 0.45 m, 2) 315 g in fractional form is \(\frac{315}{1000}\) kg because 1 kg = 1000 g and in decimal form it is 0.315 g, 3) 455 ml in fractional form is \(\frac{455}{1000}\) ml because 1 l = 1000 ml and in decimal form it is 0.455 l, 4) \(\frac{5}{100}\) m in decimal form is 0.05 m while in measurement it is 5 centimetres because 1 m = 100 cm, 5) \(\frac{42}{1000}\) kg in decimal form is 0.042 kg and in measurement it is 42 grams because 1 kg = 1000 g, 6) 0.035 l in fractional form is \(\frac{35}{1000}\) l and in measurement it is 35 ml because 1l = 1000 ml, 7) 3kg 5g in fractional form can be written as \(\frac{3005}{1000}\) kg because 1kg = 1000 g so 3kg 5g = [(3 x 1000) + 5] g = 3005 g and in decimal form it is 3.005 g, 8) 2l 7ml in fractional form can be written as \(\frac{2007}{1000}\) l because 1l = 1000 ml so 2l 7 ml = [(2 X 1000) + 7] ml = 2007 ml and in decimal form it is 2.007 ml, 9)3m 4cm in fractional form can be written as \(\frac{304}{100}\) m because 1m = 100 cm so 3m 4cm = [(3 X 100) + 4] = 304 cm and in decimal form it is 3.04 m, 10) 3m 4mm in fractional form can be written as \(\frac{3004}{1000}\) m because 1m = 100 cm and 1mm = 0.1 cm and in decimal form it is 3.004 m, 11) 4kg 50g in fractional form can be written as \(\frac{4050}{1000}\) kg because 1 kg = 1000 g so 4kg 50g = [(4 x 1000) + 50] g = 4050g and in decimal form it is 4.050 g, 12) 4kg 5g in fractional form can be written as \(\frac{4005}{1000}\) kg because 1 kg = 1000 g so 4kg 5g = [(4 X 1000) + 5] g = 4005g and in decimal form it is 4.005 g, 13) 4kg 5mg in fractional form can be written as \(\frac{4000005}{1000}\) kg because 1 kg = 1000 g , 1 mg = 0.001 g so 4kg 5mg = [(4 X 1000) + 0.005] g = 4000.005 g and in decimal form it is 4.000005 g,14) 2ml in fractional form is \(\frac{2}{1000}\) ml because 1 l = 1000 ml and in decimal form it is 0.002 l, 15) 0.02l in fractional form is \(\frac{20}{1000}\) l because 1 l = 1000 ml and in measurement form it is 20 ml, 16) \(\frac{200}{1000}\) l in decimal form is 0.2l and in measurement form it is 200 ml because 1l = 1000 ml.

More and less 

Sneha’s height is 1.36 metre and Meena’s height is 1.42 metre. Who is taller?
In the sports meet, Vinu jumped 3.05 metres and Anu, 3.5 metres. Who won?
Vinu jumped 3 metres, 5 centimetres and Anu jumped 3 metres, 50 centimetres, right? So who won?
Answer:
Meena is taller,
Anu Won,

Explanation:
Given Sneha’s height is 1.36 metre and Meena’s height is 1.42 metre. Now if we compare Sneha’s height and Meena’s height 1.42 metre is more or greater than 1.36 metre so Meena is taller,
Now given Vinu jumped 3 metres, 5 centimetres and Anu jumped 3 metres, 50 centimetres, Comparing Vinu and Anu jumped heights if we see 3 metres 50 centimetres is more or greater than 3 metres 5 centimetres so Anu jumped more height therefore Anu won.

Kerala Syllabus 6th Standard Maths Solutions Chapter 5 Decimal Forms 12

Largest number
Which is the largest number among 4836, 568,97? What about these? 0.4836, 0.568, 0.97
We can also look at it like this. Both numbers have 3 in one’s place. The number 3.05 has zero in the tenth’s place while 3.50 has 5 in the tenth’s place. So 3.50 is the larger number.
Similarly which is the largest among 2.400 kilogram, 2.040 kilogram, 2.004 kilogram?
What about 0.750 litre and 0.075 litre.
Answer:
2.400 kilogram is largest among the given numbers, 0.750 litre is largest among the given numbers,

Explanation:
The numbers have 4 in tenth’s place. The number 2.040 has zero in the tenth’s place while 2.004 has 4 in the thousandth place. So 2.400 is the larger number. The numbers have 7 in tenth’s place. The number 0.075 has zero in the tenth’s place . So 0.750 is the larger number.

Kerala Syllabus 6th Standard Maths Solutions Chapter 5 Decimal Forms

Textbook Page No. 84

Question 1.
Find the larger in each of the pairs given below:
i) 1.7 centimetre, 0.8 centimetre
Answer:
1.7 centimetre is larger,

Explanation:
In 1.7, 1 is in the one’s place while in 0.8, 0 is in the one’s place. So 1.7 is larger.

ii) 2.35 kilogram, 2.47 kilogram
Answer:
2.47 kilogram is larger,

Explanation:
In 2.47, 4 is in the tenth’s place while in 2.35, 3 is the tenth’s place. So 2.47 is larger.

iii) 8.050 litre, 8.500 litre
Answer:
8.500 litre is larger,

Explanation:
In 8.500 , 5 is in tenth’s place while in 8.050, 0 is in tenth’s place. So 8.500 is larger.

iv) 1.005 kilogram, 1.050 kilogram
Answer:
1.050 kilogram is larger,

Explanation:
In 1.050, 5 is in the hundredth place while in 1.005, 0 is in the hundredth place. So 1.050 is larger.

v) 2.043 kilometre, 2.430 kilometre
Answer:
2.430 kilometre is larger,

Explanation:
In 2.430, 4 is in tenth’s place while in 2.043, 0 is in tenth’s place. So 2.430 is larger.

vi) 1.40 metre, 1.04 metre
Answer:
1.40 metre is larger,

Explanation:
In 1.40, 4 is in the tenth’s place while in 1.04, 0 is in the tenth’s place. So 1.40 is larger.

vii) 3.4 centimetre, 3.04 centimetre
Answer:
3.4 centimetre is larger,

Explanation:
In 3.4, 4 is in the tenth’s place while in 3.04 , 0 is in the tenth’s place. So 3.4 is larger.

viii) 3.505 litre, 3.055 litre
Answer:
3.505 litre is larger,

Explanation:
In 3.505, 5 is in the tenth’s place while in 3.055 , 0 is in the tenth’s place. So 3.505 is larger.

Question 2.
Arrange each set of numbers below from the smallest to the largest.
i) 11.4, 11.45, 11.04, 11.48, 11.048
Answer:
11.04, 11.048, 11.4, 11.45, 11.48,

Explanation:
The number 11.4 has 4 in tenth’s place, the number 11.45 has 4 in tenth’s place, the number 11.04 has 0 in tenth’s place, the number 11.48 has 4 in tenth’s place, the number 11.048 has 0 in tenth’s place, comparing 11.04 and 11.048 we understand 11.04 has 0 in thousandth place while 11.048 has 8 so 11.04 is the smallest, among 11.4, 11.45, 11.48, 11.4 is least because 11.4 has 0 in it’s hundredth place while 11.45 and 11.48 have 5 and 8 respectively we know 8 is greater than 5 hence 11.48 is greater than 11.45. Therefore the order of numbers is 11.04, 11.048, 11.4, 11.45, 11.48.

ii) 20.675, 20.47, 20.743, 20.074, 20.74
Answer:
20.074, 20.47, 20.675, 20.74, 20.743,

Explanation:
The number 20.074 has 0 in it’s tenth place while the other given numbers don’t have so it the smallest of all given numbers, 20.47 is the next smallest number because it has 4 in it’s tenth place while others have 6 and 7 in tenth place we know 7 is greater than 6 therefore 20.675 is the next least number, among both 20.74 and 20.743 we find 4 in hundredth place while 3 in thousandth place in one number and 0 in other so 20.743 is greater than 20.74. Therefore the order of numbers is 20.074, 20.47, 20.675, 20.74, 20.743.

iii) 0.0675, 0.064, 0.08, 0.09, 0.94
Answer:
0.064, 0.0675, 0.08, 0.09, 0.94,

Explanation:
All the given numbers contain 0 in their tenth place so we need to check the hundredth place 0.064 and 0.0675 are the numbers with least hundredth place value when we check the thousandth place 7 is greater than 4 so 0.064 is the least and 0.0675 is the next least, among 0.08, 0.09, 0.094; 0.08 is least as 8 is less than 9 and when we compare 0.09 and 0.094 we find 0.09 has 0 in it’s thousandth place while 0.094 has 4 so 0.09 is least compared to 0.094. Therefore the order of numbers is 0.064, 0.0675, 0.08, 0.09, 0.94.

Which of 11.4, 11.47, 11.465 the largest?
We can write 11.4 as 11.400 and 11.47 as 11.470.
Now can’t we find the largest?
Answer:
11.4, 11.465, 11.47,

Explanation:
All the given numbers have 4 in their tenth place so 11.4 is least because the hundredth place of the number is 0 when hundredth place is compared between  11.465 and 11.47 then 11.47 is greater than 11.465 as 7 is greater than 6. Therefore the order of numbers is 11.4, 11.465, 11.47.

Addition and subtraction

A 4.3 centimetre long line is drawn and then extended by 2.5 centimetres.
Kerala Syllabus 6th Standard Maths Solutions Chapter 5 Decimal Forms 23
What is the length of the line now?
We can put the length in millimetres and add
4.3 cm = 43 mm
2.5 cm = 25 mm
Kerala Syllabus 6th Standard Maths Solutions Chapter 5 Decimal Forms 13
Total length 43 +25 = 68mm
Turning this back into centimetres, we get 6.8 centimetres.
We can do this directly, without changing to millimetres.
Kerala Syllabus 6th Standard Maths Solutions Chapter 5 Decimal Forms 14
What if we want to add 4.3 centimetres and 2.8 centimetres?
If we change into millimetres and add, we get 71 millimetres.
And turing back into centiemtes, it becomes 7.1 centimetres.

Can we do this also directly, without changing to millimetres?
Let’s add in terms of place value.
Kerala Syllabus 6th Standard Maths Solutions Chapter 5 Decimal Forms 15
The answer is 6 ones and 11 tenths; that is, 7 ones and
1 tenth. This we can write 7.1
Kerala Syllabus 6th Standard Maths Solutions Chapter 5 Decimal Forms 16
How do we add 4.3 metres and 2.56 metres?
We can change both to centimetres and add
Kerala Syllabus 6th Standard Maths Solutions Chapter 5 Decimal Forms 17
4.3 m = 430 cm
2.56 m = 256 cm
The length is 430 + 256 = 686 centimetres.
Changing back to metres, it is 6.86 metres.
Kerala Syllabus 6th Standard Maths Solutions Chapter 5 Decimal Forms 18
We can add directly, without changing to centimetres 4.30
(when we do this, it is convenient to write 4.3 as 4.30)
What if we want to add 4.3 metres and 2.564 metre?
We can change both to millimetres and add
4300 mm + 2564 mm = 6864 mm
6864 mm = 6.864 mm = 6864
Kerala Syllabus 6th Standard Maths Solutions Chapter 5 Decimal Forms 19
Or directly add.
Kerala Syllabus 6th Standard Maths Solutions Chapter 5 Decimal Forms 20

Generally speaking, to add measurements given in decimal form, it is better to make the number of digits in the decimal parts same; for this, we need only add as many zeros as needed.
Now look at this; if from a 12.4 centimetre long stick, a 3.2 centimetre piece is cut off, what is the length of the remaining part?

3 centimetres subtracted from 12 centimetres is 9 centimetres.
2 millimetres subtracted from 4 millimetres is 2 millimetres.
We can write it like this;
Kerala Syllabus 6th Standard Maths Solutions Chapter 5 Decimal Forms 21
How do we subtract 3.9 centimetres from 15.6 centimetres?
We cannot subtract 9 millimetres from 6 millimetres. So we look at 15.6 centimetres as 14 centimetres and 16 millimetre. 9 millimetres subtracted from 16 millimetres gives 7 millimetres.
Let’s write according to place values and subtract.
Kerala Syllabus 6th Standard Maths Solutions Chapter 5 Decimal Forms 22

Another example: A sack contains 16.8 kilograms sugar. From this, 3.750 kilogram is put in a bag. How much sugar remains in the sack? Write 16.8 kilogram as 16.8000 kilograms and try it.
Answer:
Sugar remains in the sack is 13.050 kilograms,

Explanation:
Given a sack contains 16.8 kilograms sugar. From this, 3.750 kilogram is put in a bag. So sugar remains in the sack we wrote 16.8 kilogram as 16.8000 kilograms and subtracted 3.750 kilograms we get 13.050 kilograms.
Kerala Syllabus 6th Standard Maths Solutions Chapter 5 Decimal Forms- 33

Kerala Syllabus 6th Standard Maths Solutions Chapter 5 Decimal Forms

Textbook Page No. 87

Question 1.
Sunitha and Suneera divided a ribbon between them. Sunitha got 4.85 metre and Suneera got 3.75 metre. What was the length of the original ribbon?
Answer:
Total length of the ribbon is 8.60 metre,

Explanation:
Length of ribbon Sunitha has = 4.85 metre, Length of ribbon Suneera has = 3.75 metre, Total length of ribbon = 4.85 metre + 3.75 metre = 8.60 metre.
Kerala-Syllabus-6th-Standard-Maths-Solutions-Chapter-5-Decimal-Forms-18

Question 2.
The sides of a triangle are of lengths 12.4 centimetre, 16.8 centimetre, 13.7 centimetre. What is the perimeter of the triangle?
Answer:
Perimeter of the triangle is 42.9 centimetre,

Explanation:
Given sides of a triangle are 12.4 centimetre, 16.8 centimetre, 13.7 centimetre, Perimeter of the triangle = 12.4 centimetre + 16.8 centimetre + 13.7 centimetre = 42.9 centimetre.
Kerala-Syllabus-6th-Standard-Maths-Solutions-Chapter-5-Decimal-Forms-20
Question 3.
A sack has 48.75 kilograms of rice in it. From this 16.5 kilograms was given to Venu and 12.48 kilograms to Thomas. How much rice is now in the sack’?
Answer:
19.77 kilograms of rice is left in the sack,

Explanation:
Total quantity of rice the sack contain = 48.75 kilograms, Quantity of rice Venu was given = 16.5 kilograms, Quantity of rice Thomas was given = 12.48 kilograms, Quantity of rice left in the sack = Total quantity of rice the sack contain – ( Quantity of rice Venu was given + Quantity of rice Thomas was given) = 48.75 kilograms – (16.5 kilograms + 12.48 kilograms) = 19.77 kilograms.
Kerala-Syllabus-6th-Standard-Maths-Solutions-Chapter-5-Decimal-Forms-22

Question 4.
Which number added to 16.254 gives 30?
Answer:
13.746,

Explanation:
The number added to 16.254 to get 30 = 30 – 16.254 = 13.746.
Kerala-Syllabus-6th-Standard-Maths-Solutions-Chapter-5-Decimal-Forms-23

Question 5.
Faisal travelled 3.75 kilometres on bicycle, 12.5 kilometres in a bus and the remaining distance on foot. He travelled 17 kilometres in all. What distance did he walk?
Answer:
Distance walked by Faisal is 0.75 kilometres,

Explanation:
Total distance travelled by Faisal = 17 kilometres, Distance travelled by Faisal on bicycle = 3.75 kilometres, Distance travelled by Faisal on bus = 12.5 kilometres, Distance walked by Faisal = Total distance travelled by Faisal – (Distance travelled by Faisal on bicycle + Distance travelled by Faisal on bus) = 17 kilometres – (3.75 kilometres + 12.5 kilometres) = 0.75 kilometres.
Kerala-Syllabus-6th-Standard-Maths-Solutions-Chapter-5-Decimal-Forms-24

Quantities of some items are written using fraction.
Onion 1\(\frac{2}{5}\) kilogram
Tomato 1\(\frac{3}{4}\) kilogram
Chilly \(\frac{1}{4}\) kilogram
How much is the total weight? Do it by writing in decimal form which way is easier?
Answer:
Total weight is 3.4 kilogram,

Explanation:
1kg = 1000 grams, \(\frac{2}{5}\) X 1000 = 400 grams = 0.4 kilogram, \(\frac{3}{4}\) X 1000 = 750 grams = 0.75 kilogram, \(\frac{1}{4}\) = 250 grams = 0.25 kilogram, Weight of onions = 1\(\frac{2}{5}\) kilogram = 1.4 kilogram, Weight of tomatoes = 1\(\frac{3}{4}\) kilogram = 1.75 kilogram, Weight of chilly = \(\frac{1}{4}\) kilogram = 0.25 kilogram,
Total weight of vegetables = 1.4 kilogram + 1.75 kilogram + 0.25 kilogram = 3.4 kilogram,
Kerala-Syllabus-6th-Standard-Maths-Solutions-Chapter-5-Decimal-Forms-28

Question 6.
Mahadevan’s home is 4 kilometre from the school. He travels 2.75 kilometre of this distance in a bus and the remaining on foot. What distance does he walk?
Answer:
Distance walked by Mahadevan is 1.25 kilometre,

Explanation:
Total distance from Mahadevan’s home to school = 4 kilometre, Distance travelled by bus = 2.75 kilometre, Distance walked = Total distance from Mahadevan’s home to school – Distance travelled by bus = 4 – 2.75 = 1.25 kilometre.
Kerala-Syllabus-6th-Standard-Maths-Solutions-Chapter-5-Decimal-Forms-25

Question 7.
Susan bought a bangle weighing 7.4 grams and a necklace weighing 10.8 grams. She bought a ring also and the total weight of all three is 20 grams. What is the weight of the ring?
Answer:
1.8 grams is the weight of the ring,

Explanation:
Total weight of three objects = 20 grams, Weight of bangle = 7.4 grams, Weight of necklace = 10.8 grams, Weight of ring = Total weight of three objects – (Weight of bangle + Weight of necklace) = 20 – (7.4 + 10.8) = 1.8 grams.
Kerala Syllabus 6th Standard Maths Solutions Chapter 5 Decimal Forms 26

Question 8.
From a 10.5 metre rod, an 8.05 centimetre piece is cut off. What is the length of the remaining piece?
Answer:
Length of the remaining piece is 10.4195 metre,

Explanation:
Total length of the rod = 10.5 metre, Length of piece cut off = 8.05 centimetre = 0.0805 metre ( since 1 metre = 100 centimetre), Length of remaining piece = Total length of the rod – Length of piece cut off = 10.5 – 0.0805 = 10.4195 metre.
Kerala-Syllabus-6th-Standard-Maths-Solutions-Chapter-5-Decimal-Forms-27

Question 9.
We add 10.864 and the number got by interchanging the digits in its tenth’s and thousand’s this place. What do we get? What is the difference of these two numbers?
Answer:
Sum of the given number with the interchanged number is 21.332, Difference between these two numbers is 0.396,

Explanation:
The given number = 10.864, The number obtained by interchanging the digits in its tenth’s and thousand’s place = 10.468, Sum of these numbers = 21.332,
Kerala Syllabus 6th Standard-Maths-Solutions-Chapter 5 Decimal Forms 30
Difference of these two numbers = 0.396,
Kerala Syllabus 6th Standard Maths Solutions-Chapter 5 Decimal Forms 29

Question 10.
When 12.45 is added to a number and then 8.75 subtracted, the result was 7.34. What is the original number?
Answer:
The original number is 3.64,

Explanation:
Let the original number be x, when 12.45 is added to x and then subtracted by 8.75 we get 7.34, The original number ‘x’ is 12.45 + x – 8.75 = 7.34 then x = (7.34 + 8.75) – 12.45 = 16.09 – 12.45 = 3.64, Therefore the original number is 3.64.
Kerala Syllabus 6th Standard Maths Solutions Chapter 5 Decimal Forms 31

Kerala Syllabus 6th Standard Maths Solutions Chapter 10 Letter Math

You can Download Letter Math Questions and Answers, Activity, Notes, Kerala Syllabus 6th Standard Maths Solutions Chapter 10 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 6th Standard Maths Solutions Chapter 10 Letter Math

Letter Math Text Book Questions and Answers

Addition and Subtraction Textbook Page No. 149

Mary is now 4 years old; and her brother Johny is 8.
What would be Mary’s age after 2 years?
And Johny’s age?
What were their ages 3 years ago?
Can you fill up the blanks in the table.
Kerala Syllabus 6th Standard Maths Solutions Chapter 10 Letter Math 1
Kerala Syllabus 6th Standard Maths Solutions Chapter 10 Letter Math 2
In this, how do we compute Johny’s age from Mary’s age?
Adding 4 to Mary’s age gives Johny’s age, right?
We can shorten it like this:
Johny’s age = Mary’s age +4
There’s a trick to shorten it further. Let’s write m for Mary’s age and
j for Johny’s age. Then we can write
j = m + 4
Answer:
Kerala-State-Syllabus-6th-Standard-Maths-Solutions-Chapter-10-Letter-Math-Addition and Subtraction Textbook Page No. 149

Explanation:
Difference between Johny’s age and Mary’s age = Johny’s age – Mary’s age
= 8 – 4
= 4.
Johny’s age = Difference between Johny’s age and Mary’s age + Mary’s age
=> 1 + 4
=> 5 years.
Johny’s age – Difference between Johny’s age and Mary’s age = Mary’s age
=> 6 – 4
=> 2 years.
Johny’s age – Difference between Johny’s age and Mary’s age = Mary’s age
=> 7 – 4
=> 3 years.
Johny’s age = Difference between Johny’s age and Mary’s age + Mary’s age
=> 5 + 4
=> 9 years.

Here, the letter m which stands for Mary’s age can be any of the numbers 1, 2, 3 and so on. Accordingly we get
the numbers 5, 6, 7 and so on as j.
Another problem:
See this figure;
Kerala Syllabus 6th Standard Maths Solutions Chapter 10 Letter Math 3

However we draw the slanted line, what is the relation connecting the angles on the left and right?
We can write it like this:

The sum of the angles on the left and right is 180°.
What if we write the measure of the angle on the left as l° and the measure of the angle on the right as r°, then we can shorten this as:
l + r = 180

Kerala Syllabus 6th Standard Maths Solutions Chapter 10 Letter Math

One fact different ways Textbook Page No. 150

We can say the same fact in different ways.

  1. Johny is 4 years older than Mary.
  2. Mary is 4 years younger than Johny.
  3. The difference in ages between Johny and his younger sister Mary is 4.

When we write such relations using letters also, we can put them in different ways. If we write j for Johny’s age and m for Mary’s age, the statements above become

  1. j = m + 4
  2. m = j – 4
  3. j – m = 4

In what all ways can we state the relationship between the angles a line makes on the two sides of
another line?
Can you write each of these using letters?7*

Now look at this figure:

Kerala Syllabus 6th Standard Maths Solutions Chapter 10 Letter Math 4

A figure of four sides. Drawing a line from one corner to the opposite comer, we can split it into two triangles;
Kerala Syllabus 6th Standard Maths Solutions Chapter 10 Letter Math 5
What about a five sided figure?
Kerala Syllabus 6th Standard Maths Solutions Chapter 10 Letter Math 6
We can draw lines from one comer to two other corners to get three triangles.
Kerala Syllabus 6th Standard Maths Solutions Chapter 10 Letter Math 7
And a six sided figure?
Kerala Syllabus 6th Standard Maths Solutions Chapter 10 Letter Math 8
Answer:
Kerala-State-Syllabus-6th-Standard-Maths-Solutions-Chapter-10-Letter-Math-Addition and Subtraction Textbook Page No. 149-And a six sided figure

Explanation:
A six sided figure can be drawn by joining the edges and making two straight lines in between the lines.

Kerala Syllabus 6th Standard Maths Solutions Chapter 10 Letter Math

Changing angles 

We can use a slider to make an angle which can be changed as we like
Choose slider and click on the Graphic view. In the window which opens up, choose Integer and give Min = 0, Max = 180. Click Apply. We get a slider named n.
Kerala Syllabus 6th Standard Maths Solutions Chapter 10 Letter Math 9
Mark two points A, B.
Kerala Syllabus 6th Standard Maths Solutions Chapter 10 Letter Math 10
Choose Angle with Given size and click on A and B in order. In the window which opens up, give n° as the size of the angle and click OK. We get a new point A’.
Draw the lines BA and BA’. As we move the slider, the size of angle B changes.

Draw seven sided and eight sided figures like this. From one particular comer, draw lines to other corners to split them into triangles. Make a table like this:
Kerala Syllabus 6th Standard Maths Solutions Chapter 10 Letter Math 11
Answer:
Kerala-State-Syllabus-6th-Standard-Maths-Solutions-Chapter-10-Letter-Math-Changing angles 

Explanation:
Number of lines + 1 = Number of Triangles.
=> 3 + 1 = 4.
=> 4 + 1 = 5.
=> 5 + 1 = 6.

In a 12 sided figure, how many such lines can be drawn from one particular comer?
Answer:
9 lines can be drawn from one particular comer.

Explanation:
Number of Sides – 3 = Number of lines
=> 12 – 3
=> 9.

• What is the relation connecting the number of sides and number of lines in general?
Answer:
The relation connecting the number of sides and number of lines in general is they are directly connected to eachother.

Explanation:
The number of lines and number of sides are directly related to eachother in forming the number of shapes.

• What is the relation connecting the number of sides and the number of triangles?
Answer:
The number of lines is the relation connecting the number of sides and the number of triangles.

Explanation:
The relation connecting the number of sides and the number of triangles is the number of lines used.

• What is the relation connecting the number of lines and the number of triangles?
Answer:
Number of sides used to be formed is the relation connecting the number of lines and the number of triangles.

Explanation:
Number of sides connects the number of lines and the number of triangles.

Writing the number of sides as s, the number of lines as l and the number of triangles as t, how do we write these relations?
This is how Sneha wrote them:

  • s – 3 = l
  • t + 2 = s
  • t – 1 = l

In what other ways can we write them?
Try!
Answer:
Other ways can we write them are:
t = s – 2.
t = l + 1.

Explanation:
t + 2 = s
=> t = s – 2.
t – 1 = l
=> t = l + 1.

Kerala Syllabus 6th Standard Maths Solutions Chapter 10 Letter Math

Regular polygons Textbook Page No. 152

Polygons with equal sides and equal angles are called regular polygons.
Kerala Syllabus 6th Standard Maths Solutions Chapter 10 Letter Math 12
We can easily draw such figures using Geogebra. We use Regular polygon to do this.

Choose this and click on two positions. In the window, which opens up give the number of sides and click OK.
Kerala Syllabus 6th Standard Maths Solutions Chapter 10 Letter Math 13

Now look at this problem:

A shopkeeper decides to sell something for a price 1oo rupees more than what he bought it for. If the had bought it for 500 rupees, at what price would he sell it? What if the price at which he bought is 600 rupees? Here what is the relation between the prices at which he bought and the price at which he sells?

This hundred rupee increase in price is called the profit in the sale. If he wants a profit of 150 rupees, what would be the relation between how much the shopkeeper bought it for and how much he sells it for? What if he wants a profit of 200 rupees? Write down these relations using letters.

How do we write in general, the relation between the price at which a shopkeeper buys something, the price at which he sells it and the profit? Adding the profit to the price at which it is bought gives the price at which it is sold.
Writing the price at which it is bought as b, the profit as p and the price at which it is sold as s, we get
s = b + p

In what other ways can you write this relation?
Answer:
In a transaction, the selling price is greater than the cost price, it means we earn a profit.

Explanation:
The selling price of it as S, the profit bought of it as profit P, the cost price for what it is purchased as C.
=> Profit = Selling price -Cost price.
=> P = SP – CP.

Question 1.
For getting books through post, 25 rupees has to be added to the price of the book, as postage. Fill in the blanks of the table below:
Kerala Syllabus 6th Standard Maths Solutions Chapter 10 Letter Math 14
In what all ways can we say the relation between the price of a book and the total cost? Write these using letters. What if the postage is 30 rupees? What if it is 35 rupees?
Kerala Syllabus 6th Standard Maths Solutions Chapter 10 Letter Math 15
Now if the postage also changes according to the price of the book, in what all ways can we say this relation?
Write them using letters.
Answer:
Profit = Selling price – Cost price.

Explanation:
If the selling price(SP) is lesser than the cost price (CP), whatever difference you get between the two is the loss suffered. Similarly, Loss is C.P. – S.P. Always remember that you calculate profit or loss on the cost price.
Total cost (TC) = Price of the postage(CP) + Postage (P)
=> Total cost (TC) – Price of the postage(CP) = Postage (P)
Kerala-State-Syllabus-6th-Standard-Maths-Solutions-Chapter-10-Letter-Math-Regular polygons Textbook Page No. 152-1

Question 2.
Make a table showing the number of girls, the number of boys and the total number of children
in each class of your school. What are the different ways of stating the relation between these numbers? Write these using letters.
Answer:
C = G + B = 63.

Explanation:
Number of girls in a class (G)= 31.
Number of boys in a class (B) = 32.
Total number of students in class (C) = Number of girls in a class (G)+ Number of boys in a class (B)
= 31 + 32
= 63.

Question 3.
The sides of a triangle are 4 centimetres, 6 centimetres and 8 centimetres. What is its perimeter?
Denoting the length of the sides as a, b, c and the perimeter as p, how do we write the relation between them?
Answer:
Perimeter of the triangle (p) = 18 cm.

Explanation:
Side of the triangle (a) = 4 cm.
Side of the triangle (b) = 6 cm.
Side of the triangle (c) = 8 cm.
Perimeter of the triangle (p) = Side of the triangle (a) + Side of the triangle (b) + Side of the triangle (c)
= 4 + 6 + 8
= 10 + 8
= 18 cm.

Kerala Syllabus 6th Standard Maths Solutions Chapter 10 Letter Math

Textbook Page No. 154

We can draw regular polygon with as many sides as we like, using a slider.
Choose slider and click. In the window which opens up, choose Integer and type 3 for min. We get a slider named n.
Kerala Syllabus 6th Standard Maths Solutions Chapter 10 Letter Math 16
Choose Regular polygon and click on two positions. In the window which opens up, give n as the number of sides instead of a specific number

Click and drag the dot on the slider to change the number n The number of sides of the polygon changes accordingly.

Letter multiplication

Rani is making triangles with matchsticks:
Kerala Syllabus 6th Standard Maths Solutions Chapter 10 Letter Math 17
How many triangles are there in the picture?
Kerala Syllabus 6th Standard Maths Solutions Chapter 10 Letter Math 18
How many matchsticks are used to make them?
How did you calculate it?
Did you add repeatedly as 3 + 3 + 3 + 3 = 12?
Or multiply as 3 × 4 = 12?

How many matchsticks do we need to make 10 triangles like this?
In general, the number of matchsticks is three times the number of triangles.
How about writing this in shorthand, using letters?
If we take the number of triangles as t and the number of matchsticks as m, what is the relation between the numbers t and m?
m = 3 × t

When we write numbers as letters, we don’t usually write the multiplication sign; that is, we omit the multiplication in 3 × t and write it simply as 3t. Thus if we take the number of matchsticks Rani needs to make t triangles as m, then we usually write the relation between the numbers m and t as
m = 3t
Now let’s see how many triangles can be made with 45 matchsticks. The number of matchsticks is three times the number of triangles. So the number of triangles is a third of the number of matchsticks.
So with 45 matchsticks, we can make \(\frac{45}{3}\) = 15 triangles.
In general, the number of triangles is the number of matchsticks divided by 3.

We can write this also using letters.
t = n ÷ 3.
We usually wnte this as t = \(\frac{m}{3}\).

Inside a circle Textbook Page No. 155

We have seen in the lesson Angles how regular polygons can be drawn by dividing a circle into equal parts.
Let’s see how we can do this in GeoGebra.
Make an Integer slider named n. Draw a circle centred at a point A and mark a point B on it. Choose Angle with
Given Size and click on B, A in order.
In the window which opens up. give the angles as (360/n)°. We get a new point B on the circle. Choose Regular
Polygon and click on B, B’. Give the number of sides as n.
Kerala Syllabus 6th Standard Maths Solutions Chapter 10 Letter Math 19

Question 1.
How many squares are there in the picture? How many matchsticks are used to make it? How many matchsticks are needed to make five squares like this? In what different ways can we state relation between the number of squares and number of matchsticks?
Write them using letters.
Kerala Syllabus 6th Standard Maths Solutions Chapter 10 Letter Math 20
Answer:
Total number of matchsticks used = Number of squares in the figure × Number of matchsticks used in a square.
Number of matchsticks used to make 5 square = 20.

Explanation:
Number of squares in the figure = 3.
Number of matchsticks used in a square = 4.
=> Total number of matchsticks used = Number of squares in the figure × Number of matchsticks used in a square
= 3 × 4
= 12.
Number of squares needed to make = 5.
Number of matchsticks used to make 5 square = Number of squares needed to make × Number of matchsticks used in a square
= 5 × 4
= 20.

Kerala Syllabus 6th Standard Maths Solutions Chapter 10 Letter Math

Question 2.
All children in the school bought pens from the co-operative store, at 5 rupees each. Write in the table, how much the children of various classes paid.
Kerala Syllabus 6th Standard Maths Solutions Chapter 10 Letter Math 21
In what all ways can we say the relation between the number of children and the amount they paid? Write them using letters.
Answer:
Total amount 6A childen paid = 170 rupees.
Total amount 6B childen paid = 160 rupees.
Total amount 6C childen paid = 180 rupees.

Explanation:
Number of children 6A = 34.
Number of children 6B = 32.
Number of children 6C = 36.
Cost of each pen children bought = 5 rupees.
Total amount 6A childen paid = Number of children 6A × Cost of each pen children bought
= 34 × 5
= 170 rupees.
Total amount 6B childen paid = Number of children 6A × Cost of each pen children bought
= 32 × 5
= 160 rupees.
Total amount 6C childen paid = Number of children 6A × Cost of each pen children bought
= 36 × 5
= 180 rupees.

Question 3.
What is the perimeter of a square of side 5 centimetres. What about a square of perimeter 6 centimetres? In what all ways can we say the relation between the length of a side and perimetre of a square? Write all these using letters.
Answer:
Perimeter of a square (p)= 4 × Side of the square (s)

Explanation:
Side of the square (s)= 5 cm.
Perimeter of a square (p)= 4 × Side of the square (s)
= 4 × 5
= 20 cm.
Perimeter of a square (p) = 6 cm.
Perimeter of a square (p)= 4 × Side of the square (s)
=> 6 = 4 × Side of the square (s)
=> 6 ÷ 4 = Side of the square (s)
=> 1.5 cm = Side of the square (s)

Question 4.
How much money does 5 ten rupee notes make? What about 7 ten rupee notes? In what all ways can we say the number of ten rupee notes and the total amount? Using t to denote the number of ten rupee notes and a to denote the total amount, in what all ways can we write this relation?
Answer:
Total amount 5 ten rupee notes make (a) = 50 rupees.
Total amount 7 ten rupee notes make (a) = 70 rupees.

Explanation:
One rupee note = 1 rupees.
Ten rupee note (t) = 10 rupees.
Total amount 5 ten rupee notes make (a) = 5 × Ten rupee note (t)
= 5 × 10
= 50 rupees.
Total amount 7 ten rupee notes make (a) = 7 × Ten rupee note (t)
= 7 × 10
= 70 rupees.

Multiplication again

What is the area of a rectangle of length 5 centimetres and breadth 3 centimetres?
How about a rectangle of length 5\(\frac{1}{2}\) centimetres and breadth 3\(\frac{1}{4}\) metres?
Whatever be the length and breadth, area is their product, right?
How do we write this using letters?

Taking the length as l centimetres, breadth as b centimetres and area as a square centimetres,
a = l × b = lb
See how we have omitted the multiplication sign here also.

Like this, the volume of a rectangular block is the product of its length, breadth and height.
This also we can write using letters. Taking length as l centimetres, breadth as b centimetres, height as h centimetres and volume as v cubic centimetres, we can write
v = lbh

Kerala Syllabus 6th Standard Maths Solutions Chapter 10 Letter Math

Textbook Page No. 157

Using the angles drawn in changing angle, we can draw a nice picture.
First change the values of slider n from 0 to 360. Use Circle with Centre through Point to draw a circle centred at A’ through A. Right click on the circle and choose trace on.
Kerala Syllabus 6th Standard Maths Solutions Chapter 10 Letter Math 22
Now right click on the slider and choose Animation on. Don’t you get a picture like this?
Kerala Syllabus 6th Standard Maths Solutions Chapter 10 Letter Math 23

Question 1.
What is the total price of 5 pens, each of price 8 rupees? What is the price of 10 notebooks, each of price 12 rupees?

i. In what all ways can we say the relation between the price of something, the number bought and the total price?
Answer:
Total price (t) = Cost of each pen × Number of pens = 40 rupees.
Total price (t) = Cost of each notebook × Number of notebooks = 120 rupees.

Explanation:
Cost of each pen (p) = 8 rupees.
Number of pens (n) = 5.
Total price (t) = Cost of each pen × Number of pens
= 8 × 5
= 40 rupees.
Cost of each notebook (p) = 12 rupees.
Number of notebooks (n) = 10.
Total price (t) = Cost of each notebook × Number of notebooks
= 12 × 10
= 120 rupees.

ii. Taking the price of an article as p, their number as n and the total price as t, in what all ways can we write the relations between p, n and t ?
Answer:
Total price of the articles (t) = Cost of each article (p) × Number of articles (n)

Explanation:
Total price of the articles (t) = Cost of each article (p) × Number of articles (n)
=> Total price of the articles (t) – Cost of each article (p)= Number of articles (n)
=> Total price of the articles (t) ÷ Number of articles (n) = Cost of each article (p)

Question 2.
One litre of kerosene weighs 800 grams.

i. What is the weight of 2 litres of keroscene?
Answer:
Total weigh of kerosene = 1,600 liters.

Explanation:
Weigh of one litre of kerosene = 800 grams.
Number of liters of kerosene = 2.
Total weigh of kerosene = Weigh of one litre of kerosene × Number of liters of kerosene
= 800 × 2
= 1,600 liters.

ii. What is the weight of \(\frac{1}{2}\) litre of keroscene?
Answer:
Total weigh of kerosene = 400 liters.

Explanation:
Weigh of one litre of kerosene = 800 grams.
Number of liters of kerosene = \(\frac{1}{2}\)
Total weigh of kerosene = Weigh of one litre of kerosene × Number of liters of kerosene
= 800 × \(\frac{1}{2}\)
= 400 liters.

iii. What is the weight of 1 millilitre of keroscene?
Answer:
Weight of 1 millilitre of keroscene = 0.8.

Explanation:
Weigh of one litre of kerosene = 800 grams.
1 milliliters of kerosene = ??
Conversation:
1 liter = 1000 milliliters.
=> ?? = 1 milliliter
=> 1 × 1 = ?? × 1000
=> 1 ÷ 1000 = ??
=> 800 ÷ 1000 = ??
=> 0.8 milliliters= ??

iv. Taking the weight of v millilitres of kersone as w grams, write a relation between v and w.
Answer:
Weight of millilitres of kersone (v) = Weight of kersone ÷ 1000.

Explanation:
Weight of millilitres of kersone = (v)
Weight of kersone = w grams.
Conversation:
1 liter = 1000 milliliters.
=> Weight of millilitres of kersone (v) = Weight of kersone ÷ 1000.

Kerala Syllabus 6th Standard Maths Solutions Chapter 10 Letter Math

Question 3.
One cubic centimetre of iron weighs 7.8 grams.

i. Taking the volume of an iron object as v cubic centimetres and weight w grams, write a relation between v and w.
Answer:
Density = w/v.

Explanation:
Weigh of one cubic centimetre of iron = 7.8 grams.
volume of an iron object = v cubic centimetres.
Weight of iron = w grams.
Weight is the amount of matter an object contains, while volume is how much space it takes up.

ii. Taking the length, breadth and height of a rectangular block of iron as l, b, h and its weight as w, write a relation between l, b, h and w.
Answer:
A relation between l, b, h and w is that we get to know the area of the rectangular block of iron.

Explanation:
Length of the rectangular block of iron = l.
Breadth of the rectangular block of iron = b.
Height of the rectangular block of iron = h.
Weight of the rectangular block of iron = w.
=> Area of four walls =

Sequence rule

Look at these numbers:
1, 1, 2, 3, 5, 8, ………….
Can you say what the next number is?
For any three consecutive numbers a, b, c of this sequence, we must have a + b = c
Now try writing some more numbers of this. It is called Fibonacci sequence.

Multiplication and addition

Ravi has 3 ten rupee notes and a one rupee coin; Lissy has 5 ten rupee notes and a one rupee coin.
How much money does Ravi have?
And Lissy?
How did you calculate?
Similarly how much does 25 ten rupee notes and a one rupee coin make?
(10 × 25) + 1 = 251
In general, how much money does some ten rupee notes and a one rupee coin make?
We have to multiply the number of notes by 10 and add 1, right?

Let’s write it using letters.
Take the number of ten rupee notes as t.
So how much money is t ten rupee notes and a one rupee coin?
What about 8 ten rupee notes and 7 one rupee coins?
What is the general method to calculate the amount of money, some ten rupee notes and some one rupee coins make?
Multiply the number of notes by 10 and add the number of coins.
How do we write this using letters?
t ten rupee notes and c coins make 10t + c rupees.

Question 1.
How much money does 8 ten rupee notes and 2 five rupee notes make? What about 7 ten rupee notes and 4 five rupee notes?

i. How do we say the relation between the number of ten rupee notes, the number of five rupee notes and the total amount?
Answer:
The relation between the number of ten rupee notes, the number of five rupee notes and the total amount is ten rupee notes are 4 times greater than five rupee notes.

Explanation:
Number of ten rupee notes = 8.
Number of five rupee notes = 2.
Total amount = (Number of ten rupee notes × 10) + (Number of five rupee notes × 2)
= (8 × 10) + (5 × 2)
= 80 + 10
= 90 rupees.
Number of seven rupee notes = 7.
Number of five rupee notes = 4.
Total amount = (Number of ten rupee notes × 10) + (Number of five rupee notes × 2)
= (7 × 7) + (5 × 4)
= 49 + 20
= 69 rupees.

ii. Taking the number of ten rupee notes as t, the number of five rupee notes as f and the total amount as a how do we write the relation between t, f and a?
Answer:
The relation between t, f and a is a direct relationship.

Explanation:
Number of ten rupee notes = t.
Number of five rupee notes = f.
Total amount = a.
Total amount = (Number of ten rupee notes × 10) + (Number of five rupee notes × 5)
=> a = (t × 10) + (f × 5)

Question 2.
The price of a pen is 7 rupees and the price of a notebook is 12 rupees.

i. What is the total price of 5 pens and 6 notebooks?
Answer:
Total price of pens = 35 rupees.
Total price of notebooks = 72 rupees.

Explanation:
Price of a pen = 7 rupees.
Number of pens = 5.
Total price of pens = Price of a pen × Number of pens
= 7 × 5
= 35 rupees.
Price of a notebook = 12 rupees.
Number of notebooks = 6.
Total price of notebooks = Price of a notebook × Number of notebooks
= 12 × 6
= 72 rupees.

ii. What about 12 pens and 7 notebooks?
Answer:
Total price of pens = 84 rupees.
Total price of notebooks = 84 rupees.

Explanation:
Price of a pen = 7 rupees.
Number of pens = 12.
Total price of pens = Price of a pen × Number of pens
= 7 × 12
= 84 rupees.
Price of a notebook = 12 rupees.
Number of notebooks = 7.
Total price of notebooks = Price of a notebook × Number of notebooks
= 12 × 7
= 84 rupees.

iii. What is the relation between the number of pens, the number of books and the total price?
Answer:
The number of pens, the number of books and the total price is a direct relationship.

Explanation:
The number of pens, the number of books and the total price is a direct relationship as in number of pens increases the total price increases and same with the number of books.

iv. Taking the numbers of pens as p, the number of notebooks as n and total price as t, how do we write the relation between them?
Answer:
The relation between them is that if any one increases among them automaticaaly the other one increases.

Explanation:
Numbers of pens = p.
Number of notebooks = n.
Total price = t.
If number of pens increases the total price of pens will increase and same with the number of notebooks.

Kerala Syllabus 6th Standard Maths Solutions Chapter 10 Letter Math

Question 3.
What length of wire is neeed to make a triangle of sides 10 centimetres each? To make a square of side 10 centimetres?

i. What is the total length needed to make 5 such triangles and 6 such squares?
Answer:
Total length of wire is needed to make triangles = 50 cm.
Total length of wire is needed to make squares = 60 cm.

Explanation:
Length of wire is needed to make a triangle = 10 centimetres each.
Number of triangles = 5.
Total length of wire is needed = Length of wire is needed to make a triangle × Number of triangles
= 10 × 5
= 50 cm.
Length of wire is needed to make a square = 10 centimetres each.
Number of squares= 6.
Total length of wire is needed = Length of wire is needed to make a square × Number of squares
= 10 × 6
= 60 cm.

ii. What about 4 triangles and 3 squares?
Answer:
Total length of wire is needed to make triangles = 40 cm.
Total length of wire is needed to make squares = 30 cm.

Explanation:
Length of wire is needed to make a triangle = 10 centimetres each.
Number of triangles = 4.
Total length of wire is needed = Length of wire is needed to make a triangle × Number of triangles
= 10 × 4
= 40 cm.
Length of wire is needed to make a square = 10 centimetres each.
Number of squares= 3.
Total length of wire is needed = Length of wire is needed to make a square × Number of squares
= 10 × 3
= 30 cm.

iii. What is the relation between the number of triangles, the number of squares and the total length of wire?
Answer:
The number of triangles, the number of squares and the total length of wire is direct relationship.

Explanation:
The number of triangles, the number of squares and the total length of wire is that if number of triangles or squares increases the total length of them will increases.

iv. Taking the number of triangles as t, the number of squares as s and the total length as l , how do we write the relation between t, s and l?
Answer:
The relation between t, s and l is that they are directly propotionate to eachother.

Explanation:
Number of triangles = t.
Number of squares = s.
Total length = l.
If number of triangles or squares increases the total length also increases.

Addition and multiplication

Four friends went to buy pens and note books. The price of a pen is 8 rupees and the price of a notebook is 12 rupees. The shopkeeper calculated like this:
Price of 4 pens is 8 × 4 = 32 rupees
Price of 4 notebooks is 12 × 4 = 48 rupees
Total 80 rupees
The friends calculated like this:
The amount one has to spend is 8 + 12 = 20 rupees
Total expense 20 × 4 = 80 rupees
Another problem: we want to make a rectangle with eerkkil bits; of length 5\(\frac{1}{2}\) centimetres and breadth 3\(\frac{1}{2}\) centimetres. What is the total length of eerkkil needed?
Kerala Syllabus 6th Standard Maths Solutions Chapter 10 Letter Math 24
We can calculate the total length as
5\(\frac{1}{2}\) + 3\(\frac{1}{2}\) + 5\(\frac{1}{2}\) + 3\(\frac{1}{2}\) = 18
Or, we can calculate it as two eerkkil bits of length 5\(\frac{1}{2}\) centimetres and two of length 3\(\frac{1}{2}\) centimetres;
(2 × 5\(\frac{1}{2}\)) + (2 × 3\(\frac{1}{2}\)) = 11 + 7 = 18
Kerala Syllabus 6th Standard Maths Solutions Chapter 10 Letter Math 25
There is a third way: take it as two eerkkil bits of length 5\(\frac{1}{2}\) + 3\(\frac{1}{2}\) centimetres:
(2 × 5\(\frac{1}{2}\) + 3\(\frac{1}{2}\)) = 2 × 9 = 18
Kerala Syllabus 6th Standard Maths Solutions Chapter 10 Letter Math 26
Which is the easiest?

So if we take the length of a rectangle as l, breadth as b and perimeter as p, the relation between l, b andp can be written in various ways as
p = l + b + l + b
p = 2l + 2b
p = 2 (l + b)
Usually, the last one is the easiest to use.

For example, we can quickly calculate the perimeter of a rectangle of length 27 centimetres and breadth 43 centimetres as 2 × (27 + 43) = 140 centimetres.

Question 1.
There are 25 children in one room and 35 in another. 5 biscuits are to be given to each. How many biscuits are needed?
i. What if the number of children are 20 and 40?
Answer:
Total number of biscuits needed = 300.

Explanation:
Number of children on one room = 20.
Number of children on another room = 40.
Number of biscuits given to each = 5.
Total number of biscuits needed = (Number of children on one room + Number of children on another room) × Number of biscuits given to each
= (20 + 40) × 5
= 60 × 5
= 300.

ii. If we take the number of children in the first room as f in the second room as s and the total number of biscuits as t, in what all ways can we write the relation between f s and t? What if each is given 6 biscuits instead of 5?
Answer:
If there is a increase in the price of each biscuit the total number of biscuits needed also increases.

Explanation:
Number of children in the first room = f.
Number of children in the second room = s.
Total number of biscuits = t.
Total number of biscuits will be effected if there is a increase in the price of each biscuit increases.

iii. If the number of biscuits given to each is taken as b, in what all ways can we write the relation between f s, t and b?
Answer:
t = (f + s) × b.

Explanation:
Number of biscuits given to each = b.
Number of children in the first room = f.
Number of children in the second room = s.
Total number of biscuits = t.
Total number of biscuits needed = (Number of children on one room + Number of children on another room) × Number of biscuits given to each

Kerala Syllabus 6th Standard Maths Solutions Chapter 10 Letter Math

Question 2.
In the picture, M is the point right at the middle of AC.
Kerala Syllabus 6th Standard Maths Solutions Chapter 10 Letter Math 27
What is the length of AM?
Answer:
length of AM = 4 cm.

Explanation:
Length of AC = AB + BC
=> 6 + 2
= 8 cm.
M is the point right at the middle of AC.
=> Length of AM = Length of AC ÷ 2
=> 8 ÷ 2
= 4 cm.

i. If a 5 centimetre long line is extended by 4 centimetres, at what distance from an end point of the longer line is the point exactly at its middle?
Answer:
Length of the line from an end point of the longer line is the point exactly at its middle =4.5 cm.

Explanation:
Length of long line = 5 cm.
Length of the line extended = 4 cm.
Total length of the line = Length of long line + Length of the line extended
= 5 + 4
= 9 cm
Length of the line from an end point of the longer line is the point exactly at its middle = Total length of the line ÷ 2
= 9 ÷ 2
= 4.5 cm.

ii. What if a 7\(\frac{1}{2}\) centimetre long line is extended by 2\(\frac{1}{2}\) centimetres?
Answer:
Length of the line from an end point of the longer line is the point exactly at its middle = 5 cm.

Explanation:
Length of long line = 7\(\frac{1}{2}\) cm.
Length of the line extended = 2\(\frac{1}{2}\) cm.
Total length of the line = Length of long line + Length of the line extended
= 7\(\frac{1}{2}\) + 2\(\frac{1}{2}\)
= {[(7 × 2) + 1] ÷ 2} + {[(2 × 2) + 1] ÷ 2}
= [(14 + 1) ÷ 2] + [(4 + 1) ÷ 2]
= \(\frac{15}{2}\) + \(\frac{5}{2}\)
= (15 + 5) ÷ 2
= \(\frac{20}{2}\)
= 10 cm.
Length of the line from an end point of the longer line is the point exactly at its middle = Total length of the line ÷ 2
= 10 ÷ 2
= 5 cm.

iii. A line of length l centimetres is extended by e centimetres. The mid point of the long line, is m centimetres away from one of its end points. What is the relation between l, e and m?
Answer:
There is a direct relationship between l, e and m.

Explanation:
Line of length = l centimetres.
Line of length is extended = e centimetres.
Mid point of the long line = m centimetres.
The relation between if line length increases the mid point distance to the end point also increases.

Question 3.
The length of a rectangle is 4 centimetres and its breadth is 3 centimetres. The length is increased by 2 centimetres to make a large rectangle.
Kerala Syllabus 6th Standard Maths Solutions Chapter 10 Letter Math 28

i. What is the area of the large rectangle? If the length is increased by 3 centimetres, what is the area of the large rectangle?
Answer:
Area of large rectangle = 42 square cm.

Explanation:
Length of the rectangle = 4 cm.
Breadth of the rectangle = 3 cm.
Area of rectangle = Length of the rectangle  × Breadth of the rectangle
= 4 × 3
= 12 square cm.
length is increased by 3 centimetres
=> Length of the rectangle = 4 cm + 3 cm = 7 cm.
Breadth of the rectangle = 3 cm + 3 cm = 6 cm.
Area of large rectangle = Length of the rectangle  × Breadth of the rectangle
= 7 × 6
= 42 square cm.

ii. Taking the length and breadth of the original rectangle as l centimetres and b centimetres, the increase in length as i centimetres and the area of the large rectangle as a square centimetres, in what all ways can we write the relation between l, b, i and a?
Answer:
The relation between l, b, i and a is that if there is any change in length and breadth of rectangle will automatically increases the area of the rectangle.

Explanation:
Length of the original rectangle = l centimetres.
Breadth of the original rectangle = b centimetres.
Increase in length
=> length increased = i centimetres.
Area of the large rectangle = a square centimetres.
If there is any decrease or increase in the length and breadth, the area of the rectangle also changes.

Kerala Syllabus 6th Standard Maths Solutions Chapter 6 Numbers

You can Download Numbers Questions and Answers, Activity, Notes, Kerala Syllabus 6th Standard Maths Solutions Chapter 6 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 6th Standard Maths Solutions Chapter 6 Numbers

Numbers Text Book Questions and Answers

Let’s make a rectangle Textbook Page No. 95

A rectangle with 20 dots:
Kerala Syllabus 6th Standard Maths Solutions Chapter 6 Numbers 1
5 dots wide, 4 dots high.
Can we make other rectangles, rearranging the dots?
Kerala Syllabus 6th Standard Maths Solutions Chapter 6 Numbers 4
How about this?
Kerala Syllabus 6th Standard Maths Solutions Chapter 6 Numbers 2
Also like this:
Kerala Syllabus 6th Standard Maths Solutions Chapter 6 Numbers 3
Any other?
What about the number of dots along the width and height?
Their product must be 20, right?
In what all ways can we write 20 as the product of two natural numbers?
Answer:
4 and 5 are the two natural numbers.
Explanation:
4 x 5 = 20
5 x 4 = 20

Now make different rectangles with 24 dots. Also write down the number of dots along the width and height.
Kerala Syllabus 6th Standard Maths Solutions Chapter 6 Numbers 5
Answer:

Explanation:
Total number of 24 dots,
2, 4, 6 and 12 are factors of 24,
and the product of 4 x 6 = 24 or 6 x 4 =24
or 2 x 12  or 12 x 2 = 24

What about 30 dots?

Let’s think about it, without actually making rectangles. What are the possible number of dots along the width and height?
The product of numbers in every row of the table is 30.
Kerala Syllabus 6th Standard Maths Solutions Chapter 6 Numbers 6
Answer:

Explanation:
5 x 6 = 30 or 6 x 5 = 30
5 dots horizontal by 6 dots vertical.
There’s another way of stating this; all these numbers are factors of 30.

Now can you write down the different rectangles with 40 dots?
How about 45 dots?
And 60 dots?
What about 61 dots?
Answer:

Explanation:
9 x 5 = 45
5 x 9 = 45
10 x 6 = 60
6 x 10= 60
61 not possible

Kerala Syllabus 6th Standard Maths Solutions Chapter 5 Decimal Forms

Factor pairs

What are the factors of 72?
Two quick ones are 1 and 72.
We can divide 72 by 2 without any remainder. That is 2 is also a factor of 72.And 72 divided by 2 gives 36.
72 = 2 × 36
So 36 is also a factor of 72.
Thus we can find factors in pairs.
Since
72 ÷ 3 = 24
We have
72 = 3 × 24
This gives 3 and 24 as another pair of factors.

Can’t we find other pairs like this?

(1, 72)

(2, 36)

(3, 24)

(4, 18)

(6, 12)

(8, 9)

Now try to find the factors of 90, 99, 120 as pairs.
Answer:
Factors of 90 = 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, and 90.
Factors of 99 = 1, 3, 9, 11, 33, and 99.
Factors of 120 = 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60 and 120.
Explanation:
A factor is a number that divides the given number without any remainder.
The factors of a number can either be positive or negative.
factor pairs of 90
(1, 90)
(2, 45)
(3, 30)
(5, 18)
(6, 15)
(9, 10)
factor pairs of 99
(1, 99)
(3, 33)
(9, 11)
factor pairs of 120
(1, 120)
(2, 60)
(3, 40)
(4, 30)
(5, 24)
(6, 20)
(8, 15)
(10, 12)

• If 2 and 3 are factors of a number, should 6 also be a factor of that number’?
Answer:
yes,
Explanation:
Yes, 6 also be a factor of that number.
2 x 3 = 6
6 x 1 = 6
3 x 2 = 6

• If 3 and 5 are factors of a number, should 15 also be a factor of that number’?
Answer:
yes,
Explanation:
5 x 3 = 15
15 x 1 =15
3 x 5 = 15
3, 5 are the factors of 15.

• If 4 and 6 are factors of a number, should 24 also be a factor of that number’?
Answer:
Yes,
Explanation:
4 x 6 = 24
6 x 4 = 24
4 , 6 are the factors of 24

• If 4 and 6 are factors of a number, what is the largest number we can say for sure is a factor of that number’?
Answer:
12,
Explanation:
12 is the largest number is a factor of 24.
The largest number is 12.
The numbers whose factors and 4 and 6 are common multiples of 4 and 6 .
These numbers are also multiples of common multiple of 12.

• Given Two factors of a number, under what conditions can we say for sure that the product of these factors is also a factor’?
Answer:
2 and 12.
Explanation:
2 x 12 = 24
The numbers whose factors and 2 and 12 are common multiples of 2 and 12.
These numbers are also multiples of common multiple of 24.

Odd and even

We have found the factors of many numbers like 20, 24, 30, 40, 45, 60, 61, 72, 90, 99, 120.
See how many factors each has.
All of them have an even number of factors, right?
Why is this so?
Is it true for all numbers?
Write the factor pairs of 36.
(1, 36), (2, 18), (3, 12), (4, 9), (6, 6)
So what are the factors of 36?
1, 2, 3, 4, 6, 9, 12, 18, 36
9 factors in all.

Why is the number of factors odd in the case?
Can you find any other number with an odd number of factors?
Take 16, for example.
How about 25?
What is the specialty of numbers with an odd number of factors?
Answer:
1, 5, 25
Explanation:
1 x 25 = 25
5 x 5 =25
1, 5, 25 are the factors of 25.

Can you find all numbers between 1 and 100; which have an odd number of factors?
Answer:
Yes,
Explanation:
To find an odd factor, exclude the even prime factor 2.
The odd numbers from 1 to 100 are:
1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99

Kerala Syllabus 6th Standard Maths Solutions Chapter 5 Decimal Forms

Repeated multiplication Textbook Page No. 97

How many factors does 5 have?
How about 17?
5 and 17 are prime numbers, aren’t they?
And a prime has only two factors, right? 1 and the number itself.

All composite numbers have more than two factors.
For example, let’s have a look at 32.
Kerala Syllabus 6th Standard Maths Solutions Chapter 6 Numbers 7
32 = 2 × 2 × 2 × 2 × 2
Taking the first 2 alone and all the other 2’s together, we can write
32 = 2 × 16
How about taking the first two 2 s together and then other 2’s together?
32 = 4 × 8
Taking all the 2’s together can be written as
32 = 1 × 32
Thus, the factors of 32 are the 6 numbers
1, 2, 4, 8, 16, 32
Let’s look at the factors of 81 like this:
Writing 81 as a product of prime numbers, we get
81 = 3 × 3 × 3 × 3
So we can write 81 as
3 × 27
9 × 9
1 × 81
Thus we have five factors.
1, 3, 9, 27, 81
We can put this in a different way.
Taking 3’s in groups we get the factors.
3
3 × 3 = 9
3 × 3 × 3 = 27
3 × 3 × 3 × 3 = 81
and find the 5 factors of 81 as 1, 3, 9, 27 and 81.
In these examples, 32 is a product of 2’s; and 81 is a product of 3’s.

Like this, can’t we easily find the factors of a number, which can be factorized as repeated product of a single prime?
Answer:
This is true as any composite number can be expressed in terms of prime factors.
Explanation:
The prime factor for number 2 is same for the numbers.
for example;
12 = 2 x 2 x 3
18 = 2 x 3 x 3
both have prime factors 2 and 3 but the combination 12 is different from that of 18.

We can split 216 as
216 = 6 × 6 × 6
Can we say that the only factors of 216 are the 4 numbers 1, 6, 36, 24. What are the other factors of 216?
Kerala Syllabus 6th Standard Maths Solutions Chapter 6 Numbers 16
Answer:
Therefore, the factors of 216 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 27, 36, 54, 72, 108, and 216.
Explanation:
The factor of a number is a number that divides the given number without any remainder.
To find the factors of given number, divide the number with the least prime number, i.e. 2.
1 x 216 =216
2 x 108 = 216
3 x 72 = 216
4 x 54 = 216
6 x 36 = 216
8 x 27 = 216
9 x 24 = 216
12 x 18 = 216

Question 1.
Find all the factors of the numbers below:
(i) 256
Answer:
The factors of 256 are 1, 2, 4, 8, 16, 32, 64, 128 and 256.
Explanation:
The factors of 256 are 1, 2, 4, 8, 16, 32, 64, 128 and 256.
1 x 256 =256
2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 2 x 128 = 256
2 x 2 x 64 = 4 x 64 = 256
2 x 2 x 2 x 64 = 8 x 32 = 256
2 x 2 x 2 x 2 x 16 = 16 x 16 = 256

(ii) 625
Answer:
The factors of 625 are 1, 5, 25, 125 and 625.
Explanation:
The factors of 625 are 1, 5, 25, 125 and 625.
1 x 625 = 625
5 x 125 = 625
25 x 25 = 625

(iii) 243
Answer:
The factors of 243 are 1, 3, 9, 27, 81 and 243.
Explanation:
The factors of 243 are 1, 3, 9, 27, 81 and 243.
3 x 81 = 243
9 x 27 = 243

(iv) 343
Answer:
The factors of 343 are 1, 7, 49 and 343.
Explanation:
The factors of 343 are 1, 7, 49 and 343.
1 x 343 = 343
7 x 49 = 343

(v) 121
Answer:
The factors of 121 are 1, 11 and 121.
Explanation:
The factors of 121 are 1, 11 and 121.
1 x 121 = 121
11 x 11 = 121

Question 2.
Which are the numbers between 1 and 100 having exactly three factors?
Answer:
4, 9 , 25 and 49.
Explanation:
1, 2 and 4 three factor for 4
1, 3 and 9 three factor for 9
1, 5 and 25 three factor for 25
1, 7 and 49 three factor for 49
We know that the numbers between 1 and 100,
which have exactly three factors are 4, 9, 25 and 49.

Prime factors Textbook Page No. 99

How do we find the factors of 16?
The only prime factor of 16 is 2. Writing
16 = 2 × 2 × 2 × 2
We see that the factors of 16, except 1, are products of 2’ s.
2
2 × 2 = 4
2 × 2 × 2 = 8
2 × 2 × 2 × 2 = 16
Taking 1 also, we get all the factors of 16 as 1,2, 4, 8, 16.
Now let’s try 16 × 3 = 48.
48 = (2 × 2 × 2 × 2) × 3
To get its factors, we can multiply some of the 2’s only; or some 2 ‘s and 3.
Taking only 2’s, what we get are the factors of 16.
2, 4, 8, 16
What if we take 2’s and 3?
(2 × 3) = 6
(2 × 2) × 3 = 4 × 3 = 12
(2 × 2 × 2) × 3 = 8 × 3 = 24
(2 × 2 × 2 × 2) × 3 = 48
Thus we get also the factors.
6, 12, 24, 48

3 alone is also a factor. Also 1,which is a factor of every number. We can separate these factors like this;
Kerala Syllabus 6th Standard Maths Solutions Chapter 6 Numbers 8
What is the relation between each number in the first row with the number below it.
Now let’s take 48 × 3 = 144
144 = (2 × 2 × 2 × 2) × (3 × 3)
The factors can be got by taking only some 2’ s, some 2’ s and one 3 or some 2’s and two 3’s.
Taking 3 ’s only we get 3 and 9.
And 1 also is a factor.
Kerala Syllabus 6th Standard Maths Solutions Chapter 6 Numbers 9
These can also be written in a table like this:
The numbers in the first row, multiplied by 3, give the numbers in the second row.
And numbers in the second row, multiplied by 3, give the numbers in the third row.

Let’s look at the table along the columns.
First column is 1, 3, 9; these numbers do not have 2 as a factor.
Second column is 2, 6, 18; these have a single 2 as a factor.
What about the third and fourth columns?
Kerala Syllabus 6th Standard Maths Solutions Chapter 6 Numbers 10
Thus the numbers in each column, multiplied by 2, give the numbers in the next column.
So, a factor of 144 can be found like this:

Multiply some 2’s and 3’s . The number of 2’s must be less than or equal to 4 (we can also choose to take no 2 at all). The number of 3’s must be less than or equal to 2 (or no 3 at all). Such factors, together with lgive all the factors.

For example, 24 is the product of three 2’s and one 3.
24 = 2 × 2 × 2 × 3
And 18 is the product of a single 2 and two 3’s.
Can you find the factors of 200 like this?
200 = 2 × 2 × 2 × 5 × 5
Make a table like this:
Kerala Syllabus 6th Standard Maths Solutions Chapter 6 Numbers 11
Answer:

Explanation:
factors of 200 = 2 × 2 × 2 × 5 × 5
Thus the numbers in each column, multiplied by 5, to give the numbers in the next column.

Find all the factors of the numbers below:

(i) 242
Answer:
1, 2 and 11 are the factors of 242
Explanation:
2 × 11 = 22
2 × (11 × 11) = 22 × 11 = 242

(ii) 225
Answer:
1, 3 and 5 are the factors of 225
Explanation:
5 × 5 = 25
(5 x 5) × (3 × 3) = 25 × 9 = 225

(iii) 400
Answer:
1, 2 and 5 are the factors of 400
Explanation:
5 × 5 = 25
2 x 2 x 2 x 2 = 16
(5 x 5) × (2 x 2 x 2 x 2) = 25 × 16 = 400

(iv) 1000
Answer:
1, 2 and 5 are the factors of 400
Explanation:
5 × 5 x 5 = 125
(5 x 5 x 5) × (2 x 2 x 2) = 125 × 8 = 400

We have found the factors of 144.
Now let’s try 144 × 5 = 720
720 = 2 × 2 × 2 × 2 × 3 × 3 × 5
We can separate the factors as those without 5 and those with 5.
The factors without 5 are factors of 144.
And these can be found as before.
Kerala Syllabus 6th Standard Maths Solutions Chapter 6 Numbers 12
Multiplying all these by 5 gives the factors with 5.
Kerala Syllabus 6th Standard Maths Solutions Chapter 6 Numbers 13

Let’s write all these factors of 720 in a single table:
Kerala Syllabus 6th Standard Maths Solutions Chapter 6 Numbers 14
What about 144 × 25 = 3600?
We can expand the factor table of 720 like this:
Kerala Syllabus 6th Standard Maths Solutions Chapter 6 Numbers 15

Factorize each of the numbers below as the product of primes and write all factors in a table.
Write also the number of factors of each.

(i) 72
Answer:
1, 2 and 3 are factors of 72

Explanation:
factorization of 72 is,
1 x 2 = 2
3 × 3 = 9
2 x 2 x 2  = 8
(3 x 3) × (2 x 2 x 2) = 9 × 8 = 72
So, 1, 2 and 3 are factors of 72.

(ii) 108
Answer:

Explanation:
factorization of 108 is,
2 × 2 = 4
(3 x 3 x 3) × (2 x 2) = 27 × 4 = 108
So, 1, 2, 3 are factors of 108

(iii) 300
Answer:

Explanation:
Factorization of 300 is,
3 x 2 x 2 x 5 x 5 x 1
= 3 x 4 x 25
= 300

(iv) 96
Answer:
1, 2 and  3 are factors of 96

Explanation:
factorization of 96 is 3 x 2 x 2 x 2 x 2 x 2 x 1.
factors of 96 are 1, 2 and 3.

(v) 160
Answer:

Explanation:
factorization of 160 is,
2 x 2 x 2 x 2 x 2 x 5 x 1 = 160

(vi) 486
Answer:
2 x 3 x 3 x 3 x 3 x 3 = 486
Explanation:
factorization of 486 is,
2 x 3 x 3 x 3 x 3 x 3 = 486
LCM of 486 is as follows,

(vii) 60
Answer:
1, 2 ,3 and 5 are the factors of 60

Explanation:
factorization of 60 is,
2 x 2 x 3 x 5 = 160
LCM of 60 is shown below,

(viii) 90
Answer:

Explanation:
factorization of 90 is,
2 x 3 x 3 x 5 = 90
factor of 90 are 1, 2, 3 and 5
LCM of 90 is shown below,

(ix) 150
Answer:

Explanation:
factorization of 150 is,
2 x 3 x 5 x 5 = 150
LCM of 150 is shown below,

(i) Find the number of factors of 6, 10, 15, 14, 21. Find some other numbers with exactly four factors.
Answer:
Factors of 6 are 1, 2, 3 and 6.
Factors of 10 are 1, 2, 5 and 10.
Factors of 15 are 1, 3, 5 and 15.
Factors of 14 are 1, 2, 7 and 14.
Factors of 21 are 1, 3, 7 and 21.
Explanation:
A factor is a number which divides the number without remainder.
Factors of 6 are 1, 2, 3 and 6.
1 x 6 = 6
2 x 3 = 6
3 x 2 = 6
6 x 1 = 6
Factors of 10 are 1, 2, 5 and 10.
1 x 10 = 10
2 x 5 = 10
5 x 2 = 10
10 x 1 = 10
Factors of 15 are 1, 3, 5 and 15.
1 x 15 = 15
3 x 5 = 15
5 x 3 = 15
15 x 1 = 15
Factors of 14 are 1, 2, 7 and 14.
1 x 14 = 14
2 x 7 = 14
7 x 2 = 14
14 x 1 = 14
Factors of 21 are 1, 3, 7 and 21.
1 x 21 = 21
3 x 7 = 21
7 x 3 = 21
21 x 1 = 21

(ii) Is it correct to say that any number with exactly four factors is a product of two distinct primes?
Answer:
No, this is not correct.
Explanation:
If any prime number then its 3rd power has exactly 4 divisors,
and is obviously not the product of two distinct primes.
Take 36, it has two prime factors, 2 and 3.

Number of factors Textbook Page No. 104

We know how to find all the factors of 64.
Without writing down all the factors, can we just find the number of factors?
64 = 2 × 2 × 2 × 2 × 2 × 2
We can take one 2, two 2’s, three 2’s and so on to get factors. How many such factors are there?
Here there are six 2’s. So we can take one to six 2’s, and 1 is also a factor.
6 + 1 = 7 factors in all.
Can we find the number of factors of243 like this?
243 = 3 × 3 × 3 × 3 × 3
How many 3’s?
Taking one 3, two 3’s, three 3’s and so on, how many factors do we get?
Together with 1?
5 + 1 = 6 factors in all.
If a number can be split as the repeated product of a single prime, how do we find the number of factors of that number quickly?
What if we have two primes?
For example, let’s take 64 × 3 = 192
192 = (2 × 2 × 2 × 2 × 2 × 2) × 3
1 and products of 2’s give 7 factors as above; these factors multiplied by 3 give another 7 factors. Altogether 14 factors.
How about one more 3?
So how many factors does 192 × 3 = 576 have?
576 = (2 × 2 × 2 × 2 × 2 × 2) × (3 × 3)

We can separate the factors of 576 like this.

(i) Factors without 3
1 2 4 8 16 32 64

(ii) Product of these by 3
3 6 12 24 48 96 192

(iii) Product of the first factors by two 3’s
9 18 36 72 144 288 576

7 of each type. 7 × 3 = 21 in all.
We can put this in a different way. Take the products of 2’s and 3’s separately.
576 = 64 × 9

Look at the three types of factors of 576 again

(i) 1, 2, 4, 8, 16, 32, 64 – Factors of 64
(ii) 3, 6, 12, 24, 48, 96, 192 – Products of the factors of 64 by the factor 3 of 9
(iii) 9, 18, 36, 72, 144, 288, 576 – Products of the factors of 64 by the factor 9 of 9

We can also say that the factors we write first are the product of the factors of 64 by the factor 1 of 9.

Thus the factors of 576 are the product of each factor of 64 by each factor of 9.
64 has 7 factors and 9 has 3 factors. So 64 × 9 = 576 has 3 groups of 7 factors.
That is 7 × 3 = 21 factors.
Like this, can we find how many factors 1000 has?
1000 = (2 × 2 × 2) × (5 × 5 × 5)

In this, 2 × 2 × 2 = 8 has 4 factors; and 5 × 5 × 5 = 125 also has 4 factors.
We can multiply each of the 4 factors of 8 by each of the 4 factors of 125 to get all factors of 1000. That is 4 groups of 4 factors, making 4 × 4 = 16 in all.
Now let’s see how many factors 3600 has.
3600 = (2 × 2 × 2 × 2) × (3 × 3) × (5 × 5)
2 × 2 × 2 × 2 = 16 has 5 factors, 3 × 3 = 9 and 5 × 5 = 25 have 3 factors each.

Multiplying each factor of 16 by each factor of 9 gives 5 × 3 = 15 factors of 16 × 9.
Multiplying each of these by factors of 25 give all factors of 16 × 9 × 25 = 3600.
That means 15 × 3 = 45 factors
(Look once more the factor table of 3600 done earlier).
Kerala Syllabus 6th Standard Maths Solutions Chapter 6 Numbers 17

The number 4 has 3 factors and number 6 has 4 factors. Can we say that 4 × 6 = 24 has 3 × 4 = 12 factors’? Multiply each factor of 4 by each factor of 6. Why did we get the number of factors wrong?
Answer:
Yes, 4 × 6 = 24 has 3 × 4 = 12 factors’.
Explanation:
factors of 4 = 1, 2, 4
factors of 6 = 1, 2, 3, 6
Let 48 be the number,
factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 48.
4 is one of the factor of 48,
6 is one of the factor of 48.
4 x 6 = 24 is also a factor of 48.
let x be the number,
if 4 and 6 are 2 of the factors of x,
then 4 x 6 = 24 is also one of the factor of x.

Kerala Syllabus 6th Standard Maths Solutions Chapter 5 Decimal Forms

Textbook Page No. 107

Question 1.
The factor table of a number is given below. Some of the factors are written.
Kerala Syllabus 6th Standard Maths Solutions Chapter 6 Numbers 19
Kerala Syllabus 6th Standard Maths Solutions Chapter 6 Numbers 18
Answer:

(i) What is the number with this factor table?
Answer:
1, 2 , 14, 100, 245, 490, 4900
Explanation:
1 x 1 = 1
1 x 2 = 2
7 x 2 = 14
7 x 7 x 5 = 245
2 x 2 x 5 x 5 x 7 x 7 = 4900

(ii) Fill in the numbers in the circles
Answer:
1, 14, 245, 4900
Explanation:
1 x 1 = 1
7 x 2 = 14
7 x 7 x 5 = 245
2 x 2 x 5 x 5 x 7 x 7 = 4900

(iii) Write the numbers below in the correct cells
4, 25, 140, 200
Answer:

Explanation:
Take the LCM of given numbers 4, 25, 140 and 200
or find the factors of all the numbers above to fit in the cells as shown below.

(iv) Which of the numbers below cannot be in the table?
32, 40, 50, 200, 300, 350
Answer:
32 and 300
can not be in the table
Explanation:
Due to non availability of factor Five 2’s and 3 in the table 32 and 300 can not be in the table

Question 2.
Find the number of factors of each of these numbers.
(i) 500
Answer:
12
Explanation:
find the LCM of 500
500 = 2x 53.
= (2 + 1) x (3 + 1)
= 3 x 4
= 12

(ii) 600
Answer:
24
Explanation:
find the LCM of 600
600 = 23 x 31 x 52.
= (3 + 1) x (1 + 1) x (1 + 2)
= 4 x 2 x 3
= 24

(iii) 700
Answer:
12
Explanation:
first take the LCM of 700
700 = 22 x 52 x 71.
= (2 + 1) x (1 + 2) x (1 + 1)
= 3 x 3 x 2
= 12

(iv) 800
Answer:
18
Explanation:
find the LCM of 800
800 = 25 x 52.
= (5 + 1) x (1 + 2)
= 6 x 3
= 18

(v) 900
Answer:
27
Explanation:
Find the LCM of 900
900 = 22 x 32 x 52.
= (2 + 1) x (2 + 1) x (2 + 1)
= 3 x 3 x 3
= 27

Question 3.
How many factors does a product of three distinct primes have? What about a product of 4 distinct primes?
Answer:
Product of 3 distinct primes have 8 factors.
Product of 4 distinct primes have 12 factors.
Explanation:
To find the factors of product of three distinct primes have and four distinct primes have.
The factors of product of three distinct primes have four factors that include 1.
Let us consider a product of the primes 2, 3 and 5.
2 × 3 × 5 = 30
Factors of 30 are 1 , 2 , 3, 5 , 6 , 10 , 30 where the prime factors are 2, 3 and 5.
Hence there are 8 factors in total for a number that is the product of three distinct primes.
Let us consider a product of the primes 2, 3, 5 and 7.
2 × 3 × 5 x 7 = 210
The factors of product of four distinct primes have five factors that include 1.
The factors of 210 are 1, 2, 3, 5, 6, 7, 15, 30, 42, 70, 105 and 210.
So, it has twelve factors.
Hence there are 8 factors in total for a number that is the product of three distinct primes.

Question 4.
i) Find two numbers with exactly five factors.
Answer:
16 and 81
Explanation:
A factor is a number which divides the number without remainder.
factors of 16 = 1, 2, 4, 8 and 16.
factors of 81 = 1, 3, 9, 27, 81.

ii) What is the smallest number with exactly five factors?
Answer:
16
Explanation:
A factor is a number which divides the number without remainder.
factors of 16 = 1, 2, 4, 8 and 16.

Question 5.
How many even factors does 3600 have?
Answer:
36 factors.
Explanation:
Number of even factors = no. of total factors – no. of odd factors.
Prime Factorization of 3600 =  24 x 32 x 52
Total factors of 3600 = 45;
Total number of odd factors of 3600 is (2 + 1)(2 + 1) = 3 × 3 = 9
Even factors = 45 – 9 = 36

Kerala Syllabus 6th Standard Maths Solutions Chapter 7 Decimal Operations

You can Download Decimal Operations Questions and Answers, Activity, Notes, Kerala Syllabus 6th Standard Maths Solutions Chapter 7 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 6th Standard Maths Solutions Chapter 7 Decimal Operations

Decimal Operations Text Book Questions and Answers

Triangle Problem Textbook Page No. 109

Anup made a triangle with three sticks of length 4 centimetres each. What is the perimeter of this triangle?
How did you do it?
Suma made a triangle using 4.3 centimetre sticks instead.
What is the perimeter?
4.3 + 4.3 + 4.3 = 12.9 cm.
Kerala Syllabus 6th Standard Maths Solutions Chapter 7 Decimal Operations 1
Instead of adding again and again, we only compute 3 times 4.3
How do we find it?
4.3 centimetres mean 43 millimetres. And 3 times 43 millimetres is 43 × 3 = 129 millimetres.
This is 12.9 millimeters.
There’s another way of doing this:
4.3 = 4\(\frac{3}{10}\) = \(\frac{43}{10}\)
So, 3 times \(\frac{43}{10}\) is
\(\frac{43}{10}\) × 3 = \(\frac{129}{10}\) = 12.9 cm.
That is, 4.3 × 3 = 12.9

Kerala Syllabus 6th Standard Maths Solutions Chapter 7 Decimal Operations

Cloth problem

To make a shirt for a boy in the class, 1.45 metres of cloth is needed, on average.
How much cloth is needed to make shirts for the 34 boys in the class?
We must calculate 34 times 1.45.
1.45 metres mean 145 centimetres; And 34 times 145 is
145 × 34 = 4930
How much metres is 4930 centimetres?
\(\frac{4930}{100}\) metre = 49.30 metres
Kerala Syllabus 6th Standard Maths Solutions Chapter 7 Decimal Operations 2

How about writing all measurements as tractions?
1.45 = 1\(\frac{45}{100}\) = \(\frac{145}{100}\)
1.45 × 34 = 1\(\frac{45}{100}\) × 34 = \(\frac{145}{100}\) × 34 = \(\frac{4930}{100}\)
We can write it as a decimal.
\(\frac{4930}{100}\) = 49.30 = 49.3
Thus 1.45 × 34 = 49.3

The area of a square of side 1 centimetre 1 square centimetre and the area of a square of side 1 millimetre is 1 square millimetre. 1 centimetre is 10 millimetres. So in the bigger square we can stack 10 smaller squares each along the length and breadth 10 × 10 = 100 small squares in all. So the smaller square is \(\frac{1}{100}\) of the bigger square. That means 1 sq.mm. = \(\frac{1}{100}\) sq.cm

Area

We know how to calculate the area of a rectangle of length 8 centimetres and height 6 centimetres. What about a rectangle of length 8.5 centimetres and breadth 6.5 centimetres? The lengths in millimetres are 85 and 65. So area is 85 × 65 = 5525 square millimetres. How do we change it into square centimetres?
1 square millimetre = \(\frac{1}{100}\) square centimetre.
5525 square millimetres = \(\frac{5525}{100}\) = 55.25 square centimetres.

How about writing all measurements as fractions?
8.5 centimetres = 8\(\frac{5}{10}\) centimetres = \(\frac{85}{10}\) centimetres
6.5 centimetres = 6\(\frac{5}{10}\) centimetres = \(\frac{65}{10}\) centimetres
Area is \(\frac{85}{10}\) × \(\frac{65}{10}\) square centimetres.
\(\frac{85}{10}\) × \(\frac{65}{10}\) = \(\frac{5525}{100}\) = 55.25
Thus area is 55.25 square centimetres.
Let’s write the computation using numbers only.
8.5 × 6.5 = 55.25

Kerala Syllabus 6th Standard Maths Solutions Chapter 7 Decimal Operations

Textbook Page No. 111

Question 1.
The sides of a square are of length 6.4 centimeters. What is its perimeter?
Answer:
25.6 centimeters.
Explanation:
6.4 centimeter is length of one side,
The perimeter of a square is P = 6.4 + 6.4 + 6.4 + 6.4 = 25.6 centimeters.
There’s another way of doing this is,
Perimeter of a square is P = 4 x side
P = 4 x 6.4
P = 25.6 centimeters.

Question 2.
3 rods of length 6.45 meters each are laid end to end. What is the total length?
Answer:
19.62 meters.
Explanation:
6.45 centimeter is length of one side.
The perimeter of a triangle is P = 6.45 + 6.45 + 6.45 = 19.62 meters.
There’s another way of doing this is,
Perimeter of a square is P = 3 x side
P = 3 x 6.45
P = 19.62 centimeters

Question 3.
A bag can be filled with 4.575 kilograms of sugar. How much sugar can be filled in 8 such bags?
Answer:
36.6 kg
36.6 kilograms
Explanation:
A bag can be filled with 4.575 kilograms of sugar.
8 bags of sugar = 8 x 4.575 = 36.6 kg.

Question 4.
The price of one kilogram of rice is 34.50 rupees. How much money do we need to buy 16 kilograms?
Answer:
552 rupees.
Explanation:
The price of one kilogram of rice is 34.50 rupees.
Cost of 16 kilograms = 16 x 34.50 = 552 rupees.

Question 5.
6 bottles are filled with the coconut oil in a can. Each bottle contains 0.478 liters. How much oil was in the can, in liters?
Answer:
2.868 liters.
Explanation:
6 bottles are filled with the coconut oil in a can.
Each bottle contains 0.478 liters.
Total oil in the can, in liters was = 6 x 0.478 = 2.868 liters

Question 6.
The length and breadth of a rectangular room are 8.35 meters and 3.2 meters. What is the area of that room?
Answer:
26.72 meter square.
Explanation:
Area of rectangle = length x breadth
length of a rectangle room = 8.35 meters
breadth of a rectangle room = 3.2 meters
Area = 8.35 x 3.2 = 26.72 meter square.

Multiplication

What is the meaning of 4.23 × 2.4?
4.23 × 2.4 = \(\frac{423}{100}\) × \(\frac{24}{10}\) = \(\frac{423 \times 24}{1000}\)
To compute this, we have to multiply 423 by 24 and then divide by 1000.
423 × 24 = 10152
\(\frac{423 \times 24}{1000}\) × \(\frac{10152}{1000}\) = 10.152

In the answer, how many digits are there after the decimal point? Why three?
Look at the fraction form of the answer. The denominator is 1000, right?
How did we get this 1000?

Look at the denominator of the fractions we multiplied.
So how do we complete 4.23 × 0.24?
First find 423 × 24 = 10152.

Now how many digits are there after the decimal point in the product?
If we write 4.23 × 0.24 as a fraction, what would be the denominator of the product?
4.23 as a fraction has denominator 100.
0.24 as a fraction has denominator 100.What about the denominator of the product?
So, 4.23 × 0.24 = \(\frac{10152}{10000}\) = 1.0152

Like this, how do we do 2.45 × 3.72?
First calculate 245 × 372.
Now we must find out the number of digits after the decimal point.
What is the denominator of 2.45 as a fraction.
And of 3.72?
What is the denominator of the product?
So,
2.45 × 3.72 = 9.1140 = 9.114
Kerala Syllabus 6th Standard Maths Solutions Chapter 7 Decimal Operations 3
Answer:
0.1 × 0.1 = 0.01
0.01 × 0.01 = 0.0001
0.001 × 0.001 = 0.000001
0.0001 × 0.0001 = 0.00000001
Explanation:
To multiply a decimal number by a decimal number,
we first multiply the two numbers ignoring the decimal points.
Then place the decimal point in the product, in such a way that decimal places in the product is equal to the sum of the decimal places in the given numbers as shown above.

Kerala Syllabus 6th Standard Maths Solutions Chapter 7 Decimal Operations

Textbook Page No. 113

Question 1.
Calculate the products below:

i) 46.2 × 0.23
Answer:
10.626
Explanation:
If we write 46.2 × 0.23 as a fraction,
46.2 as a fraction has denominator 10.
0.23 as a fraction has denominator 100.
= \(\frac{462}{10}\) × \(\frac{23}{100}\)
= \(\frac{462 \times 23}{1000}\)
To compute this, we have to multiply 462 by 23 and then divide by 1000.
462 × 23 = 10,626
= \(\frac{10626}{1000}\) = 10.626
So, 46.2 × 0.23 = 10.626

ii) 57.52 × 31.2
Answer:
1794.624
Explanation:
If we write 57.52 × 31.2 as a fraction,
57.52 as a fraction has denominator 100.
31.2 as a fraction has denominator 10.
= \(\frac{5752}{100}\) × \(\frac{312}{10}\)
= \(\frac{5752 \times 312}{1000}\)
To compute this, we have to multiply 5752 by 312 and then divide by 1000.
5752 × 312 = 17,694,624
= \(\frac{17,94,624}{1000}\) = 1794.624
So, 57.52 × 31.2 = 1794.624

iii) 0.01 × 0.01
Answer:
Explanation:
If we write 0.01 × 0.01 as a fraction,
0.01 as a fraction has denominator 100.
0.01 as a fraction has denominator 100.
= \(\frac{1}{100}\) × \(\frac{1}{100}\)
= \(\frac{1 \times 1}{10000}\)
To compute this, we have to multiply 1 by 1 and then divide by 10000.
1 × 1 = 1
= \(\frac{1}{10000}\) = 0.0001
So, 0.01× 0.01 = 0.0001

iv) 2.04 × 2.4
Answer:
4.896
Explanation:
If we write 2.04 × 2.4 as a fraction,
2.04 as a fraction has denominator 100.
2.4 as a fraction has denominator 10.
= \(\frac{204}{100}\) × \(\frac{24}{10}\)
= \(\frac{204 \times 24}{1000}\)
To compute this, we have to multiply 204 by 24 and then divide by 1000.
204 × 24 = 4896
= \(\frac{4896}{1000}\) = 4.896
So, 2.04 x 2.4 = 4.896

v) 2.5 × 3.72
Answer:
9.3
Explanation:
If we write 2.5 × 3.72 as a fraction,
2.5 as a fraction has denominator 10.
3.72 as a fraction has denominator 100.
= \(\frac{25}{10}\) × \(\frac{372}{100}\)
= \(\frac{25 \times 372}{1000}\)
To compute this, we have to multiply 25 by 372 and then divide by 1000.
25 × 372 = 9300
= \(\frac{9300}{1000}\) = 9.3
So, 2.5× 3.72 = 9.3

vi) 0.2 × 0.002
Answer:
0.0004
Explanation:
If we write 0.2 × 0.002 as a fraction,
0.2 as a fraction has denominator 10.
0.002 as a fraction has denominator 1000.
= \(\frac{2}{10}\) × \(\frac{2}{1000}\)
= \(\frac{2 \times 2}{10000}\)
To compute this, we have to multiply 2 by 2 and then divide by 10000.
2 × 2 = 4
= \(\frac{4}{10000}\) = 0.0004
So, 0.2 × 0.002 = 0.0004

Question 2.
Given that 3212 × 23 = 73876, find the products below, without actually multiplying?

i) 321.2 × 23 = _____
Answer:
7387.6,
Explanation:
Given 3212 × 23 = 73876,
In 321.2 × 23 first find out the number of digits after the decimal point.
(1 + 0) = 1
So, 321.2 × 23 = 7387.6

ii) 0.3212 × 23 = _____
Answer:
7.3876
Explanation:
Given 3212 × 23 = 73876,
In 0.3212 × 23 first find out the number of digits after the decimal point.
(4 + 0) = 4
So, 0.3212 × 23 = 7.3876

iii) 32.12 × 23 = ____
Answer:
738.76
Explanation:
Given 3212 × 23 = 73876,
In 32.12 × 23 first find out the number of digits after the decimal point.
(2 + 0) = 2
So, 32.12 × 23 = 738.76

iv) 32.12 × 0.23 = ____
Answer:
7.3867
Explanation:
Given 3212 × 23 = 73876,
In 32.12 × 0.23 first find out the number of digits after the decimal point.
(2 + 2) = 4
So, 32.12 × 0.23 = 7.3876

v) 3.212 × 23 = ____
Answer:
73.876
Explanation:
Given 3212 × 23 = 73876,
In 3.212 × 23 first find out the number of digits after the decimal point.
(3 + 0) = 3
So, 3.212 × 23 = 73.876

vi) 321.2 × 0.23 = _____
Answer:
73.876
Explanation:
Given 3212 × 23 = 73876,
In 321.2 × 0.23 first find out the number of digits after the decimal point.
(1 + 2) = 3
So, 321.2 × 0.23 = 73.876

Question 3.
Which of the products below is equal to 1.47 × 3.7?
i) 14.7 × 3.7
ii) 147 × 0.37
iii) 1.47 × 0.37
iv) 0.147 × 37
v) 14.7 × 0.37
vi) 0.0147 × 370
vii) 1.47 × 3.70
Answer:
Option iv, v and vii has the equal products.
Explanation:
first find out the number of digits after the decimal point, then add to the product.
i) 14.7 × 3.7 = 54.39 (1 + 1 = 2)
ii) 147 × 0.37 = 54.39 (0 + 2 = 2)
iii) 1.47 × 0.37 = 0.5439 (2 + 2 = 4)
iv) 0.147 × 37 = 5.439 (3 + 0 = 3)
v) 14.7 × 0.37 = 5.439 (1 + 2 = 3)
vi) 0.0147 × 370 = 0.5439 (4 + 0 = 4)
vii) 1.47 × 3.70 = 0.5439 (2 + 2 = 4)
So, iv, v and vii has the equal products.

Question 4.
A rectangular plot is of length 45.8 meters and breadth 39.5 meters .What is its area?
Answer:
1809.10 meter square.
Explanation:
Area of rectangle = length x breadth
length of a rectangle plot= 45.8 meters
breadth of a rectangle plot = 39.5 meters
Area = 45.8 x 39.5 = 1809.10 meter square.

Question 5.
The price of petrol is 68.50 rupees per liter. What is the price of 8.5 liters?
Answer:
582.25 liters.
Explanation:
The price of petrol is 68.50 rupees per liter.
The price of 8.5 liters = 68.50 x 8.5 = 582.25 L

Kerala Syllabus 6th Standard Maths Solutions Chapter 7 Decimal Operations

Question 6.
Which is the largest product among those below.
i) 0.01 × .001
ii) 0.101 × 0.01
iii) 0.101 × 0.001
iv) 0.10 × 0.001
Answer:
Option (ii)
Explanation:
The largest product among those below are,
i) 0.01 × .001 = 0.00001 (2 + 3 = 5)
ii) 0.101 × 0.01 = 0.00101 (3 + 2 = 5)
iii) 0.101 × 0.001 = 0.000101 (3 + 3 = 6)
iv) 0.10 × 0.001 = 0.00010 (2 + 3 = 5)
So, the largest product is 0.101 x 0.01 = 0.00101

It is easy to calculate these products;
384 × 10
230 × 100

Now calculate these products:

• 125 × 10
Answer:
1250
Explanation:
First multiply the number by ignoring zeros.
125 x 1 = 125
Then add zero to the product.
125 + 0 = 1250
So, 125 x 10 = 1250

• 4.2 × 10
Answer:
42
Explanation:
To multiply a decimal by 10,
move the decimal point in the multiplication by one place to the right.
4.2 x 10 = 42

• 13.752 × 10
Answer:
137.52
Explanation:
To multiply a decimal by 10,
move the decimal point in the multiplication by one place to the right.
13.752 x 10 = 137.52

• 4.765 × 100
Answer:
476.5
Explanation:
To multiply a decimal by 100,
move the decimal point in the multiplication by two places to the right.
So, 4.765 x 100 = 476.5

• 3.45 × 100
Answer:
345
Explanation:
To multiply a decimal by 100,
move the decimal point in the multiplication by two places to the right.
3.45 x 100 = 345

• 14.572 × 100
Answer:
1457.2
Explanation:
To multiply a decimal by 100,
move the decimal point in the multiplication by two places to the right.
14.572 x 100 = 1457.2

• 1.345 × 1000
Answer:
1345
Explanation:
To multiply a decimal by 1000,
move the decimal point in the multiplication by three places to the right.
1345 x 1000 = 1345

• 2.36 × 1000
Answer:
0.236
Explanation:
To multiply a decimal by 1000,
move the decimal point in the multiplication by three places to the right.
2.36 x 1000 = 0.236

• 1.523 × 1000
Answer:
1523
Explanation:
To multiply a decimal by 1000,
move the decimal point in the multiplication by three places to the right.
1.523 x 1000 = 1523

Have you found out an easy way to multiply decimals by numbers 10,100,1000 and so on?
Answer:
Yes, just by moving the places towards the right.
Explanation:
When a decimal number is multiplied by 10, 100 or 1000,
the digits in the product are the same as in the decimal number,
but the decimal point in the product is shifted to the right as many places as there are zeros.

Kerala Syllabus 6th Standard Maths Solutions Chapter 7 Decimal Operations

Let’s divide! Textbook Page No. 114

4 girls divided a 12 meter long ribbon among them. What length did each get?
It is not difficult to calculate this.
How about a 13 meter long ribbon?
12 meter divided into 4 equal parts give 3 meter long pieces; the remaining 1 meter divided into 4 gives \(\frac{1}{4}\) meter. Altogether 3\(\frac{1}{4}\) meters.
Kerala Syllabus 6th Standard Maths Solutions Chapter 7 Decimal Operations 4
So, each gets 3\(\frac{1}{4}\) meters
We can write this as 13 ÷ 4 = 3\(\frac{1}{4}\)
We can also write it as a decimal.
\(\frac{1}{4}\) meter means 25 centimeter; that is, 0.25 meters.
So, instead of 3\(\frac{1}{4}\) metrer, we can write 3.25 meters.

Look at this problem;

A square is made with a 24.8 centimeter long rope. What is the length of its side?
To find the length of a side, 24.8 must be divided into four equal parts.
24.8 centimeters means 24 centimeters and 8 millimeters.
24 centimeters divided into four equal parts give 6 centimeters each.
The remaining 8 millimeters divided into four equal parts give 2 millimeters each.
Thus the length of a side is 6 centimeters and 4 millimeters, that is 6.2 centimeters.
This problem also we can write using numbers only.
24.8 ÷ 4
The way we found the answer can also be written using just numbers.
Kerala Syllabus 6th Standard Maths Solutions Chapter 7 Decimal Operations 5
24.8 mean 24 and 8 tenths. Dividing each by 4 gives 6 and 2 tenths; that is 6.2
These operations can be written in short hand as shown on the right.

A line of length 13.2 centimeters is divided into 3 equal parts .What is the length of each part?
We first divide 12 centimeters of 13.2 centimeters into 3 equal parts, getting 4 centimeter long parts; 1 centimeter and 2 millimeters remaining.
That is, 12 millimeters are left.

Dividing this into 3 equal parts gives 4 millimeters each. So, 13.2 centimeters divided into 3 equal parts give 4 centimeters and 4 millimeters as the length of a part.
That is 4.4 centimeters.
How about writing this as a division of numbers?
13.2 ÷ 3 = 4.4

How did we do this?
13.2 mean 13 and 2 tenths. in this, dividing 13 by 3 gives quotient 4 and remainder 1.
Kerala Syllabus 6th Standard Maths Solutions Chapter 7 Decimal Operations 6
Changing this 1 to tenths and adding them to the 2 tenths already there, we get 12 tenths. 12 divided by 3 gives 4.
Thus we get 4 and 4 tenths; that is 4.4.
These operations also we can write in shorthand.

Kerala Syllabus 6th Standard Maths Solutions Chapter 7 Decimal Operations

Let’s look at another problem:

4 people shared 16.28 kilograms of rice. How much does each get?
If 16 kilograms is divided in to 4 equal parts, how much is each part?
0.28 kilograms means 280 grams.
What if we divide 280 grams into 4?
So, how much does each get?
How about writing this using only numbers?
16.28 ÷ 4 = 4.07
16.28 means 16 and 2 tenths and 8 hundredths.
16 divided by 4 gives 4.

Changing 2 tenths to 20 hundredths and adding to the original 8 hundredths give 28 hundredths.
28 divided by 4 gives 7
So the total quotient is 4 and 7 hundredths.
That is 4.07.
The operation can be written like this:
Kerala Syllabus 6th Standard Maths Solutions Chapter 7 Decimal Operations 7
25.5 kilograms of sugar is packed into 6 bags of the same size. How much is in each bag?
24 kilograms divided into 6 equal parts give 4 kilograms each. The remaining 1.5 kilograms, changed to grams are 1500 grams.
Dividing this into 6 equal parts gives 1500 ÷ 6 = 250 grams.

So one bag contains 4 kilograms and 250 grams; that is 4.250 kilograms.
We usually write this as 4.25 kilograms.
As numbers, we find
25.5 ÷ 6 = 4.25
The method of finding the answer can also be written using only numbers.
25.5 means 25 and 5 tenths.
25 divided by 6 gives 4 and remainder 1.
The remaining 1, changed to tenths and added to the original 5 tenths give 15 tenths; divided this by 6 gives 2 tenths and remainder 3 tenths.

These 3 tenths can be changed into 30 hundredths ; and this divided by 6 gives 5 hundredths.
What then is the total quotient?
4 and 2 tenths and 5 hundredths
That is ,4.25
Let’s write these operations in shorthand.
Kerala Syllabus 6th Standard Maths Solutions Chapter 7 Decimal Operations 8

Textbook Page No. 118

Question 1.
The total amount of milk given to the children in a school for the 5 days of last week is 132.575 liters. How much was given on average each day?
Answer:
26.515 liters
Explanation:
Total milk given in last 5 days = 132.575 liters
Number of days = 5
Average milk given on each day = Total milk given by number of days.
= 132.575 ÷ 5
= 26.515 liters

Question 2.
8 people shared 33.6 kilograms of rice. Sujitha divided her share into three equal parts and gave one part to Razia. How much did Razia get?
Answer:
1.4 kg
Explanation:
Number of people = 8
Total rice shared = 33.6 kilograms.
No. of kgs each person got = 33.6 ÷ 8 = 4.2 kg
Sujitha divided her share into three equal parts = 4.2 ÷ 3 = 1.4 kg
Razia gets 1.4 kg share of Sujitha.

Question 3.
A ribbon of length 0.8 meters is divided into 16 equal parts. What is the length of each part’?
Answer: 5 cm
Explanation:
A ribbon of length 0.8 meters is divided into 16 equal parts.
1 m = 100 cm
0.8 m = 100 x 0.8 = 80 cm
The length of each part = 80 ÷ 16 = 5 cm

Question 4.
Do the problems below:
i) 54.5 ÷ 5
Answer:
10.9
Explanation:
Place the decimal point in the quotient directly above the decimal point in the dividend.
Divide the same way you would divide with whole numbers.
Divide until there is no remainder, or until the quotient begins to repeat in a pattern.

ii) 14.24 ÷ 8
Answer:
1.78
Explanation:
Place the decimal point in the quotient directly above the decimal point in the dividend.
Divide the same way you would divide with whole numbers.
Divide until there is no remainder, or until the quotient begins to repeat in a pattern

iii) 56.87 ÷ 11
Answer:
5.17
Explanation:
Place the decimal point in the quotient directly above the decimal point in the dividend.
Divide the same way you would divide with whole numbers.
Divide until there is no remainder, or until the quotient begins to repeat in a pattern

iv) 3.1 ÷ 2
Answer:
1.55
Explanation:
Place the decimal point in the quotient directly above the decimal point in the dividend.
Divide the same way you would divide with whole numbers.
Divide until there is no remainder, or until the quotient begins to repeat in a pattern

v) 35.523 ÷ 3
Answer:
11.841
Explanation:
Place the decimal point in the quotient directly above the decimal point in the dividend.
Divide the same way you would divide with whole numbers.
Divide until there is no remainder, or until the quotient begins to repeat in a pattern

vi) 36.48 ÷ 12
Answer:
3.4
Explanation:
Place the decimal point in the quotient directly above the decimal point in the dividend.
Divide the same way you would divide with whole numbers.
Divide until there is no remainder, or until the quotient begins to repeat in a patter

vii) 16.56 ÷ 9
Answer:
1.84
Explanation:
Place the decimal point in the quotient directly above the decimal point in the dividend.
Divide the same way you would divide with whole numbers.
Divide until there is no remainder, or until the quotient begins to repeat in a pattern

viii) 32.454 ÷ 4
Answer:
8.1135
Explanation:
Place the decimal point in the quotient directly above the decimal point in the dividend.
Divide the same way you would divide with whole numbers.
Divide until there is no remainder, or until the quotient begins to repeat in a pattern

ix) 425.75 ÷ 25
Answer:
17.03
Explanation:
Place the decimal point in the quotient directly above the decimal point in the dividend.
Divide the same way you would divide with whole numbers.
Divide until there is no remainder, or until the quotient begins to repeat in a pattern

Kerala Syllabus 6th Standard Maths Solutions Chapter 7 Decimal Operations

Question 5.
Given 105.728 ÷ 7 = 15.104, find the answer to the problems below, with out actual division.
i) 1057.28 ÷ 7
Answer:
151.04
Explanation:
1057.28 mean 1057 and 2 tenths, 8 hundredths.
in this, dividing 1057 by 7 gives quotient 151.
Changing this 2 tenths to 20 hundredths and adding to the original 8 hundredths gives 28 hundredths.
Then, we get 28 ÷ 7 = 4
Thus we get 151 and 2 tenths and hundredths; that is 151.04.

ii) 1.05728 ÷ 7
Answer:
0.15104
Explanation:
1.05728 mean 1 and 0 tenths, 5 hundredths, 7 thousandths, 2 ten thousandths and 8 lakhs.
count the number of decimals and move the decimals from right in the quotient.
we get 1.057281 ÷ 7 0.15104

Question 6.
A number multiplied by 9 gives 145.71.  What is the number?
Answer:
16.19
Explanation:
Let the number be x.
9x = 145.71
x = \(\frac{145.71}{9}\)
= \(\frac{145.71 × 100}{9 × 100}\)
= \(\frac{14571}{900}\)
= 16.19
So, 16.19 x 9 = 145.71

16.34 ÷ 10 = 163.4
25.765 ÷ 100 = _____.
347.5 ÷ 100 = ______.
238.4 ÷ 1000 = _____.
What have you found out about dividing a number in decimal form by 10, 100, 1000 and so on?
Answer:
When we divide a decimal by 10, 100 and 1000,
the place value of the digits decreases.
The digits move to the right since the number gets smaller,
but the decimal point does not move.
Explanation:
When we observe the below division, there is no change in decimal places.
16.34 ÷ 10 = 163.4
25.765 ÷ 100 = 0.25765
347.5 ÷ 100 = 3.475
238.4 ÷ 1000 = 0.2384

Other Divisions

A rope of length 8.4 meters is cut into 0.4 meter long pieces. How many pieces can we make?
8.4 meters is 840 centimeters and 0.4 meter is 40 centimeters. So the number of pieces is 840 ÷ 40 = 21
We can write this as
8.4 ÷ 0.4 = 21
What does this mean?
8.4 is 21times 0.4
How about doing this with fractions?
84 = \(\frac{84}{10}\), 0.4 = \(\frac{4}{10}\)
\(\frac{84}{10}\) ÷ \(\frac{4}{10}\) means, finding out the number, \(\frac{4}{10}\) of which is \(\frac{84}{10}\).

And we know that it is \(\frac{10}{4}\) times \(\frac{84}{10}\).
That is \(\frac{84}{10}\) ÷ \(\frac{4}{10}\) = \(\frac{84}{10}\) × \(\frac{10}{4}\) = 21
That is, \(\frac{84}{10}\) ÷ \(\frac{4}{10}\) = \(\frac{84}{10}\) ÷ \(\frac{10}{4}\) = 21
Can we compute 36.75 ÷ 0.5 like this?
36.75 = \(\frac{3675}{100}\), 0.5 = \(\frac{5}{10}\)
\(\frac{3675}{100}\) ÷ \(\frac{5}{10}\) = \(\frac{3675}{100}\) × \(\frac{10}{5}\) = \(\frac{735}{10}\)
That is, 36.75 ÷ 0.5 = 73.5
We can also write \(\frac{36.75}{0.5}\) = 73.5
So how do we find \(\frac{48.72}{0.12}\)?
\(\frac{48.72}{0.12}\) = 48.72 ÷ 0.12 = \(\frac{4872}{100}\) ÷ \(\frac{12}{100}\)
= \(\frac{4872}{100}\) × \(\frac{100}{12}\)
= \(\frac{4872}{12}\)
= 406

Textbook Page No. 119

Question 1.
The area of a rectangle is 3.25 square meters and its length is 2.5 centimeters. What is its breadth’?
Answer:
1.3 meters
Explanation:
Area of rectangle = length x breadth
The area of a rectangle is 3.25 square meters,
length is 2.5 centimeters.
breadth = \(\frac{3.25}{2.5}\)
b = 1.3 meters

Question 2.
A can contains 4.05 liters of coconut oil. It must be filled in to 0.45 liter bottles. How many bottles are needed?
Answer:
9 bottles.
Explanation:
A can contains 4.05 liters of coconut oil.
Capacity of one bottle = 0.45 liters
Number of bottles required = \(\frac{4.05}{0.45}\)
Divided the numerator and denominator by 100
= \(\frac{405}{45}\) = \(\frac{81}{9}\)
= 9 bottles.

Kerala Syllabus 6th Standard Maths Solutions Chapter 7 Decimal Operations

Question 3.
Calculate the quotients below:

i) \(\frac{35.37}{0.03}\)
Answer:
1,179
Explanation:
\(\frac{35.37}{0.03}\) = 35.37 ÷ 0.03
= \(\frac{3537}{100}\) ÷ \(\frac{3}{100}\)
= \(\frac{3537}{100}\) × \(\frac{100}{3}\)
= \(\frac{3537}{3}\)
= 1179

ii) \(\frac{10.92}{2.1}\)
Answer:
52
Explanation:
\(\frac{10.92}{2.1}\) = 10.92 ÷ 2.1
= \(\frac{1092}{100}\) ÷ \(\frac{21}{10}\)
= \(\frac{1092}{100}\) × \(\frac{10}{21}\)
= \(\frac{10920}{210}\)
= 52

iii) \(\frac{40.48}{1.1}\)
Answer:
3,680
Explanation:
\(\frac{40.48}{1.1}\) = 4048 ÷ 11
= \(\frac{4048}{100}\) ÷ \(\frac{11}{10}\)
= \(\frac{4048}{100}\) × \(\frac{10}{11}\)
= \(\frac{40480}{11}\)
= 3680

iv) \(\frac{0.045}{0.05}\)
Answer:
0.9
Explanation:
\(\frac{0.045}{0.05}\) = 0.045 ÷ 0.05
= \(\frac{45}{1000}\) ÷ \(\frac{5}{100}\)
= \(\frac{45}{1000}\) × \(\frac{100}{5}\)
= \(\frac{45}{50}\)
= 0.9

v) 0.001 ÷ 0.1
Answer:
0.01
Explanation:
0.001 ÷ 0.1
= \(\frac{1}{1000}\) ÷ \(\frac{1}{10}\)
= \(\frac{1}{1000}\) × \(\frac{10}{1}\)
= \(\frac{1}{100}\)
= 0.01

vi) 5.356 ÷ 0.13
Answer:
41.2
Explanation:
5.356 ÷ 0.13
= \(\frac{5356}{1000}\) ÷ \(\frac{13}{100}\)
= \(\frac{5356}{1000}\) × \(\frac{100}{13}\)
= \(\frac{5356}{130}\)
= 41.2

vii) \(\frac{0.2 \times 0.4}{0.02}\)
Answer:
4
Explanation:
\(\frac{0.2 \times 0.4}{0.02}\)
\(\frac{0.08}{0.02}\) = 4

viii) \(\frac{0.01 \times 0.01}{0.001 \times 0.1}\)
Answer:
1
Explanation:
\(\frac{0.01 \times 0.01}{0.001 \times 0.1}\)
\(\frac{0.0001}{0.0001}\) = 1

Question 4.
12125 divided by which number gives 1.2125?
Answer:
10,000
Explanation:
Let number be divided by x.
\(\frac{12125}{x}\) = 1.2125
x = \(\frac{12125}{1.2125}\)
x = 10,000
12125 divided by 10,000 gives 1.212

Question 5.
0.01 multiplied by which number gives 0.00001?
Answer:
0.001
Explanation:
Let the number be multiplied by x.
0.01x = 0.00001
x = \(\frac{0.00001}{0.01}\)
multiply both numerator and denominator with 100
x = \(\frac{0.00001}{0.01}\) x \(\frac{100}{100}\)
x = 0.001
0.01 x 0.001 = 0.00001

Kerala Syllabus 6th Standard Maths Solutions Chapter 7 Decimal Operations

Fractions and decimals

Fractions written as decimals are of denominators 10,100, 1000 and so on.
For some fractions, we can first change the denominator into one of these and then write in decimal form. For example,
\(\frac{1}{2}\) = \(\frac{5}{10}\) = 0.5
\(\frac{1}{4}\) = \(\frac{25}{100}\) = 0.25
\(\frac{3}{4}\) = \(\frac{75}{100}\) = 0.75

How do we write \(\frac{1}{8}\) in decimal form?
8 = 2 × 2 × 2
So, multiplying 8 by three 5’s we can make it a product of 10’s.
8 × (5 × 5 × 5) = (2 × 2 × 2) × (5 × 5 × 5)
= (2 × 5) × (2 × 5) × (2 × 5)
= 10 × 10 × 10 = 1000
5 × 5 × 5 = 125, right? So
\(\frac{1}{8}\) = \(\frac{125}{8 \times 125}\) = \(\frac{125}{1000}\) = 0.125
In much the same way,
\(\frac{5}{8}\) = \(\frac{5 \times 125}{8 \times 125}\) = \(\frac{625}{1000}\) = 0.625

How about \(\frac{1}{40}\) ?
40 = (2 × 2 × 2) × 5
To get a product of 10’s we have to multiply 40 by two 5’s; that is
40 × 25 = (2 × 2 × 2 × 5) × (5 × 5)
= (2 × 5) × (2 × 5) × (2 × 5)
= 10 × 10 × 10
= 1000
So,
\(\frac{1}{40}\) = \(\frac{25}{40 \times 25}\) = \(\frac{25}{1000}\) = 0.025
And \(\frac{21}{40}\)?
\(\frac{21}{40}\) = \(\frac{21 \times 25}{40 \times 25}\) = \(\frac{525}{1000}\) = 0.525

Similarly, since 125 × 8 = 1000, we can write
\(\frac{121}{125}\) = \(\frac{121 \times 8}{125 \times 8}\) = \(\frac{968}{1000}\) = 0.968
Thus we can find the decimal form of any fraction whose denominator is a multiple of 2’s and 5’s.

Now look at this problem:
24 kilograms of sugar are packed into 25 packets of the same size. How much does each packet contain?
24 kilograms means 24000 grams. So each packet contains \(\frac{24000}{25}\) grams.
\(\frac{24000}{25}\) = 960

Thus each packet contains 960 grams or 0.96 kilograms.
We can do this in a different way. Each packet contains \(\frac{24}{25}\) kilograms.
\(\frac{24}{25}\) = \(\frac{24 \times 4}{25 \times 4}\) = \(\frac{96}{100}\) = 0.96
So, one packet contains 0.96 kilograms.

Textbook Page No. 121

Question 1.
Find the decimal forms of the fractions below:

i) \(\frac{3}{5}\)
Answer:
0.6
Explanation:
To find the decimal of a fraction,
divide the numerator by the denominator.
Because the number in the numerator is smaller than the number in the denominator,
place the decimal point after it and add zeros.
\(\frac{3}{5}\) = 0.6

ii) \(\frac{7}{8}\)
Answer:
0.875
Explanation:
To find the decimal of a fraction,
divide the numerator by the denominator.
Because the number in the numerator is smaller than the number in the denominator,
place the decimal point after it and add zeros.

iii) \(\frac{5}{16}\)
Answer:
0.3125
Explanation:
To find the decimal of a fraction,
divide the numerator by the denominator.
Because the number in the numerator is smaller than the number in the denominator,
place the decimal point after it and add zeros.

iv) \(\frac{3}{40}\)
Answer:
0.075
Explanation:
To find the decimal of a fraction,
divide the numerator by the denominator.
Because the number in the numerator is smaller than the number in the denominator,
place the decimal point after it and add zeros.

v) \(\frac{3}{32}\)
Answer:
0.09375
Explanation:
To find the decimal of a fraction,
divide the numerator by the denominator.
Because the number in the numerator is smaller than the number in the denominator,
place the decimal point after it and add zeros.

vi) \(\frac{61}{125}\)
Answer:
0.488
Explanation:
To find the decimal of a fraction,
divide the numerator by the denominator.
Because the number in the numerator is smaller than the number in the denominator,
place the decimal point after it and add zeros.

Kerala Syllabus 6th Standard Maths Solutions Chapter 7 Decimal Operations

Question 2.
Write the answer to the questions below in decimal form.
(i) 3 liters of milk is used to fill 8 identical bottles. How much does each bottle contain?
Answer:
0.375 liters.
Explanation:
3 liters of milk is used to fill 8 identical bottles.
Each bottle contains = \(\frac{3}{8}\)
= 0.375 liters.

(ii) A 17 meter long string is cut into 25 equal parts. What is the length of each part?
Answer:
0.68 meters
Explanation:
A 17 meter long string is cut into 25 equal parts.
The length of each part = \(\frac{17}{25}\)
= 0.68 meters

(iii) 19 kilograms of rice is divided among 20 people. How much does each get?
Answer:
0.95 kilograms.
Explanation:
19 kilograms of rice is divided among 20 people.
Total kilograms of rice each get = \(\frac{19}{20}\)
= 0.95 kg

Question 3.
What is the decimal form of \(\frac{1}{2}\) + \(\frac{1}{4}\) + \(\frac{1}{8}\) + \(\frac{1}{16}\)?
Answer:
0.9375
Explanation:
\(\frac{1}{2}\) + \(\frac{1}{4}\) + \(\frac{1}{8}\) + \(\frac{1}{16}\)
numerators are same denominators are different, so find the LCM of denominators
= \(\frac{(1 ×16) + (1 × 8) + (1 × 4) + (1×  2)}{32}\)
= \(\frac{16 + 8 + 4 + 2}{32}\)
= \(\frac{30}{32}\)
= 0.9375

Question 4.
A two digit number divided by another two digit number gives 4.375.What are the numbers’?
Answer:
The two digit numbers are 70 and 16.
Explanation:
Given number is 4.375
convert the decimal number to whole number,
\(\frac{4375}{1000}\)
find the factors of above fraction,
factors of 4375 = 5 x 5 x 5 x 5 x 7
factors of 1000 = 2 x 2 x 2 x 5 x 5 x 5
Divide the common factors in both the numbers,
\(\frac{35}{8}\)
multiply both numerator and denominator with 2,
\(\frac{35 × 2}{8 × 2}\) = \(\frac{70}{16}\) = 4.375

Question 1.
What is the volume of a rectangular block of length 25.5 centimeters, breadth 20.4 centimeters and height 10.8 centimeters?
Answer:
5618.16 cm3
Explanation:
Given, length 25.5 centimeters, breadth 20.4 centimeters and height 10.8 centimeters.
Volume of a cuboid = l x b x h
V = 25.5 x 20.4 x 10.8 = 5618.16 cm3

Question 2.
The heights of three boys sitting on a bench are 130.5 centimeters 128.7 centimeters and 134.6 centimeters .What is the average height’?
Answer:
131.26 centimeters.
Explanation:
The heights of three boys sitting on a bench are,
130.5 centimeters 128.7 centimeters and 134.6 centimeters.
The average height of 3 boys = \(\frac{130.5 + 128.7 + 134.6}{3}\)
= 393.8 ÷ 3 = 131.26

Question 3.
Calculate \(\frac{4 \times 3.06}{3}\).
Answer:
4.08
Explanation:
\(\frac{4 \times 3.06}{3}\)
\(\frac{12.24}{3}\) = 4.08

Question 4.
The price of 22 pencils is 79.20 rupees. What is the price of 10 pencils’?
Answer:
36 rupees
Explanation:
The price of 22 pencils is 79.20 rupees.
Price of each pencil = 79.20 ÷ 22 = 3.60 rupees
The price of 10 pencils = 3.60 x 10 = 36 rupees.

Question 5.
Calculate the following:

i) \(\frac{2.3 \times 3.2}{0.4}\)
Answer:
18.4
Explanation:
\(\frac{2.3 \times 3.2}{0.4}\)
\(\frac{7.36}{0.4}\) = 18.4

ii) \(\frac{0.01 \times 0.001}{0.1 \times 0.01}\)
Answer:
0.01
Explanation:
\(\frac{0.01 \times 0.001}{0.1 \times 0.01}\)
\(\frac{0.00001}{0.001}\) = 0.01

Question 6.
Dividing 0.1 by which number gives 0.001?
Answer:
100
Explanation:
Let the number to be divided is x
0.1 ÷  x = 0.001
0.1 = 0.001x
x = \(\frac{0.1}{0.001}\)
multiply both numerator and denominator with 1000.
x = \(\frac{100}{1}\)
x = 100

Kerala Syllabus 6th Standard Maths Solutions Chapter 8 Joining Angles

You can Download Joining Angles Questions and Answers, Activity, Notes, Kerala Syllabus 6th Standard Maths Solutions Chapter 8 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 6th Standard Maths Solutions Chapter 8 Joining Angles

Joining Angles Text Book Questions and Answers

Joining Angles Textbook Page No. 123

Your geometry box has two set squares. Each of them has three angles.
Kerala Syllabus 6th Standard Maths Solutions Chapter 8 Joining Angles 1
Look at an angle drawn with a corner of a set square.
Kerala Syllabus 6th Standard Maths Solutions Chapter 8 Joining Angles 2
How much is ∠CAB?
What if we draw another angle on top of this, using a corner of the other set square?
Answer:
∠CAB = 30
Explanation:
By observing the figure triangle ABC, we came to conclude that,
as ∠CAB is 30 degrees, ∠ABC is 90 degrees and angle ∠BCA is 60 degrees
These set squares come in two usual forms,
both right triangles one with 90-45-45 degree angles,
the other with 30-60-90 degree angles.
Combining the two forms by placing the hypotenuses together will get 15° and 75° angles.

Kerala Syllabus 6th Standard Maths Solutions Chapter 8 Joining Angles 3
How much is ∠DAC?
And ∠DAB?
Answer:
75°

Explanation:
By observing set square, the figure angle ∠DAB,
we conclude that, as ∠DAB is 75 degrees, ∠CAB is 30 degrees and angle ∠DAC is 45 degrees.
∠DAC = 45° And ∠DAB = 75°

 

Kerala Syllabus 6th Standard Maths Solutions Chapter 8 Joining Angles 4
Answer:
∠DAB = 75°

Explanation:
By observing set square, the figure angle ∠DAB,
we conclude that, as ∠DAB is 75 degrees, ∠CAB is 30 degrees and angle ∠DAC is 45 degrees.
∠DAC = 45° And ∠DAB = 75°

Now suppose we draw angles as shown below:
Kerala Syllabus 6th Standard Maths Solutions Chapter 8 Joining Angles 5
How much ∠DAC?
Like this, angles of what different measures can we draw using the two set squares?
Answer:
15°
Explanation:
These set squares come in two usual forms, both right triangles
one with 90-45-45 degree angles, the other with 30-60-90 degree angles.
Combining the two forms by placing the hypotenuses together will get 15° and 75° angles.

In each picture below, two angles are given. Calculate the third angle as a sum or difference;

Kerala Syllabus 6th Standard Maths Solutions Chapter 8 Joining Angles 6
∠DAC = ____ – ____ = _____
Answer:
40°
Explanation:
∠DAC = 40°
∠CAB = 60°
∠DAB = 40° + 60° = 100°

Kerala Syllabus 6th Standard Maths Solutions Chapter 8 Joining Angles 7
∠DAB = ____ – ____ = _____
Answer:
75°
Explanation:
∠DAB = 30° + 60° = 90°

Kerala Syllabus 6th Standard Maths Solutions Chapter 8 Joining Angles 8
∠DAC = ____ – ____ = _____
Answer:
50°
Explanation:
Given;
∠DAC = 80°; ∠CAB = 30°
∠DAC = 80° – 30° = 50°

Kerala Syllabus 6th Standard Maths Solutions Chapter 8 Joining Angles 9
∠DAC = ____ – ____ = _____
Answer:
75°
Explanation:
Given;
∠DAB = 120°; ∠CAB = 45°
∠DAC = 120° – 45° = 75°

Kerala Syllabus 6th Standard Maths Solutions Chapter 8 Joining Angles

Two sides Textbook Page No. 126

Draw a line and then a perpendicular at one end.
Kerala Syllabus 6th Standard Maths Solutions Chapter 8 Joining Angles 10
We have noted that such an angle measures 90°.
Now extend the horizontal line a bit to the left;
Kerala Syllabus 6th Standard Maths Solutions Chapter 8 Joining Angles 11
Now there is another angle on the left of the vertical line also. What is its measure?
Perpendicular means straight up, not leaning to the left or right.
So the angle on the left is also 90°.
Kerala Syllabus 6th Standard Maths Solutions Chapter 8 Joining Angles 12
Now let’s draw a slanted line through the foot of the perpendicular.
Kerala Syllabus 6th Standard Maths Solutions Chapter 8 Joining Angles 13
How much is the angle on the left of this slanted line’?
A bit more than 90°, right’?
How much more’?
Kerala Syllabus 6th Standard Maths Solutions Chapter 8 Joining Angles 14
By how much is the angle on the right less than 90°?
Kerala Syllabus 6th Standard Maths Solutions Chapter 8 Joining Angles 15
Now can’t we calculate the angle on the left also?
Kerala Syllabus 6th Standard Maths Solutions Chapter 8 Joining Angles 16
see this picture
Kerala Syllabus 6th Standard Maths Solutions Chapter 8 Joining Angles 17

How much is the angle on the left of the slanted line? Imagine a perpendicular through the point where the lines meet.
By how much is the angle on the right less than 90°?
So, by how much is the angle on the left more than 90°?
Thus the angle on the left is 90° + 40° = 130°.

Kerala Syllabus 6th Standard Maths Solutions Chapter 8 Joining Angles

Joining setsquares Textbook Page No. 128

The picture shows two identical setsquares placed together;
Kerala Syllabus 6th Standard Maths Solutions Chapter 8 Joining Angles 18
What are the measures of the angles of this triangle?
Answer:
60°
Explanation:
The sum of the equilateral triangle is 120°
60° + 60° + 60° = 120°
60° in three corners as shown in the below figure

In the pictures below, two angles are marked and one of them is given. Find the other:

Kerala Syllabus 6th Standard Maths Solutions Chapter 8 Joining Angles 19
Answer:
80°
Explanation:

As we know a straight line is having 180° angle.
∠AOB = 180°
∠AOC = ∠AOB – ∠COB
∠AOC = 180° – 100° = 80°

Kerala Syllabus 6th Standard Maths Solutions Chapter 8 Joining Angles 20
Answer:
140°
Explanation:

As we know a straight line is having 180° angle.
∠AOB = 180°
∠AOC = ∠AOB – ∠COB
∠AOC = 180° – 40° = 140°

 

Kerala Syllabus 6th Standard Maths Solutions Chapter 8 Joining Angles 21
Answer:
130°
Explanation:

as we know a straight line is having 180° angle
∠AOB = 180°
∠AOC = ∠AOB – ∠COB
∠AOC = 180° – 50° = 130°

 

Kerala Syllabus 6th Standard Maths Solutions Chapter 8 Joining Angles 22
Answer:
100°
Explanation:

As we know a straight line is having 180° angle
∠AOB = 180°
∠AOC = ∠AOB – ∠COB
∠AOC = 180° – 80° = 100°

Meeting lines

See these pictures:

Kerala Syllabus 6th Standard Maths Solutions Chapter 8 Joining Angles 23

All of them show two lines meeting ; and each has two angles, one on the left and one on the right.
In the first picture, both angles are 90°. In the second ,the angle on the right is less than 90° and the angle on the left is greater than 90°; in the third picture , it is the other way round.

In the second and third picture, the angle on one side is that much more than 90° as the angle on the other side is less than 90°.
So, the sum of the angles on either side is 90° + 90° = 180°, right?

Thus we can write this as a general principle:
When two lines meet, the sum of the angles on either side is 180°.

Two such angles, made by two lines meeting is called a liner pair. So we can state this principle like this also:
The sum of the angles in a linear pair is 180°.

Kerala Syllabus 6th Standard Maths Solutions Chapter 8 Joining Angles

Textbook Page No. 130

Question 1.
How much is ∠ACE in the picture below?
Kerala Syllabus 6th Standard Maths Solutions Chapter 8 Joining Angles 26
Answer:
105°
Explanation:
as we know a straight line is having ∠ACB 180° angle
∠ACB = 180°
∠ACE = ∠ACB – ∠ECB
∠ACE = 180° – 75° = 105°

Question 2.
What is the angle between the lines in this picture?
Kerala Syllabus 6th Standard Maths Solutions Chapter 8 Joining Angles 27
Answer:
75°
Explanation:
Sum of the angles = 180°
with the help of a protractor we can measure the angles as 45° and 60°
180° – (45° + 60°)
180° – 105° = 75°

Question 3.
In the picture below, ∠ACE = ∠BCD. Find the measure of each.
Kerala Syllabus 6th Standard Maths Solutions Chapter 8 Joining Angles 25
Answer:
150°
Explanation:
Sum of the angles = 180°
∠ACE + ∠BCD = 180° – ∠DCE
180° – 30° = 150°
∠ACE + ∠BCD = 150°
∠ACE = 75°
∠ACE = ∠BCD = 75°

Question 4.
One angle of a linear pair is twice the other. How much is each?
Answer:
60° and 120°
Explanation:
Sum of the angles = 180°
If one angle of linear is 60°
twice the other is 60° + 60° = 120°

Question 5.
The angles in a linear pair are consecutive odd numbers. How much is each’?
Answer:
89° and 91°
Explanation
A linear pair are consecutive odd number.
x + x + 2 = 180°
2x + 1 = 180°
2x = 178°
x = 89°
The two consecutive odd numbers are 89° and and 91°

So, required two angles that are consecutive odd numbers and are in a linear pair are 89° and 91°.

Crossing lines

See the picture:

Kerala Syllabus 6th Standard Maths Solutions Chapter 8 Joining Angles 28
How much is the angle on the left?
What if we extend the top line downwards crossing the horizontal line?
Kerala Syllabus 6th Standard Maths Solutions Chapter 8 Joining Angles 29
Now we have two angles below also. What are their measures?
The angles above and below, on the right of the slanted line, form a linear pair, right’?
Kerala Syllabus 6th Standard Maths Solutions Chapter 8 Joining Angles 30
Thus don’t we get one angle below?
Like this, the angle above and below on the left also form a linear pair.
Kerala Syllabus 6th Standard Maths Solutions Chapter 8 Joining Angles 31
Thus we get the angle below on the left also. Let’s look at all the angles together:
Kerala Syllabus 6th Standard Maths Solutions Chapter 8 Joining Angles 32

Kerala Syllabus 6th Standard Maths Solutions Chapter 8 Joining Angles

Linear pair Textbook Page No. 131

Draw the line AB and a point C on it. Draw a circle centered at C. Mark a point D on the circle.
Kerala Syllabus 6th Standard Maths Solutions Chapter 8 Joining Angles 33
Join CD. Now we can hide the circle. By choosing Angle and clicking on B, C, D in order, we get the measure of ∠BCD. In the same way, click on A, C, D to get ∠ACD.

Using Move change the position of D. How do the angles change? Look at the sum of BCD and ACD.
Answer:
Yes,
Explanation:
There is no change in the angles after the move position of D.
The sum of the angle ∠BCD + ∠ACD = 180° degrees

Textbook Page No. 132

Some pictures showing two lines crossing each other are given below. One of the four angles so formed is given. Calculate the other three and write them in the pictures.

Kerala Syllabus 6th Standard Maths Solutions Chapter 8 Joining Angles 34
Answer:
The angles three are 135°, 45°, 135° as shown in the below figure,

Explanation:
∠AOB is straight angle is equal to 180°
∠BOD and ∠AOD  are adjacent angles,
the sum of ∠BOD and ∠AOD is 180°
∠BOD = 45°
∠AOD = 180° – 45°
∠AOD = 135°
As opposite angle of a bisecting line at one center point is equal.
∠COB and ∠AOD are equal to 135°
∠BOD and ∠AOC are equal to 45°

Kerala Syllabus 6th Standard Maths Solutions Chapter 8 Joining Angles 35
Answer:
The three angles are 60°, 120°, 60° as shown in the below figure,

Explanation:
∠AOB is straight angle is equal to 180°
∠BOD and ∠AOD  are adjacent angles,
the sum of ∠BOD and ∠AOD is 180°
∠AOD = 120°
∠BOD = 180° – 120°
∠BOD = 60°
as opposite angle of a bisecting line at one center point is equal.
∠COB and ∠AOD are equal to 120°
∠BOD and ∠AOC are equal to 60°

Kerala Syllabus 6th Standard Maths Solutions Chapter 8 Joining Angles 36
Answer:
90 degrees as shown below,

Explanation:
∠AOB is straight angle is equal to 180°
∠BOD and ∠AOD  are adjacent angles and the sum of ∠BOD and ∠AOD is 180° degrees
∠BOD = 90°
∠AOD = 180° – 90°
∠AOD = 90°
as opposite angle of a bisecting line at one center point is equal
∠COB and ∠AOD are equal to 90°
∠BOD and ∠AOC are equal to 90°

Near and opposite

The picture shows the four angles made by the line CD crossing the line AB:
Kerala Syllabus 6th Standard Maths Solutions Chapter 8 Joining Angles 37
We can pair these four angles in various ways. Of these, four are linear pairs. which are they?
Answer:

• ∠APC
• ∠BPC
•∠APD
•∠DPB
These are nearby angles in the picture. What about the other two pairs?
Answer:

•∠APC, ∠BPD are opposite angles
•∠APC, ∠BPC are opposite angles

  • ∠APC, ∠BPD
  • ∠APD, ∠BPC

They are not nearby angles; they are opposite angles. What is the relation between them?

Look at ∠APC and ∠BPD. If we add ∠BPC to any of these, we get 180°. In other words, each of these is ∠BPC subtracted from 180°.
So, ∠APC = ∠BPD.
Similarly, can’t you see that the other pair of opposite angles are also equal?

This we write as a general principle;
The opposite angles formed by two lines crossing each other are equal.

We can combine the general result on nearby and opposite angles.
Of the four angles formed by two lines crossing each other, the sum of the nearby angles is 1800, the opposite angles are equal.

Draw a circle center at a point A. Mark four points B, C, D, E on the circle.
Kerala Syllabus 6th Standard Maths Solutions Chapter 8 Joining Angles 38
Draw the lines BD and CE. Now hide the circle.
Use Angle to mark the four angles in the picture.
Use Move to change of B, C, D, E.
Observe what happens to the opposite angles.

Kerala Syllabus 6th Standard Maths Solutions Chapter 8 Joining Angles 39
In the picture above the sum of the green and red angles is 180° and the sum of the green and blue angles is also 180°. So the red and blue angles are equal. Can you see that the green and yellow angles are equal?
Answer:
Yes, the red and blue angles are equal.
Explanation:
As opposite angles of a bisecting two lines at one point, the opposite angles are equal.
So, the green and yellow angles are equal.
Angles that are opposite to each other when two lines cross, are also known as vertical angles, because the two angles share the same corner.
Opposite angles are also congruent angles, when they are equal or have the same measurement.

Question 1.
Two picture of lines passing through a point are given below. Some of the angles are given. Calculate the other angles marked and write in the figure:
Kerala Syllabus 6th Standard Maths Solutions Chapter 8 Joining Angles 40
Answer:

Explanation:
As A and A’ and X and X’ are straight lines and the straight angle is 180 degrees.
With reference to the center point and bisecting lines of BB’ and CC’ angles are noted.
Same as in the figure XX’ bisects the lines YY’ and ZZ’ at common center the angles are noted. Opposite angles are opposite to each other when two lines cross, are also called vertical angles, because the two angles share the same corner.
Opposite angles are also congruent angles, because they are equal or have the same measurement.

Kerala Syllabus 6th Standard Maths Solutions Chapter 8 Joining Angles

Question 2.
Of the four angles made by two lines crossing each other, one angle is half of another angle. Calculate all four angles.
Answer:
The four angles are 60°, 120°, 60°, 120°.
Explanation:
Linear pair,
∠a + ∠b = 180°.
∠c + ∠d = 180°.
∠a = ∠c
∠b = ∠d
Let ∠a = 2x
∠b = half of ∠a
∠b = x
∠a + ∠b = 180°.
2x + x = 180°.
3x = 180°.
x = 180/3
x = 60°.
∠b = 60°, ∠c = 60°.
∠a = 2x
∠a = 2 x 60
∠a = 120°.
So, the four angles are 60°, 120°, 60°, 120°.

Question 3.
Of the four angles formed by two lines crossing each other, the sum of two angles is 100°. Calculate all four angles.
Answer:
Four angles are 50°, 130°, 50°, 130°
Explanation:
Linear pairs lie on the same line.
Sum of the four angles = 180°.
∠A + ∠C = 100°.
x + x = 100°.
2x = 100°.
x = 100/2 = 50°.
Linear pair,
∠A + ∠B = 180°.
50° + ∠B = 180°.
∠B = 180°- 50°.
∠B = 130°
Linear pair,
∠C + ∠D = 180°.
50° + ∠D = 180°.
∠D = 180°- 50°.
∠D = 130°
So, the four angles are 50°, 130°, 50°, 130°

Kerala Syllabus 6th Standard Maths Solutions Chapter 11 Statistics

You can Download Statistics Questions and Answers, Activity, Notes, Kerala Syllabus 6th Standard Maths Solutions Chapter 11 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 6th Standard Maths Solutions Chapter 11 Statistics

Statistics Text Book Questions and Answers

Bar graphs Textbook Page No. 165

Remember how we drew bar graphs to show numerical information? See the bar graph below.
It shows the amount of money each class in a school donated to the Snehasparsham medical aid fund.
Kerala Syllabus 6th Standard Maths Solutions Chapter 11 Statistics 1

  • What is the total amount donated?
  • Which class gave the most?
  • And the least?

What other things should we get from this table?
Answer:
The other things we get from the data given are What is the average amount denoted? and
How many classes gave the funds?

Explanation:
From the data given above, the other things we get:
What is the average amount denoted?
How many classes gave the funds?

Kerala Syllabus 6th Standard Maths Solutions Chapter 11 Statistics

The table below shows the number of children in each class in a school.
Kerala Syllabus 6th Standard Maths Solutions Chapter 11 Statistics 2

Let’s draw a bar graph of this.
The height of each bar should be according to the number of children. If we take it as 1 centimetre per child, what would be the heights?
So, what length per child would be a convenient scale?
What other things should we decide to draw the graph?

  • The width of a bar
  • Distance between bars
    Kerala Syllabus 6th Standard Maths Solutions Chapter 11 Statistics 3

Now draw a graph in your notebook.
We get the number of children in each class from this graph and we get the amount each class donated from the first graph.

For example, in class VA, there are 25 children and they donated 600 rupees. So how much did each child of this class give on average?
Calculate the average amount for other classes also.

• Which class has the highest average?
Answer:
VIIA class has the highest average.

Explanation:
Number of students of VA = 25.
Total amount donated by VA = 600 rupees.
Amount denoted by each student = Total amount donated by VA ÷ Number of students of VA
= 600 ÷ 25
= 24.
Number of students of VB = 30.
Total amount donated by VB = Amount denoted by each student × Number of students of VB
= 30 × 24
= 720.
Number of students of VIA = 30.
Total amount donated by VIA = Amount denoted by each student × Number of students of VIA
= 30 × 24
= 720.
Number of students of VIB = 20.
Total amount donated by VIB = Amount denoted by each student × Number of students of VIB
= 20 × 24
= 480.
Number of students of VIIA = 40.
Total amount donated by VIIA = Amount denoted by each student × Number of students of VIIA
= 40 × 24
= 960.
Number of students of VIIB = 35.
Total amount donated by VIIB = Amount denoted by each student × Number of students of VIIB
= 35 × 24
= 840.
Total amount donated = 600 + 720 + 480 + 960 + 840
= 3600.
Total number of students = 25 + 30 + 30 + 20 + 40 + 35
= 180.
Average amount donated = Total amount donated ÷ Total number of students
= 3600 ÷ 180
= 20.
Average of VA = 600 ÷ 20 = 30.
Average of VB = 720 ÷ 20 = 36.
Average of VIA = 720 ÷ 20 = 36.
Average of VIB = 480 ÷ 20 = 24.
Average of VIIA = 960 ÷ 20 = 48.
Average of VIIB = 840 ÷ 20 = 42.

• And the lowest?
Answer:
VIB Class the lowest.

Explanation:
Number of students of VA = 25.
Total amount donated by VA = 600 rupees.
Amount denoted by each student = Total amount donated by VA ÷ Number of students of VA
= 600 ÷ 25
= 24.
Number of students of VB = 30.
Total amount donated by VB = Amount denoted by each student × Number of students of VB
= 30 × 24
= 720.
Number of students of VIA = 30.
Total amount donated by VIA = Amount denoted by each student × Number of students of VIA
= 30 × 24
= 720.
Number of students of VIB = 20.
Total amount donated by VIB = Amount denoted by each student × Number of students of VIB
= 20 × 24
= 480.
Number of students of VIIA = 40.
Total amount donated by VIIA = Amount denoted by each student × Number of students of VIIA
= 40 × 24
= 960.
Number of students of VIIB = 35.
Total amount donated by VIIB = Amount denoted by each student × Number of students of VIIB
= 35 × 24
= 840.
Total amount donated = 600 + 720 + 480 + 960 + 840
= 3600.
Total number of students = 25 + 30 + 30 + 20 + 40 + 35
= 180.
Average amount donated = Total amount donated ÷ Total number of students
= 3600 ÷ 180
= 20.
Average of VA = 600 ÷ 20 = 30.
Average of VB = 720 ÷ 20 = 36.
Average of VIA = 720 ÷ 20 = 36.
Average of VIB = 480 ÷ 20 = 24.
Average of VIIA = 960 ÷ 20 = 48.
Average of VIIB = 840 ÷ 20 = 42.

In class VI, 20 children got A grade, 50 got B grade, 20 got C grade, 15 got D grade and 5 got E grade. Draw a bar graph showing this.
Answer:
Kerala-State-Syllabus-6th-Standard-Maths-Solutions-Chapter-11-Statistics-Double bar-Bar graph

Explanation:
Number of students of VI – Y-axis representation.
Types of Grades – X- axis representation.
20 children got A grade,
50 children got B grade,
20 children got C grade,
15 children got D grade,
5 children got E grade.

Kerala Syllabus 6th Standard Maths Solutions Chapter 11 Statistics

Double bar

The bar graph below shows the number of boys and girls present in class 5 of a school, from 1st to 5th of June.
Kerala Syllabus 6th Standard Maths Solutions Chapter 11 Statistics 4
Complete the table, based on this.
Kerala Syllabus 6th Standard Maths Solutions Chapter 11 Statistics 5
Answer:
Kerala-State-Syllabus-6th-Standard-Maths-Solutions-Chapter-11-Statistics-Double bar

Explanation:
1/6/15 – Boys = 23    Girls = 26     Total = 23 + 26 = 49.
2/6/15 – Boys = 26   Girls = 26      Total = 26 + 26 = 52.
3/6/15 – Boys = 25   Girls =  22     Total = 25 + 22 = 47.
4/6/15 – Boys = 24    Girls = 27      Total = 24 + 27 = 51.
5/6/15 – Boys = 28    Girls = 26      Total = 28 + 26 = 54.

• On which day was the least number of children present?
Answer:
On 3/6/15 was the least number of children present.

Explanation:
1/6/15 – Boys = 23    Girls = 26     Total = 23 + 26 = 49.
2/6/15 – Boys = 26   Girls = 26      Total = 26 + 26 = 52.
3/6/15 – Boys = 25   Girls =  22     Total = 25 + 22 = 47.
4/6/15 – Boys = 24    Girls = 27      Total = 24 + 27 = 51.
5/6/15 – Boys = 28    Girls = 26      Total = 28 + 26 = 54.

• On which day was the most number of boys present? And the least?
Answer:
On 5/6/15 was the most number of boys present and 1/6/15 the least.

Explanation:
1/6/15 – Boys = 23    Girls = 26     Total = 23 + 26 = 49.
2/6/15 – Boys = 26   Girls = 26      Total = 26 + 26 = 52.
3/6/15 – Boys = 25   Girls =  22     Total = 25 + 22 = 47.
4/6/15 – Boys = 24    Girls = 27      Total = 24 + 27 = 51.
5/6/15 – Boys = 28    Girls = 26      Total = 28 + 26 = 54.
Most boys – 5/6/15.
Least boys – 1/6/15

• What about girls?
Answer:
On 4/6/15 was the most number of girls present and 3/6/15 the least.

Explanation:
1/6/15 – Boys = 23    Girls = 26     Total = 23 + 26 = 49.
2/6/15 – Boys = 26   Girls = 26      Total = 26 + 26 = 52.
3/6/15 – Boys = 25   Girls =  22     Total = 25 + 22 = 47.
4/6/15 – Boys = 24    Girls = 27      Total = 24 + 27 = 51.
5/6/15 – Boys = 28    Girls = 26      Total = 28 + 26 = 54.
Most girls – 4/6/15.
Least girls – 3/6/15

• On which day was the difference in number between boys and girls the most?
Answer:
1/6/15, 3/6/15 and 4/6/15 these days the difference in number between boys and girls the most.

Explanation:
1/6/15 – Boys = 23    Girls = 26     Total = 23 + 26 = 49.
2/6/15 – Boys = 26   Girls = 26      Total = 26 + 26 = 52.
3/6/15 – Boys = 25   Girls =  22     Total = 25 + 22 = 47.
4/6/15 – Boys = 24    Girls = 27      Total = 24 + 27 = 51.
5/6/15 – Boys = 28    Girls = 26      Total = 28 + 26 = 54.
Total boys and girls = 49 + 52 + 47 + 51 + 54 = 253.
Difference on 1/6/15:
26 – 23 = 3.
Difference on 2/6/15:
26 – 26 = 0.
Difference on 3/6/15:
25 – 22 = 3.
Difference on 4/6/15:
27 – 24 = 3.
Difference on 5/6/15:
28 – 26 = 2.
There is no specific date for the difference in number between boys and girls the most. Major difference of 3 is on three days of 3.

100 grams of rice is taken per child for lunch. How much rice was used each day during this week?
Answer:
Total quantity of rice children nedded for a week = 1,77,100 grams.

Explanation:
Number of grams of rice is taken per child for lunch = 100.
Number of days in a week = 7.
Total number of boys and girls = 49 + 52 + 47 + 51 + 54 = 253.
Number of grams each children eats for lunch = Number of grams of rice is taken per child for lunch × Total number of boys and girls
= 100 × 253
= 25,300 grams.
Total quantity of rice children nedded for a week = Number of grams each children eats for lunch × Number of days in a week
= 25,300 × 7
= 1,77,100 grams.

Textbook Page No. 168

Question 1.
The table shows the number of notebooks sold at the school store during six months.
Kerala Syllabus 6th Standard Maths Solutions Chapter 11 Statistics 6
Draw a bar graph of this.
Answer:
Kerala-State-Syllabus-6th-Standard-Maths-Solutions-Chapter-11-Statistics-Double bar-Textbook Page No. 168-1

Explanation:
June – 140 note books.
July – 130 note books.
August – 150 note books.
September – 160 note books.
October – 120 note books.
November – 150 note books.

Question 2.
The table shows the various expenses of George’s family for the last month.
Kerala Syllabus 6th Standard Maths Solutions Chapter 11 Statistics 7
Draw a bar graphs of this. Write down some facts we get from the graph.
Answer:
Kerala-State-Syllabus-6th-Standard-Maths-Solutions-Chapter-11-Statistics-Double bar-Textbook Page No. 168-2

Explanation:
Less expense the family invests on Travelling.
Most of their expenses is consumption of food.

Question 3.
The table below shows the amount of electricity used at Soumya’s home during last year.
Kerala Syllabus 6th Standard Maths Solutions Chapter 11 Statistics 8

i. How many units were used in all last year?
Answer:
Total number of units used in the last year = 2100 KW.

Explanation:
Number of units used in:
Month = January, February = 340 KW.
Month = March, April = 440 KW.
Month = May, June = 410 KW.
Month = July, August = 290 KW.
Month = September, October = 300 KW.
Month = November, December = 320 KW.
Total number of units used in the last year = 340 KW + 440 KW + 410 KW + 290 KW + 300 KW + 320 KW = 2100 KW.

ii. What is the average use every two months?
Answer:
Average use of January, February = 340 KW ÷ 2 = 170 KW.
Average use of  March, April = 440 KW ÷ 2 = 220 KW.
Average use of May, June = 410 KW ÷ 2 = 205 KW.
Average use of July, August = 290 KW ÷ 2 = 145 KW.
Average use of September, October = 300 KW ÷ 2 = 150 KW.
Average use of November, December = 320 KW ÷ 2 = 160 KW.

Explanation:
Number of units used in:
Month = January, February = 340 KW.
Average use of January, February = 340 KW ÷ 2 = 170 KW.
Month = March, April = 440 KW.
Average use of  March, April = 440 KW ÷ 2 = 220 KW.
Month = May, June = 410 KW.
Average use of May, June = 410 KW ÷ 2 = 205 KW.
Month = July, August = 290 KW.
Average use of July, August = 290 KW ÷ 2 = 145 KW.
Month = September, October = 300 KW.
Average use of September, October = 300 KW ÷ 2 = 150 KW.
Month = November, December = 320 KW.
Average use of November, December = 320 KW ÷ 2 = 160 KW.

iii. During which two months was the use closest to the average?
Answer:
Average use of January, February of 170 KW is the use closest to the average.

Explanation:
Total number of units used in the last year = 2100 KW.
Average use in last year = 2100 KW ÷ 12 = 175.
Average use of January, February = 340 KW ÷ 2 = 170 KW.
Average use of  March, April = 440 KW ÷ 2 = 220 KW.
Average use of May, June = 410 KW ÷ 2 = 205 KW.
Average use of July, August = 290 KW ÷ 2 = 145 KW.
Average use of September, October = 300 KW ÷ 2 = 150 KW.
Average use of November, December = 320 KW ÷ 2 = 160 KW.

Kerala Syllabus 6th Standard Maths Solutions Chapter 11 Statistics

Question 4.
The bar graph shows the percent of those who voted in some wards in a panchayath election.
Kerala Syllabus 6th Standard Maths Solutions Chapter 11 Statistics 10
The table shows the total number of voters.
Kerala Syllabus 6th Standard Maths Solutions Chapter 11 Statistics 9
Calculate the actual number of men and women who voted in each ward.
Answer:
Ward 1: Actual number of total = 558 + 456 = 1,014.
Ward 2: Actual number of total = 632 + 756 = 1,388.
Ward 3: Actual number of total = 640 + 532 = 1,172.
Ward 4: Actual number of total = 680 + 810 = 1,490.
Ward 5: Actual number of total =576 + 666 = 1,242.

Explanation:
Ward 1:
Number of men = 620 × 90%
= 620 × 90 ÷ 100
= 62 × 9
= 558.
Number of Women = 570 × 80%
= 570 × 80 ÷ 100
= 57 × 8
= 456.
Actual number of total = 558 + 456 = 1,014.
Ward 2:
Number of men = 790 × 80%
= 790 × 80 ÷ 100
= 79 × 8
= 632.
Number of Women = 840 × 90%
= 840 × 90 ÷ 100
= 84 × 9
= 756.
Actual number of total = 632 + 756 = 1,388.
Ward 3:
Number of men = 800 × 80%
= 800 × 80 ÷ 100
= 80 × 8
= 640.
Number of Women = 760 × 70%
= 760 × 70 ÷ 100
= 76 × 7
= 532.
Actual number of total = 640 + 532 = 1,172.
Ward 4:
Number of men = 850 × 80%
= 850 × 80 ÷ 100
= 85 × 8
= 680.
Number of Women = 900 × 90%
= 900 × 90 ÷ 100
= 90 × 9
= 810.
Actual number of total = 680 + 810 = 1,490.
Ward 5:
Number of men = 720 × 80%
= 720 × 80 ÷ 100
= 72 × 8
= 576.
Number of Women = 740 × 90%
= 740 × 90 ÷ 100
= 74 × 9
= 666.
Actual number of total =576 + 666 = 1,242.

Kerala Syllabus 6th Standard Maths Solutions Chapter 2 Average

You can Download Average Questions and Answers, Activity, Notes, Kerala Syllabus 6th Standard Maths Solutions Chapter 2 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 6th Standard Maths Solutions Chapter 2 Average

Average Text Book Questions and Answers

Donation math Textbook Page No. 19

Children in class 6A decided to raise 1000 rupees to buy books for the library. There are 40 children in the class. And they decided that all of them should give the same amount. How much should each give?
To compute this, we need only divide 1000 by 40, right?
Kerala Syllabus 6th Standard Maths Solutions Chapter 2 Average 1
There are 30 children in class 6B. They raised 1200 rupees for a medical fund. Can you calculate how much each gave?
Kerala Syllabus 6th Standard Maths Solutions Chapter 2 Average 2
Here, the amounts given may not be the same. So, we cannot say exactly how much each gave.

Yet, we can say something about the amount each gave.
If all had given the same amount, each would have given 40 rupees.
If all had given less than 40 rupees, they could not have raised 1200 rupees.
In the same way, all could not have given more than 40 rupees.
So, we can say something like these:

If all had given the same amount, then each would have given 40 rupees; if some had given less than 40 rupees, then some others must have given more than 40 rupees.
Here we say that the average amount each kid gave is 40 rupees.

Kerala Syllabus 6th Standard Maths Solutions Chapter 2 Average

Average problems

Manikuttan supplies milk to the society each day. Last week, he supplied 56 litres in all. How much did he give each day of this week on average?

The amount given on all days may not be the same. Average per day means how much he would have given each day, had he given the same amount on all days. So, the average in this case is 56 ÷ 7 = 8 litres.

As mentioned at the beginning, this does not mean he supplied exactly 8 litres each day.
It might be 7 litres one day and 9 litres another day. It might be slightly more or less than 8 litres each day.
But it is very much unlikely that he gave only 1 litre one day and 15 litres another day.

On five days, a man spends 300 rupees, 250 rupees, 270 rupees, 280 rupees and 290 rupees. How much did he spend each day on average?
Answer:
Given that: A man had spent 300 rupees, 250 rupees, 270 rupees, 280 rupees and 290 rupees in 5 days.
The average amount spent by the man can be calculated by dividing the total amount he had spent in 5 days.
Total amount=300+250+270+280+290=1390 rupees.
So, the average amount spent each day will be 1390÷5=278 rupees.
This does not mean that he had spent 278 rupees daily. This is the average of his total expenditure.

How much did he spent in all?
Answer:
The total amount spent in five days will be sum of all the amounts spent during this period.
Total amount=300+250+270+280+290=1390 rupees.

And in how many days?
Answer:
The man had spent the total amount,1390 rupees in 5 days.

To get the average expenditure per day, we have to divide the total amount spent by the number of days.
See the length of cloth needed to make shirts for some kids in Sudheer’s class.
Kerala Syllabus 6th Standard Maths Solutions Chapter 2 Average 3
There are 23 boys in the class. How much should be bought to make shirts for all?
How do we compute this?

If the length of cloth is the same for all, we can calculate exactly how much is for 23 kids.
By the table, the total length needed for five is 600 centimetres.
If it is the same amount for all, we can say each needs 120 centimetres.
In other words, the average length needed for each is 120 centimetres.

We can use the average length of 120 centimetre to calculate the total length of cloth needed. But would it be right to cut out a 120 centimetre piece for each?

Since they are all in the same class, there wouldn’t be much difference among the lengths needed for them.
So, we can estimate the total length of cloth needed as 23 × 120 centimetres = 2760 centimetres. or 27 metres and 60 centimetres.

Kerala Syllabus 6th Standard Maths Solutions Chapter 2 Average

Textbook Page No. 22

Question 1.
The number of children who attended class from Monday to Friday are 34, 35, 32, 33, 31. What is the average number of children who attended classes each day?
Answer:
Given that: The number of children attended class from Monday to Friday are 34, 35, 32, 33, 31.
Step 1: The total strength who attended class from Monday to Friday will be 34+35+32+33+31=165
Step 2: The average number of students for this week can be calculated by dividing the total strength by number of days. Therefore, the average will be 165÷5=33.
The number of children attending class may not be the same on all the days. Average per day means how many children would have attended each day.So, this does not mean 33 children attended each day.It might be 30 students attending on monday while 25 on tuesday. It might be slightly more or less than 33 students each day.

Question 2.
The table shows the amount of electricity used in Majeed’s house for some months. What is the average amount of electricity used per month? Which are the months on which usage is more than the average?
Kerala Syllabus 6th Standard Maths Solutions Chapter 2 Average 4
Answer:
Given that: The amount of electricity spent in 5 months.
Step 1: The total amount of electricity spent can be calculated by adding the units spent in all the months.So, it will be 85+90+75+82+78=410 units.
Step 2: The average amount of electricity spent in 5 months can be calculated by dividing the total units spent in 5 months.Therefore, the average amount of electricity used per month will be 410÷5=82 units.
Step 3: In January and February the amount of electricity usage is more than the average.
In April, it is same as of the average units.While in March and May, it is less than the average units.

Question 3.
The weights of players in a team are 68 kilograms, 72 kilograms, 80 kilograms, 70 kilograms, 60 kilograms, 70 kilograms. What is the average weight of a player in the team?
Answer:
Given that: The weights of players in a team are 68 kilograms, 72 kilograms, 80 kilograms, 70 kilograms, 60 kilograms, 70 kilograms.
Step 1:The total weight of all the players will be 68+72+80+70+60+70=420 kilograms.
Step 2: To get the average weight, we need to divide the total weight by number of players.Therefore, it will be 420÷6=70 kilograms.

Question 4.
The total income of a man in 8 days is 1840 rupees. What is his average income per day?
Answer:
GIven that: The total income of a man in 8 days is 1840 rupees.
Step 1: Total income for 8 days is 1840 rupees.
Step 2: To calculate the average income per day, we need to divide the total income by number of days.Therefore, the average income will be 1840÷8=230 rupees.
The income earned may or may not be the same on all the days. It might be slightly more or less than 230 rupees each day or can also be 230 rupees each day.

Kerala Syllabus 6th Standard Maths Solutions Chapter 2 Average

Which is better?

Ouseph and Abu grow different types of coconut trees. Ouseph has 20 trees and Abu has 18. See how many coconuts each got last year:
Kerala Syllabus 6th Standard Maths Solutions Chapter 2 Average 5
Which kind gives more coconuts?
Answer:
Given that: The number of coconuts got in january,april,august and november.
Step 1: The total number of coconuts Ouseph got will be 160+280+200+260=900 coconuts.
Step 2: The total number of coconuts Abu got will be 200+264+240+160=864 coconuts.
By observing the figures above, we can say that Ouseph got more coconuts than Abu.

Can we decide this by comparing just the total numbers each got?
Answer:
Yes, by observing the total number of coconuts, we can decide who got more coconuts.
In this scenario Ouseph got 36 more coconuts than Abu.

So, how do we decide?
Let’s compute the average number of coconuts per tree for each kind.
Kerala Syllabus 6th Standard Maths Solutions Chapter 2 Average 6
How many coconuts per tree did Ouseph got on average?
Answer:
Given that: The total number of coconuts got from 20 Ouseph kind is 900.
Step 1:To calculate the average number of coconuts per tree, we need to divide the total number of coconuts with the number of trees.
Therefore, the average number of coconuts got per tree will be 900÷20=45.

And Abu?
Answer:
Given that: The total number of coconuts got from 18 Abu kind is 864.
Step 1:To calculate the average number of coconuts per tree, we need to divide the total number of coconuts with number of trees.
Therefore, the average number of coconuts got per tree will be 864÷18=48.

Computing like this, we can decide which kind of tree gives better yield.
Answer:
Yes, from the average number of coconuts we can decide which kind of tree gives better yield.
Here Abu kind yields more than the Ouseph.

Question 1.
During the Forest Fest celebration, two divisions of class V decided to plant trees as a Haritha Club activity. 35 children of class VA planted 245 saplings and 30 children of class VB planted 240. On the basis of average number of saplings planted per kid, which division did a better job?
Answer:
Given that: 35 children of class VA planted 245 saplings and 30 children of class VB planted 240.
Step 1: To calculate the average saplings of VA, we need to divide the total number of saplings with the total number of children.
Therefore, the average saplings of VA children will be 245÷35=7.
Step 2:To calculate the average saplings of VB, we need to divide the total number of saplings with the total number of children.
Therefore, the average saplings of VB children will be 240÷30=8.
From the above figures, we can say that VB division did better job by sapling more than VA.

Question 2.
The table shows the number of members and the amount of water used in a month of three households:
Kerala Syllabus 6th Standard Maths Solutions Chapter 2 Average 7
How much water did one person in the first household use on average?
Answer:
Given that: In the first household 18000 litres of water was used in a month by 6 members.
Step 1: To calculate the average usage of water by 1 person, we need to divide the total water used by the members.
Therefore, the average will be 18000÷6=3000 litres.
The water used may not be the same by each family member. It might be slightly more or less than 3000 litres per person in a month.

What about the other households?
Answer:
Given that: In the second household 16000 litres of water was used in a month by 4 members.
Step 1: To calculate the average usage of water by 1 person, we need to divide the total water used by the members.
Therefore, the average will be 16000÷4=4000 litres.
The water used may not be the same by each family member. It might be slightly more or less than 4000 litres per person in a month.

Given that: In the third household 16500 litres of water was used in a month by 5 members.
Step 1: To calculate the average usage of water by 1 person, we need to divide the total water used by the members.
Therefore, the average will be 16500÷5=3300 litres.
The water used may not be the same by each family member. It might be slightly more or less than 3300 litres per person in a month.

According to these figures, in which household used most water per person?
Answer:
The average amount of water used by each member of first household is 3000 litres.
The average amount of water used by each member of second household is 4000 litres.
The average amount of water used by each member of third household is 3300 litres.
From the above figures, we can say that the second household has used most water per person.

Kerala Syllabus 6th Standard Maths Solutions Chapter 2 Average

Some other problems

Milk math

Ramu checked his sale of milk for some days and calculated the average income to be 150 rupees per day. If he continues like this, how much can he expect from the sale of milk in June?
There are 30 days in June. So if we get 150 rupees per day on average during the month, he would get 150 × 30 = 4500 rupees.

Trade math Textbook Page No. 24

The incomes for five days of a trade are 6435 rupees, 6927 rupees, 6855 rupees, 7230 rupees and 6562 rupees. After the sixth day, he calculated the average income as 6500 rupees per day. How much did he get on the sixth day?
The amount got each day for the first 5 days is given. Adding all these, we can get the total income for these 5 days. Since the average income is 6500 rupees per day for the first six days, the total income can be calculated by multiplying by 6. Now can’t we find the income on the sixth day?
Answer:
Given that:The incomes for five days of a trade are 6435 rupees, 6927 rupees, 6855 rupees, 7230 rupees and 6562 rupees and the average income is 6500 rupees per day for the first six days.
Step 1: The total income for first five days will be 6435+6927+6855+7230+6562=34009 rupees.
Step 2: To calculate the total income for six days, we need to multiply the average income with the total number of days.Therefore, the total income for first six days will be 6500×6=39000 rupees.
Step 3: To get the sixth day income, we need to subtract the total income of five days from the total income of six days. So, the sixth day income will be 39000-34009=4991 rupees.

Question 1.
Children were asked to donate books to the school library. Using the given details, fill up the table below.
Kerala Syllabus 6th Standard Maths Solutions Chapter 2 Average 8
Answer:
Kerala-Syllabus-6th-Standard-Maths-Solutions-Chapter-2-Average-8
6B:
To calculate the average of 6B students, we need to divide the number of books by number of children.
Therefore, the average books donated by 6B children will be 240÷40=6.
6C:
To calculate the number of children in 6C, we need to divide the number of books by the average number of books.
Therefore, the average books donated by 6C children will be 175÷5=35.
6D:
To calculate the number of books in 6D, we need to multiply the number of children with the average number of books.
Therefore, the average books donated by 6D children will be 32×10=320.

The sum of 7 consecutive natural numbers is 70. What are the numbers? The sum of 8 consecutive natural number is 92. What are the numbers? Can the sum of 9 consecutive numbers be 58?
Answer:
Case 1:
Given that: The sum of 7 consecutive natural numbers is 70.
Step 1: Let us assume the first natural number be ‘a’.Now, the other six consecutive numbers will be a+1,a+2,a+3,a+4,a+5 and a+6.
Given that the sum is 70. So, a+a+1+a+2+a+3+a+4+a+5+a+6=70
7a+21=70
7a=70-21
7a=49
a=\(\frac{49}{7}\)
a=7
Therefore, the required number is 7 and the other consecutive numbers are 8,9,10,11,12 and 13.

Case 2:
Given that: The sum of 8 consecutive natural numbers is 92.
Step 1: Let us assume the first natural number be ‘a’.Now, the other six consecutive numbers will be a+1,a+2,a+3,a+4,a+5,a+6 and a+7.
Given that the sum is 92. So, a+a+1+a+2+a+3+a+4+a+5+a+6+a+7=92
8a+28=92
8a=92-28
8a=64
a=\(\frac{64}{8}\)
a=8
Therefore, the required number is 8 and the other consecutive numbers are 9,10,11,12,13,14 and 15.

Case 3:
Given that: The sum of 9 consecutive natural numbers is 58.
Step 1: Let us assume the first natural number be ‘a’.Now, the other six consecutive numbers will be a+1,a+2,a+3,a+4,a+5,a+6,a+7 and a+8.
Given that the sum is 92. So, a+a+1+a+2+a+3+a+4+a+5+a+6+a+7+a+8=58
9a+36=58
9a=58-36
9a=22
a=\(\frac{22}{9}\)
a=2.44
The sum of 9 consecutive natural numbers cannot be 58 as the numbers obtained will not give the sum 58.

Question 2.
The average age of a child in a class of 35 is 11. The average age, including the teacher is 12. How old is the teacher?
Answer:
Given that:The average age of a child in a class of 35 is 11.The average age, including the teacher is 12.
Step 1: The number of students in a class are 35 and their average age is 11.So, the age of class will be 35×11=385
Step 2: The age of 35 students and a teacher i.e. 36 members is 12. So, the age of class will be 36×12=432
Step 3: Therefore, the age of a teacher can be calculated by subtracting the students age from total. So, it will 432-385=47.
The teacher is 47 years old.

Question 3.
The average weight of a kid in a group of 10 is 35 kilograms. When Sonu also joined them, the average became 36 kilograms. How much does Sonu weigh?
Answer:
Given that:The average weight of a kid in a group of 10 is 35 kilograms.When Sonu joined, the average of 11 kids became 36.
Step 1: The total weight of 10 kids will be 10×35=350 kilograms.
Step 2: After sonu joined, the total weight of 11 kids will be 11×36=396 kilograms.
Step 3: So, the weight of Sonu can be calculated by subtracting the total weight of 10 kids from the total weight of 11 kids. Therefore, the total weight of sonu is 396-350=46 kilograms.

Question 4.
There are 8 teachers in a school. When a 35 year old teacher was transferred and another teacher joined, the average age ws increased by 2 years. How old is the new teacher?
Answer:
Given that:There are 8 teachers in a school. When a 35 year old teacher was transferred and another teacher joined, the average age ws increased by 2 years.
Explanation:
Step 1:
Let us assume the age of 7 teachers be ‘a’ and a 35 year old teacher was transferred. The total age of 8 teachers will be a+35.
Let us assume the average age of 8 teachers be x.
x = \(\frac{(a+35)}{8}\)
8x=a+35
8x-35=a                                          (1)
Step 2:
After the new teacher joined, the average age was increased by 2 years which is x+2.
The new average age will be \(\frac{a+age of New teacher}{8}\)=x+2
a+New teacher’s age=8(x+2)
a+New teacher’s age=8x+16         (2)
Step 3:
Substitute the value of a from (1) in (2)
8x-35+New teacher’s age=8x+16
New teacher’s age=8x+16-8x+35
New teacher’s age=8x-8x+16+35
New teacher’s age=0+16+35
New teacher’s age=51
Therefore, the age of new teacher will be 51 years old.

Question 5.
The average rainfall per month during 2014 in a place was calculated to be 23 centimetres. The total rainfall there during June, July and August was 150 centimetres.

i) What is the average rainfall per month during these three months?
Answer:
Given that: The total rainfall during June, July and August was 150 centimetres.
Explanation:
The average rainfall during these three months will be 150÷3=50 centimetres.

ii) What was the total rainfall during the entire year of 2014?
Answer:
The total rainfall during the entire year or 12 months will be 23×12=276 centimetres.

iii) What is the average rainfall per month during the other 9 months?
Answer:
Step 1: The total rainfall of 9 months can be calculated by subtracting the total rainfall of 3 months from the total rainfall of 12 months.
Therefore, the total rainfall per month during the other 9 months will be 276-50=226 centimetres.
Step 2: The average rainfall per month during the other 9 months will be 226÷9=75.33 centimetres.

Question 6.
When a person calculated his expenses from Sunday to Thursday, he found the average expenditure to be 400 rupees per day. Including Friday, the average increased to 430 rupees per day. How much did he spent on Friday?
Answer:
Given that:
The average expenditure from Sunday to Thursday is 400 rupees per day.
The total sum for five days will be 400×5=2000 rupees.
The average increased to 430 rupees per day. Therefore, the total sum for six days will be 430×5=2580 rupees.
Therefore, the money he spent on friday will be the amount obtained after subtracting the total five days sum from the total six days sum,which is 2580-2500=580 rupees.
So, he had spent 580 rupees on friday.

Including Saturday, the average decreased to 390 rupees per day. How much did he spend on Saturday?
Answer:
The total expenditure from sunday to saturday will be 390×7=2730 rupees.
The total amount spent on saturday will be 2730-2580=150 rupees.

Kerala Syllabus 6th Standard Maths Solutions Chapter 2 Average

Question 7.
40 children of class VI donated 50 rupees on average to the Mutual Aid Fund. 30 children of class V donated 800 rupees in all. If we consider both classes together, how much did each donate on average?
Answer:
Given that: 40 children of class VI donated 50 rupees on average to the Mutual Aid Fund. 30 children of class V donated 800 rupees in all.
Step 1: 40 students of class VI donated 50 rupees on average. So, the total amount donated will be 50×40=2000 rupees.
Step 2: 30 student of class V donated 800 rupees in all. Therefore, the total amount spent by both class V and class VI students will be 2000+800=2800 rupees.
Step 3: To calculate the average amount of money donated by each student, we need to divide the total sum by the total number of students.
Therefore, the average amount will be 2800÷70=40.
On an average each children donated 40 rupees. If all had given the same amount, then each would have given 40 rupees; if some had given less than 40 rupees, then some others must have given more than 40 rupees.

Question 8.
Three groups of 10 kids. The average weight of a kid in each group is 35 kilograms. One more kid joined each group.

i) The average weight of a kid in the first group is still 35 kilograms.
Answer:
Total number of kids in 3 groups will be 10×3=30 students.
The average weight of 30 kids in each group will be 10×35=350 kilograms.
One more kid joined in first group. So, the total weight of 31 stuudents will be 31×35=385 kilograms.

ii) The average weight of a kid in the second group is now 36 kilograms.
Answer:
One more kid joined in second group. So, the total weight of 31 stuudents will be 31×36=396 kilograms.

iii) The average weight of a kid in the third group is now 34 kilograms.
Answer:
One more kid joined in third group. So, the total weight of 31 stuudents will be 31×34=374 kilograms.

Compute the weight of the new kid in each group.
Answer:
The weight of the new kid in first group will be 385-350=35 kilograms.
The weight of the new kid in second group will be 396-350=46 kilograms.
The weight of the new kid in third group will be 374-350=24 kilograms.

In your class, are the boys or girls taller on average? Calculate the average height, considering all kids in the class. Compare it with the average height of boys and girls.
Answer:
Assumed data:
Height of class VIA girls is 100cm,102cm,96cm,98cm,104cm,75cm,85cm,95cm,97cm,83cm and 115cm.
Height of class VIA boys is 101cm,92cm,99cm,85cm,98cm,80cm,95cm,89cm,121cm,115cm and 125cm.
Explanation:
The average height of class VIA will be the total sum of student’s height divided by the total number of students. Therefore, the average the height of VIA class will be 2150÷20=107.5cm
The average height of class VIA girls will be 1050÷10=105cm
The average height of class VIA boys will be 1100÷10=110cm
From the figures we can say that the average height of boys is more than that of girls.

Write any five consecutive natural numbers and find their sum. Does the middle number have any relation with the sum? What if we take 9 consecutive natural numbers? Does the same relation hold for any odd number of numbers?
What if the number of numbers is even?
How about taking consecutive odd numbers or consecutive even numbers?
Answer:
Assumed data: Let us assume the five consecutive natural numbers be 2,3,4,5 and 6.
Case 1: The sum of the assumed five consecutive natural numbers will be 2+3+4+5+6=20.
The average of the five consecutive natural numbers will be 20÷5=4.
The average of five consecutive natural numbers is equal to the middle number.
Thus, the middle number is equal to the average of the consecutive odd numbers.

Case 2: The sum of the assumed nine consecutive natural numbers will be 1+2+3+4+5+6+7+8+9=45.
The average of the nine consecutive natural numbers will be 45÷9=5.
The average of nine consecutive natural numbers is equal to the middle number.
Thus, the middle number is equal to the average of the consecutive odd numbers.

Case 3: The sum of the assumed four consecutive natural numbers will be 1+2+3+4=10.
The average of the four consecutive natural numbers will be 10÷4=2.5.
In case of consecutive even numbers, the average of four consecutive natural numbers is not equal to the middle number.

Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles

You can Download Angles Questions and Answers, Activity, Notes, Kerala Syllabus 6th Standard Maths Solutions Chapter 1 to help you to revise the complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 6th Standard Maths Solutions Chapter 1 Angles

Angles Text Book Questions and Answers

Angles in a circle Textbook Page No. 7

Remember dividing a circle into equal parts using set squares?
(The lesson, Part Number of the class 5 textbook)
See these pictures:
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles 1
You also know how to divide a circle into equal parts using the corners of the other set square, don’t you?
Look at the angles made at the centre of the circle in each case. If we make the angle larger, does the number of parts increase or decrease?
Answer: If we make the angle larger, the number of parts decreases because the space is getting smaller.

Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles

Measure of an angle

We have seen in class 5 how we can divide a circle into three equal parts, using a set square (The section, Three parts, of the lesson, Part Number)
So using the corners of a set square, we can divide a circle into three, four or six parts.
Can we make five equal parts?
Answer: No, we can’t make five equal parts using set square.

Using corners of a set of squares, we cannot make such an angle at the centre.
We need some other method to draw and measure angles of different sizes. Measure lengths of lines starting with small lengths such as millimetres and centimetres.
In the same way, we use a small angle to measure all other angles. This angle is got by dividing a circle into 360 equal parts.

The measure of this angle is said to be 1 degree, and it is written 1°. An angle twice as large as this is said to have measured 2°, thrice as large to have measured 3° and so on.
There is a device in your geometry box to measure angles of different sizes.
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles 2
It is called a protractor.
See the lines drawn on it?

At the end of each line, we see two numbers, one below the other. Look at the numbers below:
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles 3
These numbers are the degree measure of the angles made by these lines with the bottom line marked 0.
For example, the angle made by this bottom line and the line just above it is of measure at 10° (10 degrees). The angle made by the bottom line and the line marked 40 is of measure at 40°.
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles 4
In other words, an angle of 40° is made by 4 angles of 10° each.
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles 5
Another round of numbers is given on top to draw and measure angles on the left.
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles 6
How can we measure an angle using a protractor?
Look at this picture.
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles 7
Note also how the angle is marked.

Answer: It is marked with degrees from 0 to 180 degrees. It can be directly used to measure any angle from 0 to 360 degrees. The markings are made in two ways, 0 to 180 degrees from the right to left and vice versa.

Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles

Here are some more examples:
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles 8

Now draw a line and at one end, thaw a line straight up using the square comer of a set square (The section, Let’s draw of the lesson, When lines Join of the class 5 textbook).
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles 9
Next, measure this angle using a protractor.
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles 10
So the angle at a square comer is 90° Such an angle is also called a right angle.

In a figure, we mark a right angle like this:
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles 11
Without actually measuring the angles below, can you say which are less than 90°, which are more than 90° and which are exactly 90°?
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles 12
Answer:  In the given question, some figures are given.
Now, we will find the figures which are less than 90°, which are more than 90°.

The above figure angle is Less than 90°.
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles _image(i)
The above figure angle is more than 90°.
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles _image(ii)
The above figure angle is 90°.
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles _image(iii)
The above figure angle is less than 90°.

The above figure angle is 90°.
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles _image(v)
The above figure angle is more than 90°.

Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles

Textbook Page No. 8

Let’s see how we measure an angle in GeoGebra. First mark three points A, B, C. Select the Angle tool and click on B, A, C in this order. (Test what happens if you click in some other order)
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles 13
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles 14
We can get the measure of the angle by clicking on the lines AB. AC also.
To get the measure of the angle shown below, in what order should we click?
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles 15
Answer: To get the measure of the angle by clicking on the lines EF and ED also.

Textbook Page No. 11

Question 1.
Measure all angles below and write names and sizes in degrees below each:
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles 16
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles 17
Answer: Given the figures,
Now, we will find the angles and names.
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles _Protractor(i)
The angle name and size is ∠ABC = 60°.

Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles _Protractor(ii)

The below figure angle name and size is ∠EDF= 40°.

Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles _Protractor(iii)

The below figure angle name and size is ∠GHI= 110°.


The below figure angle name and size is ∠IKJ= 50°.

Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles _Protractor(v)

The below figure angle name and size is ∠MNO= 115°.
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles _Protractor(vi)

The below figure angle name and size is ∠PQR= 30°.

Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles _Protractor(vii)

The below figure angle name and size is ∠YOB= 30°.

Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles _Protractor(viii)

The below figure angle name and size is ∠XYZ= 50°, ∠XYT= 20° and ∠TYZ= 30°.

Question 2.
Measure and write all angles of the figure below:
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles 18
Answer: As given in the question, four figures are given.
Now, we will find the angles of all the angles of figures.
The names of the figures are triangle, square, pentagon and hexagon.
Now, in the triangle, all angles are 60°.
The square all angles are 90°.
In the pentagon, all angles are 108°.
In the Hexagon all angles are 120°.

Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles

Drawing an angle

Let’s see how we can draw an angle like this:
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles 19
Now can you draw this angle?
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles 20

Answer: As given in the question, the figure is given.
Now, we can draw that given figure step by step.
First, draw a line with the names A and B.
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles _Angle(i)
The next step is, to take a protractor. Place the protractor at B point and mark a 40° angle.
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles _Angle(ii)
Now, take a scale. Draw a line from the 40° angle point marked to B. The line name is marked as C.
The final angle figure is below,
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles _Angle(iii)

Question 1.
A rectangle has four angles. What is the degree measure of each?
Answer: The degree measure of each angle of a rectangle is 90°.

Question 2.
Draw a rectangle of length 5 centimetres and breadth 3 centimetres using a ruler and a protractor.

Answer: Given the length and breadth are 5cm and 3cm.
Using the protractor with an angle of 90°, we can draw the rectangle.
Below is the rectangle,

Question 3.
Draw the figures shown below with the specified measures, in your notebook:
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles 21
Answer: Given the figures in the question.
Now, we can draw the given figures in a notebook.
Using Protractor, we can easily draw.
So, the figure is,
(i)
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles _Que(i)

(ii)
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles _Que(ii)

(iii)
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles _Que(iii)

(iv)
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles _Que(iv)

Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles

Drawing angles

Draw a line AB in GeoGebra. Select the Angle with the Giving size tool and click on B, and A in this order. In the dialogue box gives the measure of the angle needed and click OK.
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles 22
We get a new point B’. Join A and B’.

Circle division Textbook Page No. 14

An angle of 1° is got by dividing a circle into 360 equal parts. Putting this the other way round, by drawing 1° degree angles at the centre, we can divide a circle into 360 equal parts.
If we take two of these angles together, we get 2° angles; and 180 equal parts of the circle.
What if take the angles three at a time?
How much would be each angle?
And how many equal parts of the circle?
So, to divide the circle into 30 equal parts, how many of the 360 parts should we take together?

Answer: If we take three of these angles together, we get 3° angles and 120 equal parts of the circle. So, we can divide the circle into 30 equal parts, we get an angle is 12°.

We have divided a circle into equal parts using other corners of set squares. How many equal parts did we get in each case?
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles 23
360 ÷ 4 = 90
We have divided a circle into equal parts using other corners of set squares. How many equal parts did we get in each case?

Answer: Given the circle,
Now, we can divide the circle into equal parts with each angle.
First, if we divide the circle into 4 equal parts, each angle will be 90°.
If we divide the circle into 6 equal parts, each angle will be 60°.
If we divide the circle into 8 equal parts, each angle will be 45°.
If we divide the circle into 9 equal parts, each angle will be 40°.

Look at this picture:
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles 24
Using this corner of the set square, we have divided the circle into 8 equal parts.
So, how much is each angle at the centre?
360 ÷ 8 =45
Thus the angle at this corner of the set square is 45°.
In the same way, the angle at the other non-square corner of this set square is also 45°.

Now find the angles at the corners of the other set square.
Now let’s look at our earlier problem of dividing a circle into five equal parts.
To divide a circle into 5 equal parts, what should be the size of the angle at the centre?
360 ÷ 5 = 72
Answer: The size of the angle at the centre is 72°.

Draw angles of 72° at the centre of a circle:
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles 25
Continuing like this, can’t we divide a circle into five equal parts?
Now can you draw this figure?
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles 26
What is the name of this figure?
Like this, draw figures of 6, 8, 9, 10, and 12 sides inside a circle.

Answer: Given the figure,
The name of the figure is Pentagon.
(i) The figure of 6 sides inside a circle is below,
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles _Fig(i)
The name of the figure is Hexagon.

(ii) The figure of 8 sides inside a circle is below,
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles _Fig(ii)
The name of the figure is Octagon.

(iii) The figure of 9 sides inside a circle is below,
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles _Fig(iii)
The name of the figure is Nonagon.

(iv) The figure of 10 sides inside a circle is below,
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles _Fig(iv)
The name of the figure is Decagon.

(v) The figure of 12 sides inside a circle is below,
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles _Fig(v)

Textbook Page No. 17

Question 1.
Can you draw angles of the sizes given below, using a set square? (See the section, Joining angles, of the lesson, When Lines Join, in the class 5 textbook)
(i) 75°
Answer: Given the angle is 75°
We can draw an angle of 30°+45°.
In the figure shown below,
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles _(i)

(ii) 105°
Answer: Given the angle is 105°
We can draw angles with 60°+45°.
In the figure shown below,
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles _(ii)

(iii) 135°
Answer: Given the angle is 135°
We can draw an angle of 90°+45°.
In the figure shown below,
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles _(iii)

(iv) 15°
Answer: Given the angle is 15°
We can draw an angle of 45°- 30° = 15°
In the figure shown below,
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles _(iv)

Question 2.
In the figures below, calculate the fraction of the whole circle the green and yellow parts are:
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles 27
Answer: Given the figures,
Now, we will find the green parts and yellow parts fractions.
(i) In the first figure, the yellow part is 20° and the total circle angle is 360°
So, the yellow part fraction is, 20°/360° = 1/18
The green part fraction is 340°/360° = 17/18

(ii) In the second figure, the yellow part is 24° and the total circle angle is 360°
So, the yellow part fraction is, 24°/360° = 12/180 = 6/90 = 2/30 = 1/15
The green part fraction is 336°/360° = 14/15

(iii) In the third figure, the yellow part is 54° and the total circle angle is 360°
So, the yellow part fraction is, 54°/360° = 3/20
The green part fraction is 306°/360° = 17/20

(iv) In the fourth figure, the yellow part is 80° and the total circle angle is 360°
So, the yellow part fraction is, 80°/360° = 2/9
The green part fraction is 280°/360° = 7/9

(v) In the fifth figure, the yellow part is 108° and the total circle angle is 360°
So, the yellow part fraction is, 108°/360° = 3/10
The green part fraction is 252°/360° = 7/10

(vi) In the sixth figure, the yellow part is 150° and the total circle angle is 360°
So, the yellow part fraction is, 150°/360° = 5/12
The green part fraction is 210°/360° = 7/12

Question 3.
Draw circles and mark the fractional parts given below. Colour them also:

(i) \(\frac{3}{8}\)

Answer: Given the value is \(\frac{3}{8}\)
Now, we will draw the circle and mark the fractional parts.
We know that the total angle of a circle is 360°.
In a given question, the circle total parts are 8.
So, 360°/8 = 45°
Now, we will fill the colour up to 45°.
The circle with colour is as shown below,
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles _Problem 3(i)

(ii) \(\frac{2}{5}\)

Answer: Given the value is \(\frac{2}{5}\)
Now, we will draw the circle and mark the fractional parts.
We know that the total angle of a circle is 360°.
In a given question, the circle total parts are 5.
So, 360°/5 = 72°
Now, we will fill the colour up to 72°.
The circle with colour is as shown below,
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles _Problem 3(ii)

(iii) \(\frac{4}{9}\)

Answer: Given the value is \(\frac{4}{9}\)
Now, we will draw the circle and mark the fractional parts.
We know that the total angle of a circle is 360°.
In a given question, the circle total parts are 9.
So, 360°/9 = 40°
Now, we will fill the colour up to 40°.
The circle with colour is as shown below,
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles _Problem 3(iii)

(iv) \(\frac{5}{12}\)

Answer: Given the value is \(\frac{5}{12}\)
Now, we will draw the circle and mark the fractional parts.
We know that the total angle of a circle is 360°.
In a given question, the circle total parts are 12.
So, 360°/12 = 30°
Now, we will fill the colour up to 30°.
The circle with colour is as shown below,
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles _Problem 3(iv)

(v) \(\frac{5}{24}\)

Answer: Given the value is \(\frac{5}{24}\)
Now, we will draw the circle and mark the fractional parts.
We know that the total angle of a circle is 360°.
In a given question, the circle total parts are 24.
So, 360°/24 = 15°
Now, we will fill the colour up to 15°.
The circle with colour is as shown below,
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles _Problem 3(v)

Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles

Angles in a clock

The hands of a clock make different angles at different times: What is the angle between them at 3 o’clock? And at 9 o’clock? The hour hand rotates through 360° in 12 hours. So it rotates through 360° ÷ 12 = 30° in an hour. So at 1 o’clock, the angle between the hands is 30°. What is the angle at 2 o’clock?
And at 4 0’ clock?
Kerala Syllabus 6th Standard Maths Solutions Chapter 1 Angles 28

Answer: Given the clock figure,
Now, we will find the angle at 2 o’clock and 4 o’clock.
It rotates 30° in an hour. So, the angle between the hands is 30°.
So, the angle at 2 o’clock is 30°+30° = 60°.
Next, the angle at 4 o’clock is 30°+30°+30°+30° = 120°.