Kerala Syllabus 6th Standard Basic Science Solutions Guide

Expert Teachers at HSSLive.Guru has created Kerala Syllabus 6th Standard Basic Science Solutions Guide Pdf Free Download in both English Medium and Malayalam Medium of Chapter wise Questions and Answers, Notes are part of Kerala Syllabus 6th Standard Textbooks Solutions. Here HSSLive.Guru has given SCERT Kerala State Board Syllabus 6th Standard Basic Science Textbooks Solutions Pdf of Kerala Class 6 Part 1 and Part 2.

Kerala State Syllabus 6th Standard Basic Science Textbooks Solutions

Kerala Syllabus 6th Standard Basic Science Guide

Kerala State Syllabus 6th Standard Basic Science Textbooks Solutions Part 1

  • Chapter 1 Caskets of Life
  • Chapter 2 The Essence of Change
  • Chapter 3 Flower to Flower
  • Chapter 4 Along with Motion
  • Chapter 5 Food for Health

Kerala State Syllabus 6th Standard Basic Science Textbooks Solutions Part 2

  • Chapter 6 Living in Harmony
  • Chapter 7 Attraction and Repulsion
  • Chapter 8 Moon and Stars
  • Chapter 9 Mix and Separate
  • Chapter 10 For Shape and Strength

We hope the given Kerala Syllabus 6th Standard Basic Science Solutions Guide Pdf Free Download in both English Medium and Malayalam Medium of Chapter wise Questions and Answers, Notes will help you. If you have any queries regarding SCERT Kerala State Board Syllabus Class 6th Basic Science Textbooks Answers Guide Pdf of Part 1 and Part 2, drop a comment below and we will get back to you at the earliest.

Kerala Syllabus 6th Standard Social Science Solutions Guide

Expert Teachers at HSSLive.Guru has created Kerala Syllabus 6th Standard Social Science Solutions Guide Pdf Free Download in both English Medium and Malayalam Medium of Chapter wise Questions and Answers, Notes are part of Kerala Syllabus 6th Standard Textbooks Solutions. Here HSSLive.Guru has given SCERT Kerala State Board Syllabus 6th Standard Social Science Textbooks Solutions Pdf of Kerala Class 6 Part 1 and Part 2.

Kerala State Syllabus 6th Standard Social Science Textbooks Solutions

Kerala Syllabus 6th Standard Social Science Guide

Kerala State Syllabus 6th Standard Social Science Textbooks Solutions Part 1

  • Chapter 1 Medieval India: The Centres of Power
  • Chapter 2 Medieval India: Society, Resource, and Trade
  • Chapter 3 Kerala: The Land, the Rain, and the People
  • Chapter 4 Production Process
  • Chapter 5 The Earth: Myth and Reality
  • Chapter 6 World of Diversities

Kerala State Syllabus 6th Standard Social Science Textbooks Solutions Part 2

  • Chapter 7 Medieval India: Art and Literature
  • Chapter 8 Medieval World
  • Chapter 9 Medieval Kerala
  • Chapter 10 Democracy and Rights
  • Chapter 11 Diversity ¡n Social Life
  • Chapter 12 Gift of Nature

We hope the given Kerala Syllabus 6th Standard Social Science Solutions Guide Pdf Free Download in both English Medium and Malayalam Medium of Chapter wise Questions and Answers, Notes will help you. If you have any queries regarding SCERT Kerala State Board Syllabus Class 6th Social Science Textbooks Answers Guide Pdf of Part 1 and Part 2, drop a comment below and we will get back to you at the earliest.

Kerala Syllabus 6th Standard English Solutions Guide

Expert Teachers at HSSLive.Guru has created Kerala Syllabus 6th Standard English Solutions Guide Pdf Free Download of Textbook Questions and Answers, Chapter Wise Notes, English Textbook Activity Answers, Chapters Summary in Malayalam, English Study Material, Kerala Reader English Book Answers, English Teachers Hand Book are part of Kerala Syllabus 6th Standard Textbooks Solutions. Here HSSLive.Guru has given SCERT Kerala State Board Syllabus 6th Standard English Textbooks Solutions Pdf of Kerala Class 6 Part 1 and Part 2.

Kerala State Syllabus 6th Standard English Textbooks Solutions

Kerala Syllabus 6th Standard English Guide

Unit 1 Rain of Love

  • Chapter 1 Life with Grandfather (Kesavan Sankara Pillai)
  • Chapter 2 The Little Boy and The Old Man (Sheldon Allan)
  • Chapter 3 Making a Mango Pickle (Bibhuti Bhushan Bandopadhyay)

Unit 2 Still We Rise

  • Chapter 1 Cinderella
  • Chapter 2 Woman Work (Maya Angelou)
  • Chapter 3 One Child, One Teacher, One Pen and One Book can Change the World (Malala Yousafzai)

Unit 3 Glimpses of Nature

  • Chapter 1 The Mountain and The Squirrel (Ralph Waldo Emerson)
  • Chapter 2 The Rightful Inheritors of the Earth (Vaikom Muhammad Basheer)
  • Chapter 3 The Book of Nature (Jawaharlal Nehru)

Unit 4 Work is Worship

  • Chapter 1 The Grain as Big as a Hen’s Egg (Leo Tolstoy)
  • Chapter 2 Clever Carla (Robert Scotellaro)
  • Chapter 3 The Ploughman (Oliver Wendell Holmes)

Unit 5 Helping Hands

  • Chapter 1 The Champ (Girija Rani Asthana)
  • Chapter 2 A Glass of Milk
  • Chapter 3 If I Can Stop One Heart from Breaking (Emily Dickinson)

We hope the given Kerala Syllabus 6th Standard English Solutions Guide Pdf Free Download of Textbook Questions and Answers, Chapter Wise Notes, English Textbook Activity Answers, Chapters Summary in Malayalam, English Study Material, Kerala Reader English Book Answers, English Teachers Hand Book will help you. If you have any queries regarding SCERT Kerala State Board Syllabus Class 6th English Textbooks Answers Guide Pdf of Part 1 and Part 2, drop a comment below and we will get back to you at the earliest.

Kerala Syllabus 6th Standard Hindi Solutions Guide

Expert Teachers at HSSLive.Guru has created Kerala Syllabus 6th Standard Hindi Solutions Guide Pdf Free Download of Textbook Questions and Answers, Chapter Wise Notes, Activity Answers, Chapters Summary in Malayalam, Hindi Study Material, Teachers Hand Book are part of Kerala Syllabus 6th Standard Textbooks Solutions. Here HSSLive.Guru has given SCERT Kerala State Board Syllabus 6th Standard Hindi Textbooks Solutions Pdf of Kerala Class 6 Part 1 and Part 2.

Kerala State Syllabus 6th Standard Hindi Textbooks Solutions

Kerala Syllabus 6th Standard Hindi Guide

इकाई 1

  • Chapter 1 बादल दानी (बालगीत)
  • Chapter 2 बारिश कैसे मिली (चित्र कहानी)

इकाई 2

  • Chapter 1 भारत देश हमारा (लेख)
  • Chapter 2 जगमग तारा (कविता)
  • Chapter 3 हे मेरे वतन के लोगो (देशभक्ति गान)

इकाई 3

  • Chapter 1 इंसानियत की मिसाल (समाचार)
  • Chapter 2 दो भाई (कहानी)

इकाई 4

  • Chapter 1 नन्हा उल्लू (कहानी)
  • Chapter 2 बबुआ और बूढ़ा (कविता)

इकाई 5

  • Chapter 1 ज़मीं को जादू आता है (कविता)
  • Chapter 2 मार्च (संस्मरणात्मक लेख)
  • Chapter 3 धूप की संदूक (कविता)

We hope the given Kerala Syllabus 6th Standard Hindi Solutions Guide Pdf Free Download of Textbook Questions and Answers, Chapter Wise Notes, Activity Answers, Chapters Summary in Malayalam, Hindi Study Material, Teachers Hand Book will help you. If you have any queries regarding SCERT Kerala State Board Syllabus Class 6th Hindi Textbooks Answers Guide Pdf of Part 1 and Part 2, drop a comment below and we will get back to you at the earliest.

Kerala Syllabus 6th Standard Textbooks Solutions Guide

Expert Teachers at HSSLive.Guru has created Kerala Syllabus 6th Standard Textbooks Solutions Guide Pdf Free Download all Subjects in both English Medium and Malayalam Medium of Chapter wise Questions and Answers, Notes are part of Kerala Class 6 Solutions. Here HSSLive.Guru has given SCERT Kerala State Board Syllabus 6th Standard Textbooks Solutions Pdf Part 1 and Part 2.

Kerala State Syllabus 6th Standard Textbooks Solutions

We hope the given Kerala Syllabus 6th Standard Textbooks Solutions Guide Pdf Free Download all Subjects in both English Medium and Malayalam Medium of Chapter wise Questions and Answers, Notes will help you. If you have any queries regarding SCERT Kerala State Board Syllabus Class 6th Textbooks Answers Guide Pdf of Part 1 and Part 2, drop a comment below and we will get back to you at the earliest.

Kerala Syllabus 6th Standard Maths Solutions Guide

Expert Teachers at HSSLive.Guru has created Kerala Syllabus 6th Standard Maths Solutions Guide Pdf Free Download in English Medium and Malayalam Medium of Chapter wise Questions and Answers, Notes are part of Kerala Syllabus 6th Standard Textbooks Solutions. Here HSSLive.Guru has given SCERT Kerala State Board Syllabus 6th Standard Maths Textbooks Solutions Pdf of Kerala Class 6 Part 1 and Part 2.

Kerala State Syllabus 6th Standard Maths Textbooks Solutions

Kerala Syllabus 6th Standard Maths Guide

Kerala State Syllabus 6th Standard Maths Textbooks Solutions Part 1

Kerala State Syllabus 6th Standard Maths Textbooks Solutions Part 2

We hope the given Kerala Syllabus 6th Standard Maths Solutions Guide Pdf Free Download in both English Medium and Malayalam Medium of Chapter wise Questions and Answers, Notes will help you. If you have any queries regarding SCERT Kerala State Board Syllabus Class 6th Maths Textbooks Answers Guide Pdf of Part 1 and Part 2, drop a comment below and we will get back to you at the earliest.

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 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

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