Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions

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Kerala State Syllabus 6th Standard Maths Solutions Chapter 3 Fractions

Fractions Text Book Questions and Answers

Multiplication and times Questio

A bottle can hold 250 millilitres. How much water do we need to fill three such bottles?
3 × 250 millilitres = 750 millilitres
We can say this in words
750 millilitres is 3 time as much as 250 millilitres.
Using only numbers, we have
3 times as much as 250 = 3 × 250 = 750
A packet can hold 500 grams of sugar. How much sugar do we need to fill four packets?
4 × 500 grams = 2000 grams
How about stating this as we did earlier?
4 times as much as 500 grams is 2000 grams.
And using only numbers?
4 times as much as 500 = 4 × 500 = 2000
2000 grams means 2 kilograms, right?
And 500 grams is \(\frac{1}{2}\) kilograms. So,
4 times as much as \(\frac{1}{2}\) kilogram is 2 kilograms
Using only numbers
4 times as much as \(\frac{1}{2}\) is 2.

Just as we write 4 × 500 for 4 times 500, we can write 4 times \(\frac{1}{2}\) as
4 × \(\frac{1}{2}\)
That is,
4 × \(\frac{1}{2}\) = 4 times \(\frac{1}{2}\) = 2
Let’s say our water problem in litres instead of millilitres.
250 millilitres means a quarter litre; thrice a quarter makes a three-quarter. So,
3 times \(\frac{1}{4}\) litre is \(\frac{3}{4}\) litre
Using only numbers,
3 time \(\frac{1}{4}\) is \(\frac{3}{4}\)
How about writing this as multiplication’?
3 × \(\frac{1}{4}\) = 3 times \(\frac{1}{4}\) = \(\frac{3}{4}\)

Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions

What part?

In the picture, what fraction of the triangle is colored red?
Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions 9
The large triangle is divided into how many small triangles?
Of these, how many are coloured red?
So the red coloured part is \(\frac{18}{25}\) of the large triangle.

Let’s look at it this way. The red coloured portion is made up of 3 equal parts; and each of these has 6 small triangles. So the part coloured red is
3 × \(\frac{6}{25}\) = \(\frac{18}{25}\)

Another problem:
If five strings \(\frac{1}{4}\) metre long are laid end to end, what would be the total length?
Four quarter metres make one metre.
One more quarter makes one and a quarter metre.
Let’s say this in words and as multiplication
5 times \(\frac{1}{4}\) is, 1\(\frac{1}{4}\)
And as multiplication?
5 × \(\frac{1}{4}\) = 1\(\frac{1}{4}\)

Textbook Page No. 28

Like this, find the answer to each of the problems below and then write it in words and as multiplication of numbers

Question 1.
i) What is the total weight of two pieces of pumpkin, each weighing 250 grams?
Answer:
Total weight of two pieces of pumpkin = 500 grams.

Explanation:
Weight of each pumpkin = 250 grams.
Number of pumpkins = 2.
Total weight of two pieces of pumpkin = Weight of each pumpkin × Number of pumpkins
= 250 × 2
= 500 grams.

ii) What if the weight is put in kilograms?
Answer:
Total weight of two pieces of pumpkin = \(\frac{1}{2}\) kg.

Explanation:
Total weight of two pieces of pumpkin = 500 grams.
Conversion:
1 kilogram = 1,000 grams.
?? kg = 500 g
=> 1 × 500 = 1000 × ??
=> 500 ÷ 1000 = ??
=> 1 ÷ 2 = ??
=> \(\frac{1}{2}\) = ?? kg

Question 2.
i) What is the total length of four pieces of ribbon, each of length 75 centimetres.
Answer:
Total length of four pieces of ribbon = 300 cm.

Explanation:
Number of ribbons = 4.
Length of each ribbon = 75 cm.
Total length of four pieces of ribbon = Number of ribbons × Length of each ribbon
= 4 × 75
= 300 cm.

ii) What if the length is put in metres?
Answer:
Total length of four pieces of ribbon = 30,000 m.

Explanation:
Total length of four pieces of ribbon = 300 cm.
Conversion:
1 cm = 100 m.
300 cm = ?? m
=> 1 × ?? = 100 × 300
=> ?? = 30,000 m.

Question 3.
(i) One cup can hold \(\frac{1}{3}\) litre of milk. How much milk can we pour in two cups?
Answer:
Number of liters milk can we pour in two cups = \(\frac{2}{3}\)

Explanation:
Number of liters of milk a cup can hold =  \(\frac{1}{3}\)
Number of cups = 2.
Number of liters milk can we pour in two cups = Number of liters of milk a cup can hold × Number of cups
=   \(\frac{1}{3}\) × 2
=   \(\frac{2}{3}\)

(ii) In four cups?
Answer:
Number of liters milk can we pour in two cups = \(\frac{4}{3}\)

Explanation:
Number of liters of milk a cup can hold =  \(\frac{1}{3}\)
Number of cups = 4.
Number of liters milk can we pour in two cups = Number of liters of milk a cup can hold × Number of cups
=   \(\frac{1}{3}\) × 4
=   \(\frac{4}{3}\)

Multiplication and part Textbook Page No. 29

A six metre long string is cut into two equal pieces. How long is each peace?
Half of six metres is three metres.
Half is written \(\frac{1}{2}\). So,
\(\frac{1}{2}\) of six metres is 3 metres.
Using only numbers, we can say
\(\frac{1}{3}\) of 6 is 3
Part of numbers is also written as multiplication; that is
\(\frac{1}{2}\) × 6 = \(\frac{1}{2}\) of 6 = 3

What about cutting a 2 metre long ribbon into three equal parts?
The length of each piece is \(\frac{2}{3}\) metre
(The section, Measuring parts, of the lesson, Part Number, in the class 5 textbook).
Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions 10

That is,
\(\frac{1}{3}\) of 2 is \(\frac{2}{3}\)
This also we write as multiplication
\(\frac{1}{3}\) × 2 = \(\frac{1}{3}\) of 2 = \(\frac{2}{3}\)
What is a quarter of five kilograms?
A quarter of four kilograms is one kilogram; a quarter of the remaining one kilogram is a quarter kilogram. One
and a quarter kilogram in all.
That is,
\(\frac{1}{4}\) of 5 kilogram is 1\(\frac{1}{4}\) kilogram.
Writing this as multiplication,
\(\frac{1}{4}\) × 5 = \(\frac{1}{4}\) of 5 = 1\(\frac{1}{4}\)

Different ways Textbook Page No. 30

If three litres of milk is equally divided among four persons, how much would each get?
One-fourth of three litres, which is three quarters of a litre.
We can also think like this.
If one litre is divided among four, each gets a quarter litre.
With three litres, this division can be done thrice. So, each gets three times a quarter litre, which is three quarters of a litre.
Thus a quarter of three litres and three times a quarter litre are equal.
As multiplication.
\(\frac{1}{4}\) × 3 = 3 × \(\frac{1}{4}\)

Textbook Page No. 30

Like this, find the answer to each of the problem below and then write it in words and as multiplication of numbers.

Question 1.
(i) Nine litres of milk is equally shared by four kids. How much does each get?
Answer:
Number of liters of milk each kid gets = \(\frac{9}{4}\) or 2\(\frac{1}{4}\)

Explanation:
Number of kids = 4.
Number of liters of milk shared = 9.
Number of liters of milk each kid gets = Number of liters of milk shared ÷ Number of kids
= 9 ÷ 4
= \(\frac{9}{4}\) or 2\(\frac{1}{4}\)

(ii) What if it were shared by three?
Answer:
Number of liters of milk each kid gets = 3.

Explanation:
Number of kids = 3.
Number of liters of milk shared = 9.
Number of liters of milk each kid gets = Number of liters of milk shared ÷ Number of kids
= 9 ÷ 3
= 3.

Question 2.
(i) Six kilograms of rice is packed into four identical bags. How much rice is in each bag?
Answer:
Number of kilograms each rice bag =  \(\frac{3}{2}\) or 1\(\frac{1}{2}\)

Explanation:
Number of kilograms rice = 6.
Number of identical bags the rice is packed = 4.
Number of kilograms each rice bag =  Number of kilograms rice ÷ Number of identical bags the rice is packed
= 6 ÷ 4
= 3 ÷ 2
= \(\frac{3}{2}\) or 1\(\frac{1}{2}\)

(ii) What if it were packed into two bags?
Answer:
Numb\(\frac{1}{2}\)er of kilograms each rice bag has = 3.

Explanation:
Number of kilograms rice = 6.
Number of identical bags the rice is packed = 2.
Number of kilograms each rice bag has =  Number of kilograms rice ÷ Number of identical bags the rice is packed
= 6 ÷ 2
= 3.

Question 3.
(i) An eight metre long string is cut into three equal parts. What is the length of each piece?
Answer:
Length of each piece = \(\frac{8}{3}\) or 2\(\frac{2}{3}\)

Explanation:
Length of the string = 8 m.
Number of parts its cut = 3.
Length of each piece = Length of the string ÷ Number of parts its cut
= 8 ÷ 3
= \(\frac{8}{3}\) or 2\(\frac{2}{3}\)

(ii) What if it were cut into six equal parts?
Answer:
Length of each piece =  \(\frac{4}{3}\) or 1\(\frac{1}{3}\)

Explanation:
Length of the string = 8 m.
Number of parts its cut = 6.
Length of each piece = Length of the string ÷ Number of parts its cut
= 8 ÷ 6
= 4 ÷ 3
= \(\frac{4}{3}\) or 1\(\frac{1}{3}\)

Question 4.
(i) A rectangle of area seven square centimetres is cut into three equal rectangles. What is the area of each?
Answer:
Area of each rectangle = \(\frac{7}{3}\) or 2\(\frac{1}{3}\)

Explanation:
Area of rectangle = 7 square cm.
Number of pieces its cut = 3.
Area of each rectangle = Area of rectangle ÷ Number of pieces its cut
= 7 ÷ 3
= \(\frac{7}{3}\) or 2\(\frac{1}{3}\)

(ii) What if it were cut into four?
Answer:
Area of each rectangle = \(\frac{7}{4}\) or 1\(\frac{3}{4}\)

Explanation:
Area of rectangle = 7 square cm.
Number of pieces its cut = 4.
Area of each rectangle = Area of rectangle ÷ Number of pieces its cut
= 7 ÷ 4
= \(\frac{7}{4}\) or 1\(\frac{3}{4}\)

Question 5.
(i) Twelve children are divided into four equal groups. How many children are there in each group?
Answer:
Number of children in each group = 3.

nation:
Number of  children = 12.
Number of groups they divided = 4.
Number of children in each group = Number of children ÷ Number of groups they divided
= 12 ÷ 4
= 3.

(ii) What if they were divided into three groups?
Answer:
Number of children in each group = 4.

Explanation:
Number of children = 12.
Number of groups they divided = 3.
Number of children in each group = Number of children ÷ Number of groups they divided
= 12 ÷ 3
= 4.

Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions

Multiplication method Textbook Page No. 31

If 4 strings of length \(\frac{1}{3}\) metre were laid end to end, what would be the total length?
3 strings of \(\frac{1}{3}\) metre make I metre; one more string makes 1\(\frac{1}{3}\) metres.
Thus, 4 times \(\frac{1}{3}\) metre is 1\(\frac{1}{3}\) metres.
Using only numbers, 4 times \(\frac{1}{3}\) is \(\frac{1}{3}\).
As multiplication,
4 × \(\frac{1}{3}\) = 1\(\frac{1}{3}\)
We can also think like this: 4 times \(\frac{1}{3}\) metre means,
4 pieces of \(\frac{1}{3}\) metres.
\(\frac{1}{3}\) + \(\frac{1}{3}\) + \(\frac{1}{3}\) + \(\frac{1}{3}\) = \(\frac{1+1+1+1}{3}\) = \(\frac{4}{3}\) = 1\(\frac{1}{3}\)

Similarly, how do we calculate 4 times \(\frac{2}{3}\)?
\(\frac{2}{3}\) + \(\frac{2}{3}\) + \(\frac{2}{3}\) + \(\frac{2}{3}\) = \(\frac{4 \times 2}{3}\) = \(\frac{8}{3}\) = 2\(\frac{2}{3}\)
How do we calculate 10 times \(\frac{2}{3}\) like this?
\(\frac{2}{3}\) × 10 = \(\frac{2 \times 10}{3}\) = \(\frac{20}{3}\) = 6\(\frac{2}{3}\)

Now look at this problem:
A bottle can hold \(\frac{3}{4}\) litre of milk. How many litres of milk is there in 7 such bottles?
We want to calculate 7 times \(\frac{3}{4}\).
7 × \(\frac{3}{4}\) = \(\frac{3 \times 7}{4}\) = \(\frac{21}{4}\)
How do we split \(\frac{21}{4}\)?
We divide 21 by 4 and write like this:
21 = (5 × 4) + 1 So,
\(\frac{21}{4}\) = \(\frac{(5 \times 4)+1}{4}\) = \(\frac{5 \times 4}{4}\) + \(\frac{1}{4}\) = 5 + \(\frac{1}{4}\) = 5\(\frac{1}{4}\)
Thus 7 bottles contain 5\(\frac{1}{4}\) litres in all.

Textbook Page No. 32

Question 1.
An iron block weighs \(\frac{1}{4}\) kilogram.

(i) What is the weight of 15 such blocks?
Answer:
Weight of 15 iron blocks = \(\frac{15}{4}\)  or 3\(\frac{3}{4}\) kilograms.

Explanation:
Weight of one iron block = \(\frac{1}{4}\) kilogram.
Number of iron blocks = 15.
Weight of 15 iron blocks = Weight of one iron block × Number of iron blocks
= \(\frac{1}{4}\) × 15
= \(\frac{15}{4}\)  or 3\(\frac{3}{4}\) kilograms.

(ii) What about 16 blocks?
Answer:
Weight of 16 iron blocks =  4 kilograms.

Explanation:
Weight of one iron block = \(\frac{1}{4}\) kilogram.
Number of iron blocks = 16.
Weight of 16 iron blocks = Weight of one iron block × Number of iron blocks
= \(\frac{1}{4}\) × 16
= \(\frac{1}{1}\)  × 4
= 4 kilograms.

Question 2.
Each of some iron rods, of length 2 meters is cut into five equal pieces.
(i) What is the length of each piece?
Answer:
Length of each piece of iron rod = \(\frac{2}{5}\) meters.

Explanation:
Length of each iron rod = 2 m.
Number of pieces the rod is cut = 5.
Length of each piece of iron rod = Length of each iron rod ÷ Number of pieces the rod is cut
= 2 ÷ 5
= \(\frac{2}{5}\) meters.

(ii) What is the total length of 4 such pieces?
Answer:
Length of each piece of iron rod = \(\frac{1}{2}\) meters.

Explanation:
Length of each iron rod = 2 m.
Number of pieces the rod is cut = 4.
Length of each piece of iron rod = Length of each iron rod ÷ Number of pieces the rod is cut
= 2 ÷ 4
= \(\frac{1}{2}\) meters.

(iii) What about 10 such pieces?
Answer:
Length of each piece of iron rod =  \(\frac{1}{5}\) meters.

Explanation:
Length of each iron rod = 2 m.
Number of pieces the rod is cut = 10.
Length of each piece of iron rod = Length of each iron rod ÷ Number of pieces the rod is cut
= 2 ÷ 10
= \(\frac{1}{5}\) meters.

Question 3.
There are some cans, each containing 5 litres of milk. The milk in each vessel is used to fill 6 identical bottles.
(i) How much milk is there in each bottle?
Answer:
Number of liters of milk in each bottle = \(\frac{5}{6}\).

Explanation:
Number of liters each can contains = 5.
Number of identical bottles the milk is used = 6.
Number of liters of milk in each bottle = Number of liters each can contains ÷ Number of identical bottles the milk is used
= 5 ÷ 6
= \(\frac{5}{6}\)

(ii) How much milk in 3 such bottles?
Answer:
Number of liters of milk in 3 bottle = \(\frac{5}{2}\) or 2\(\frac{1}{2}\)

Explanation:
Number of liters each can contains = 5.
Number of bottles = 3.
Number of liters of milk in each bottle = \(\frac{5}{6}\).
Number of liters of milk in 3 bottle = Number of bottles × Number of liters of milk in each bottle
= 3 × \(\frac{5}{6}\)
= 1 × \(\frac{5}{2}\)
= \(\frac{5}{2}\) or 2\(\frac{1}{2}\)

(iii) In 12 bottles?
Answer:
Number of liters of milk in 12 bottles = 10.

Explanation:
Number of liters each can contains = 5.
Number of bottles = 12.
Number of liters of milk in each bottle = \(\frac{5}{6}\).
Number of liters of milk in 12 bottles = Number of bottles × Number of liters of milk in each bottle
= 12 × \(\frac{5}{6}\)
= 2 × \(\frac{5}{1}\)
= 10.

Part of part Textbook Page No. 33

Suhara has a 1 metre long silk ribbon. She gave half of it to Soumya. She in term gave half of this to Reena. What is the length of the piece Reena got?
Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions 11
Suhara got half of a metre. What is half of this?
Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions 12
If both halves are halved, we can see this quickly:
Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions 13

Reena got \(\frac{1}{4}\) metre. That is, half of half is a quarter.
How about writing this as multiplication?
\(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{1}{4}\)
Like this, if a metre is cut into three equal pieces, then each piece is \(\frac{1}{3}\) metre.
Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions 14

What is half of \(\frac{1}{3}\) metre’?
Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions 15
The picture shows four pieces; but they are not of the same length. To get equal pieces, let’s divide the other two one-third metres also into two equal parts:
Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions 16
Now we have six equal pieces. And we need the length of one piece. It is \(\frac{1}{6}\) metre, right’? So, half of a third is a sixth.

As multiplication,
\(\frac{1}{2}\) × \(\frac{1}{3}\) = \(\frac{1}{6}\)
Like this, what \(\frac{1}{4}\) of \(\frac{1}{3}\) metre?
Let’s think without any picture.
To get \(\frac{1}{3}\) metre, one metre must be divided into 3 equal parts. And we want \(\frac{1}{4}\) of one such part.

To get equal parts, we must divide each of the first three parts into how many equal parts?
So how many parts in all?
What is the length of one part?

So what is \(\frac{1}{4}\) of \(\frac{1}{3}\) ?

As multiplication,
\(\frac{1}{3}\) × \(\frac{1}{4}\) = \(\frac{1}{12}\)
Look at the way we got the answer. How did we get 12 here?
Including that also, we write
\(\frac{1}{3}\) × \(\frac{1}{4}\) = \(\frac{1}{3 \times 4}\) = \(\frac{1}{12}\)

Can you calculate \(\frac{1}{6}\) of \(\frac{1}{4}\) in head, like this?
Answer:
Yes,  \(\frac{1}{6}\) of \(\frac{1}{4}\) = \(\frac{1}{12}\).

Explanation:
\(\frac{1}{6}\) of \(\frac{1}{4}\) = \(\frac{1}{6 \times 4}\) =
\(\frac{1}{12}\)

Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions

Rectangle division Textbook Page No. 34

A rectangle is vertically divided into two equal parts.
Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions 17
Now suppose we divide this horizontally into three equal parts.
Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions 18
The green part is a third of the yellow part. That is, a third of half.
It is also a sixth of the whole rectangle.
A third half is a sixth.
\(\frac{1}{2}\) × \(\frac{1}{3}\) = \(\frac{1}{6}\)

Textbook Page No. 35

Question 1.
A string of length one metre is cut into five equal parts. What is the length of half of one such piece? How many centimetres?
Answer:
Length of half of one such piece = 10 cm.

Explanation:
Length of the string = 1 m.
Number equal parts it is cut = 5.
Length of each part = Length of the string  ÷ Number of equal parts it is cut
= 1 ÷ 5
= \(\frac{1}{5}\) m
Length of half of one such piece = \(\frac{1}{2}\) × \(\frac{1}{5}\) = \(\frac{1}{10}\) m.
Conversion:
1 meter = 100 cm.
\(\frac{1}{10}\) m = ?? cm
=> 1 × ?? = 100 × \(\frac{1}{10}\)
=> ?? = 10 × \(\frac{1}{1}\)
=> ?? = 10 cm.

Question 2.
One litre of milk fills two identical bottles. A quarter of the milk in one bottle is used to make a cup of tea. How much milk was used to make tea? How many millilitres?
Answer:
Quantity of milk used to make a cup of tea = 250 ml.

Explanation:
Number of liters of milk = 1.
Number of identical bottles = 2.
Quantity of milk used to make a cup of tea = \(\frac{1}{4}\) of 1 liter.
= \(\frac{1}{4}\) × 1
= \(\frac{1}{4}\) liter.
Conversion:
1 liter = 1000 ml.
\(\frac{1}{4}\) liter = ?? ml
=> 1 × ?? = 1000 × \(\frac{1}{4}\)
=> ?? = 250 × \(\frac{1}{1}\)
=> ?? = 250 ml.

Vertically and Horizontally
\(\frac{1}{3}\) of \(\frac{1}{2}\) is \(\frac{1}{6}\)
Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions 19
And in reverse?
Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions 20
\(\frac{1}{2}\) of \(\frac{1}{3}\) is also \(\frac{1}{6}\)

Question 3.
One kilogram yam is cut into three equal pieces. One of the pieces is halved. What is the weight of this piece’?
Answer:
Weight of this piece = \(\frac{1}{6}\) kg.

Explanation:
Number of kilogram of yam = 1.
Number of pieces yam is cut = 3.
One of the pieces is halved.
Weight of one piece = Number of kilogram of yam ÷ Number of pieces yam is cut
= 1 ÷ 3
= \(\frac{1}{3}\) kg.
Weight of this piece = Weight of one piece  ÷ Halve of 1 kg
= \(\frac{1}{3}\) ÷ 2
= \(\frac{1}{3}\) × \(\frac{1}{2}\)
= \(\frac{1}{6}\) kg.

Question 4.
Half the children in a class are girls. A thrid of them are in the Math Club. What fraction of the total class are they’?
Answer:
Fraction of the total class are they = \(\frac{1}{6}\)

Explanation:
Number of children are girls =  \(\frac{1}{2}\)
Number of them are in Math Club = \(\frac{1}{3}\)
Fraction of the total class are they = Number of children are girls × Number of them are in Math Club
= \(\frac{1}{2}\) × \(\frac{1}{3}\)
= \(\frac{1}{6}\)

Question 5.
Calculate the following in your head. Write them as products.
(i) \(\frac{1}{4}\) of \(\frac{1}{2}\)
Answer:
Product of \(\frac{1}{4}\) of \(\frac{1}{2}\), we get \(\frac{1}{8}\)

Explanation:
\(\frac{1}{4}\) of \(\frac{1}{2}\)
= (1 × 1) ÷ (4 × 2)
= \(\frac{1}{8}\)

(ii) \(\frac{1}{2}\) of \(\frac{1}{4}\)
Answer:
Product of \(\frac{1}{2}\) of \(\frac{1}{4}\), we get \(\frac{1}{8}\).

Explanation:
\(\frac{1}{2}\) of \(\frac{1}{4}\)
= (1 × 1) ÷ (2 × 4)
= \(\frac{1}{8}\)

(iii) \(\frac{1}{5}\) of \(\frac{1}{3}\)
Answer:
Product of  \(\frac{1}{5}\) of \(\frac{1}{3}\), we get \(\frac{1}{15}\).

Explanation
\(\frac{1}{5}\) of \(\frac{1}{3}\)
= (1 × 1) ÷ (5 × 3)
= \(\frac{1}{15}\)

(iv) \(\frac{1}{3}\) of \(\frac{1}{5}\)
Answer:
Product of \(\frac{1}{3}\) of \(\frac{1}{5}\), we get

Explanation:
\(\frac{1}{3}\) of \(\frac{1}{5}\)
= (1 × 1) ÷ (3 × 5)
= \(\frac{1}{15}\)

(v) \(\frac{1}{6}\) of \(\frac{1}{3}\)
Answer:
Product of \(\frac{1}{6}\) of \(\frac{1}{3}\), we get \(\frac{1}{18}\)

Explanation:
\(\frac{1}{6}\) of \(\frac{1}{3}\)
= (1 × 1) ÷ (6 × 3)
= \(\frac{1}{18}\)

(vi) \(\frac{1}{3}\) of \(\frac{1}{6}\)
Answer:
Product of \(\frac{1}{3}\) of \(\frac{1}{6}\), we get \(\frac{1}{18}\)

Explanation:
\(\frac{1}{3}\) of \(\frac{1}{6}\)
= (1 × 1) ÷ (3 × 6)
= \(\frac{1}{18}\)

Times in part Textbook Page No. 36

Two litres of milk is used to fill three bottles of the same size. A quarter of one such bottle is poured into a glass. How much milk is in the glass?

Each bottle of \(\frac{1}{3}\) has 2 litres.
That is, \(\frac{2}{3}\) litre.
The glass contains \(\frac{1}{4}\) of this.
That is \(\frac{1}{4}\) of \(\frac{2}{3}\) litre.
How do we calculate this’?
\(\frac{2}{3}\) means \(\frac{1}{3}\) of 2.
So, \(\frac{1}{4}\) of \(\frac{2}{3}\) means of \(\frac{1}{4}\) of \(\frac{1}{3}\) of 2.

\(\frac{1}{4}\) of \(\frac{1}{3}\) is \(\frac{1}{4}\) × \(\frac{1}{3}\) = \(\frac{1}{4 \times 3}\) = \(\frac{1}{12}\)
So, \(\frac{1}{4}\) of \(\frac{2}{3}\) is \(\frac{1}{12}\) of 2.
That is,
\(\frac{1}{12}\) × 2 = \(\frac{2}{12}\) = \(\frac{1}{6}\)
Thus the glass has \(\frac{1}{6}\) litre of milk.
Here, we calculated \(\frac{1}{4}\) of \(\frac{2}{3}\).
Thus we can write \(\frac{1}{4}\) × \(\frac{2}{3}\)
So,
\(\frac{1}{4}\) × \(\frac{2}{3}\) = \(\frac{2}{12}\) = \(\frac{1}{6}\)
And how did we calculate this?
\(\frac{2}{3}\) × \(\frac{1}{4}\) = 2 × \(\frac{1}{3}\) × \(\frac{1}{4}\)
= 2 × \(\frac{1}{3 \times 4}\)
= 2 × \(\frac{1}{12}\)
= \(\frac{2}{12}\) = \(\frac{1}{6}\)

Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions

Milk d\(\frac{1}{6}\)istribution 

A can is full of milk. It is used to fill three identical bottles. Each bottle is used to fill four cups. What fraction of the milk in the can does each cup contain?
Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions 21
Answer:
Fraction of the milk in the can does each cup contain = \(\frac{1}{12}\)

Explanation:
Quantity of milk in a can = 1.
Number of identical bottles milk is filled = 3.
Number of cups used to fill the bottle with milk = 4.
Fraction of the milk in the can does each bottle contain = Quantity of milk in a can ÷ Number of identical bottles milk is filled
= 1 ÷ 3
= \(\frac{1}{3}\)
Fraction of the milk in the bottle does each cup contain = Quantity of milk in a bottle ÷ Number of cups used in a bottle
= 1 ÷ 4
= \(\frac{1}{4}\)
Fraction of the milk in the can does each cup contain = Fraction of the milk in the can does each bottle contain × Fraction of the milk in the bottle does each cup contain
= \(\frac{1}{3}\) × \(\frac{1}{4}\)
= \(\frac{1}{12}\)

Square math

A square is vertically divided into three equal parts.
Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions 22
Again, it is horizontally divided into five equal parts.
Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions 23
The part of the picture coloured green is \(\frac{2}{15}\) of the whole square. And also \(\frac{1}{5}\) of the part coloured yellow.
Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions 24
Thus \(\frac{1}{5}\) of \(\frac{2}{3}\) is,
\(\frac{2}{3}\) × \(\frac{1}{5}\) = \(\frac{2}{15}\)

Another problem:

\(\frac{1}{2}\) kilogram of rice is equally filled in 4 bags. If we take 3 such bags together, how much rice do we get?
Each bag has \(\frac{1}{4}\) of \(\frac{1}{2}\) kilograms.
That is. \(\frac{1}{4}\) × \(\frac{1}{2}\) = \(\frac{1}{4 \times 2}\) = \(\frac{1}{8}\)
3 bags contain 3 times as much; 3 × \(\frac{1}{8}\) = \(\frac{3}{8}\)
Three bags contain \(\frac{3}{8}\) kilograms (375 grams).
One bag contains a quarter of half kilogram. Three such bags make three quarters of half kilogram.
Thus we see that \(\frac{3}{4}\) of \(\frac{1}{2}\) is \(\frac{3}{8}\).
As multiplication, \(\frac{3}{4}\) × \(\frac{1}{2}\) = \(\frac{3}{8}\)
Look again at the way we calculated this:
\(\frac{3}{4}\) × \(\frac{1}{2}\) = 3 × \(\frac{1}{4}\) × \(\frac{1}{2}\)
= 3 × \(\frac{1}{4 \times 2}\)
= 3 × \(\frac{1}{8}\) = \(\frac{3}{8}\)

Can you find \(\frac{2}{5}\) of \(\frac{1}{3}\) metre like this?
We divide \(\frac{1}{3}\) metre into 5 equal parts and put 2 such pieces end to end. What is this length’?
2 times \(\frac{1}{5}\) of \(\frac{1}{3}\) metre.
As multiplication,
\(\frac{2}{5}\) × \(\frac{1}{3}\) = 2 × \(\frac{1}{5}\) × \(\frac{1}{3}\)
= 2 × \(\frac{1}{5}\) × \(\frac{1}{3}\)
= 2 × \(\frac{1}{15}\)
= \(\frac{2}{15}\)

Now how do we find \(\frac{4}{5}\) of \(\frac{2}{3}\)?
First find \(\frac{1}{5}\) of \(\frac{2}{3}\) and then 4 times this.
How do we find the \(\frac{1}{5}\) of \(\frac{2}{3}\) in this?
We have to find \(\frac{1}{5}\) of \(\frac{1}{3}\) of 2.
\(\frac{1}{5}\) × \(\frac{2}{3}\) = \(\frac{1}{5}\) × \(\frac{1}{3}\) × 2 = \(\frac{1}{5 \times 3}\) × 2 = \(\frac{1}{15}\) × 2 = \(\frac{2}{15}\)

Now we need only find 4 times \(\frac{2}{15}\).
4 × \(\frac{2}{15}\) = \(\frac{8}{15}\)
If we do all multiplication only at the end, we write
\(\frac{4}{5}\) × \(\frac{2}{3}\) = 4 × \(\frac{1}{5}\) × \(\frac{1}{3}\) × 2
= 4 × \(\frac{1}{5}\) × \(\frac{1}{3}\) × 2
= 4 × \(\frac{1}{5 \times 3}\) × 2
= \(\frac{4 \times 2}{5 \times 3}\)
= \(\frac{8}{15}\)
Can’t we find \(\frac{4}{9}\) of \(\frac{3}{5}\) like this?
\(\frac{4}{9}\) × \(\frac{3}{5}\) = \(\frac{4 \times 3}{9 \times 5}\) = \(\frac{12}{45}\) = \(\frac{4}{15}\)

Another way

We can do \(\frac{4}{9}\) × \(\frac{3}{5}\) like this also.
\(\frac{4}{9}\) × \(\frac{3}{5}\) = \(\frac{4 \times 3}{9 \times 5}\) = \(\frac{4 \times 3}{3 \times 3 \times 5}\)
= \(\frac{4}{3 \times 5}\) = \(\frac{4}{15}\)

Textbook Page No. 40

Question 1.
Draw the line AB, 12 centimetres long. Mark C on it such that AC is \(\frac{2}{3}\) of AB. Mark D such that AD is \(\frac{1}{4}\) of AC. What fraction of AB is AD?
Answer:
Fraction of AB is AD = \(\frac{1}{6}\) cm.

Explanation:
Length of AB = 12 cm.
Length\(\frac{1}{4}\) of AC = \(\frac{2}{3}\)cm of AB.
= \(\frac{2}{3}\) × 12
= \(\frac{2}{1}\) × 4
= 8 cm.
Length of AD = \(\frac{1}{4}\) cm of AC.
= \(\frac{1}{4}\) × 8
= \(\frac{1}{1}\) × 2
= 2 cm.
Fraction of AB is AD = Length of AD ÷ Length of AB
= 2 ÷ 12
= 1 ÷ 6
= \(\frac{1}{6}\) cm.
Kerala-State-Syllabus-6th-Standard-Maths-Solutions-Chapter-3-Fractions-Textbook Page No. 40

Question 2.
A two metre long rope is cut into five equal pieces. What is the length of three quarters of one piece? How many centimetres is this?
Answer:
Length of three quarters of one piece = 30 cm.

Explanation:
Length of the rope = 2m.
Number of pieces the rope is cut = 5.
Length of each piece of rope = Length of the rope ÷ Number of pieces the rope is cut
= 2 ÷ 5
= \(\frac{2}{5}\) m.
Three quarters of one piece of cord is three quarters of \(\frac{2}{5}\) metres,
=> 3/4 × \(\frac{2}{5}\)
=> \(\frac{6}{20}\) ÷ \(\frac{2}{2}\)
=> \(\frac{3}{10}\)  meters.
Conversion:
1 cm = 100 m.
?? cm = \(\frac{3}{10}\) m
=> 1 × \(\frac{3}{10}\) = ?? × 100
=> \(\frac{3}{10}\) ÷ 100 = ??
=> \(\frac{3}{10}\) × \(\frac{100}{1}\) = ??
=> \(\frac{3}{1}\)  × \(\frac{10}{1}\) = ??
=> 30 cm = ??

Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions

Question 3.
Three liters of water is used to fill four identical bottles. One bottle is used to fill five identical cups. How much water is in one cup? How many milliliters is this?
Answer:
Quantity of water in a cup = 50 milliliters.

Explanation:
Number of liters of water = 3.
Number of identical bottles the water is filled = 4.
Number of identical cups used to fill one bottle = 5.
Number of liters of water filled in one bottle = Number of liters of water ÷ Number of identical bottles the water is filled
= 3 ÷ 4
= \(\frac{3}{4}\)
One bottle is used to fill five identical cups.
=> Number of cups of water filled in one bottle = Number of liters of water filled in one bottle ÷ 5
=> \(\frac{3}{4}\) ÷ 5
=> \(\frac{3}{4}\) × \(\frac{1}{5}\)
=> \(\frac{3}{20}\) liters.
Conversion:
1 liter = 1000 milliliters.
\(\frac{3}{20}\) liters = ?? ml
=> 1 × ?? = 1000 × \(\frac{3}{20}\)
=> ?? = 50 × \(\frac{3}{1}\)
=> ?? = 50 milliliters.

Question 4.
A five kilogram pumpkin is cut into five equal pieces. Each piece is further cut into two. What is the weight of each such piece? How many grams is this?
Answer:
Weight of each such piece = 500 grams.

Explanation:
Number of kilograms the pumpkin = 5.
Number of pieces the pumpkin is cut = 5.
Number of parts the piece is cut = 2.
Weight of each piece of pumpkin = Number of kilograms the pumpkin ÷ Number of parts the piece is cut
= 5 ÷ 5
= 1 kg.
Weight of each such piece = Weight of each piece of pumpkin ÷ Number of parts the piece is cut
= 1 ÷ 2
= \(\frac{1}{2}\) kg.
Conversion:
1 kg = 1,000 g.
\(\frac{1}{2}\) kg = ?? g
=> 1 × ?? = 1000 × \(\frac{1}{2}\)
=> ?? = 500 × \(\frac{1}{1}\)
=> ?? = 500 g.

Question 5.
Calculate each of the following using multiplication.
(i) \(\frac{3}{7}\) of \(\frac{2}{5}\)
Answer:
Product of \(\frac{3}{7}\) of \(\frac{2}{5}\), we get \(\frac{6}{35}\).

Explanation:
\(\frac{3}{7}\) of \(\frac{2}{5}\)
= (3 × 2) ÷ (7 × 5)
= \(\frac{6}{35}\)

(ii) \(\frac{3}{5}\) of \(\frac{2}{7}\)
Answer:
Product of \(\frac{3}{5}\) of \(\frac{2}{7}\), we get \(\frac{6}{35}\).

Explanation:
\(\frac{3}{5}\) of \(\frac{2}{7}\)
= (3 × 2) ÷ (5 × 7)
= \(\frac{6}{35}\)

(iii) \(\frac{2}{3}\) of \(\frac{3}{4}\)
Answer:
Product of \(\frac{2}{3}\) of \(\frac{3}{4}\), we get \(\frac{1}{2}\).

Explanation:
\(\frac{2}{3}\) of \(\frac{3}{4}\)
= (2 × 3) ÷ (3 × 4)
= \(\frac{6}{12}\) ÷ \(\frac{2}{2}\)
= \(\frac{1}{2}\)

(iv) \(\frac{5}{6}\) of \(\frac{3}{10}\)
Answer:
Product of \(\frac{5}{6}\) of \(\frac{3}{10}\), we get \(\frac{1}{4}\).

Explanation:
\(\frac{5}{6}\) of \(\frac{3}{10}\)
= (5 × 3) ÷ (6 × 10)
= \(\frac{15}{60}\) ÷ \(\frac{5}{5}\)
= \(\frac{3}{12}\) ÷ \(\frac{3}{3}\)
= \(\frac{1}{4}\)

Part in times

A bottle can hold one and a half litres of water. Four such bottles of water is poured into a jar. How much water is there in the jar?
Two bottles make three litres; four bottles, six litres.
Here, what we calculate is 4 time 1\(\frac{1}{2}\).

As multiplication, we write
4 × 1\(\frac{1}{2}\) = 6
Suppose we pour in 3 bottles containing 2\(\frac{1}{4}\) litres each.
If they were 2 litre bottles, we would have got 6 litres. Here each bottle contain \(\frac{1}{4}\) litre more.
So, we have to add \(\frac{3}{4}\) litres more; that is 6\(\frac{3}{4}\) litres.
How about writing this as multiplication?
3 × 2\(\frac{1}{4}\) = 3 × (2 + \(\frac{1}{4}\))
= (3 × 2) + (3 × \(\frac{1}{4}\))
= 6 + \(\frac{3}{4}\) = 6\(\frac{3}{4}\)
We can do this in another way. We can write 2\(\frac{1}{4}\) litres as \(\frac{9}{4}\) litres; that is, \(\frac{1}{4}\) of 9 litres. We want to calculate 3 times this.
So,
3 × 2\(\frac{1}{4}\) = 3 × \(\frac{9}{4}\)
= \(\frac{27}{4}\) = 6\(\frac{3}{4}\)
In this way, we can calculate 5 times 3\(\frac{1}{2}\):
5 × 3\(\frac{1}{2}\) = 5 × \(\frac{7}{2}\)
= \(\frac{5 \times 7}{2}\)
= \(\frac{35}{2}\) = 17\(\frac{1}{2}\)

Let’s look at another thing.
Six metres is three times two metres.
What about seven metres?

Three times two metres and then one more metre. In other words, three times two metres and then half of two metres.
So, we can say seven metres is three and a half times two metres.
Written as multiplication,
3\(\frac{1}{2}\) × 2 = (3 + \(\frac{1}{2}\)) × 2 = (3 × 2) + (\(\frac{1}{2}\) × 2) = 6 + 1 = 7
In the same way, two and a quarter times five means, two times five and a quarter of five together; that is ten with one and a quarter, which makes eleven and a quarter.
(2\(\frac{1}{4}\) × 5) = (2 + \(\frac{1}{4}\)) × 5
= (2 × 5) + (\(\frac{1}{4}\) × 5)
= 10 + 1\(\frac{1}{4}\)
= 11\(\frac{1}{4}\)

We can also do it like this:
2\(\frac{1}{4}\) × 5 = \(\frac{9}{4}\) × 5
= \(\frac{9 \times 5}{4}\)
= \(\frac{45}{4}\) = 11\(\frac{1}{4}\)
Now let’s see how we can calculate 3\(\frac{1}{2}\) times 2\(\frac{1}{4}\)
3\(\frac{1}{2}\) × 2\(\frac{1}{4}\) = \(\frac{7}{2}\) × \(\frac{9}{4}\) = \(\frac{63}{8}\) = 7\(\frac{7}{8}\)
we can calculate 3 times 2\(\frac{1}{4}\) and \(\frac{1}{2}\) of 2\(\frac{1}{4}\) seperately and add.

Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions

Textbook Page No. 43

Question 1.
1\(\frac{1}{2}\) metres of cloth is needed for a shirt. How much cloth is needed for five such shirts?
Answer:
Length of cloth needed to make 5 shirts = \(\frac{15}{2}\)   or 7\(\frac{1}{2}\)   meters.

Explanation:
Length of the cloth needed to make a shirt = 1\(\fr meters.ac{1}{2}\) meters.
Number of shirts = 5.
Length of cloth needed to make 5 shirts = Length of the cloth needed to make a shirt × Number of shirts
= 1\(\frac{1}{2}\)  × 5
= [(1 × 2) + 1] ÷ 2} × 5
= \(\frac{3}{2}\)  × 5
= \(\frac{15}{2}\)   or 7\(\frac{1}{2}\) meters.

Question 2.
The price of one kilogram of okra is 30 rupees. What is the price of 2\(\frac{1}{2}\) kilograms?
Answer:
Cost of 2\(\frac{1}{2}\) kilograms of okra = 75 rupees.

Explanation:
Cost of 1 kilogram of okra = 30 rupees.
Number of kilograms is the okra = 2\(\frac{1}{2}\)
Cost of 2\(\frac{1}{2}\) kilograms of okra = Cost of 1 kilogram of okra × Number of kilograms is the okra
= 30 × 2\(\frac{1}{2}\)
= 30 × {[(2 × 2) + 1] ÷ 2}
= 30 × [(4 + 1) ÷ 2]
= 30 × \(\frac{5}{2}\)
= 15 × \(\frac{5}{1}\)
= 75  rupees.

Question 3.
A man walks one and a half kilometres in one hour. How many kilometres does he walk in one and a half hours at this speed?
Answer:
Number of kilometers a man walks in half hours = \(\frac{3}{4}\)

Explanation:
Number of kilometers a man walks in an hour = 1\(\frac{1}{2}\)
Number of kilometers a man walks in half hours = Number of kilometers a man walks in an hour ÷ 2
= 1\(\frac{1}{2}\) ÷ 2
= {[(1 × 2) + 1] ÷ } ÷ 2
= [(2 + 1) ÷ 2] ÷ 2
= \(\frac{3}{2}\) ÷ 2
= \(\frac{3}{2}\) × \(\frac{1}{2}\)
= (3 × 1) ÷ (2 × 2)
= \(\frac{3}{4}\)

Question 4.
Rony has 36 stamps. Zaheera says she has 2\(\frac{1}{4}\) times as much. How many stamps does she have?
Answer:
Number of stamps Zaheera has = 81.

Explanation:
Number of stamps Rony has = 36.
Number of times Zaheera has stamps = 2\(\frac{1}{4}\)
Number of stamps Zaheera has = Number of stamps Rony has  × Number of times Zaheera has stamps
= 36 × 2\(\frac{1}{4}\)
= 36 × {[(2 × 4) + 1] ÷ 4}
= 36 × [(8 + 1) ÷ 4]
= 36 × \(\frac{9}{4}\)
= 9 × \(\frac{9}{1}\)
= 81.

Question 5.
Calculate the following:

(i) 4 times 5\(\frac{1}{3}\)
Answer:
4 times of 5\(\frac{1}{3}\), we get the result \(\frac{64}{3}\)  or 21\(\frac{1}{3}\).

Explanation:
4 times 5\(\frac{1}{3}\)
= 4 × 5\(\frac{1}{3}\)
= 4 × {[(5 × 3) + 1] ÷ 3}
= 4 × [(15 + 1) ÷ 3]
= 4 × \(\frac{16}{3}\)
= \(\frac{64}{3}\)  or 21\(\frac{1}{3}\)

(ii) 4\(\frac{1}{3}\) times 5
Answer:
4\(\frac{1}{3}\) times of 5, we get the result \(\frac{65}{3}\) or 21\(\frac{2}{3}\).

Explanation:
4\(\frac{1}{3}\) times  × 5
= {[(4 × 3) + 1] ÷ 3] × 5
= [(12 + 1) ÷ 3] × 5
= \(\frac{13}{3}\) × 5
= (13 × 5) ÷ 3
= \(\frac{65}{3}\) or 21\(\frac{2}{3}\)

(iii) 1\(\frac{1}{2}\) times \(\frac{2}{3}\)
Answer:
1\(\frac{1}{2}\) times of \(\frac{2}{3}\), we get the result 1.

Explanation:
1\(\frac{1}{2}\) times × \(\frac{2}{3}\)
= {[(1 × 2) + 1] ÷ 2}  × \(\frac{2}{3}\)
= [(2 + 1) ÷ 2] × \(\frac{2}{3}\)
= \(\frac{3}{2}\) × \(\frac{2}{3}\)
= \(\frac{1}{1}\) × \(\frac{1}{1}\)
= 1.

(iv) \(\frac{2}{5}\) of 2\(\frac{1}{2}\)
Answer:
\(\frac{2}{5}\) of 2\(\frac{1}{2}\), we get the result 1.

Explanation:
\(\frac{2}{5}\) ×  2\(\frac{1}{2}\)
= \(\frac{2}{5}\) × {[(2 × 2) + 1] ÷ 2}
= \(\frac{2}{5}\) × [(4 + 1) ÷ 2]
= \(\frac{2}{5}\) × \(\frac{5}{2}\)
= \(\frac{1}{1}\) × \(\frac{1}{1}\)
= 1.

(v) 2\(\frac{1}{2}\) times 5\(\frac{1}{2}\)
Answer:
2\(\frac{1}{2}\) times of 5\(\frac{1}{2}[/late, we get the result [latex]\frac{55}{4}\) or 13\(\frac{3}{4}\).

Explanation:
2\(\frac{1}{2}\) times × 5\(\frac{1}{2}\)
= {[(2 × 2) + 1] ÷ 2} × {[(5 × 2) + 1] ÷ 2}
= [(4 + 1) ÷ 2] × [(10 + 1) ÷ 2]
= \(\frac{5}{2}\) × \(\frac{11}{2}\)
= (5 × 11) ÷ (2 × 2)
= \(\frac{55}{4}\) or 13\(\frac{3}{4}\)

(vi) 4\(\frac{1}{3}\) of 4\(\frac{1}{2}\)
Answer:
4\(\frac{1}{3}\) of 4\(\frac{1}{2}\), we get the result \(\frac{39}{2}\) or 19\(\frac{1}{2}\).

Explanation:
4\(\frac{1}{3}\) × 4\(\frac{1}{2}\)
= {[(4 × 3) + 1] ÷ 3} × {[(4 × 2)  + 1] ÷ 2}
= [(12 + 1) ÷ 3] × [(8 + 1) ÷ 2]
= \(\frac{13}{3}\) × \(\frac{9}{2}\)
= \(\frac{13}{1}\) × \(\frac{3}{2}\)
= (13 × 3) ÷ (1 × 2)
= (39 ÷ 2)
= \(\frac{39}{2}\) or 19\(\frac{1}{2}\)

Fractional area

You have learnt about area of rectangles in class 5.1
What is the area of a rectangle of length 5 centimetres and breadth 3 centimetres, in square centimetres.
Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions 25
The area of a square of side one centimetre is one square centimetre, isn’t it’? How do we measure the area of smaller rectangles?

Look at this picture:
A square of side one centimetre is divided into two equal parts. Each such rectangle is of the square.
Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions 26

So, we can say each has area \(\frac{1}{2}\) square centimetre.
What are the lengths of its sides?
Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions 27
Now suppose we divide our square again into three equal parts.
Each rectangle is \(\frac{1}{6}\) of the whole square; its area is \(\frac{1}{6}\) square centimetre. Thus the area of a rectangle of sides \(\frac{1}{2}\) centimetre and \(\frac{1}{3}\) centimetre are \(\frac{1}{6}\) square centimetre.
We can see this in another way. We can make a square of side 1 centimetre, by stacking 6 rectangles of sides \(\frac{1}{2}\) centimetre and \(\frac{1}{3}\) centimetre.
Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions 28
What is the area of a rectangle of length 5\(\frac{1}{2}\) centimetre and breadth 3\(\frac{1}{3}\) centimetre?
Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions 29
We first divide the bottom side into \(\frac{1}{2}\) centimetre parts.
How many parts do we get?

10 lines of \(\frac{1}{2}\) centimetres make 5 centimetres; to get 5\(\frac{1}{2}\) centimetres, we need one more
Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions 30
Next, we divide the left side of the rectangle into \(\frac{1}{3}\) centimetre parts.
9 lines of \(\frac{1}{3}\) centimetre make 3 centimetres; to get 3\(\frac{1}{3}\) centimetre, we need one more
Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions 31
Thus we can fill part of the rectangle using rectangles of sides \(\frac{1}{2}\) centimetre and \(\frac{1}{3}\) centimetre.
Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions 32

How many such rectangles do we need to fill the whole large rectangle?
Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions 33
11 × 10 = 110 rectangles in all; the area of each is \(\frac{1}{6}\) square centimetre. Total area is
\(\frac{1}{6}\) × 110 = \(\frac{55}{3}\) = 18\(\frac{1}{3}\) square centimeter
Here, we calculated \(\frac{1}{6}\) × 11 × 10. This can write like this;
\(\frac{1}{6}\) × 11 × 10 = \(\frac{1}{2}\) × \(\frac{1}{3}\) × 11 × 10 = \(\frac{11}{2}\) × \(\frac{10}{3}\) = 5\(\frac{1}{2}\) × 3\(\frac{1}{3}\)
So, the area of a rectangle is the product of length and breadth, even if they are fractions.

Area again! Textbook Page No. 45

What is the area of a rectangle of sides \(\frac{3}{4}\) centimetre and \(\frac{2}{5}\) centimetre?
Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions 34
Placing 4 of them side by side we get a rectangle like this;
Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions 35
Stacking 5 such rectangles one over another, we get a larger rectangle;
Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions 36
What is the area of this rectangle?
How many small rectangles are in it?
So, what fraction of the whole rectangle is each small rectangle?
Thus we see that the area of each small rectangle is \(\frac{1}{20}\) of 6 square centimetres. How much is it?
6 sq.cm × \(\frac{1}{20}\) = \(\frac{6}{20}\) sq.cm.
This we can simplify as \(\frac{3}{10}\) square centimetre.
Any way, the area is \(\frac{3}{4}\) × \(\frac{2}{5}\) isn’t it?
Answer:
Area of a rectangle of sides = \(\frac{3}{10}\) square centimeter.
Number of small rectangles = 20.

Explanation:
Length of the rectangle = \(\frac{3}{4}\) centimetre.
Breadth of the rectangle = \(\frac{2}{5}\) centimetre.
Area of a rectangle of sides = Length of the rectangle ×  Breadth of the rectangle
= \(\frac{3}{4}\) × \(\frac{2}{5}\)
= (3 × 2) ÷ (4 × 5)
= (6 ÷ 20) ÷ (2÷ 2)
= 3 ÷ 10
= \(\frac{3}{10}\) square centimeter.
Number of small rectangles = 5 × 4
= 20.

Textbook Page No. 46

Question 1.
The length and breadth of some rectangles are given below. Calculate their areas.

(i) 4\(\frac{1}{2}\) cm, 3\(\frac{1}{4}\) cm
Answer:
Area of the rectangle =  \(\frac{117}{8}\)  or 14\(\frac{5}{8}\) square cm.

Explanation:
Length of the rectangle = 4\(\frac{1}{2}\) cm
Breadth of the rectangle = 3\(\frac{117}{8}\)
Area of the rectangle = Length of the rectangle × Breadth of the rectangle
= 4\(\frac{1}{2}\) × 3\(\frac{1}{4}\)
= {[(4 × 2) + 1] ÷ 2} × {[(3 × 4) + 1] ÷ 4}
= [(8 + 1) ÷ 2] × [(12 + 1) ÷ 4]
= \(\frac{9}{2}\)  × \(\frac{13}{4}\)
= (9 × 13) ÷ (2 × 4)
= \(\frac{117}{8}\)  or 14\(\frac{5}{8}\) square cm.

(ii) 6\(\frac{3}{4}\) m, 5\(\frac{1}{3}\) m
Answer:
Area of the rectangle = 36 square meters.

Explanation:
Length of the rectangle = 6\(\frac{3}{4}\) m.
Breadth of the rectangle = 5\(\frac{1}{3}\) m.
Area of the rectangle = Length of the rectangle × Breadth of the rectangle 6\(\frac{3}{4}\)  × 5\(\frac{1}{3\)
= {[(6 × 4) + 3] ÷ 4} × {[(5 × 3) + 1] ÷ 3}
= [(24 + 3) ÷ 4] × [(15 + 1) ÷ 3}
= \(\frac{27}{4}\)  × \(\frac{16}{3}\)
= \(\frac{9}{1}/latex]  × [latex]\frac{4}{1}\)
= (9 × 4) ÷ (1 × 1)
= \(\frac{36}{1}\)
= 36 square meters.

(iii) 1\(\frac{1}{3}\) m, \(\frac{3}{4}\) m
Answer:
Area of the rectangle = 1 square meter.

Explanation:
Length of the rectangle = 1\(\frac{1}{3}\) m
Breadth of the rectangle = \(\frac{3}{4}\) m
Area of the rectangle = Length of the rectangle  × Breadth of the rectangle
= 1\(\frac{1}{3}\)  × \(\frac{3}{4}\)
= {[(1 × 3) + 1] ÷ 3} × \(\frac{3}{4}\)
= [(3 + 1) ÷ 3] × \(\frac{3}{4}\)
= \(\frac{4}{3}\) × \(\frac{3}{4}\)
= \(\frac{1}{1}\) × \(\frac{1}{1}\)
= 1 square meter.

 

Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions

Question 2.
What is the area of a square of side 1\(\frac{1}{2}\) metre?
Answer:
Area of the square = \(\frac{9}{4}\)  or 2\(\frac{1}{4}\)  square meter.

Explanation:
Side of the square = 1\(\frac{1}{2}\) meter.
Area of the square =  Side of the square  × Side of the square
= 1\(\frac{1}{2}\)  × 1\(\frac{1}{2}\)
= {[(1 × 2) + 1] ÷ 2} × {[(1 × 2) + 1] ÷ 2}
= [(2 + 1) ÷ 2] × [(2 + 1) ÷ 2]
= \(\frac{3}{2}\)  × \(\frac{3}{2}\)
= (3 × 3) ÷ (2 × 2)
= \(\frac{9}{4}\)  or 2\(\frac{1}{4}\)  square meter.
\(\frac{1}{4}\)

Question 3.
The perimeter of a square is 14 metres. What is its area?
Answer:
Area of the square = \(\frac{49}{4}\) or 12\(\frac{1}{4}\) square meter.

Explanation:
Perimeter of a square = 14 metres.
Perimeter of a square =  4 × Side of the square
=> 14 = 4 × Side of the square
=> 14 ÷ 4 = Side of the square
=> 7 ÷ 2 = Side of the square
=> \(\frac{7}{2}\)  or 3\(\frac{1}{2}\) metre = Side of the square
Area of the square = Side of the square × Side of the square
= \(\frac{7}{2}\)  × \(\frac{7}{2}\)
= (7 × 7) ÷ (2 × 2)
= \(\frac{49}{4}\)  or 12\(\frac{1}{4}\) square meter.

Back and forth Textbook Page No. 47

A small jar can hold 2 litres of water and a large jar, 6 litres. So, the large jar can hold 3 times as much as the small one.
Saying it in reverse, the small jar can hold \(\frac{1}{3}\) as much as the large one.
That is 3 times 2 is 6; on the other hand, \(\frac{1}{3}\) of 6 is 2.
Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions 37
What about a jar of 4 litres and a jar of 6 litres?
How much times 4 is 6?
4 and its half 2 make 6; that is 1
6 is 1\(\frac{1}{2}\) times 4.
How do we say it in reverse?
Let’s think like this.
1\(\frac{1}{2}\) means 3 times 1. So 1\(\frac{1}{2}\) times 4 means times \(\frac{1}{2}\) of 4.
\(\frac{1}{2}\) of 4 is 2, and 3 times 2 is 6.
Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions 38
So in reverse, we can take \(\frac{1}{3}\) of 6 to get 2 and then 2 times 2 to get 4.
Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions 39
Thus 4 is 2 times \(\frac{1}{3}\) of 6; that is \(\frac{2}{3}\) of 6.
How about combining all these?
3 times \(\frac{1}{2}\) is \(\frac{3}{2}\) times; 2 times \(\frac{1}{3}\) to \(\frac{2}{3}\).
Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions 40
Writing these as multiplication,
\(\frac{3}{2}\) × 4 = 6 \(\frac{2}{3}\) × 6 = 4
Let’s look at another such calculation: 6 and its one third, that is 2, make 8; that is, 8 is 1\(\frac{1}{3}\) of 6.
We can also say like this: \(\frac{1}{3}\) of 6 is 2 and 4 times 2 is 8.
Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions 41

How do we say in reverse?
\(\frac{1}{4}\) of 8 is 2 and 3 times 2 is 6.
Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions 42
Let’s put these together:
Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions 43
And we can write the multiplications
\(\frac{4}{3}\) × 6 = 8 \(\frac{3}{4}\) × 8 = 6

One more calculation: 8 and its \(\frac{1}{4}\), that is 2, make 10; that is, 10 is 1\(\frac{1}{4}\) times 8.
We can say \(\frac{5}{4}\) instead of 1\(\frac{1}{4}\). Thus 10 is \(\frac{5}{4}\) times 8.
In reverse, 8 is 4 times \(\frac{1}{5}\) of 10.
Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions 44
\(\frac{5}{4}\) × 8 = 10, \(\frac{4}{5}\) × 10 = 8
Did you notice something in all these?
To say in reverse, we need only turn the fraction upside down.
Instead of saying “turn the fraction upside d
=> We noticed interchanging the numerator and denominator”.own”, we can say, “interchanging the numerator and denominator”. The fraction thus got is called the reciprocal.

Textbook Page No. 47

Take a strip of paper and cut it into 5 equal parts.
Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions 55
Take two such pieces and put them together?
Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions 53
It is \(\frac{2}{5}\) of the strip. Take two more pieces and put them together with the first two.
Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions 54
Now we have two \(\frac{2}{5}\)‘s. That is, two times \(\frac{2}{5}\). What remain is half of \(\frac{2}{5}\). Put it also with the others. Then we have 2 times \(\frac{2}{5}\) and \(\frac{1}{2}\) of \(\frac{2}{5}\); that is 2\(\frac{1}{2}\) times. This is the whole strip. What do we see here?
Answer:
2 times \(\frac{2}{5}\) and 1 times \(\frac{1}{5}\)
We have noticed the interchanging of denominators.

Explanation:
2 times \(\frac{2}{5}\) and \(\frac{1}{2}\) of \(\frac{2}{5}\)
=> 2 × \(\frac{2}{5}\) = (2 × 2) ÷ 5 = \(\frac{4}{5}\)
\(\frac{1}{2}\) of \(\frac{2}{5}\)
=> \(\frac{1}{2}\) × \(\frac{2}{5}\) =  \(\frac{1}{1}\) × \(\frac{1}{5}\) = \(\frac{1}{5}\)
=> \(\frac{4}{5}\) + \(\frac{1}{5}\) = \(\frac{5}{5}\) = 1.
=> 2\(\frac{1}{2}\) times. This is the whole strip.

Textbook Page No. 49

A square of side 1 metre can \(\frac{1}{5}\) be divided vertically into 3 equal parts and horizontally into 4 equal parts.
Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions 45
Now cut out the top three pieces.
Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions 46

Stack these pieces on the left as below:
Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions 47
What are the length and breadth of this rectangle? And its area?
Answer:
Area of the rectangle =  \(\frac{1}{12}\)  square meters.

Explanation:
Length of the rectangle = \(\frac{1}{4}\)  m.
Breadth of the rectangle = \(\frac{1}{3}\)  m.
Area of the rectangle =  Length of the rectangle ×  Breadth of the rectangle
= \(\frac{1}{4}\)  × \(\frac{1}{3}\)
= (1 × 1) ÷ (4 × 3)
= \(\frac{1}{12}\)  square meters.

Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions

Now look at this problem:

The price of 1\(\frac{1}{2}\) kilogram of tomatoes is 30 rupees. What is the price of one kilogram?
There are several ways to do this. One way is this

Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions 48

  • 1\(\frac{1}{2}\) is 3 times \(\frac{1}{2}\)
  • The price of 1\(\frac{1}{2}\) kilogram of tomatoes is 3 times that of \(\frac{1}{2}\) kilograms.
  • The price of \(\frac{1}{2}\) kilogram is 30 ÷ 3 = 10 rupees.
  • The price of 1 kilogram is 2 × 10 = 20 rupees.

We can also think like this.

  • 2 times 1\(\frac{1}{2}\) is 3.
  • The price of 3 kilograms is 2 × 30 = 60 rupees.
  • The price of 1 kilogram is 60 ÷ 3 = 20 rupees.

We can do this directly, using reciprocal:

  • 30 rupees is \(\frac{3}{2}\) times the price of 1 kilogram.
  • The price of 1 kilogram is \(\frac{2}{3}\) of 30 rupees.
  • 30 × \(\frac{2}{3}\) = 20 rupees.

Textbook Page No. 51

Question 1.
The length of one string is 4 metres and the length of another string is 14 metres.

(i) What fraction of the longer string is the shorter string?
Answer:
Fraction of the longer string is the shorter string = \(\frac{7}{2}\) meters.

Explanation:
Length of one string = 4 meters.
Length of another string = 14 meters.
Fraction of the longer string is the shorter string = Length of another string ÷ Length of one string
= 14 ÷ 4
= (14 ÷ 4) ÷ (2 ÷ 2)
=  \(\frac{7}{2}\) meters.

(ii) How much times of the shorter string is the longer string?
Answer:
7 times of shorter string is the longer string.

Explanation:
Length of one string = 4 meters.
Length of another string = 14 meters.
Fraction of the longer string is the shorter string = \(\frac{7}{2}\) meters.
=> 7 times of shorter string is the longer string.

There is one block weighs 6 kilogram and another, 26 kilogram.
(i) What fraction of the weight of the heavier block is the lighter block?
Answer:
Fraction of the weight of the heavier block is the lighter block = \(\frac{13}{3}\) kilogram.

Explanation:
Weight of one block = 6 kilograms.
Weight of another block = 26 kilograms.
Fraction of the weight of the heavier block is the lighter block = Weight of another block ÷ Weight of one block
= 26 ÷ 6
= (26 ÷ 6) ÷ (2 ÷ 2)
= 13 ÷ 3 or \(\frac{13}{3}\) kilogram.

(ii) How much times of the weight of the lighter block is the heavier block?
Answer:
13 times of the weight of the lighter block is the heavier block.

Explanation:
Weight of one block = 6 kilograms.
Weight of another block = 26 kilograms.
Fraction of the weight of the heavier block is the lighter block = \(\frac{13}{3}\) kilograms.
=> 13 times o or \(\frac{1}{3}\)f the weight of the lighter block is the heavier block.

Question 3.
A pumpkin is cut into three equal pieces. Two pieces together weigh one kilogram. What is the weight of the whole pumpkin?
Answer:
Weight of the whole pumpkin = 1 kilogram.

Explanation:
Number of pieces a pumpkin is cut = 3.
Two pieces together weigh one kilogram.
Weight of one piece = Number of pumpkins ÷ Number of pieces a pumpkin is cut
= 1 ÷ 3 or \(\frac{1}{3}\)
Weight of the whole pumpkin = Number of pieces a pumpkin is cut × Weight of one piece
= 3 × \(\frac{1}{3}\)
= 1 × \(\frac{1}{1}\)
= 1 kilogram.

Question 4.
1\(\frac{1}{2}\) litres of water is needed to fill \(\frac{3}{4}\) of a can. How much water is needed to fill it completely?
Answer:
Water is needed to fill it completely = \(\frac{1}{2}\) liters.

Explanation:
Weight of the can = \(\frac{3}{4}\)
Number of liters of water is needed to fill a can = 1\(\frac{1}{2}\)  = \(\frac{3}{2}\)
water is needed to fill it completely:
\(\frac{3}{4}\) – \(\frac{3}{2}\) liters.
\(\frac{1}{4}\) – ?? liters
=> \(\frac{3}{4}\) × ?? = \(\frac{3}{2}\) × \(\frac{1}{4}\)
=> \(\frac{3}{4}\) × ?? = (3 × 1) ÷ (2 × 4)
=> \(\frac{3}{4}\) × ?? = \(\frac{3}{8}\)
=> ?? = \(\frac{3}{8}\)  ÷ \(\frac{3}{4}\)
=> ?? = \(\frac{3}{8}\)  × \(\frac{4}{3}\)
=> ?? = \(\frac{1}{2}\)  × \(\frac{1}{1}\)
=> ?? = \(\frac{1}{2}\) liters.

Question 5.
There are three pieces of ribbon. Two of the pieces and half the third piece, laid end to end, make one metre. What is the length of a piece, in centimetres?
Answer:
Length of each piece of ribbon  = 40 cm.

Explanation:
Number of pieces of ribbon = 3.
Two of the pieces and half the third piece, laid end to end, make one metre.
Let the one piece of the ribbon be x cm.
Conversion:
1 m = 100 cm.
=> x + x + \(\frac{1}{2}\) x = 100 cm.
=> 2x + \(\frac{x}{2}\)  = 100 cm.
=> [2x(2) + x ] ÷ 2 = 100 cm.
=> (4x +
Length of each piece of ribbon  = 40 cm.x) ÷ 2 = 100 cm.
=> 4x + x = 100 × 2
=> 5x = 200
=>x = 200 ÷ 5
=> x= 40 cm.

Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions

Fraction division

The area of a rectangle is 85 square metre and the length of one of its sides is 5 metres. What is the length of the other side?
We want to find out which number should be multiplied by 5 to get 85.
For that, we should divide 85 by 5.
85 ÷ 5 = 17
So, the other side is of length 17 metres.
Suppose the question is like this:

The area of a rectangle is \(\frac{1}{2}\) square metre and the length of one side is \(\frac{3}{4}\) metre. What is the length of the other side?

We have seen that area of a rectangle is the product of the length of the sides, even if they are fractions. So in this problem, \(\frac{3}{4}\) multiplied by some number gives \(\frac{1}{2}\). What is this number?
Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions 49
We can use reciprocal to reverse this. The number is \(\frac{4}{3}\) times \(\frac{1}{2}\).
Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions 50
That is,
\(\frac{1}{2}\) × \(\frac{4}{3}\) = \(\frac{2}{3}\)
Thus the other side is \(\frac{2}{3}\) metre.
Here we see that \(\frac{2}{3}\) is the number which should be multiplied by \(\frac{3}{4}\) to get \(\frac{1}{2}\).
As in the case of natural numbers, we write this also as division.
\(\frac{1}{2}\) ÷ \(\frac{3}{4}\) = \(\frac{2}{3}\)
Let’s look at another problem:
To fill \(\frac{3}{4}\) of a vessel, 1\(\frac{1}{2}\) litres of water is needed.
How much water would it be, if it is completely filled?

We an think in terms of reciprocals. \(\frac{3}{4}\) of the vessel is 1\(\frac{1}{2}\) litres. The fall vessel is \(\frac{4}{3}\) times 1\(\frac{1}{2}\) litre.
1\(\frac{1}{2}\) × \(\frac{4}{3}\) = 2
We can also think like this: The problem says the capacity of the vessel, multiplied by \(\frac{3}{4}\) gives 1\(\frac{1}{2}\) . So the question is, by what number \(\frac{3}{4}\) should be multiplied to get 1\(\frac{1}{2}\). And it is easy to see that it is 2.
This also, we can write as division:
1\(\frac{1}{2}\) ÷ \(\frac{3}{4}\) = 2

Generally speaking, division by a fraction is just multiplication by the reciprocal.

Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions 51
Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions 52

Let’s look at a few more examples:
A 10 metre string can be cut into how many \(\frac{1}{2}\) metre pieces?
The question is, how many times \(\frac{1}{2}\) is 10.
We can easily see that it is 20 times.
Thus there are 20 pieces of string.
We can also think like this: One piece is \(\frac{1}{2}\) metre; so the total length of 10 metres is half the number of pieces.

So the question is this:
\(\frac{1}{2}\) of a number is 10: what is the number?
The number is 2 times 10, that is 20.
We can also put it as a division:
Which number multiplied by \(\frac{1}{2}\) gives 10?
To get the number, 10 must be divided by \(\frac{1}{2}\).
10 ÷ \(\frac{1}{2}\) = 10 × \(\frac{2}{1}\) = 20

Now look at this problem:
12 litres of oil is to be stored in \(\frac{3}{4}\) litre bottles. How many bottles are needed?
Each bottle is \(\frac{3}{4}\) litre. So 12 litres is \(\frac{3}{4}\) of the number of bottles. If we think like this, the question becomes this:
\(\frac{3}{4}\) of a number 12: what is the number?
We can use reciprocal to find the number.
12 × \(\frac{4}{3}\) = 16
We can also think in terms of division?
What number multiplied by \(\frac{3}{4}\) gives 12?
And the method of finding the number like this:
12 ÷ \(\frac{3}{4}\) = 12 × \(\frac{4}{3}\) =16

Now describe each of these problems either using division or reciprocals and find the answer.

Textbook Page No. 55

Question 1.
A 16 metre rod is cut into \(\frac{2}{3}\) metre pieces. How many pieces do we get?
Answer:
Number of pieces the rod is cut = 24.

Explanation:
Length of the rod = 16 meter.
Length of  each piece cut = \(\frac{2}{3}\) metre.
Number of pieces the rod is cut = Length of the rod ÷ Length of each piece
= 16 ÷  \(\frac{2}{3}\)
= 16 × \(\frac{3}{2}\)
= 8 × \(\frac{3}{1}\)
= 8 × 3
= 24.

Question 2.
5\(\frac{1}{4}\) litres of water is to be stored in \(\frac{3}{4}\) litre bottles. How many bottles do we need?
Answer:
Number of bottles needed to fill = 7.

Explanation:
Quantity of water =  5\(\frac{1}{4}\) litres.
Quantity of bottles the water is stored = \(\frac{3}{4}\) litre.
Number of bottles needed to fill = Quantity of water ÷ Quantity of bottles the water is stored
= 5\(\frac{1}{4}\) ÷ \(\frac{3}{4}\)
= {[(5 × 4) + 1] ÷4}  ÷ \(\frac{3}{4}\)
= [(20 + 1) ÷ 4] ÷ \(\frac{3}{4}\)
= \(\frac{21}{4}\) × \(\frac{4}{3}\)
= \(\frac{7}{1}\) × \(\frac{1}{1}\)
= 7.

Kerala Syllabus 6th Standard Maths Solutions Chapter 3 Fractions

Question 3.
12\(\frac{1}{2}\) kilograms of sugar is to be packed in 2\(\frac{1}{2}\) kilogram bags. How many bags do we need?
Answer:
Number of bags to be needed for packing = 5.

Explanation:
Quantity of sugar = 12\(\frac{1}{2}\) kilograms.
Quantity of bag to be packed the sugar = 2\(\frac{1}{2}\) kilogram
Number of bags to be needed for packing = Quantity of sugar ÷ Quantity of bag to be packed the sugar
= 12\(\frac{1}{2}\) ÷ 2\(\frac{1}{2}\)
= {[(12 × 2) + 1] ÷ 4} ÷ {[(2 × 2) + 1] ÷ 2}
= [(24 + 1) ÷ 4] ÷ [(4 + 1) ÷ 2]
= \(\frac{25}{2}\) ÷ \(\frac{5}{2}\)
= \(\frac{25}{2}\) × \(\frac{2}{5}\)
= \(\frac{5}{1}\) × \(\frac{1}{1}\)
= 5.

Question 4.
The area of a rectangle is 12\(\frac{1}{2}\) square centimetres and the length of one of its sides is 3\(\frac{3}{4}\) centimetres. What is the length of the other side?
Answer:
Length of another side of rectangle = \(\frac{10}{3}\) centimeters.

Explanation:
Area of a rectangle = 12\(\frac{1}{2}\) square centimetres.
Length of
one side of rectangle =  3\(\frac{3}{4}\) centimetres.
Length of another side of rectangle = ?? centimeters.
Area of a rectangle =  Length of one side of rectangle × Length of another side of rectangle
=>  12\(\frac{1}{2}\) centimeters.x =3\(\frac{3}{4}\)  × ??
=> {[(12 × 2) + 1] ÷ 2} = {[(3 × 4) + 3] ÷ 4} × ??
=> \(\frac{25}{2}\) = \(\frac{15}{4}\) × ??
=> \(\frac{25}{2}\) ÷ \(\frac{15}{4}\) = ??
=> \(\frac{25}{2}\) × \(\frac{4}{15}\) = ??
=>\(\frac{5}{1}\) × \(\frac{2}{3}\) = ??
=> (5 × 2) ÷ (1 × 3) = ??
=>\(\frac{10}{3}\) centimeters = ??

Question 5.
From 11\(\frac{1}{2}\) metre rope, 2\(\frac{1}{2}\) metre pieces are cut out. How many pieces do we get? What is the length of the remaining piece?
Answer:
Length of the remaining piece =   \(\frac{23}{5}\)  meter.
Number of pieces the rope is cut = 9.

Explanation:
Total length of rope = 11\(\frac{1}{2}\) metre
Length of the each piece of rope cut =  2\(\frac{1}{2}\) metre
Length of the remaining piece = Total length of rope ÷  Length of the each piece of rope cut
= {[(11 × 2) + 1] ÷ 2} ÷ {[(2 × 2) + 1] ÷ 2}
= [(22 + 1) ÷ 2] ÷ [(4 + 1) ÷ 2]
= (23 ÷ 2) ÷ (5 ÷ 2)
= \(\frac{23}{2}\) × \(\frac{2}{5}\)
= \(\frac{23}{1}\) × \(\frac{1}{5}\)
=  \(\frac{23}{5}\)  meter.
Number of pieces the rope is cut = 11\(\frac{1}{2}\)  – 2\(\frac{1}{2}\)
= {[(11 × 2) + 1] ÷ 2} – {[(2 × 2) + 1] ÷ 2}
= [(22 + 1) ÷ 2] – [(4 + 1) ÷ 2]
= (23 ÷ 2) – (5 ÷ 2)
= (23 – 5) ÷ 2
= (18 ÷ 2) ÷ (2 ÷ 2)
= 9 ÷ 1
=9.

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HSSLive Plus Two

Plus Two Zoology Previous Year Question Papers and Answers Kerala

HSE Kerala Board Syllabus Plus Two Zoology Previous Year Model Question Papers and Answers Pdf HSSLive Free Download in both English medium and Malayalam medium are part of SCERT Kerala Plus Two Previous Year Question Papers and Answers. Here HSSLive.Guru have given Higher Secondary Kerala Plus Two Zoology Previous Year Sample Question Papers with Answers based on CBSE NCERT syllabus.

Board SCERT, Kerala Board
Textbook NCERT Based
Class Plus Two
Subject Zoology
Papers Previous Papers, Model Papers, Sample Papers
Category Kerala Plus Two

Kerala Plus Two Zoology Previous Year Question Papers and Answers

We hope the given HSE Kerala Board Syllabus Plus Two Zoology Previous Year Model Question Papers and Answers Pdf HSSLive Free Download in both English medium and Malayalam medium will help you. If you have any query regarding HSS Live Kerala Plus Two Zoology Previous Year Sample Question Papers with Answers based on CBSE NCERT syllabus, drop a comment below and we will get back to you at the earliest.

HSSLive Plus Two

Plus Two Sociology Chapter Wise Questions and Answers Kerala

HSE Kerala Board Syllabus HSSLive Plus Two Sociology Chapter Wise Questions and Answers Pdf Free Download in both English Medium and Malayalam Medium are part of SCERT Kerala Plus Two Chapter Wise Questions and Answers. Here HSSLive.Guru has given Higher Secondary Kerala Plus Two Sociology Chapter Wise Questions and Answers based on CBSE NCERT syllabus.

Board SCERT, Kerala
Text Book NCERT Based
Class Plus Two
Subject Sociology
Chapter All Chapters
Category Kerala Plus Two

Kerala Plus Two Sociology Chapter Wise Questions and Answers

Plus Two Sociology Chapter Wise Questions and Answers Part A: Indian Society

Plus Two Sociology Chapter Wise Questions and Answers Part B: Social Change and Development in India

We hope the given HSE Kerala Board Syllabus HSSLive Plus Two Sociology Chapter Wise Questions and Answers Pdf Free Download in both English Medium and Malayalam Medium will help you. If you have any queries regarding Higher Secondary Kerala Plus Two Sociology Chapter Wise Questions and Answers based on CBSE NCERT syllabus, drop a comment below and we will get back to you at the earliest.

HSSLive Plus Two

Plus Two Maths Chapter Wise Questions and Answers Kerala

HSE Kerala Board Syllabus HSSLive Plus Two Maths Chapter Wise Questions and Answers Pdf Free Download in both English Medium and Malayalam Medium are part of SCERT Kerala Plus Two Chapter Wise Questions and Answers. Here HSSLive.Guru has given Higher Secondary Kerala Plus Two Maths Chapter Wise Questions and Answers based on CBSE NCERT syllabus. Students can also read NCERT Solution.

Board SCERT, Kerala
Text Book NCERT Based
Class Plus Two
Subject Maths
Chapter All Chapters
Category Kerala Plus Two

Kerala Plus Two Maths Chapter Wise Questions and Answers

We hope the given HSE Kerala Board Syllabus HSSLive Plus Two Maths Chapter Wise Questions and Answers Pdf Free Download in both English Medium and Malayalam Medium will help you. If you have any query regarding Higher Secondary Kerala Plus Two Maths Chapter Wise Questions and Answers based on CBSE NCERT syllabus, drop a comment below and we will get back to you at the earliest.

HSSLive Plus Two

Plus Two Computer Science Notes Chapter Wise HSSLive Kerala

HSE Kerala Board Syllabus HSSLive Plus Two Computer Science Notes Chapter Wise Pdf Free Download in both English Medium and Malayalam Medium are part of SCERT Kerala HSSLive Plus Two Notes. Here HSSLive.Guru has given Higher Secondary Kerala Plus Two Computer Science Chapter Wise Quick Revision Notes based on CBSE NCERT syllabus.

Board SCERT, Kerala
Text Book NCERT Based
Class Plus Two
Subject Computer Science
Chapter All Chapters
Category Kerala Plus Two

Kerala Plus Two Computer Science Notes Chapter Wise

We hope the given HSE Kerala Board Syllabus HSSLive Plus Two Computer Science Notes Chapter Wise Pdf Free Download in both English Medium and Malayalam Medium will help you. If you have any query regarding Higher Secondary Kerala Plus Two Computer Science Chapter Wise Quick Revision Notes based on CBSE NCERT syllabus, drop a comment below and we will get back to you at the earliest.

HSSLive Plus Two

Plus Two Political Science Chapter Wise Questions and Answers Kerala

HSE Kerala Board Syllabus HSSLive Plus Two Political Science Chapter Wise Questions and Answers Pdf Free Download in both English Medium and Malayalam Medium are part of SCERT Kerala Plus Two Chapter Wise Questions and Answers. Here HSSLive.Guru has given Higher Secondary Kerala Plus Two Political Science Chapter Wise Questions and Answers based on CBSE NCERT syllabus.

Board SCERT, Kerala
Text Book NCERT Based
Class Plus Two
Subject Political Science
Chapter All Chapters
Category Kerala Plus Two

Kerala Plus Two Political Science Chapter Wise Questions and Answers

Plus Two Political Science Part A – Contemporary World Politics

Plus Two Political Science Part B – Politics In India Since Independence

We hope the given HSE Kerala Board Syllabus HSSLive Plus Two Political Science Chapter Wise Questions and Answers Pdf Free Download in both English Medium and Malayalam Medium will help you. If you have any query regarding Higher Secondary Kerala Plus Two Political Science Chapter Wise Questions and Answers based on CBSE NCERT syllabus, drop a comment below and we will get back to you at the earliest.

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