Fractions Class 6 Solutions | NCERT Ganita Prakash Chapter 7 Questions and Answers

 NCERT Solutions for Class 6 Maths Ganita Prakash Chapter 7 Fractions

7.1 Fractional Units and Equal Shares Figure it Out (Page No. 152 – 153)

Fill in the blanks with fractions.

Question 1.

Three guavas together weigh 1 kg. If they are roughly of the same size, each guava will roughly weigh ____kg.

Solution:

13

Question 2.

A wholesale merchant packed 1 kg of rice in four packets of equal weight. The weight of each packet is _________ kg.

Solution:

14

Question 3.

Four friends ordered 3 glasses of sugarcane juice and shared it equally. Each one drank ____________ glass of sugarcane juice.

Solution:

34

As total quantity is 3 which is to be divided into four equal parts. So, the required fraction is 

34

.

Question 4.

The bis fish weighs 

12

 kg. The small one weighs 14

 kg. Together they weigh ____ kg.

Solution:

Given the weighs of big fish = 

12

kg and the weighs of small fish = 14

kg

Total weight of both fish = 

12

+

14

=

2+14

 kg

= 

34

 kg

Question 5.

Arrange these fraction words in order of size from the smallest to the biggest in the empty box below: One and a half, three quarters, one and a quarter, half, quarter, two and a half.

Solution:

∴ The fractions from smallest to the biggest are as follows: quarter, half, three quarters one and a quarter, one and a half, two and a half.

7.2 Fractional Units as Parts of a Whole Figure it Out (Page No. 155)

Question 1.

The figure below shows different fractional units of a whole chikki. How much of a whole chikki is each piece?

Solution:

(a)

Total no. of pieces formed of given size = 12

Required fraction = 

112

(b)

Total no. of pieces formed of given size = 4

Required fraction = 

14

(c)

Total no. of pieces formed of given size = 8

Required fraction = 

18

(d)

Total no. of pieces formed of given size = 6

Required fraction = 

16

(e)

Total no. of pieces formed of given size = 8

Required fraction = 

18

(f)

Total no. of pieces formed of given size = 8

Required fraction = 

18

(g)

Total no. of pieces formed of given size = 24

Required fraction = 

124

(h)

Total no. of pieces formed of given size = 12

Required fraction = 

112

7.3 Measuring Using Fractional Units Figure it Out (Page No. 158)

Question 1.

Continue this table of 

12

 for 2 more steps.

Solution:

Question 2.

Can you create a similar table for 

14

 ?

Solution:

Yes.

Question 3.

Make 

13

 using a paper strip. Can you use this to also make 16

?

Solution:

Take a strip of paper.

Fold the strip into three equal parts and then open up.

Yes, we can also make 

16

 using a paper strip by folding 6 again the above strip.

Question 4.

Draw a picture and write an addition statement as above to show:

(a) 5 times 

14

 of a roti

Solution:

5 times 

14

 of a roti

= 

14

+

14

+

14

+

14

+

14

(b) 9 times 

14

 of a roti

Solution:

9 times 

14

 of a roti

= 

14

+

14

+

14

+

14

+

14

+

14

+

14

+

14

+

14

Question 5.

Match each fractional unit with the correct picture:

Solution:

7.4 Marking Fraction Lengths on the Number Line Figure it Out (Page No. 160)

Question 1.

On a number line, draw lines of length 

110

, 310

, and 45

.

Solution:

Divide the unit into 10 equal parts and point A represents 

110

.

Divide a unit into 10 equal parts and point B represents 

310

.

Divide a unit into 5 equal parts and point C represents 

45

.

Question 2.

Write five more fractions of your choice and mark them on the number line.

Solution:

The fractions are 

35

,

13

,

57

,

25

 and 18

.

Their number line representations are:

Question 3.

How many fractions lie between 0 and 1? Think, discuss with your classmates, and write your answer.

Solution:

There are an infinite number of fractions between 0 and 1.

Example: 

35

,

45

,

710

12

 etc.

Question 4.

What is the length of the pink line and black line shown below? The distance between 0 and 1 is 1 unit long, and it is divided into two equal parts. The length of each part is 

12

. So the pink line is y units long. Write the fraction that gives the length of the black line in the box.

Solution:

Length of black line is 

12

;

Length of black line is 

12

 + 12

 + 12

Fraction that gives length of black line = 

32

Question 5.

Write the fraction that gives the lengths of the black lines in the respective boxes.

Solution:

Intext Questions

Question 1.

Here, the fractional unit is dividing a length of 1 unit into three equal parts. Write the fraction that gives the length of the pink line in the box or in your notebook. (Page 159)

Solution:

Here number line OR is divided into three equal parts OP, PQ and QR.

Hence length of pink line = OP + PQ = 

13

+

13

=

23

Question 2.

Here, a unit is divided into 5 equal parts. Write the fraction that gives the length of the pink lines in the respective boxes or in your notebook.

Solution:

Here number line OT = 1 unit is divided into five equal parts OP, PQ, QR, RS and ST.

Hence length of pink line OQ = OP + PQ = 

15

+

15

=

25

Now, length of pink line OS = OP + PQ + QR + RS = 

15

+

15

+

15

+

15

=

45

Hence, OQ = 

25

 OS = 45

 

Question 3.

Now, a unit is divided into 8 equal parts. Write the appropriate fractions in your notebook Solution:Here number line OH is divided into 8 equal parts OA, AB, BC, CD, DE, EF, FG and GH.

Solution:

Also, OA = 

18

, OB = 28

, OC = 38

, OH = 88

 = 1

7.5 Mixed Fractions Figure it Out (Page No. 162)

Question 1.

How many whole units are there in 

72

?

Solution:

72

=

12

+

12

+

12

+

12

+

12

+

12

+

12

=3+

12

So, there are 3 whole units in 

72

.

Question 2.

How many whole units are there in 

43

 and in 73

?

Solution:

43

=

13

+

13

+

13

+

13

=1+

13

So, there are 1 whole unit in 

43

.
73

=

13

+

13

+

13

+

13

+

13

+

13

+

13

=2+

13

So, there are 2 whole units in 

73

.

7.5 Mixed Fractions Figure it Out (Page No. 162)

Question 1.

Figure out the number of whole units in each of the following fractions:

(a) 

83

(b) 

115

(c) 

94

Solution:

(a) 2

(b) 2

(c) 2

Question 2.

Can all fractions greater than 1 be written as such mixed numbers?

Solution:

Yes.

Question 3.

Write the following fractions as mixed fractions (e.g. 

92

 = 412

)

(a) 

92

Solution:

= 4

12

(b) 

95

Solution:

= 1

45

(c) 

2119

Solution:

= 1

219

(d) 

479

Solution:

= 5

29

(e) 

1211

Solution:

= 1

111

(f) 

196

Solution:

= 3

16

7.5 Mixed Fractions Figure it Out (Page No. 163)

Question 1.

Write the following mixed numbers as fractions:

(a) 3

14

(b) 7

23

(c) 9

49

(d) 3

16

(e) 2

311

(f) 3

910

Solution:

7.6 Equivalent Fractions Figure it Out 7.7 Simplest form of a Fraction Figure it Out (Page No. 165)

Question 1.

Are 

36

,

48

,

510

 equivalent fractions? Why?

Solution:

Here, simplest form of 

36

=

3÷36÷3

=

12

 [HCF of 3 and 6 is 3]

and simplest form of 

48

 is 4÷48÷4

=

12

 [HCF of 4 and 8 is 4]

and simplest form of 

510

 is 5÷510÷5

=

12

 [HCF of 5 and 10 is 5]

Hence, 

36

,

48

,

510

 are equivalent fractions.

Question 2.

Write two equivalent fractions for 

26

.

Solution:

From the fractional wall we can choose any two fractions that denote the same length as 

26

⋅

26

=

13

=

39

Question 3.

46

 = ___________ = ___________ = ___________ = ___________________

(Write as many as you can)

Solution:

Here,

Intext Questions

Answer the following questions after looking at the fraction wall: [Page 164]

Question 1.

Are the lengths 

12

 and 36

 equal?

Solution:

Yes, here lengths 

12

 and 36

 = 12

Lengths are equal.

Question 2.

Are 

23

 and 46

 equivalent fractions? Why?

Solution:

Yes, lengths 

23

 and 46

 = 13

 are equivalent fraction, as they have same length.

Question 3.

How many pieces of length 

16

 will make a length of 12

?

Solution:

Total no.of pieces = 

12

16

=

12

×

61

=

62

 = 3

Hence three pieces of length 

16

 will make a length of 12

Question 4.

How many pieces of length 

16

 will make a length of 13

?

Solution:

Total no. of pieces = 

13

16

=

13

×

61

=

63

 = 2

Hence two pieces of length 

16

 will make a length of 13

.

7.7 Simplest form of a Fraction Figure it Out (Page No. 166)

Question 1.

Three rotis are shared equally by four children, show the division in the picture and write a fraction of how much each child gets. Also, write the corresponding division facts, addition facts, and, multiplication facts.

The fraction of roti each child gets is ___________

Division fact:

Addition fact:

Multiplication fact:

Compare your picture and answer with your classmates!

Solution:

One roti is shared as shown in the figure below:

The four shares must be equal to each other!

Similar distribution will be done for the second and third roti also.

So, each child will get 

34

 a piece of roti.

The division fact is 3 ÷ 4 = 

34

The addition fact is 3 = 

34

+

34

+

34

+

34

The multiplication fact is 3 = 4 × 

34

Question 2.

Draw a picture to show how much each child gets when 2 rotis are shared equally by 4 children. Also, write the corresponding division facts, addition facts, and multiplication facts.

Solution:

One roti is shared as shown in the figure below:

The four shares must be equal to each other!

A similar distribution will be done for the second roti also.

So, each child will get 

14

 part from a rod.

So, the total fraction of roti received by each child from 2 rotis = 

24

 = 12

The division fact is 2 ÷ 4 = 

24

The addition fact is = 

24

+

24

+

24

+

24

The multiplication fact is 2 = 4 × 

24

Question 3.

Anil was in a group where 2 cakes were divided equally among 5 children. How much cake would Anil get?

Solution:

Anil would get 

25

 part of the cake.

7.7 Simplest form of a Fraction Figure it Out (Page No. 168 – 169)

Question 1.

Find the missing numbers:

(a) 5 glasses of juice shared equally among 4 friends is the same as ____________ glasses of juice shared equally among 8 friends. So, 

54

 = ?8

.

(b) 4 kg of potatoes divided equally in 3 bags is the same as 12 kgs of potatoes divided equally in ____________ bags. So, 

43

 = 12?

.

(c) 7 rods divided among 5 children is the same as rods divided among children. So, 

75

 = ____________

Solution:

(a) Here, the amount of juice each friend gets when 5 glasses are shared among 4 friends = 

 number of glasses  number of friends 

=

54

Now to determine how many glasses of juice would be needed to give each of the 8 friends the same amount = 8 × 

54

= 10 glasses

So, 10 glasses of juice shared equally among 8 friends is the same as 5 glasses of juice shared equally among 4 friends.

∴ 

54

=

108

(b) Here 4 kg of potatoes divided equally in 3 bags then amount of potatoes per bag = 

4 kg3 bags 

=

43

 kg per bag

Let x is the number of bags for 12 kg of potatoes, where each bag has the same amount of potatoes then

12 kgx bags 

=

43

 kg per bag

⇒ 12 × 3 = 4 × x

⇒ 36 = 4x

⇒ x = 

364

⇒ x = 9

∴ 

43

=

129

(c) Dividing 7 rotis among 4 children gives 7 each child = 

75

 of a roti. We can find an

equivalent fraction by multiplying both the numerator and the denominator by the same number. For example, multiplying both by 2.

7×25×2

=

1410

So, 7 rotis divided among 5 children is the same as 14 rotis divided among 10 children

∴ 

75

=

1410

Intext Questions

Question 1.

Find equivalent fractions for the given pairs of fractions such that the fractional units are the same. (Page 172)

(a) 

72

 and 35

Solution:

Given fractions are 

72

 and 35

Here, the denominators are 2 and 5.

And least common multiple of 2 and 5 is 10.

Hence for both fractions let’s have same denominator of 10.

Now for 

72

 multiply both the numerator and the denominator by 5.
72

=

7×52×5

=

3510

And for 

35

 multiply both the numerator and the denominator by 2, we get,
3×25×2

=

610

Hence, the equivalent fractions with the same denominator are:

3510

 and 610

(b) 

83

 and 56

Solution:

Given fractions are 

83

 and 56

Here, the denominators are 3 and 6.

And least common multiple of 3 and 6 is 6.

Now for 

83

 multiply both the numerator and the denominator by 2.
83

=

8×23×2

=

166

56

 already have a denominator 6.

Hence, the equivalent fractions with the same denominator are:

166

 and 56

(c) 

34

 and 35

Solution:

Given fractions are 

34

 and 35

Here, the denominators are 4 and 5.

And least common multiple of 4 and 5 is 20.

Now for 

34

 multiply both the numerator and the denominator by 5.
34

=

3×54×5

=

1520

And for 

35

 multiply both the numerator and the denominator by 4, we get
35

=

3×45×4

=

1220

So, the equivalent fractions with the same denominator are:

1520

 and 1220

(d) 

67

 and 85

Solution:

Given fractions are 

67

 and 85

Here, the denominators are 7 and 5.

And least common multiple of 7 and 5 is 35.

Now for 

67

 multiply both the numerator and the denominator by 5.
67

=

6×57×5

=

3035

And for 

85

 multiply both the numerator and the denominator by 7, we get
85

=

8×75×7

=

5635

So, the equivalent fractions with the same denominator are:

3035

 and 5635

(e) 

94

 and 52

Solution:

Given fractions are 

94

 and 52

Here, the denominators are 4 and 2.

And least common multiple of 4 and 2 is 4.

Now for 

52

 multiply both the numerator and the denominator by 2.
52

=

5×22×2

=

104

and 

94

 already have a denominator 4

So, the equivalent fractions with the same denominator are:

94

 and 104

(f) 

110

 and 29

Solution:

Given fractions are and 

110

 and 29

Here, the denominators are 10 and 9.

And least common multiple of 10 and 9 is 90.

Now for 

110

 multiply both the numerator and the denominator by 9.
110

=

1×910×9

=

990

And for 2 multiply both the numerator and the denominator by 10, we get

29

=

2×109×10

=

2090

So, the equivalent fractions with the same denominator are:’

990

 and 2090

(g) 

83

 and 114

Solution:

Given fractions are 

83

 and 114

Here, the denominators are 3 and 4.

And least common multiple of 3 and 4 is 12.

Now for 

83

 multiply both the numerator and the denominator by 4.
83

=

8×43×4

=

3212

And for 

114

 multiply both the numerator and the denominator by 3, we get
114

=

11×34×3

=

3312

So, the equivalent fractions with the same denominator are:

3212

 and 3312

(h) 

136

 and 19

Solution:

Given fractions are 

136

 and 19

Here, the denominators are 6 and 9.

And least common multiple of 6 and 9 is 18.

Now for 

136

 multiply both the numerator and the denominator by 3.
136

=

13×36×3

=

3918

And for 

19

 multiply both the numerator and the denominator by 2, we get
19

=

1×29×2

=

218

So, the equivalent fractions with the same denominator are:

3918

 and 218

7.7 Simplest form of a Fraction Figure it Out (Page No. 173)

Question 1.

Express the following fractions in lowest terms:

(a) 

1751

Solution:

13

(b) 

64144

Solution:

49

(c) 

126147

Solution:

67

(d) 

525112

Solution:

7516

7.8 Comparing Fractions Figure it Out (Page No. 174)

Question 1.

Compare the following fractions and justify your answers:

(a) 

83

,

52

(b) 

49

,

37

(c) 

710

,

914

(d) 

125

,

85

(e) 

94

,

52

Solution:

Question 2.

Write following fractions ascending order.

(a) 

710

,

1115

,

25

Solution:

The given fractions are 

710

,

1115

,

25

Let us find LCM of denominator 10, 15, 5

∴ LCM of 10, 15 and 5 = 2 × 3 × 5 = 30

Now let us make denominator of each fractions as LCM

Hence given fractions in ascending order are: 

25

,

710

115

(b) 

1924

,

56

,

712

Solution:

The given fractions are 

1924

,

56

,

712

Here LCM of 24, 6, 12 is 24.

On arranging in ascending Order, we get

1424

,

1924

,

2024

⇒ 

712

,

1924

,

56

Question 3.

Write the following fractions in descending order.

(a) 

2516

,

78

,

134

,

1732

Solution:

(b) 

34

,

125

,

712

,

54

Solution:

7.9 Relation to Number Sequences Figure it Out (Page No. 179)

Question 1.

Add the following fractions using Brahmagupta’s method:

(a) 

27

+

57

+

67

Solution:

27

+

57

+

67

= 

2+5+67

=

137

=1 

67

(b) 

34

+

13

Solution:

34

+

13

=

34

×

33

+

13

×

44

= 

912

+

412

=

9+412

= 

1312

= 1 

112

(c) 

23

+

56

Solution:

23

+

56

=

23

×

22

+

56

46

+

56

=

96

=

32

= 1 

12

(d) 

23

+

27

Solution:

23

+

27

=

23

×

77

+

27

×

33

= 

1421

+

621

=

2021

(e) 

34

+

13

+

15

Solution:

4560

+

2060

+

1260

= 

7760

= 1

1760

(f) 

23

+

45

Solution:

1015

+

1215

=

2215

= 1

715

(g) 

45

+

23

Solution:

1215

+

1015

=

2215

= 1

715

(h) 

35

+

58

Solution:

2440

+

2540

=

4940

= 

940

(i) 

92

+

54

Solution:

184

+

54

=

234

= 5

34

(j) 

83

+

27

Solution:

5621

+

621

=

6221

= 2

2021

(k) 

34

+

13

+

15

Solution:

4560

+

2060

+

1260

=

7760

= 1

1760

(l) 

23

+

45

+

37

Solution:

70105

+

84105

+

45105

=

199105

= 1

94105

(m) 

92

+

54

+

76

Solution:

5412

+

1512

+

1412

=

8312

= 

1112

Question 2.

Rahim mixes 

23

 liters of yellow paint with 34

 liters of blue paint to make green paint What is the volume of green paint he has made?

Solution:

Quantity of yellow paint added = 

23

 litres

Quantity of blue paint added = 

34

 litres

Total quantity of green paint made = 

23

 + 34

LCM of 3 and 4 is 12.

23

=

23

×

44

=

812

34

=

34

×

33

=

912

812

+

912

=

8+912

=

1712

So, the total quantity of paint made is 

1712

 liters.

Question 3.

Geeta bought 

25

 meter of lace and Shamim bought 34

 meter of the same lace to put a complete border on a table cloth whose perimeter is 1 meter long. Find the total length of the lace they both have bought. Will the lace be sufficient to cover the whole border?

Solution:

Length of lace bought by Geeta = 

25

 m

Length of lace bought by Shamim = 

34

 m

Total length of lace bought = 

25

 + 34

LCM of 5 and 4 is 20.

25

=

25

×

44

=

820

34

=

34

×

55

=

1520

820

+

1520

=

2320

=1

320

This length is more than 1 m. So, lace is more than sufficient or will be left extra after covering the border.

7.9 Relation to Number Sequences Figure it Out (Page No. 181)

Question 1.

58

−

38

Solution:

Given 

58

−

38

As fractional unit is same i.e., 

18

 we shall simply subtract numerators keeping fractional unit as 18

Then 

58

−

38

=

5−38

= 

28

=

14

 (representing in simplest form)

Question 2.

79

−

59

Solution:

Given 

79

−

59

As fractional unit is same i.e., 

19

 we shall simply subtract numerators keeping fractional unit as 19

79

−

59

= 

7−59

=

29

Question 3.

1027

−

127

1027

−

127

Solution:

Here 

1027

−

127

= 

10−127

= 

927

=

13

7.9 Relation to Number Sequences Figure it Out (Page No. 182)

Question 1.

Carry out the following subtractions using Brahmagupta’s method:

(a) 

815

−

315

Solution:

Given 

815

−

315

Fractional unit for both fractions is 

115

 then
815

−

315

=

8−315

= 

515

=

13

(b) 

25

−

415

Solution:

Given 

25

−

415

Here LCM of 5 and 15 is 15. Fractional unit for both fractions should be 

115

then 

2×35×3

−

4×115×1

= 

615

−

415

= 

6−415

= 

215

(c) 

56

−

49

Solution:

Given 

56

−

49

Hence LCM of 6 and 9 is 18. Fractional unit for both fractions should be 

118

 then

(d) 

23

−

12

Solution:

Given 

23

−

12

Here LCM of 3 and 2 is 6. Fractional unit for both fractions should be 

16

Question 2.

Subtract as indicated:

(a) 

134

 from 103

Solution:

The denominators of the given fractions are 3 and 4. The LCM of 3 and 4 is 12.

Then 

134

=

13×34×3

=

3912

,

103

=

10×43×4

=

4012

Therefore, 

103

−

134

=

4012

−

3912

=

112

(b) 

185

 from 233

Solution:

The denominators of the given fractions are 3 and 5.

The LCM of 3 and 5 is 15.

Then, 

233

=

23×53×5

=

11515

,

185

=

18×35×3

=

5415

Therefore, 

233

−

185

=

11515

−

5415

=

6115

=4

115

(c) 

297

 from 457

Solution:

The denominators are same.

Therefore, 

457

−

297

=

167

=2

27

Question 3.

Solve the following problems:

(a) Jaya’s school is 

710

 km from her home. She takes an auto for 12

 km from her home daily, and then walks the remaining distance to reach her school. How much does she walk daily to reach the school?

Solution:

(a) Total distance between school and home = 

710

 km

Distance travelled in Auto = 

12

 km.

∴ Distance she walks daily to reach the school

(b) Jeevika takes 

103

 minutes to take a complete round of the park and her friend Namit takes 134

 minutes to do the same. Who takes less time and by how much?

Solution:

Time taken by Jeevika = 

103

 minutes

and time taken by Narnit = 

134

 minutes

Now, 

103

×

44

=

4012

 and 134

×

33

=

3912

Clearly, 

103

 > 134

∴ Jeevika takes less ti me by 

(

103

−

134

)

 minutes

= 

(

4012

−

3912

)

 minutes

= 

112

 minutes.