NCERT Solutions For Class 6 Maths Chapter 7 Fractions free pdf

NCERT Solutions For Class 6 Maths Chapter 7 Fractions

Here at MentorAtHome, We provide NCERT Solutions For Class 6 Maths Chapter 7 Fractions. NCERT solutions help students to improve their performance, and attain better results. In NCERT Solutions Class 6 Maths Chapter 6 there is 3 exercise.

NCERT Solutions For Class 6 Maths Chapter 7 fraction, will discuss fraction. NCERT Solutions For Class 6 Maths Chapter 7 is important for basic understanding.

NCERT Solutions For Class 6 Maths Chapter 7 Fractions Exercises 7.1

Class 6th Maths Chapter 7 Solutions ex 7.1

Q.1. Write the fraction representing the shaded portion.

Ans:-
(i) Total number of parts = 4
Number of shaded parts = 2
∴ Fraction = 2/4

(ii) Total number of parts = 9
Number of shaded parts = 8
∴ Fraction = 8/9

(iii) Total number of parts = 8
Number of shaded parts = 4
∴ Fraction = 4/8

(iv) Total number of parts = 4
Number of shaded parts = 1
∴ Fraction = 1/4

(v) Total number of parts = 7
Number of shaded parts = 3
∴ Fraction = 3/7

(vi) Total number of parts = 12
Number of shaded parts = 3
∴ Fraction = 3/1

(vii) Total number of parts = 10
Number of shaded parts = 10
∴ Fraction = 10/10

(viii) Total number of parts = 9
Number of shaded parts = 4
∴ Fraction = 4/9

(ix) Total number of parts = 8
Number of shaded parts = 4
∴ Fraction = 4/8

(x) Total number of parts = 2
Number of shaded part = 1
∴ Fraction = 1/2

Q.2. Color the part according to the given fraction.

Ans:-

Q.3. Identify the error, if any.

Ans:-
(a) the shaded part is not half.
∴ This is not 1/2.
(b) the parts are not equal.
∴ The shaded part is not 1/4
(c) the part is not equal.
∴ The shaded part is not 3/4

Q.4. What fraction of a day is 8 hours?
Ans:-
Since a day has 24 hours
we have 8 hours,
∴ Required fraction = 8/24

Q.5. What fraction of an hour is 40 minutes?
Ans:-

Since I hours = 60 minutes
we have 40 minutes
∴ Required fraction = 40/60

Q.6. Arya, Abhimanyu, and Vivek shared lunch. Arya has brought two sandwiches, one made of vegetable and one of jam. The other two boys forgot to bring their lunch. Arya agreed to share his sandwiches so that each person will have an equal share of each sandwich.
(a) How can Arya divide his sandwiches so that each person has an equal share?
Ans:-
Arya has divided his sandwich into three equal parts.
So, each of them will get one part.
(b) What part of a sandwich will each boy receive?
Ans:-
Each one of them will receive 1/3 part.
∴ Required fraction = 1/3

Q.7. Kanchan dyes dresses. She had to dye 30 dresses. She has so far finished 20 dresses. What fraction of dresses has she finished?
Ans:-
Total number of dresses to be dyed = 30
Number of dresses finished = 20
∴ Required fraction = 20/30 = 2/3A

Q.8. Write the natural numbers from 2 to 12. What fraction of them are prime numbers?
Ans:-
Natural numbers between 2 and 12 are;
2,3,4, 5, 6, 7, 8, 9, 10,11, 12
Number of given natural numbers = 11
Number of prime numbers = 5
∴ Required fraction = 5/11

Q.9. Write the natural numbers from 102 to 113. What fraction of them are prime numbers?
Ans:-
Natural numbers from 102 to 113 are;
102,103,104,105,106, 107,108, 109,110, 111, 112,113
Total number of given natural numbers = 12
Prime numbers are 103, 107, 109, 113
The prime numbers = 4
∴ Required fraction = 4/12 = 1/3

Q.10. What fraction of these circles have X’s in them?

Ans:-
Total number of circles = 8
Number of circles having X’s in them = 4
Required fraction = 4/8 = 1/2

Q.11. Kristin received a CD player for her birthday. She bought 3 CDs and received 5 others as gifts. What fraction of her total CDs did she buy and what fraction did she receive as gifts?
Ans:-
Number of CDs bought by her from the market = 3
Number of CD’s received as gifts = 5
Total number of CDs = 3 + 5 = 8
∴ Fraction of CD (bought) = 3/8
∴ the fraction of CDs (gifted) = 5/8

NCERT Solutions For Class 6 Maths Chapter 7 Fractions Exercises 7.2

NCERT Class 6 Maths Chapter 7 Solutions ex 7.2

Q.1. Draw number lines and locate the points on them:

We have divided the number line from 0 to 1 into four equal parts.
C represents 2/4 i,e., 1/2
B represents 1/4
D represents 3/4
and E represents 4/4, i.e., 1.

We have divided the number line from 0 to 1 into eight equal parts.
B represents 1/8 
C represents 2/8
D represents 3/8
and H represents 7/8

From the above number line, we have
C represents 2/5
D represents 3/5
E represents 4/5
and I represent 8/5

Q.2. Express the following as mixed fractions:

Q.3. Express the following as improper fractions:

NCERT Solutions For Class 6 Maths Chapter 7 Fractions Exercises 7.3

NCERT Class 6 Maths Chapter 7 Solutions ex 7.3

Q.1. Write the fractions. Are all these fractions equivalent?

All the fractions in their simplest form are not equal
Hence, they are not equivalent fractions.

Q.2. Write the fractions and pair up the equivalent fractions from each row.

The following pairs of fractions: represent the equivalent fractions.
(a) and (ii) = 1/2
(b) and (iv) = 2/3
(c) and (i) = 1/3
(d) and (v) = 1/4
(e) and (iii) = 3/4

Q.3. Replace ☐ in each of the following by the correct number:

Q.4. Find the equivalent fraction of 3 / 5 having
(a) denominator 20
(b) numerator 9
(c) denominator 30
(d) numerator 27
Ans:-
(a) We require denominator 20
Let N be the numerator of the fractions

(b) We require numerator 9
Let D be the denominator of the fractions

(c) We require denominator 30
Let N be the numerator of the fraction

(d) We require numerator 27
Let D be the denominator of the fraction

Q.5. Find the equivalent fraction of 36 / 48 with
(a) numerator 9
(b) denominator 4
Ans:-
(a) We require numerator 9
Let D be the denominator of the fraction

(b) We require denominator 4
Let N be the numerator of the fraction

Q.6. Check whether the given fractions are equivalent:

Ans:-
We have 5 x 54 = 270
and 9 x 30 = 270
Here 5 x 54 = 9 x 30
They are equivalent fractions

Ans:-
We have 3 x 50 = 150
and 10 x 12 = 120
Here 3 x 50 ≠ 10 x 12
They are not equivalent fractions

Ans:-
We have 7 x 11 = 77
and 5 x 13 = 65
Here 7 x 11 ≠ 5 x 13
They are not equivalent fractions

Q.7. Reduce the following fractions to simplest form:

Ans :-
(a) we know that 48 and 60 both are divisible by 12 so.
48 / 60 = (12 × 4) / (12 × 5)
hance 48 / 60 = 4 / 5
(b) we know that 150 and 60 both are divisible by 30 so
150 / 60 = (30 × 5) / (30 × 2)
hance 150 / 60 = 5 / 2
(c) we know that 84 and 98 both are divisible by 14 so
84 / 98 = (14 × 6) / (14 × 7)
hance 84 / 98 = 6 / 7
(d) we know that 12 and 52 both are divisible by 4 so
12 / 52 = (3 × 4) / (13 × 4)
hance 84 / 98 = 3 / 13
(e) we know that 7 and 28 both are divisible by 7 so
7 / 28 = 7 / (7 × 4)
hance 7 / 28 =1 / 4

NCERT Solutions For Class 6 Maths Chapter 7 Fractions Exercises 7.4

Ch 7 Class 6 Maths Solutions ex 7.4

1. Write shaded portion as a fraction. Arrange them in ascending and descending order using the correct sign ‘’ between the fractions:

(c) Show 2 / 6, 4 / 6, 8 / 6 and 6 / 6 on the number line. Put appropriate signs between the fractions given.

Ans:-
(a) (i) Total number of divisions = 8
Number of shaded parts = 3
∴ Fraction = 3 / 8
(ii) Total number of divisions = 8
Number of shaded parts = 6
∴ Fraction = 6 / 8
(iii) Total number of divisions = 8
Number of shaded parts = 4
∴ Fraction = 4 / 8
(iv) Total number of divisions = 8
Number of shaded parts = 1
∴ Fraction = 1 / 8
∴ The arranged fractions are:

Ans:-
(b) (i) Total number of divisions = 9
Number of shaded parts = 8
∴ Fraction = 8 / 9
(ii) Total number of divisions = 9
Number of shaded parts = 4
∴ Fraction = 4 / 9
(iii) Total number of divisions = 9
Number of shaded parts = 3
∴ Fraction = 3 / 9
(iv) Total number of divisions = 9
Number of shaded parts = 6
∴ Fraction = 6 / 9
∴ The arranged fractions are:

Q.2. Compare the fractions and put an appropriate sign.

Ans:-
(a) denominators of the two fractions are the same and 3 < 5.
∴ 3 / 6 < 5 / 6
(b) numerators of the fractions are the same So, the fraction having a lesser denominator is the greater
∴ 1 / 7 < 1 / 4
(c) denominators of the two fractions are the same and 4 < 5.
∴ 4 / 5 < 5 / 5
(d) numerators of the two fractions are the same So, the fraction having a lesser denominator is the greater
∴ 3 / 7 < 3 / 5

Q.3. Make five more such pairs and put appropriate signs.
Ans:-

Q.4. Look at the figures and write ‘<’ or ‘>’, ‘=’ between the given pairs of fractions

5. How quickly can you do this? Fill appropriate sign. ( ‘<’, ‘=’, ‘>’)
(a) 1 / 2 ☐ 1 / 5 (b) 2 / 4 ☐ 3 / 6
(c) 3 / 5 ☐ 2 / 3 (d) 3 / 4 ☐ 2 / 8
(e) 3 / 5 ☐ 6 / 5 (f) 7 / 9 ☐ 3 / 9
(g) 1 / 4 ☐ 2 / 8 (h) 6 / 10 ☐ 4 / 5
(i) 3 / 4 ☐ 7 / 8 (j) 6 / 10 ☐ 3 / 5
(k) 5 / 7 ☐ 15 / 21
Ans:-
(a) 1 / 2 ☐ 1 / 5
numerators of the two fractions are the same So, the fraction having a lesser denominator is the greater
∴ 1 / 2 > 1 / 5
(b) 2 / 4 ☐ 3 / 6
We have the cross multiple are
2 x 6 = 12
4 x 3 = 12
12 = 12
∴ 2 / 4 = 3 / 6
(c) 3 / 5 ☐ 2 / 3
We have the cross multiple are
3 x 3 = 9
5 x 2 = 10
9 < 10
∴ 3 / 5 < 2 / 3
(d) 3 / 4 ☐ 2 / 8
We have the cross multiple are
3 x 8 = 24
4 x 2 = 8
24 > 8
∴ 3 / 4 > 2 / 8
(e) 3 / 5 ☐ 6 / 5
denominators of the two fractions are same and 3 < 6
∴ 3 / 5 < 6 / 5.
(f) 7 / 9 ☐ 3 / 9
denominators of the two fractions are same and 7 > 3
∴ 7 / 9 > 3 / 9
(g) 1 / 4 ☐ 2 / 8
We have the cross multiple are
1 x 8 = 8
4 x 2 = 8
8 = 8
∴ 1 / 4 = 2 / 8
(h) 6 / 10 ☐ 4 / 5
We have the cross multiple are
6 x 5 = 30
10 x 4 = 40
30 < 40
∴ 6 / 10 < 4 / 5
(i) 3 / 4 ☐ 7 / 8
We have the cross multiple are
3 x 8 = 24
4 x 7 = 28
24 < 28
∴ 3 / 4 < 7 / 8
(j) 6 / 10 ☐ 3 / 5
We have the cross multiple are
6 x 5 = 30
10 x 3 = 30
30 = 30
∴ 6 / 10 = 3 / 5
(k) 5 / 7 ☐ 15 / 21
We have the cross multiple are
5 x 21 = 105
7 x 15 = 105
105 =105
∴ 5 / 7 = 15 / 21

6. The following fractions represent just three different numbers. Separate them into three groups of equivalent fractions, by changing each one to its simplest form.
(a) 2 / 12 (b) 3 / 15 (c) 8 / 50
(d) 16 / 100 (e) 10 / 60 (f) 15 / 75
(g) 12 / 60 (h) 16 / 96 (i) 12 / 75
(j) 12 / 72 (k) 3 / 18 (l) 4 / 25
Ans:-
(a) 2 / 12
we know that 2 and 12 both are divisible by 2 so.
2 / 12 = (1 × 2) / (6 × 2)
hance 2 / 12 = 1 / 6
(b) 3 / 15
we know that 3 and 15 both are divisible by 3 so.
3 / 15 = (1 × 3) / (5 × 3)
hance 3 / 15 = 1 / 5
(c) 8 / 50
we know that 8 and 50 both are divisible by 2 so
8 / 50 = (4 × 2) / (25 × 2)
hance 8 / 50 = 4 / 25
(d) 16 / 100
we know that 16 and 100 both are divisible by 4 so
16 / 100 = (4 × 4) / (25 × 4)
hance 16 / 100 = 4 / 25
(e) 10 / 60
we know that 10 and 60 both are divisible by 10 so
10 / 60 = (1 × 10) / (6 × 10)
hance 10 / 60 = 1 / 6
(f) 15 / 75
we know that 15 and 75 both are divisible by 15 so
15 / 75 = (1 × 15) / (5 × 15)
hance 15 / 75 = 1 / 5
(g) 12 / 60
we know that 12 and 60 both are divisible by 12 so
12 / 60 = (1 × 12) / (5 × 12)
hance 12 / 60 = 1 / 5
(h) 16 / 96
we know that 16 and 96 both are divisible by 16 so
16 / 96 = (1 × 16) / (6 × 16)
hance 16 / 96 = 1 / 6
(i) 12 / 75
we know that 12 and 75 both are divisible by 3 so
12 / 75 = (4 × 3) / (25 × 3)
hance 12 / 75 = 4 / 25
(j) 12 / 72
we know that 12 and 72 both are divisible by 12 so
12 / 72 = (1 × 12) / 6 × 12)
hance 12 / 72 = 1 / 6
(k) 3 / 18
we know that 3 and 18 both are divisible by 3 so
3 / 18 = (1 × 3) / (6 × 3)
hance 3 / 18 = 1 / 6
(l) 4 / 25
we know that 4 and 25 both are divisible by 1 so
4 / 25 = (1 × 4) / (1 × 25)
hance 4 / 25 = 4 / 25
Now grouping the above fractions into equivalent fractions, we have
(i) 2 /12 = 10 / 60 = 16 / 96 = 12 / 72 = 3 / 18 Each = 1 / 6
(ii) 3 / 15 = 15 / 75 = 12 / 60 Each = 1 / 5
(iii) 8 / 50 = 16 / 100 = 12 / 75 = 4 / 25 Each = 4 / 25

7. Find answers to the following. Write and indicate how you solved them.
(a) Is 5 / 9 equal to 4 / 5
(b) Is 9 / 16 equal to 5 / 9
(c) Is 4 /5 equal to 16 / 20
(d) Is 1 / 15 equal to 4 / 30
Ans:-
(a) 5 / 9 and 4 / 5
We have the cross multiple are
5 x 5 = 25
4 x 9 = 36
25 ≠ 36
∴ 25 / 45 ≠ 36 / 45
(b) 9 / 16 and 5 / 9
We have the cross multiple are
9 x 9 = 81
16 x 5 = 80
81 ≠ 80
∴ 9 / 16 ≠ 5 / 9
(c) 4 / 5 and 16 / 20
We have the cross multiple are
4 x 20 = 80
5 x 16 = 80
80 = 80
∴ 4 / 5 = 16 / 20
(d) 1 / 15 and 4 / 30
We have the cross multiple are
1 x 30 = 30
15 x 4 = 60
30 ≠ 60
∴ 1 / 15 ≠ 4 / 30

Q.8. Ila read 25 pages of a book containing 100 pages. Lalita read 2 / 5 of the same book. Who read less?
Ans:-
Total number of pages a book has = 100 pages
Lalita read = 2 / 5 × 100 = 40 pages
Ila read = 25 pages
Hence Ila reads less pages

Q.9. Rafiq exercised for 3 / 6 of an hour, while Rohit exercised for 3 / 4 of an hour. Who exercised for a longer time?
Ans:-
Rafiq exercised = 3 / 6 of an hour
Rohit exercised = 3 / 4 of an hour
3 / 6 and 3 / 4
numerators of the two fractions are the same So, the fraction having a lesser denominator is the greater
∴ 3 / 6 < 3 / 4
Therefore Rohit exercised for a longer time than Rafiq.

Q.10. In a class A of 25 students, 20 passed with 60% or more marks; in another class B of 30 students, 24 passed with 60% or more marks. In which class was a greater fraction of students getting with 60% or more marks?
Ans:-
Total number of students in Class A = 25
Students passed in first class in Class A = 20
fraction = 20 / 25 = (5 x 4) / (5 x 5)
hance = 4 / 5
Total number of students in Class B = 30
Students passed in first class in Class B = 24
fraction = 24 / 30 = (6 x 4) / 6 x 5)
hence = 4 / 5
Hence, both classes A and B have the same fractions.

NCERT Solutions For Class 6 Maths Chapter 7 Fractions Exercises 7.5

NCERT Solutions Class 6 Maths Chapter 7 Fraction Ex 7.5

Q.1. Write these fractions appropriately as additions or subtractions:

(a) Total number of parts each rectangle has = 5
Number of shaded parts in first rectangle = 1 = 1 / 5
Number of shaded parts in second rectangle = 2 = 2 / 5
Sum of first and second rectangle = 1 / 5 + 2 / 5 = 3 / 5
Number of shaded parts in third rectangle = 3 = 3 / 5

(b) Total number of parts each rectangle has = 5
Number of shaded parts in first rectangle = 5 = 5 / 5 or =1
Number of shaded parts in second rectangle = 3 = 3 / 5
Subtract of first and second rectangle = 5 / 5 – 3 / 5 = 2 / 5
Number of shaded parts in third rectangle = 2 = 2 / 5

(c) Total number of parts each rectangle has = 6
Number of shaded parts in first rectangle = 2 = 2 / 6
Number of shaded parts in second rectangle = 3 = 3 / 6
Sum of first and second rectangle = 2 / 6 + 3 / 6 = 5 / 6
Number of shaded parts in third rectangle = 5 = 5 / 6

2. Solve:
(a) 1 / 18 + 1 / 18
(b) 8 / 15 + 3 / 15
(c) 7 / 7 – 5 / 7
(d) 1 / 22 + 21 / 22
(e) 12 / 15 – 7 / 15
(f) 5 / 8 + 3 / 8
(g) 1 – 2 / 3 (1 = 3 / 3)
(h) 1 / 4 + 0 / 4
(i) 3 – 12 / 5
Ans:-
(a) 1 / 18 + 1 / 18
The LCM of denominators = 18
= (1 + 1) / 18
= 2 / 18 = 1 / 9
(b) 8 / 15 + 3 / 15
The LCM of denominators = 15
= (8 + 3) / 15
= 11 / 15
(c) 7 / 7 – 5 / 7
The LCM of denominators = 7
= (7 – 5) / 7
= 2 / 7
(d) 1 / 22 + 21 / 22
The LCM of denominators = 22
= (1 + 21) / 22
= 22 / 22 = 1
(e) 12 / 15 – 7 / 15
The LCM of denominators = 15
= (12 – 7) / 15
= 5 / 15 = 1 / 3
(f) 5 / 8 + 3 / 8
The LCM of denominators = 8
= (5 + 3) / 8
= 8 / 8 = 1
(g) 1 – 2 / 3 (1 = 3 / 3)
= 3 / 3 – 2 / 3
The LCM of denominators = 3
= (3 – 2) / 3
= 1 / 3
(h) 1 / 4 + 0 / 4
The LCM of denominators = 4
(1 + 0) / 4
= 1/ 4
(i) 3 – 12 / 5
3 / 1 – 12 / 5
The LCM of denominators = 5
= (15 – 12) / 5
= 3 / 5

Q.3. Shubham painted 2 / 3 of the wall space in his room. His sister Madhavi helped and painted 1 / 3 of the wall space. How much did they paint together?
Ans:-
wall painted by Shubham = 2 / 3
wall painted by Madhavi = 1 / 3
Total space painted by both = (2 / 3 + 1 / 3)
= (2 + 1) / 3
= 3 / 3 = 1
The together painted 1 complete wall in a room

4. Fill in the missing fractions.
(a) 7 / 10 – ▯ = 3 / 10
(b) ▯ – 3 / 21 = 5 / 21
(c) ▯ – 3 / 6 = 3 / 6
(d) ▯ + 5 / 27 = 12 / 27
Ans:-
(a) 7 / 10 – ▯ = 3 / 10
▯ = 7 / 10 – 3 / 10
▯ = (7 – 3) / 10
▯ = 4 / 10 or 2 / 5
The missing fractions = 2 / 5
(b) ▯ – 3 / 21 = 5 / 21
▯ = 5 / 21 + 3 / 21
▯ = (5 + 3) / 21
▯ = 8 / 21
The missing fractions = 8 / 21
(c) ▯ – 3 / 6 = 3 / 6
▯ = 3 / 6 + 3 / 6
▯ = (3 + 3) / 6
▯ = 6 / 6 or = 1
The missing fractions = 1
(d) ▯ + 5 / 27 = 12 / 27
▯ = 12 / 27 – 5 /27
▯ = (12 – 5) / 27
▯ = 7 /27
The missing fractions = 7 / 27

Q.5. Javed was given 5 / 7 of a basket of oranges. What fraction of oranges was left in the basket?
Ans:-
Fraction of basket as a whole can be taken as 1
Fraction of oranges given to Javed = 5 / 7
Fraction of oranges left in the basket = 1 – 5 / 7
= 7 / 7 – 5 / 7
= (7 – 5) / 7 = 2 / 7
the required fraction = 2 / 7

NCERT Solutions For Class 6 Maths Chapter 7 Fractions Exercises 7.6

Class 6th Maths Chapter 7 solutions Ex 7.6

Q.1. Solve
(a) 2 / 3 + 1 / 7
(b) 3 / 10 + 7 / 15
(c) 4 / 9 + 2 / 7
(d) 5 / 7 + 1 / 3
(e) 2 / 5 + 1 / 6
(f) 4 / 5 + 2 / 3
(g) 3 / 4 – 1 / 3
(h) 5 / 6 – 1 / 3
(i) 2 / 3 + 3 / 4 + 1 / 2
(j) 1/ 2 + 1 / 3 + 1 / 6

Ans:-
(a) 2 / 3 + 1 / 7
The LCM of 3 and 7 is =21
[(2 × 7) + (1 × 3)] / 21
= (14 + 3) / 21
= 17 / 21
(b) 3 / 10 + 7 / 15
The LCM of 10 and 15 is =30
= [(3 × 3) + (7 × 2)] / 30
= (9 + 14) / 30
= 23 / 30

(c) 4 / 9 + 2 / 7
The LCM of 9 and 7 is =63
= [(4 × 7) + (2 × 9)] / 63
= (28 + 18) / 63
= 46 / 63
(d) 5 / 7 + 1 / 3
The LCM of 7 and 3 is =21
= [(5 × 3) + (1 × 7)] / 21
= (15 + 7) / 21
= 22 / 21
(e) 2 / 5 + 1 / 6
The LCM of 5 and 6 is =30
= [(2 × 6) + (1 × 5)] / 30
= (12 + 5) / 30
= 17 / 30
(f) 4 / 5 + 2 / 3
The LCM of 5 and 3 is =15
= [(4 × 3) + (2 × 5)] / 15
= (12 + 10) / 15
= 22 / 15
(g) 3 / 4 – 1 / 3
The LCM of 4 and 3 is =12
= [(3 × 3) – (1 × 4)] / 12
= (9 – 4) / 12
= 5 / 12

(h) 5 / 6 – 1 / 3
The LCM of 6 and 3 is =6
= [(5 × 1) – (1 × 2)] / 6
= (5 – 2) / 6
= 3 / 6
(i) 2 / 3 + 3 / 4 + 1 / 2
The LCM of 3, 4, and 2 is =12
= [(2 × 4) + (3 × 3) + (1 × 6)] / 12
= (8 + 9 + 6) / 12
= 23 / 12
(j) 1/ 2 + 1 / 3 + 1 / 6
The LCM of 2, 3, and 6 is = 6
= [(1 × 3) + (1 × 2) + (1 × 1)] / 6
= (3 + 2 + 1) / 6
= 6 / 6 = 1

The LCM of 3, and 3 is =3
= [(4 x 1) + (11 x 1) / 3
= (4 + 11) / 3
= 15 / 3 = 5

The LCM of 4 and 3 is =12
= [(14 × 4) + (13 × 3)] / 12
= (56 + 39) / 12
= 95 / 12
(m) 16 / 5 – 7 / 5
The LCM of 5 and 5 is =5
= [(16 x 1) – (7 x 1) / 5
= (16 – 7) / 5
= 9 / 5
(n) 4 /3 – 1 / 2
The LCM of 3 and 2 is = 6
= [(4 × 2) – (1 × 3)] / 6
= (8 – 3) /6
= 5 / 6

Q.2. Sarita bought 2 / 5 metre of ribbon and Lalita 3 /4 metre of ribbon. What is the total length of the ribbon they bought?
Ans:-
Length of ribbon bought by Sarita = 2 / 5
Length of ribbon bought by Lalita = 3 / 4
The total length of ribbon bought by Sarita and Lalita
= 2 / 5 + 3 / 4
The LCM of 5 and 4 is =20
= [(2 × 4) + (3 × 5)] / 20
= (8 + 15) / 20
= 23 / 20 metre

Ans:-
Piece of cake given to Naina = 3 / 2
Piece of cake given to Najma = 4 / 3
Total amount of cake given to both of them = 3 / 2 + 4 / 3
The LCM of 2 and 3 is = 6
= [(3 × 3) + (4 × 2)] / 6
= (9 + 8) / 6
Hence the total amount of piece given to both = 17 / 6 or

4. Fill in the boxes:
(a) ▯ – 5 / 8 = 1 / 4
(b) ▯ – 1 / 5 = 1 / 2
(c) 1 / 2 – ▯ = 1 / 6
Ans:-
(a) ▯ – 5 / 8 = 1 / 4
▯ = 1 / 4 + 5 / 8
The LCM of 4 and 8 is = 8
▯ = [(1 × 2 + 5)] / 8
▯ = 7 / 8
The missing number is = 7 / 8

(b) ▯ – 1 / 5 = 1 / 2
▯ = 1 / 2 + 1 / 5
The LCM of 2 and 5 is = 10
▯ = [(1 × 5) + (1 × 2)] / 10
▯ = (5 + 2) / 10
▯ = 7 / 10
The missing number is = 7 / 10
(c) 1 / 2 – ▯ = 1 / 6
▯ = 1 / 2 – 1 / 6
The LCM of 2 and 6 is = 6
▯ = [(1 × 3) – (1 × 1)] / 6
▯ = (3 – 1) / 6
▯ = 2 / 6 or 1 / 3
The missing number is = 1 / 3

Q.5. Complete the addition and subtraction box.

Ans:-
(a) (i)  2 / 3 + 4 / 3
The LCM = 3
= [(1 × 2) + (1 × 4)] / 3
= (2 + 4) / 3
= 6 / 3 = 2
(ii) 1 / 3 + 2 / 3
The LCM = 3
= [(1 × 1) + (1 × 2)] / 3
= (1 + 2) / 3
= 3 / 3 = 1
(iii) 2 / 3 – 1 / 3
The LCM = 3
= [(1 × 2) – (1 × 1)] / 3
= (2 – 1) / 3
= 1 / 3

(iv) 4 / 3 – 2 / 3
The LCM = 3
= [(1 × 4) – (1 × 2)] / 3
= (4 – 2) / 3
= 2 / 3
(v) 1 / 3 + 2 / 3
The LCM = 3
= [(1 × 1) + (1 × 2)] / 3
= (1 + 2) / 3
= 3 / 3 = 1

(b) (i) 1 / 2 + 1 / 3
The LCM = 6
= [(1 × 3) + (1 × 2)] / 6
= (3 + 2) / 6
= 5 / 6
(ii) 1 / 3 + 1 / 4
The LCM = 12
= [(1 × 4) + (1 × 3)] / 12
= (4 + 3) / 12
= 7 / 12
(iii) 1 / 2 – 1 / 3
The LCM = 6
= [(1 × 3) – (1 × 2)] / 6
= (3 – 2) / 6
= 1 / 6

(iv) 1 / 3 – 1 / 4
The LCM = 12
= [(1 × 4) – (1 ×3)] / 12
= (4 – 3) / 12
= 1 / 12
(v) 1 / 6 + 1 / 12
The LCM = 12
= [(1 × 2) + 1] / 12
= (2 + 1) / 12
= 3 / 12
= 1 / 4

Q.6. A piece of wire 7 / 8 metre long broke into two pieces. One piece was 1 / 4 metre long. How long is the other piece?
Ans:-
Total length of wire = 7 / 8 metre
Length of one piece of wire = 1 / 4 metre
Length of the other piece = 7 / 8 – 1 / 4
The LCM of 8 and 4 = 8
= [(7 × 1) – (1 × 2)] / 8
= (7 – 2) / 8
= 5 / 8
∴ Length of the other piece of wire = 5 / 8 metre

Q.7. Nandini’s house is 9 / 10 km from her school. She walked some distance and then took a bus for 1 / 2 km to reach the school. How far did she walk?
Ans:-
Total distance from house to school = 9 / 10
Distance travelled by Nandini by bus = 1 / 2
∴ Distance travelled by her walk = 9 / 10 – 1 / 2
The LCM of = 10
= [(9 × 1) – (1 × 5)] / 10
= (9 – 5) / 10 = 4 / 10
= 2 / 5 km
∴ Distance travelled by her walk is 2 / 5 km

Q.8. Asha and Samuel have bookshelves of the same size partly filled with books. Asha’s shelf is 5 / 6 th full and Samuel’s shelf is 2/ 5 th full. Whose bookshelf is more full? By what fraction?
Ans:-
Asha’s bookshelf = 5 / 6
Samuel’s bookshelf = 2 / 5
Comparing 5 / 6 and 2 / 5
5 x 5 = 25
6 / 2 = 12
25 > 12
∴ 5 / 6 > 2 / 5
Hence, Asha’s shelf is full more than Samuel’s shelf
Difference = 5 / 6 – 2 / 5
The LCM of = 30
[(5 x 5) – (2 x 6)] / 30
= (25 – 12) / 30
= 13 / 30

Jaidev takes = 11 / 5 minutes
Rahul takes = 7 / 4 minutes
Comparing 11 / 5 and 7 / 4
11 x 4 = 44
5 x 7 = 35
44 > 35
∴ 11 / 5 > 7 / 4
Difference = 11 / 5 – 7 / 4
The LCM of = 20
[(11 x 4) – (5 x 7)] / 20
(44 – 35) / 20
9 / 20
Hence, Rahul walks across the school ground by 9 / 20 minutes

Class 6 Science NCERT Notes	Class 6 Complete Study Material
NCERT Solutions for Class 6 Science	Class 6 Maths Chapter 6 Solutions

NCERT Solutions For Class 6 Maths Chapter 7 Fractions are based on NCERT Books. You can download ncert book for class 6 maths chapter 7. Ch 7 class 6 maths solutions are designed by our subject expert team.

What We Learn?

1. (a) A fraction is a number representing a part of a whole. The whole may be a single
   object or a group of objects.
   (b) When expressing a situation of counting parts to write a fraction, it must be ensured
   that all parts are equal.
2. In5/7, 5 is called the numerator and 7 is called the denominator.
3. Fractions can be shown on a number line. Every fraction has a point associated with it
   on the number line.
4. In a proper fraction, the numerator is less than the denominator. The fractions, where
   the numerator is greater than the denominator are called improper fractions. An improper
   fraction can be written as a combination of a whole and a part, and such fraction then
   called mixed fractions.
5. Each proper or improper fraction has many equivalent fractions. To find an equivalent
   fraction of a given fraction, we may multiply or divide both the numerator and the
   denominator of the given fraction by the same number.
6. A fraction is said to be in the simplest (or lowest) form if its numerator and the denomi-
   nator have no common factor except 1.