FRACTIONS
EXERCISE – 7.6
EX 7.6 QUESTION 1.
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
(k) 
(l) 
(m) 16 / 5 – 7 / 5
(n) 4 / 3 – 1 / 2
Solution:
(a) 2 / 3 + 1/ 7
Taking LCM
[(2 × 7) + (1 × 3)] / 21
= (14 + 3) / 21
= 17 / 21
(b) 3 / 10 + 7 / 15
Taking LCM 30
= [(3 × 3) + (7 × 2)] / 30
= (9 + 14) / 30
= 23 / 30
(c) 4 / 9 + 2/ 7
Taking LCM 63
= [(4 × 7) + (2 × 9)] / 63
= (28 + 18) / 63
= 46 / 63
(d) 5 / 7 + 1 / 3
Taking LCM 21
= [(5 × 3) + (1 × 7)] / 21
= (15 + 7) / 21
= 22 / 21
(e) 2 / 5 + 1 / 6
Taking LCM 30
= [(2 × 6) + (1 × 5)] / 30
= (12 + 5) / 30
= 17 / 30
(f) 4 / 5 + 2 / 3
Taking LCM 15
= [(4 × 3) + (2 × 5)] / 15
= (12 + 10) / 15
= 22 / 15
(g) 3 / 4 – 1 / 3
Taking LCM 12
= [(3 × 3) – (1 × 4)] / 12
= (9 – 4) / 12
= 5 / 12
(h) 5 / 6 – 1 / 3
Taking LCM 6
= [(5 × 1) – (1 × 2)] / 6
= (5 – 2) / 6
= 3 / 6
= 1 / 2
(i) 2 / 3 + 3 / 4 + 1 / 2
Taking LCM 12
= [(2 × 4) + (3 × 3) + (1 × 6)] / 12
= (8 + 9 + 6) / 12
= 23 / 12
(j) 1 / 2 + 1 / 3 + 1 / 6
Taking LCM 6
= [(1 × 3) + (1 × 2) + (1 × 1)] / 6
= (3 + 2 + 1) / 6
= 6 / 6
= 1
(k) 
= [(3 × 1) + 1] / 3 + [(3 × 3) + 2] / 3
= (3 + 1) / 3 + (9 + 2) / 3
= 4/ / 3 + 11 / 3
= (4 + 11) / 3
= 15 / 3
= 5
(l) 
= [(3 × 4) + 2] / 3 + [(3 × 4) + 1] / 4
= 14 / 3 + 13 / 4
= [(14 × 4) + (13 × 3)] / 12
= (56 + 39) / 12
= 95 / 12
(m) 16 / 5 – 7 / 5
= (16 – 7) / 5
= 9 / 5
(n) 4 /3 – 1 / 2
Taking LCM 6
= [(4 × 2) – (1 × 3)] / 6
= (8 – 3) /6
= 5 / 6
EX 7.6 QUESTION 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?
Solution:
The length of ribbon bought by Sarita = 2 / 5 metre
The length of ribbon bought by Lalita = 3 / 4 metre
Total length of the ribbon bought by both of them = 2 / 5 + 3 / 4
Taking LCM 20
= [(2 × 4) + (3 × 5)] / 20
= (8 + 15) / 20
= 23 / 20 metre
∴ The total length of the ribbon bought by both = 23 / 20 metre
EX 7.6 QUESTION 3.
Naina was given 
piece of cake and Najma was given 
piece of cake. Find the total amount of cake was given to both of them.
Solutions:
Fraction of cake Naina got =
= 3 / 2
Fraction of cake Najma got =
= 4 / 3
Total amount of cake given to both of them = 3 / 2 + 4 / 3
= [(3 × 3) + (4 × 2)] / 6
= (9 + 8) / 6
= 17 / 6
=
EX 7.6 QUESTION 4.
Fill in the boxes:
(a) ▯ – 5 / 8 = 1 / 4
(b) ▯ – 1 / 5 = 1 / 2
(c) 1 / 2 – ▯ = 1 / 6
Solutions:
(a) ▯ – 5 / 8 = 1 / 4
▯ = 1 / 4 + 5 / 8
▯ = [(1 × 2 + 5)] / 8
▯ = 7 / 8
(b) ▯ – 1 / 5 = 1 / 2
▯ = 1 / 2 + 1 / 5
▯ = [(1 × 5) + (1 × 2)] / 10
▯ = (5 + 2) / 10
▯ = 7 / 10
(c) 1 / 2 – ▯ = 1 / 6
▯ = 1 / 2 – 1 / 6
▯ = [(1 × 3) – (1 × 1)] / 6
▯ = (3 – 1) / 6
▯ = 2 / 6
▯ 1 / 3
EX 7.6 QUESTION 5.
5. Complete the addition and subtraction box.
Solution:
(a) 2 / 3 + 4 / 3
= (2 + 4) / 3
= 6 / 3
= 2
1 / 3 + 2 / 3
= (1 + 2) / 3
= 3 / 3
= 1
2 / 3 – 1 / 3
= (2 – 1) / 3
= 1 / 3
4 / 3 – 2 / 3
= (4 – 2) / 3
= 2 / 3
1 / 3 + 2 / 3
= (1 + 2) / 3
= 3 / 3
= 1
Hence, the complete given box is
(b) 1 / 2 + 1 / 3
= [(1 × 3) + (1 × 2)] / 6
= (3 + 2) / 6
= 5 / 6
1 / 3 + 1 / 4
= [(1 × 4) + (1 × 3)] / 12
= (4 + 3) / 12
= 7 / 12
1 / 2 – 1 / 3
= [(1 × 3) – (1 × 2)] / 6
= (3 – 2) / 6
= 1 / 6
1 / 3 – 1 / 4
= [(1 × 4) – (1 ×3)] / 12
= (4 – 3) / 12
= 1 / 12
1 / 6 + 1 / 12
= [(1 × 2) + 1] / 12
= (2 + 1) / 12
= 3 / 12
= 1 / 4
Hence, the complete given box is
EX 7.6 QUESTION 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?
Solutions:
Total length of wire = 7 / 8 metre
Length of one piece of wire = 1 / 4 metre
Length of other pieces of wire = Total length of wire – one piece of wire
= 7 / 8 – 1 / 4
= [(7 × 1) – (1 × 2)] / 8
= (7 – 2) / 8
= 5 / 8
∴ Length of the other piece of wire = 5 / 8 metre
EX 7.6 QUESTION 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?
Solutions:
The distance of the school from house = 9 / 10 km
The distance she travelled by bus = 1 / 2 km
Distance walked by Nandini = Total distance of the school – Distance she travelled by bus
= 9 / 10 – 1 / 2
= [(9 × 1) – (1 × 5)] / 10
= (9 – 5) / 10
= 4 / 10
= 2 / 5 km
∴ Distance walked by Nandini is 2 / 5 km
EX 7.6 QUESTION 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?
Solution:
Fraction of Asha’s bookshelf = 5 / 6
Fraction of Samuel’s bookshelf = 2 / 5
Convert these fractions into like fractions
5 / 6 = 5 / 6 × 5 / 5
= (5 × 5) / (6 × 5)
= 25 / 30
2 / 5 = 2 / 5 × 6 / 6
= (2 × 6) / (5 × 6)
= 12 / 30
25 / 30 > 12 / 30
5 / 6 > 2 / 5
∴ Asha’s bookshelf is more full than Samuel’s bookshelf
Difference = 5 / 6 – 2 / 5
= 25 / 30 – 12 / 30
= 13 / 30
EX 7.6 QUESTION 9.
Jaidev takes 
minutes to walk across the school ground. Rahul takes 7 / 4 minutes to do the same. Who takes less time and by what fraction?
Solution:
Time taken by Jaidev to walk across the school ground = 
= 11 / 5 minutes
Time taken by Rahul to walk across the school ground = 7 / 4 minutes
Convert these fractions into like fractions
11 / 5 = 11 / 5 × 4 / 4
= (11 × 4) / (5 × 4)
= 44 / 20
7 / 4 = 7 / 4 × 5 / 5
= (7 × 5) / (4 × 5)
= 35 / 20
Clearly, 44 / 20 > 35 / 20
11 / 5 > 7 / 4
∴ Rahul takes less time than Jaidev to walk across the school ground
Difference = 11 / 5 – 7 / 4
= 44 / 20 – 35 / 20
= 9 / 20
Hence, Rahul walks across the school ground by 9 / 20 minutes
