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RD Sharma Solutions: Fractions (Exercise 6.8)

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 Page 1


 
 
 
 
 
 
Exercise 6.8                                                                               PAGE: 6.29 
1. Write these fractions appropriately as additions or subtractions: 
 
Solution: 
 
(i) It can be written as 
1/5 + 2/5 = 3/5 
 
(ii) It can be written as 
3/6 + 2/6 = 5/6 
 
2. Solve: 
(i) 5/12 + 1/12 
(ii) 3/15 + 7/15 
(iii) 3/22 + 7/22 
(iv) 1/4 + 0/4 
(v) 4/13 + 2/13 + 1/13 
(vi) 0/15 + 2/15 + 1/15 
(vii) 7/31 – 4/31 + 9/31 
(viii) 3 2/7 + 1/7 – 2 3/7 
(ix) 2 1/3 – 1 2/3 + 4 1/3 
(x) 1 – 2/3 + 7/3  
(xi) 16/7 – 5/7 + 9/7 
Solution: 
 
(i) 5/12 + 1/12 
It can be written as 
5/ 12 + 1/12 = (5 + 1)/ 12  
On further calculation 
5/ 12 + 1/12 = 6/12 = 1/2 
 
(ii) 3/15 + 7/15 
It can be written as 
3/15 + 7/15 = (3 + 7)/ 15  
On further calculation 
3/15 + 7/15 = 10/15 = 2/3 
 
(iii) 3/22 + 7/22 
It can be written as 
3/22 + 7/22 = (3 + 7)/ 22 
On further calculation 
3/22 + 7/22 = 10/22 = 5/11 
Page 2


 
 
 
 
 
 
Exercise 6.8                                                                               PAGE: 6.29 
1. Write these fractions appropriately as additions or subtractions: 
 
Solution: 
 
(i) It can be written as 
1/5 + 2/5 = 3/5 
 
(ii) It can be written as 
3/6 + 2/6 = 5/6 
 
2. Solve: 
(i) 5/12 + 1/12 
(ii) 3/15 + 7/15 
(iii) 3/22 + 7/22 
(iv) 1/4 + 0/4 
(v) 4/13 + 2/13 + 1/13 
(vi) 0/15 + 2/15 + 1/15 
(vii) 7/31 – 4/31 + 9/31 
(viii) 3 2/7 + 1/7 – 2 3/7 
(ix) 2 1/3 – 1 2/3 + 4 1/3 
(x) 1 – 2/3 + 7/3  
(xi) 16/7 – 5/7 + 9/7 
Solution: 
 
(i) 5/12 + 1/12 
It can be written as 
5/ 12 + 1/12 = (5 + 1)/ 12  
On further calculation 
5/ 12 + 1/12 = 6/12 = 1/2 
 
(ii) 3/15 + 7/15 
It can be written as 
3/15 + 7/15 = (3 + 7)/ 15  
On further calculation 
3/15 + 7/15 = 10/15 = 2/3 
 
(iii) 3/22 + 7/22 
It can be written as 
3/22 + 7/22 = (3 + 7)/ 22 
On further calculation 
3/22 + 7/22 = 10/22 = 5/11 
 
 
 
 
 
 
 
(iv) 1/4 + 0/4 
It can be written as 
1/4 + 0/4 = (1 + 0)/4 
On further calculation 
1/4 + 0/4 = ¼ 
 
(v) 4/13 + 2/13 + 1/13 
It can be written as 
4/13 + 2/13 + 1/13 = (4 + 2 + 1)/ 13 
On further calculation 
4/13 + 2/13 + 1/13 = 7/13 
 
(vi) 0/15 + 2/15 + 1/15 
It can be written as 
0/15 + 2/15 + 1/15 = (0 + 2 + 1)/ 15 
On further calculation 
0/15 + 2/15 + 1/15 = 3/15 = 1/5 
 
(vii) 7/31 – 4/31 + 9/31 
It can be written as 
7/31 – 4/31 + 9/31 = (7 – 4 + 9)/ 31 
On further calculation 
7/31 – 4/31 + 9/31 = 12/31 
 
(viii) 3 2/7 + 1/7 – 2 3/7 
It can be written as 
3 2/7 + 1/7 – 2 3/7 = (23 + 1 – 17)/ 7 
On further calculation 
3 2/7 + 1/7 – 2 3/7 = 7/7 = 1 
 
(ix) 2 1/3 – 1 2/3 + 4 1/3 
It can be written as 
2 1/3 – 1 2/3 + 4 1/3 = (7 – 5 + 13)/ 3  
On further calculation 
2 1/3 – 1 2/3 + 4 1/3 = 15/3 = 5 
 
(x) 1 – 2/3 + 7/3  
It can be written as 
1 – 2/3 + 7/3 = (3 – 2 + 7)/3 
On further calculation 
1 – 2/3 + 7/3 = 8/3 
 
(xi) 16/7 – 5/7 + 9/7 
It can be written as 
16/7 – 5/7 + 9/7 = (16 – 5 + 9)/ 7 
On further calculation 
Page 3


 
 
 
 
 
 
Exercise 6.8                                                                               PAGE: 6.29 
1. Write these fractions appropriately as additions or subtractions: 
 
Solution: 
 
(i) It can be written as 
1/5 + 2/5 = 3/5 
 
(ii) It can be written as 
3/6 + 2/6 = 5/6 
 
2. Solve: 
(i) 5/12 + 1/12 
(ii) 3/15 + 7/15 
(iii) 3/22 + 7/22 
(iv) 1/4 + 0/4 
(v) 4/13 + 2/13 + 1/13 
(vi) 0/15 + 2/15 + 1/15 
(vii) 7/31 – 4/31 + 9/31 
(viii) 3 2/7 + 1/7 – 2 3/7 
(ix) 2 1/3 – 1 2/3 + 4 1/3 
(x) 1 – 2/3 + 7/3  
(xi) 16/7 – 5/7 + 9/7 
Solution: 
 
(i) 5/12 + 1/12 
It can be written as 
5/ 12 + 1/12 = (5 + 1)/ 12  
On further calculation 
5/ 12 + 1/12 = 6/12 = 1/2 
 
(ii) 3/15 + 7/15 
It can be written as 
3/15 + 7/15 = (3 + 7)/ 15  
On further calculation 
3/15 + 7/15 = 10/15 = 2/3 
 
(iii) 3/22 + 7/22 
It can be written as 
3/22 + 7/22 = (3 + 7)/ 22 
On further calculation 
3/22 + 7/22 = 10/22 = 5/11 
 
 
 
 
 
 
 
(iv) 1/4 + 0/4 
It can be written as 
1/4 + 0/4 = (1 + 0)/4 
On further calculation 
1/4 + 0/4 = ¼ 
 
(v) 4/13 + 2/13 + 1/13 
It can be written as 
4/13 + 2/13 + 1/13 = (4 + 2 + 1)/ 13 
On further calculation 
4/13 + 2/13 + 1/13 = 7/13 
 
(vi) 0/15 + 2/15 + 1/15 
It can be written as 
0/15 + 2/15 + 1/15 = (0 + 2 + 1)/ 15 
On further calculation 
0/15 + 2/15 + 1/15 = 3/15 = 1/5 
 
(vii) 7/31 – 4/31 + 9/31 
It can be written as 
7/31 – 4/31 + 9/31 = (7 – 4 + 9)/ 31 
On further calculation 
7/31 – 4/31 + 9/31 = 12/31 
 
(viii) 3 2/7 + 1/7 – 2 3/7 
It can be written as 
3 2/7 + 1/7 – 2 3/7 = (23 + 1 – 17)/ 7 
On further calculation 
3 2/7 + 1/7 – 2 3/7 = 7/7 = 1 
 
(ix) 2 1/3 – 1 2/3 + 4 1/3 
It can be written as 
2 1/3 – 1 2/3 + 4 1/3 = (7 – 5 + 13)/ 3  
On further calculation 
2 1/3 – 1 2/3 + 4 1/3 = 15/3 = 5 
 
(x) 1 – 2/3 + 7/3  
It can be written as 
1 – 2/3 + 7/3 = (3 – 2 + 7)/3 
On further calculation 
1 – 2/3 + 7/3 = 8/3 
 
(xi) 16/7 – 5/7 + 9/7 
It can be written as 
16/7 – 5/7 + 9/7 = (16 – 5 + 9)/ 7 
On further calculation 
 
 
 
 
 
 
16/7 – 5/7 + 9/7 = 20/7 
 
3. Shikha painted 1/5 of the wall space in her room. Her brother Ravish helped and painted 3/5 of the wall 
space. How much did they paint together? How much the room is left unpainted? 
Solution: 
 
Fraction of wall space painted by Shikha = 1/5 
Fraction of wall space painted by Ravish = 3/5 
So the wall space painted by both = 1/5 + 3/5 
                                                       = (1+3)/5 
                                                       = 4/5 
We get the unpainted space = (5 – 4)/ 5 = 1/5 
 
Therefore, Shikha and Ravish painted 4/5 of the wall space together and the room space left unpainted is 1/5. 
 
4. Ramesh bought 2 ½ kg sugar whereas Rohit bought 3 ½ kg of sugar. Find the total amount of sugar 
bought by both of them. 
Solution: 
 
Sugar bought by Ramesh = 2 ½ kg 
It can be written as 
Sugar bought by Ramesh = ((2 × 2) + 1)/ 2 = 5/2 kg 
 
Sugar bought by Rohit = 3 ½ kg 
It can be written as 
Sugar bought by Rohit = ((2 × 3) + 1)/ 2 = 7/2 kg 
 
So the total sugar bought by both of them = Sugar bought by Ramesh + Sugar bought by Rohit 
By substituting the values 
Total sugar bought by both of them = 5/2 + 7/2 = 12/2 = 6kg 
 
Therefore, the total amount of sugar bought by both of them is 6kg. 
 
5. The teacher taught 3/5 of the book, Vivek revised 1/5 more on his own. How much does he still have to 
revise? 
Solution: 
 
We know that 
Fraction of book teacher taught = 3/5 
Fraction of book Vivek revised = 1/5 
So the fraction of book Vivek still have to revise = 3/5 – 1/5 
                                                                               = (3 – 1)/ 5 
                                                                               = 2/5 
 
Hence, Vivek still have to revise 2/5 of the book. 
 
6. Amit was given 5/7 of a bucket of oranges. What fraction of oranges was left in the basket? 
Solution: 
 
We know that 
Page 4


 
 
 
 
 
 
Exercise 6.8                                                                               PAGE: 6.29 
1. Write these fractions appropriately as additions or subtractions: 
 
Solution: 
 
(i) It can be written as 
1/5 + 2/5 = 3/5 
 
(ii) It can be written as 
3/6 + 2/6 = 5/6 
 
2. Solve: 
(i) 5/12 + 1/12 
(ii) 3/15 + 7/15 
(iii) 3/22 + 7/22 
(iv) 1/4 + 0/4 
(v) 4/13 + 2/13 + 1/13 
(vi) 0/15 + 2/15 + 1/15 
(vii) 7/31 – 4/31 + 9/31 
(viii) 3 2/7 + 1/7 – 2 3/7 
(ix) 2 1/3 – 1 2/3 + 4 1/3 
(x) 1 – 2/3 + 7/3  
(xi) 16/7 – 5/7 + 9/7 
Solution: 
 
(i) 5/12 + 1/12 
It can be written as 
5/ 12 + 1/12 = (5 + 1)/ 12  
On further calculation 
5/ 12 + 1/12 = 6/12 = 1/2 
 
(ii) 3/15 + 7/15 
It can be written as 
3/15 + 7/15 = (3 + 7)/ 15  
On further calculation 
3/15 + 7/15 = 10/15 = 2/3 
 
(iii) 3/22 + 7/22 
It can be written as 
3/22 + 7/22 = (3 + 7)/ 22 
On further calculation 
3/22 + 7/22 = 10/22 = 5/11 
 
 
 
 
 
 
 
(iv) 1/4 + 0/4 
It can be written as 
1/4 + 0/4 = (1 + 0)/4 
On further calculation 
1/4 + 0/4 = ¼ 
 
(v) 4/13 + 2/13 + 1/13 
It can be written as 
4/13 + 2/13 + 1/13 = (4 + 2 + 1)/ 13 
On further calculation 
4/13 + 2/13 + 1/13 = 7/13 
 
(vi) 0/15 + 2/15 + 1/15 
It can be written as 
0/15 + 2/15 + 1/15 = (0 + 2 + 1)/ 15 
On further calculation 
0/15 + 2/15 + 1/15 = 3/15 = 1/5 
 
(vii) 7/31 – 4/31 + 9/31 
It can be written as 
7/31 – 4/31 + 9/31 = (7 – 4 + 9)/ 31 
On further calculation 
7/31 – 4/31 + 9/31 = 12/31 
 
(viii) 3 2/7 + 1/7 – 2 3/7 
It can be written as 
3 2/7 + 1/7 – 2 3/7 = (23 + 1 – 17)/ 7 
On further calculation 
3 2/7 + 1/7 – 2 3/7 = 7/7 = 1 
 
(ix) 2 1/3 – 1 2/3 + 4 1/3 
It can be written as 
2 1/3 – 1 2/3 + 4 1/3 = (7 – 5 + 13)/ 3  
On further calculation 
2 1/3 – 1 2/3 + 4 1/3 = 15/3 = 5 
 
(x) 1 – 2/3 + 7/3  
It can be written as 
1 – 2/3 + 7/3 = (3 – 2 + 7)/3 
On further calculation 
1 – 2/3 + 7/3 = 8/3 
 
(xi) 16/7 – 5/7 + 9/7 
It can be written as 
16/7 – 5/7 + 9/7 = (16 – 5 + 9)/ 7 
On further calculation 
 
 
 
 
 
 
16/7 – 5/7 + 9/7 = 20/7 
 
3. Shikha painted 1/5 of the wall space in her room. Her brother Ravish helped and painted 3/5 of the wall 
space. How much did they paint together? How much the room is left unpainted? 
Solution: 
 
Fraction of wall space painted by Shikha = 1/5 
Fraction of wall space painted by Ravish = 3/5 
So the wall space painted by both = 1/5 + 3/5 
                                                       = (1+3)/5 
                                                       = 4/5 
We get the unpainted space = (5 – 4)/ 5 = 1/5 
 
Therefore, Shikha and Ravish painted 4/5 of the wall space together and the room space left unpainted is 1/5. 
 
4. Ramesh bought 2 ½ kg sugar whereas Rohit bought 3 ½ kg of sugar. Find the total amount of sugar 
bought by both of them. 
Solution: 
 
Sugar bought by Ramesh = 2 ½ kg 
It can be written as 
Sugar bought by Ramesh = ((2 × 2) + 1)/ 2 = 5/2 kg 
 
Sugar bought by Rohit = 3 ½ kg 
It can be written as 
Sugar bought by Rohit = ((2 × 3) + 1)/ 2 = 7/2 kg 
 
So the total sugar bought by both of them = Sugar bought by Ramesh + Sugar bought by Rohit 
By substituting the values 
Total sugar bought by both of them = 5/2 + 7/2 = 12/2 = 6kg 
 
Therefore, the total amount of sugar bought by both of them is 6kg. 
 
5. The teacher taught 3/5 of the book, Vivek revised 1/5 more on his own. How much does he still have to 
revise? 
Solution: 
 
We know that 
Fraction of book teacher taught = 3/5 
Fraction of book Vivek revised = 1/5 
So the fraction of book Vivek still have to revise = 3/5 – 1/5 
                                                                               = (3 – 1)/ 5 
                                                                               = 2/5 
 
Hence, Vivek still have to revise 2/5 of the book. 
 
6. Amit was given 5/7 of a bucket of oranges. What fraction of oranges was left in the basket? 
Solution: 
 
We know that 
 
 
 
 
 
 
Fraction of oranges Amit has = 5/7 
So the fraction of oranges left in the basket = 1 – 5/7 
                                                                      = (7 – 5)/ 7 
                                                                      = 2/7 
Hence, the fraction of oranges left in the basket is 2/7. 
 
7. Fill in the missing fractions: 
(i) 7/10 - ? = 3/10 
(ii) ? - 3/21 = 5/21 
(iii) ? - 3/6 = 3/6 
(iv) ? - 5/27 = 12/27 
Solution: 
 
(i) 7/10 - ? = 3/10 
It can be written as 
7/10 - 3/10 = ? 
We get 
(7 – 3)/ 10 = 2/5 
 
(ii) ? - 3/21 = 5/21 
It can be written as 
? = 5/21 + 3/21 
We get 
(5 + 3)/ 21 = 8/21 
 
(iii) ? - 3/6 = 3/6 
It can be written as 
? = 3/6 + 3/6 
We get 
(3 + 3)/ 6 = 6/6 = 1 
 
(iv) ? - 5/27 = 12/27 
It can be written as 
 ? = 12/27 + 5/27 
We get 
(12 + 5)/ 27 = 17/27
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