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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 = 17/27Read More
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