Class 6 Exam > Class 6 Notes > Mathematics (Maths) Class 6 > 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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Document Description: RD Sharma Solutions: Fractions (Exercise 6.8) for Class 6 2024 is part of Mathematics (Maths) Class 6 preparation.
The notes and questions for RD Sharma Solutions: Fractions (Exercise 6.8) have been prepared according to the Class 6 exam syllabus. Information about RD Sharma Solutions: Fractions (Exercise 6.8) covers topics
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