Gujarat Board GSEB Textbook Solutions Class 6 Maths Chapter 7 Fractions Ex 7.6 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 6 Maths Chapter 7 Fractions Ex 7.6

Question 1.
(a) (frac { 2 }{ 3 }) + (frac { 1 }{ 7 })
(b) (frac { 3 }{ 10 }) + (frac { 7 }{ 15 })
(c) (frac { 4 }{ 9 }) + (frac { 2 }{ 7 })
(d) (frac { 5 }{ 7 }) + (frac { 1 }{ 3 })
(e) (frac { 2 }{ 5 }) + (frac { 1 }{ 6 })
(f) (frac { 4 }{ 5 }) + (frac { 2 }{ 3 })
(g) (frac { 3 }{ 4 }) – (frac { 1 }{ 3 })
(h) (frac { 5 }{ 6 }) – (frac { 1 }{ 3 })
(i) (frac { 2 }{ 3 }) + (frac { 3 }{ 4 }) + (frac { 1 }{ 2 })
(j) (frac { 1 }{ 2 }) + (frac { 1 }{ 3 }) + (frac { 1 }{ 6 })
(k) 1(frac { 1 }{ 3 }) + 3(frac { 2 }{ 3 })
(l) 4(frac { 2 }{ 3 }) + 3(frac { 1 }{ 4 })
(m) (frac { 16 }{ 5 }) – (frac { 7 }{ 5 })
(n) (frac { 4 }{ 3 }) – (frac { 1 }{ 2 })
Solution:
(a) (frac { 2 }{ 3 }) + (frac { 1 }{ 7 })
We have:
(frac { 2 }{ 3 }) = (frac{2 times 7}{3 times 7}) = (frac { 14 }{ 21 }) [∵ LCM of 3 and 7 = 21]
(frac { 1 }{ 7 }) = (frac{1 times 3}{7 times 3}) = (frac { 3 }{ 21 })
Now, (frac { 2 }{ 3 }) + (frac { 1 }{ 7 }) = (frac { 14 }{ 21 }) + (frac { 3 }{ 21 }) = (frac { 14 + 3 }{ 21 }) = (frac { 17 }{ 21 })

(b) (frac { 3 }{ 10 }) + (frac { 7 }{ 15 })
We have:
(frac { 3 }{ 10 }) = (frac{3 times 3}{10 times 3}) = (frac { 9 }{ 30 }) [∵ LCM of 10 and 15 = 30]
(frac { 7 }{ 15 }) = (frac{7 times 2}{15 times 2}) = (frac { 14 }{ 30 })
Now, (frac { 3 }{ 10 }) + (frac { 7 }{ 15 }) = (frac { 9 }{ 10 }) + (frac { 14 }{ 30 }) = (frac { 9 + 14 }{ 30 }) = (frac { 23 }{ 30 })

(c) (frac { 4 }{ 9 }) + (frac { 2 }{ 7 })
We have:
(frac { 4 }{ 9 }) = (frac{4 times 7}{9 times 7}) = (frac { 28 }{ 63 }) [∵ LCM of 9 and 7 = 63]
(frac { 7 }{ 15 }) = (frac{2 times 9}{7 times 9}) = (frac { 19 }{ 63 })
Now, (frac { 4 }{ 9 }) + (frac { 2 }{ 7 }) = (frac { 28 }{ 63 }) + (frac { 18 }{ 63 }) = (frac { 28 + 8 }{ 63 }) = (frac { 46 }{ 63 })

(d) (frac { 5 }{ 7 }) + (frac { 1 }{ 3 })
We have:
(frac { 5 }{ 7 }) = (frac{5 times 3}{7 times 3}) = (frac { 15 }{ 21 }) [∵ LCM of 7 and 3 = 21]
(frac { 1 }{ 3 }) = (frac{1 times 7}{3 times 7}) = (frac { 7 }{ 21 })
Now, (frac { 5 }{ 7 }) + (frac { 1 }{ 3 }) = (frac { 15 }{ 21 }) + (frac { 7 }{ 21 }) = (frac { 15 + 7 }{ 21 }) = (frac { 22 }{ 21 })

(e) (frac { 2 }{ 5 }) + (frac { 1 }{ 6 })
We have:
(frac { 2 }{ 5 }) = (frac{2 times 6}{5 times 6}) = (frac { 12 }{ 30 }) [∵ LCM of 5 and 6 = 30]
(frac { 1 }{ 6 }) = (frac{1 times 5}{6 times 5}) = (frac { 5 }{ 30 })
Now, (frac { 2 }{ 5 }) + (frac { 1 }{ 6 }) = (frac { 12 }{ 30 }) + (frac { 5 }{ 30 }) = (frac { 12 + 5 }{ 30 }) = (frac { 17 }{ 30 })

(f) (frac { 4 }{ 5 }) + (frac { 2 }{ 3 })
We have:
(frac { 4 }{ 5 }) = (frac{4 times 3}{5 times 3}) = (frac { 12 }{ 15 }) [∵ LCM of 5 and 3 = 15]
Now, (frac { 2 }{ 3 }) = (frac{2 times 5}{3 times 5}) = (frac { 10 }{ 15 })
(frac { 4 }{ 5 }) + (frac { 2 }{ 3 }) = (frac { 12 }{ 15 }) + (frac { 10 }{ 15 }) = (frac { 22 }{ 15 })

(g) (frac { 3 }{ 4 }) – (frac { 1 }{ 3 })
We have:
(frac { 3 }{ 4 }) = (frac{3 times 3}{4 times 3}) = (frac { 9 }{ 12 }) [∵ LCM of 4 and 3 is 12]
(frac { 1 }{ 3 }) = (frac{1 times 4}{3 times 4}) = (frac { 4 }{ 12 })
Now, (frac { 3 }{ 4 }) – (frac { 1 }{ 3 }) = (frac { 9 }{ 12 }) – (frac { 4 }{ 12 }) = (frac { 9 – 4 }{ 12 }) = (frac { 5 }{ 12 })

(h) (frac { 5 }{ 6 }) – (frac { 1 }{ 3 })
We have:
(frac { 1 }{ 3 }) = (frac{1 times 2}{3 times 2}) = (frac { 2 }{ 6 }) [∵ LCM of 6 and 3 is 6]
Now, (frac { 5 }{ 6 }) – (frac { 1 }{ 3 }) = (frac { 5 }{ 6 }) – (frac { 2 }{ 6 }) = (frac { 5 – 2 }{ 6 }) = (frac { 3 }{ 6 })

(i) (frac { 2 }{ 3 }) + (frac { 3 }{ 4 }) + (frac { 1 }{ 2 })
We have:
(frac { 2 }{ 3 }) = (frac{2 times 7}{3 times 7}) = (frac { 8 }{ 12 }) [∵ LCM of 2, 3 and 4 = 12]
(frac { 3 }{ 4 }) = (frac{3 times 7}{4 times 3}) = (frac { 9 }{ 12 })
(frac { 1 }{ 2 }) = (frac{1 times 6}{2 times 6}) = (frac { 6 }{ 12 })
Now, (frac { 2 }{ 3 }) + (frac { 3 }{ 4 }) + (frac { 1 }{ 2 }) = (frac { 8 }{ 12 }) + (frac { 9 }{ 12 }) + (frac { 6 }{ 12 }) = (frac{8+9+6}{12})

(j) (frac { 1 }{ 2 }) + (frac { 1 }{ 3 }) + (frac { 1 }{ 6 })
We have:
(frac { 1 }{ 2 }) = (frac{1 times 3}{2 times 3}) = (frac { 3 }{ 6 }) [∵ LCM of 2, 3 and 6 = 6]
(frac { 1 }{ 3 }) = (frac{1 times 2}{3 times 2}) = (frac { 2 }{ 6 })
Now, (frac { 1 }{ 2 }) + (frac { 1 }{ 3 }) + (frac { 1 }{ 6 }) = (frac { 3 }{ 6 }) + (frac { 2 }{ 6 }) + (frac { 1 }{ 6 }) = (frac{3+2+1}{6}) = (frac { 6 }{ 6 }) (or 1)

(k) 1(frac { 1 }{ 3 }) + 3(frac { 2 }{ 3 })
Method I
We have: 1(frac { 1 }{ 3 }) = (frac { 4 }{ 3 }) and 3(frac { 2 }{ 3 }) = (frac { 11 }{ 3 })
Now, 1(frac { 1 }{ 3 }) + 3(frac { 2 }{ 3 }) = (frac { 4 }{ 3 }) + (frac { 11 }{ 3 }) = (frac { 4 + 11 }{ 3 }) = (frac { 15 }{ 3 }) = 5
Method II
We have: 1(frac { 1 }{ 3 }) = 1 + (frac { 1 }{ 3 }) and 3(frac { 2 }{ 3 }) = 3 + (frac { 2 }{ 3 })

(l) 4(frac { 2 }{ 3 }) + 3(frac { 1 }{ 4 })
We have: 4(frac { 2 }{ 3 }) = (frac { 14 }{ 3 }) and 3(frac { 1 }{ 4 }) = (frac { 13 }{ 4 }) [∵ LCM of 3 and 4 = 12]
∴ 4(frac { 2 }{ 3 }) + 3(frac { 1 }{ 4 }) = (frac { 14 }{ 3 }) + (frac { 13 }{ 4 })
(frac { 14 }{ 3 }) = (frac{14 times 4}{3 times 4}) = (frac { 56 }{ 12 })
and (frac { 13 }{ 4 }) = = (frac{13 times 3}{4 times 3}) = (frac { 39 }{ 12 })
∴ (frac { 14 }{ 3 }) + (frac { 13 }{ 4 }) = (frac { 56 }{ 12 }) + (frac { 39 }{ 12 })
= (frac{56+39}{12}) = (frac { 95 }{ 12 }) = 7 (frac { 11 }{ 12 })

(m) (frac { 16 }{ 5 }) – (frac { 7 }{ 5 })
We have:
(frac { 16 }{ 5 }) – (frac { 7 }{ 5 }) = (frac { 16 – 7 }{ 5 }) = (frac { 9 }{ 5 }) = 1(frac { 4 }{ 5 })

(n) (frac { 4 }{ 3 }) – (frac { 1 }{ 2 })
We have: (frac { 4 }{ 3 }) = (frac{4 times 2}{3 times 2}) = (frac { 8 }{ 6 }) [∵ LCM of 3 and 2 = 6]
(frac { 1 }{ 2 }) = (frac{1 times 3}{2 times 3}) = (frac { 3 }{ 6 })
Now, (frac { 4 }{ 3 }) – (frac { 1 }{ 2 }) = (frac { 8 }{ 6 }) – (frac { 3 }{ 6 }) = (frac { 8 – 3 }{ 6 }) = (frac { 5 }{ 6 })

Question 2.
Sarita bought (frac { 2 }{ 5 }) metre of ribbon and Lalita (frac { 3 }{ 4 }) metre of ribbon. What is the total length of the ribbon they bought?
Solution:
∵ Ribbon bought by Santa = (frac { 2 }{ 5 }) metre
Ribbon bought by Lalita = (frac { 3 }{ 4 }) metre
∴ Total ribbon bought by them = (frac { 2 }{ 5 }) m + (frac { 3 }{ 4 }) m
Now, (frac { 2 }{ 5 }) = (frac{2 times 4}{5 times 4}) = (frac { 8 }{ 20 }) and (frac { 3 }{ 4 }) = (frac{3 times 5}{4 times 5}) = (frac { 15 }{ 20 }) [∵ LCM of 4 and 5 is 20]
∴ (frac { 2 }{ 5 }) m + (frac { 3 }{ 4 }) m = (frac { 8 }{ 20 }) m + (frac { 15 }{ 20 }) m = (frac{8 m+15 m}{20}) = (frac { 23 }{ 20 }) m

Question 3.
Naina was given 1 (frac { 1 }{ 2 }) piece of cake and Najma was given 1 (frac { 1 }{ 3 }) piece of cake. Find the total amount of cake given to both of them.
Solution:
Amount of cake given to Naina = 1 (frac { 1 }{ 2 }) piece
Amount of cake given to Najma = 1 (frac { 1 }{ 3 }) piece
∴ Total amount of cake given to them = 1 (frac { 1 }{ 2 }) piece + 1 (frac { 1 }{ 3 }) piece
We have:
1 (frac { 1 }{ 2 }) = (frac { 3 }{ 2 }) and 1 (frac { 1 }{ 3 }) = (frac { 4 }{ 3 })
Also (frac { 3 }{ 2 }) = (frac{3 times 3}{2 times 3}) = (frac { 9 }{ 6 }) and (frac { 4 }{ 3 }) = (frac{4 times 2}{3 times 2}) = (frac { 8 }{ 6 })
∴ 1 (frac { 1 }{ 2 }) piece + 1 (frac { 1 }{ 3 }) = piece
= piece = (frac{9+8}{6}) piece = (frac { 17 }{ 6 }) piece

Question 4.

(a)

Here, the ‘missing fraction’ is more than (frac { 1 }{ 4 }) by (frac { 5 }{ 8 }).
∴ Missing fraction = (frac { 1 }{ 4 }) + (frac { 5 }{ 8 })
We have, (frac { 1 }{ 4 }) = (frac{1 times 2}{4 times 2}) = (frac { 2 }{ 8 })
∴ Missing fraction = (frac { 1 }{ 4 }) + (frac { 5 }{ 8 }) = (frac { 2 }{ 8 }) + (frac { 5 }{ 8 }) = (frac{2+5}{8}) = (frac { 7 }{ 8 })
∴

(b)

Here, the ‘missing fraction’ is more than (frac { 1 }{ 2 }) by (frac { 1 }{ 5 }).
∴ (frac { 1 }{ 2 }) + (frac { 1 }{ 5 }) = Missing fraction
or
[∵ LCM of 2 and 5 = 10]
or (frac { 5 }{ 10 }) + (frac { 2 }{ 10 }) = Missing fraction
or (frac{5+2}{10}) or (frac { 7 }{ 10 }) = Missing fraction
Thus,

(c)

Here, the ‘missing fraction’ is less than (frac { 1 }{ 2 }) by (frac { 1 }{ 6 })
∴ Missing fraction = (frac { 1 }{ 2 }) – (frac { 1 }{ 6 })
= (frac { 3 }{ 6 }) – (frac { 1 }{ 6 })
= (frac{3-1}{6}) = (frac { 2 }{ 6 }) = (frac { 1 }{ 3 })
∴

Question 5.
Complete the addition-subtraction box.

Solution:
(a) ∵ (frac { 2 }{ 3 }) + (frac { 4 }{ 3 }) = (frac{2+4}{3}) = (frac { 6 }{ 3 }) = 2
And (frac { 1 }{ 3 }) + (frac { 2 }{ 3 }) = (frac{1+2}{3}) + (frac { 3 }{ 3 }) = 1
Also, (frac { 2 }{ 3 }) – (frac { 1 }{ 3 }) = (frac{2-1}{3}) = (frac { 1 }{ 3 })
and (frac { 4 }{ 3 }) – (frac { 2 }{ 3 }) = (frac{4-2}{3}) = (frac { 2 }{ 3 })

We also have:
2 – 1 = 1 and (frac { 1 }{ 3 }) + (frac { 2 }{ 3 }) = 1

(b) ∵ (frac { 1 }{ 2 }) = (frac{1 times 3}{2 times 3}) = (frac { 3 }{ 6 }) and (frac { 1 }{ 3 }) = (frac{1 times 2}{3 times 2}) = (frac { 2 }{ 6 })
∵ (frac { 1 }{ 2 }) + (frac { 1 }{ 3 }) = (frac { 3 }{ 6 }) + (frac { 2 }{ 6 }) = (frac { 5 }{ 6 }) [∵ LCM of 2 and 3 = 6]

and (frac { 1 }{ 2 }) – (frac { 1 }{ 3 }) = (frac { 3 }{ 6 }) – (frac { 2 }{ 6 }) = (frac { 1 }{ 6 })
Again, (frac { 1 }{ 3 }) = (frac{1 times 4}{3 times 4}) = (frac { 4 }{ 12 }) and (frac { 1 }{ 4 }) = (frac{1 times 3}{4 times 3}) = (frac { 3 }{ 12 }) [∵ LCM of 3 and 4 = 12]
∴ (frac { 1 }{ 3 }) + (frac { 1 }{ 4 }) = (frac { 4 }{ 12 }) + (frac { 3 }{ 12 }) = (frac { 7 }{ 12 })
and (frac { 1 }{ 3 }) – (frac { 1 }{ 4 }) = (frac { 4 }{ 12 }) – (frac { 3 }{ 12 }) = (frac { 1 }{ 12 })
Again, we have (frac { 1 }{ 6 }) + (frac { 1 }{ 12 }) = (frac { 1 }{ 4 }) and (frac { 5 }{ 6 }) – (frac { 7 }{ 12 }) = (frac { 1 }{ 4 })

Question 6.
A piece of wire (frac { 7 }{ 8 }) metre long broke into two pieces. One-piece was (frac { 1 }{ 4 }) metre long. How long is the other piece?
Solution:
Length of the original piece of wire = (frac { 7 }{ 8 }) m As it is broken into two pieces, and one of the parts is (frac { 1 }{ 4 }) m,
∴ Length of the other part = m
Since, (frac { 1 }{ 4 }) = (frac{1 times 2}{4 times 2}) = (frac { 2 }{ 8 })
∴ (frac { 7 }{ 8 }) – (frac { 1 }{ 4 }) = (frac { 7 }{ 8 }) – (frac { 2 }{ 5 })
∴ m = (frac { 7 }{ 8 })m
Thus, the length of the other part is (frac { 5 }{ 8 })m

Question 7.
Nandini s house is (frac { 9 }{ 10 }) km from her school. She walked some distance and then took a bus for (frac { 1 }{ 2 }) km to reach the school. How far did she walk?
Solution:
Distance between Nandini’s house and school = (frac { 9 }{ 10 }) km
Distance covered by bus = (frac { 1 }{ 2 }) km
∴ Distance covered by walking = km

Question 8.
Asha and Samuel have bookshelves of the same size partly filled with books. Asha s shelf is (frac { 5 }{ 6 }) the full and Samuel’s shelf is (frac { 2 }{ 5 }) the full. Whose bookshelf is more full and by what fraction?
Solution:
‘ Portion of Asha’s shelf full of books = (frac { 5 }{ 6 })
Portion of Samuel’s shelf full of books = (frac { 2 }{ 5 })
Changing the fraction into equivalent fractions having the same denominator.
We have: (frac { 5 }{ 6 }) = (frac{5 times 5}{6 times 5}) = (frac { 25 }{ 30 }) and (frac { 2 }{ 5 }) = (frac{2 times 6}{5 times 6}) = (frac { 12 }{ 30 }) [ ∵ HCF of 5 and 6 is 30]
Obviously, portion of Asha’s shelf is more full.
Now, (frac { 25 }{ 30 }) – (frac { 12 }{ 30 }) = (frac { 25 – 12 }{ 30 }) = (frac { 13 }{ 30 })
∴ Fraction by which Asha’s shelf is more full = (frac { 13 }{ 30 })

Question 9.
Jaidev takes 2 (frac { 1 }{ 5 }) minutes to walk across the school ground. Rahul takes (frac { 7 }{ 4 }) minutes to do the same. Who takes less time and by what fraction?
Solution:
To walk across the school ground:
Time taken by Jaidev = 2 (frac { 1 }{ 5 }) minutes
Time taken by Rahul = (frac { 7 }{ 4 }) minutes
We have,
2 (frac { 1 }{ 5 }) = (frac { 11 }{ 5 }) = (frac{11 times 4}{5 times 4}) = (frac { 44 }{ 20 })
[∵ LCM of 4 and 5 = 20]
Also
Obviously, Rahul takes less time.
Now, (frac { 7 }{ 4 }) = (frac{7 times 5}{4 times 5}) = (frac { 35 }{ 20 })
∴ 2 (frac { 1 }{ 5 }) minutes – (frac { 7 }{ 4 }) minutes = (frac { 9 }{ 20 }) minutes
Thus, Rahul takes (frac { 9 }{ 20 }) minutes less time.