# GSEB Solutions Class 6 Maths Chapter 7 Fractions Ex 7.6

Gujarat BoardĀ 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:
$$\frac { 2 }{ 3 }$$ = $$\frac{2 \times 7}{3 \times 7}$$ = $$\frac { 14 }{ 21 }$$ [āµ 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:
$$\frac { 3 }{ 10 }$$ = $$\frac{3 \times 3}{10 \times 3}$$ = $$\frac { 9 }{ 30 }$$ [āµ 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:
$$\frac { 4 }{ 9 }$$ = $$\frac{4 \times 7}{9 \times 7}$$ = $$\frac { 28 }{ 63 }$$ [āµ 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:
$$\frac { 5 }{ 7 }$$ = $$\frac{5 \times 3}{7 \times 3}$$ = $$\frac { 15 }{ 21 }$$ [āµ 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:
$$\frac { 2 }{ 5 }$$ = $$\frac{2 \times 6}{5 \times 6}$$ = $$\frac { 12 }{ 30 }$$ [āµ 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:
$$\frac { 4 }{ 5 }$$ = $$\frac{4 \times 3}{5 \times 3}$$ = $$\frac { 12 }{ 15 }$$ [āµ 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:
$$\frac { 3 }{ 4 }$$ = $$\frac{3 \times 3}{4 \times 3}$$ = $$\frac { 9 }{ 12 }$$ [āµ 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:
$$\frac { 1 }{ 3 }$$ = $$\frac{1 \times 2}{3 \times 2}$$ = $$\frac { 2 }{ 6 }$$ [āµ 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:
$$\frac { 2 }{ 3 }$$ = $$\frac{2 \times 7}{3 \times 7}$$ = $$\frac { 8 }{ 12 }$$ [āµ 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:
$$\frac { 1 }{ 2 }$$ = $$\frac{1 \times 3}{2 \times 3}$$ = $$\frac { 3 }{ 6 }$$ [āµ 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$$\frac { 1 }{ 4 }$$ = $$\frac { 13 }{ 4 }$$ [āµ LCM of 3 and 4 = 12]
ā“ 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 $$\frac { 13 }{ 4 }$$ = = $$\frac{13 \times 3}{4 \times 3}$$ = $$\frac { 39 }{ 12 }$$
ā“ $$\frac { 14 }{ 3 }$$ + $$\frac { 13 }{ 4 }$$ = $$\frac { 56 }{ 12 }$$ + $$\frac { 39 }{ 12 }$$
= $$\frac{56+39}{12}$$ = $$\frac { 95 }{ 12 }$$ = 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: $$\frac { 4 }{ 3 }$$ = $$\frac{4 \times 2}{3 \times 2}$$ = $$\frac { 8 }{ 6 }$$ [āµ 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:
āµ Ribbon bought by Santa = $$\frac { 2 }{ 5 }$$ metre
Ribbon bought by Lalita = $$\frac { 3 }{ 4 }$$ metre
ā“ 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 $$\frac { 3 }{ 4 }$$ = $$\frac{3 \times 5}{4 \times 5}$$ = $$\frac { 15 }{ 20 }$$ [āµ LCM of 4 and 5 is 20]
ā“ $$\frac { 2 }{ 5 }$$ 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
ā“ 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 $$\frac { 4 }{ 3 }$$ = $$\frac{4 \times 2}{3 \times 2}$$ = $$\frac { 8 }{ 6 }$$
ā“ 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 $$\frac { 5 }{ 8 }$$.
ā“ Missing fraction = $$\frac { 1 }{ 4 }$$ + $$\frac { 5 }{ 8 }$$
We have, $$\frac { 1 }{ 4 }$$ = $$\frac{1 \times 2}{4 \times 2}$$ = $$\frac { 2 }{ 8 }$$
ā“ Missing fraction = $$\frac { 1 }{ 4 }$$ + $$\frac { 5 }{ 8 }$$ = $$\frac { 2 }{ 8 }$$ + $$\frac { 5 }{ 8 }$$ = $$\frac{2+5}{8}$$ = $$\frac { 7 }{ 8 }$$
ā“

(b)

Here, the ‘missing fraction’ is more than $$\frac { 1 }{ 2 }$$ by $$\frac { 1 }{ 5 }$$.
ā“Ā  $$\frac { 1 }{ 2 }$$ + $$\frac { 1 }{ 5 }$$ = Missing fraction
orĀ
[āµ 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 $$\frac { 1 }{ 6 }$$
ā“ Missing fraction = $$\frac { 1 }{ 2 }$$ –Ā  $$\frac { 1 }{ 6 }$$
= $$\frac { 3 }{ 6 }$$ –Ā  $$\frac { 1 }{ 6 }$$
= $$\frac{3-1}{6}$$ = $$\frac { 2 }{ 6 }$$ =Ā  $$\frac { 1 }{ 3 }$$
ā“

Question 5.

Solution:
(a) āµ $$\frac { 2 }{ 3 }$$ + $$\frac { 4 }{ 3 }$$ = $$\frac{2+4}{3}$$ = $$\frac { 6 }{ 3 }$$ = 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) āµ $$\frac { 1 }{ 2 }$$ = $$\frac{1 \times 3}{2 \times 3}$$ = $$\frac { 3 }{ 6 }$$ and $$\frac { 1 }{ 3 }$$ = $$\frac{1 \times 2}{3 \times 2}$$ = $$\frac { 2 }{ 6 }$$
āµ $$\frac { 1 }{ 2 }$$ + $$\frac { 1 }{ 3 }$$ = $$\frac { 3 }{ 6 }$$ + $$\frac { 2 }{ 6 }$$ = $$\frac { 5 }{ 6 }$$ [āµ 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 $$\frac { 1 }{ 4 }$$ = $$\frac{1 \times 3}{4 \times 3}$$ = $$\frac { 3 }{ 12 }$$ [āµ LCM of 3 and 4 = 12]
ā“ $$\frac { 1 }{ 3 }$$ + $$\frac { 1 }{ 4 }$$ = $$\frac { 4 }{ 12 }$$ + $$\frac { 3 }{ 12 }$$ = $$\frac { 7 }{ 12 }$$
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,
ā“ Length of the other part = m
Since, $$\frac { 1 }{ 4 }$$ = $$\frac{1 \times 2}{4 \times 2}$$ = $$\frac { 2 }{ 8 }$$
ā“ $$\frac { 7 }{ 8 }$$ – $$\frac { 1 }{ 4 }$$ = $$\frac { 7 }{ 8 }$$ – $$\frac { 2 }{ 5 }$$
ā“ 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 Nandiniās house and school = $$\frac { 9 }{ 10 }$$ km
Distance covered by bus = $$\frac { 1 }{ 2 }$$ km
ā“ 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 Samuelās shelf is $$\frac { 2 }{ 5 }$$ the full. Whose bookshelf is more full and by what fraction?
Solution:
ā Portion of Ashaās shelf full of books = $$\frac { 5 }{ 6 }$$
Portion of Samuelā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 $$\frac { 2 }{ 5 }$$ = $$\frac{2 \times 6}{5 \times 6}$$ = $$\frac { 12 }{ 30 }$$ [ āµ HCF of 5 and 6 is 30]
Obviously, portion of Ashaās shelf is more full.
Now, $$\frac { 25 }{ 30 }$$ – $$\frac { 12 }{ 30 }$$ = $$\frac { 25 – 12 }{ 30 }$$ = $$\frac { 13 }{ 30 }$$
ā“ 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 $$\frac { 1 }{ 5 }$$ = $$\frac { 11 }{ 5 }$$ = $$\frac{11 \times 4}{5 \times 4}$$ = $$\frac { 44 }{ 20 }$$
[āµ LCM of 4 and 5 = 20]
Also
Obviously, Rahul takes less time.
Now, $$\frac { 7 }{ 4 }$$ = $$\frac{7 \times 5}{4 \times 5}$$ = $$\frac { 35 }{ 20 }$$
ā“ 2 $$\frac { 1 }{ 5 }$$ minutes – $$\frac { 7 }{ 4 }$$ minutes = $$\frac { 9 }{ 20 }$$ minutes
Thus, Rahul takes $$\frac { 9 }{ 20 }$$ minutes less time.