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GSEB Solutions Class 6 Maths Chapter 7 Fractions Ex 7.6

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 } GSEB Solutions Class 6 Maths Chapter 7 Fractions Ex 7.6 img 1

(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 } GSEB Solutions Class 6 Maths Chapter 7 Fractions Ex 7.6 img 2

(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 } GSEB Solutions Class 6 Maths Chapter 7 Fractions Ex 7.6 img 3

(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 }
GSEB Solutions Class 6 Maths Chapter 7 Fractions Ex 7.6 img 4

(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 }

GSEB Solutions Class 6 Maths Chapter 7 Fractions Ex 7.6

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

GSEB Solutions Class 6 Maths Chapter 7 Fractions Ex 7.6

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 } = GSEB Solutions Class 6 Maths Chapter 7 Fractions Ex 7.6 img 5 piece
=GSEB Solutions Class 6 Maths Chapter 7 Fractions Ex 7.6 img 6 piece = \frac{9+8}{6} piece = \frac { 17 }{ 6 } piece

GSEB Solutions Class 6 Maths Chapter 7 Fractions Ex 7.6

Question 4.
GSEB Solutions Class 6 Maths Chapter 7 Fractions Ex 7.6 img 7
(a)
GSEB Solutions Class 6 Maths Chapter 7 Fractions Ex 7.6 img 8
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 }
GSEB Solutions Class 6 Maths Chapter 7 Fractions Ex 7.6 img 9

(b)
GSEB Solutions Class 6 Maths Chapter 7 Fractions Ex 7.6 img 10
Here, the ‘missing fraction’ is more than \frac { 1 }{ 2 } by \frac { 1 }{ 5 }.
∴  \frac { 1 }{ 2 } + \frac { 1 }{ 5 } = Missing fraction
or  GSEB Solutions Class 6 Maths Chapter 7 Fractions Ex 7.6 img 11
[∵ 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, GSEB Solutions Class 6 Maths Chapter 7 Fractions Ex 7.6 img 12

(c)
GSEB Solutions Class 6 Maths Chapter 7 Fractions Ex 7.6 img 13
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 }GSEB Solutions Class 6 Maths Chapter 7 Fractions Ex 7.6 img 13
= \frac{3-1}{6} = \frac { 2 }{ 6 }\frac { 1 }{ 3 }
GSEB Solutions Class 6 Maths Chapter 7 Fractions Ex 7.6 img 14

GSEB Solutions Class 6 Maths Chapter 7 Fractions Ex 7.6

Question 5.
Complete the addition-subtraction box.
GSEB Solutions Class 6 Maths Chapter 7 Fractions Ex 7.6 img 15
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 }
GSEB Solutions Class 6 Maths Chapter 7 Fractions Ex 7.6 img 16
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]
GSEB Solutions Class 6 Maths Chapter 7 Fractions Ex 7.6 img 17
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 }

GSEB Solutions Class 6 Maths Chapter 7 Fractions Ex 7.6

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 = GSEB Solutions Class 6 Maths Chapter 7 Fractions Ex 7.6 img 18 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 }
GSEB Solutions Class 6 Maths Chapter 7 Fractions Ex 7.6 img 18 m = \frac { 7 }{ 8 }m
Thus, the length of the other part is \frac { 5 }{ 8 }m

GSEB Solutions Class 6 Maths Chapter 7 Fractions Ex 7.6

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 = GSEB Solutions Class 6 Maths Chapter 7 Fractions Ex 7.6 img 19 km
GSEB Solutions Class 6 Maths Chapter 7 Fractions Ex 7.6 img 20

GSEB Solutions Class 6 Maths Chapter 7 Fractions Ex 7.6

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 }

GSEB Solutions Class 6 Maths Chapter 7 Fractions Ex 7.6

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.

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