Gujarat Board GSEB Textbook Solutions Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.3 Textbook Questions and Answers.
Gujarat Board Textbook Solutions Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.3
Question 1.
Find (i) \frac { 1 }{ 4 } of (a) \frac { 1 }{ 4 } (b) \frac { 3 }{ 4 } (c) \frac { 4 }{ 3 }
(ii) \frac { 1 }{ 7 } of (a) \frac { 2 }{ 9 } (b) \frac { 6 }{ 5 } (c) \frac { 3 }{ 10 }
Solution:
(i) (a) \frac { 1 }{ 4 } of \frac { 1 }{ 4 }
= \frac { 1 }{ 4 } x \frac { 1 }{ 4 }
= \frac { 1 ×1 }{ 4×4 } = \frac { 1 }{ 16 }
(b) \frac { 1 }{ 4 } of \frac { 3 }{ 5 }
= \frac { 1 }{ 4 } x \frac { 3 }{ 5 }
= \frac { 1 ×3 }{ 4×5 } = \frac { 3 }{ 20 }
(c) \frac { 1 }{ 4 } of \frac { 4 }{ 3 }
= \frac { 1 }{ 4 } x \frac { 4 }{ 3 }
= \frac { 1 ×4 }{ 4×3 } = \frac { 1 }{ 3 }
(ii) (a) \frac { 1 }{ 7 } of \frac { 2 }{ 9 }
= \frac { 1 }{ 7 } x \frac { 2 }{ 9 }
= \frac { 1 ×2 }{ 7×9 } = \frac { 2 }{ 63 }
(b) \frac { 1 }{ 7 } of \frac { 6 }{ 5 }
= \frac { 1 }{ 7 } x \frac { 6 }{ 5 }
= \frac { 1 ×6 }{ 7×5 } = \frac { 6 }{ 35 }
(c) \frac { 1 }{ 7 } of \frac { 3 }{ 10 }
= \frac { 1 }{ 7 } x \frac { 3 }{ 10 }
= \frac { 1 ×3 }{ 7×10 } = \frac { 3 }{ 70 }
Question 2.
Multiply and reduce to lowest form (if possible):
(i) \frac { 2 }{ 3 } x 2\frac { 2 }{ 3 }
(ii) \frac { 2 }{ 7 } x \frac { 7 }{ 9 }
(iii) \frac { 3 }{ 8 } x \frac { 6 }{ 4 }
(iv) \frac { 9 }{ 5 } x \frac { 3 }{ 5 }
(v) \frac { 1 }{ 3 } x \frac { 15 }{ 8 }
(vi) \frac { 11 }{ 2 } x \frac { 3 }{ 10 }
(vii) \frac { 4 }{ 5 } x \frac { 12 }{ 7 }
Solution:
Question 3.
Multiply and reduce to lowest form and convert into a mixed fraction:
(i) \frac { 2 }{ 5 } x 5\frac { 1 }{ 4 }
(ii) 6\frac { 2 }{ 5 } x \frac { 7}{ 9 }
(iii) \frac { 3 }{ 2 } x 5\frac { 1 }{ 3 }
(iv) \frac { 5 }{ 6 } x 2\frac { 3 }{ 7 }
(v) 3\frac { 2 }{ 5 } x 4\frac { 4 }{ 7 }
(vi) 2\frac { 3 }{ 5 } x 3
(vii) 3\frac { 4 }{ 7 } x \frac { 3 }{ 5 }
Solution:
Question 4.
Which is greater:
(i) \frac { 2 }{ 7 } of \frac { 3 }{ 4 } or \frac { 3 }{ 5 } of \frac { 5 }{ 8 }?
(ii) \frac { 1 }{ 2 } of \frac { 6 }{ 7 } or \frac { 2 }{ 3 } of \frac { 3 }{ 7 }?
Solution:
(i) Comparing \frac { 2 }{ 7 } of \frac { 3 }{ 4 } and \frac { 3 }{ 5 } of \frac { 5 }{ 8 }
∵ \frac { 2 }{ 7 } of \frac { 3 }{ 4 }
= \frac { 2 }{ 7 } x \frac { 3 }{ 4 }
= \frac { 1×3 }{ 7×2 }
= \frac { 3 }{ 14 }
and \frac { 3 }{ 5 } of \frac { 5 }{ 8 }
= \frac { 3 }{ 5 } x \frac { 5 }{ 8 }
= \frac { 3×1 }{ 1×8 }
= \frac { 3 }{ 8 }
Again \frac { 3 }{ 14 } = \frac { 3×4 }{ 14×4 }
= \frac { 12 }{ 56 } [∵ LCM of 14 and 8 is 56]
= \frac { 3 }{ 8 } = \frac { 3×7 }{ 8×7 } = \frac { 21 }{ 56 }
Now, \frac { 21 }{ 56 } > \frac { 12 }{ 56 }
∴ \frac { 3 }{ 5 } of \frac { 5 }{ 8 } is greater than \frac { 2 }{ 7 } of \frac { 3 }{ 4 }.
(ii) Comparing \frac { 1 }{ 2 } of \frac { 6 }{ 7 } or \frac { 2 }{ 3 } of \frac { 3 }{ 7 }:
∵ \frac { 1 }{ 2 } of \frac { 6 }{ 7 } = \frac { 1 }{ 2 } x \frac { 6 }{ 7 }
= \frac { 1×3 }{ 1×7 } = \frac { 3 }{ 7 }
and, \frac { 2 }{ 3 } of \frac { 3 }{ 7 }
= \frac { 2 }{ 3 } x \frac { 3 }{ 7 }
= \frac { 2×1 }{ 1×7 } = \frac { 2 }{ 7 }
Also 3 > 2 ⇒ \frac { 3 }{ 7 } > \frac { 2 }{ 7 }
∴ \frac { 1 }{ 2 } of \frac { 6 }{ 7 } is greater than \frac { 2 }{ 3 } of \frac { 3 }{ 7 }
Question 5.
Saili plants 4 saplings, in a row, in her garden. The distance between two adjacent saplings is \frac { 3 }{ 4 } m. Find the distance between the first and the last sapling.
Solution:
Number of saplings = 4
Distance between two adjacent saplings = \frac { 3 }{ 4 }m
∴ Distance between 1st and last (4th) sapling , 3
= 3 x \frac { 3 }{ 4 }m
= \frac { 3×3 }{ 4 }m
= \frac { 9 }{ 4 }m
= 2\frac { 1 }{ 4 }m
Question 6.
Lipika reads a book for 1\frac { 3 }{ 4 } hours everyday. She reads the entire book in 6 days. How many hours in all were required by her to read the book?
Solution:
Number of days = 6
Reading hours per day (for one day) = 1\frac { 3 }{ 4 } hours
∴ Total number of reading hours
= 6 x 1\frac { 3 }{ 4 }
= 6 x \frac { 7 }{ 4 } hours
= \frac { 3×7 }{ 2 } hours = \frac { 21 }{ 2 } hours = 10\frac { 1 }{ 2 } hours
Question 7.
A car runs 16 km using 1 litre of petrol. How much distance will it cover using 2\frac { 3 }{ 4 } litres of petrol.
Solution:
Distance covered in 1 litre of petrol =16 km
∴ Distance covered in 2\frac { 3 }{ 4 } litres of petrol
= 16 x 2\frac { 3 }{ 4 } km
= 16 x \frac { 11 }{ 4 } km
= \frac { 16×11 }{ 4 } km
= \frac { 4×11 }{ 1 } km
= 44 km
Question 8.
(a)
(i) Provide the number in the box 
, such that \frac { 2 }{ 3 } x = \frac { 10 }{ 30 }.
(ii) The simplest form of the number obtained in is ______.
(b)
(i) Provide the number in the box , such that \frac { 3 }{ 5 } x = \frac { 24 }{ 75 }.
(ii) The simplest form of the number obtained in is ______.
Solution:
(a)
(i) We have: \frac { 2 }{ 3 } x \frac { 5 }{ 10 } = \frac { 10 }{ 30 } , i.e. the required number = \frac { 5 }{ 10 }
(ii) Simplest form of 
(b)
(i) We have: \frac { 3 }{ 5 } x \frac { 8 }{ 15 }
= \frac { 24 }{ 75 }, i.e. the required number = \frac { 8 }{ 15 }
(ii) The simplest form of \frac { 8 }{ 15 } is \frac { 8 }{ 15 }.