# GSEB Solutions Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.3

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 $$\frac { 8 }{ 15 }$$.