RBSE Solutions for Class 7 Maths Chapter 15 Comparison of Quantities Ex 15.1

RBSE Solutions for Class 7 Maths Chapter 15 Comparison of Quantities Ex 15.1 is part of RBSE Solutions for Class 7 Maths. Here we have given Rajasthan Board RBSE Class 7 Maths Chapter 15 Comparison of Quantities Exercise 15.1.

Board	RBSE
Textbook	SIERT, Rajasthan
Class	Class 7
Subject	Maths
Chapter	Chapter 15
Chapter Name	Comparison of Quantities
Exercise	Ex 15.1
Number of Questions	10
Category	RBSE Solutions

Rajasthan Board RBSE Class 7 Maths Chapter 15 Comparison of Quantities Ex 15.1

Question 1
Find the ratio :
(i) 60 paise to ₹3
(ii) 340 cm to 4 meter
Solution:
(i) 60 paise to ₹3 = 60 paise : (3 x 100) paise
[∵ ₹1 = 100 paise]
= 60 paise : 300 paise
= [latex]\frac { 60 }{ 300 }[/latex] = [latex]\frac { 1 }{ 5 }[/latex] = 1 : 5

(ii) 340 cm to 4 meter
[∵1 meter = 100 cm]
= 340 cm : (4 x 100) cm।
= [latex]\frac { 340 }{ 400 }[/latex] = [latex]\frac { 17 }{ 20 }[/latex] = 17 : 20

Question 2
Write in (RBSESolutions.com) simplest ratios
(i) 65 : 25
(ii) 72 : 64
Solution:
(i) 65 : 25 = [latex]\frac { 65 }{ 25 }[/latex] = [latex]\frac { 13 }{ 5 }[/latex] = 13:5
(ii) 72 : 64 = [latex]\frac { 72 }{ 64 }[/latex] = [latex]\frac { 9 }{ 8 }[/latex] = 9: 8

Question 3
Find out two equivalent (RBSESolutions.com) ratios of the following ratio
(i) 3 : 5
(ii) 7 : 11
Solution:
(i) 3 : 5 = 3 x 2 : 5 x 2 = 6 : 10
and 3 : 5 = 3 x 3 : 5 x 3 = 9 : 15
(ii) 7 : 11 = 7 x 2 : 11 x 2 = 14 : 22
and 7 : 11 = 7 x 3 : 11 x 3 = 21 : 33

Question 4
The length and breadth of (RBSESolutions.com) carpet is 7 m and 35 cm. respectively. Then find the following ratio.
(i) Breadth with Lenght
(ii) Length with breadth
Solution:
Length of carpet = 7 meter = 7 x 100 cm
= 700 and breadth of carpet = 35 cm
(i) Ratio of breadth to length
= 35 : 700 = [latex]\frac { 35 }{ 700 }[/latex] = [latex]\frac { 1 }{ 20 }[/latex] = 1 : 20
(ii) Ratio of length to breadth
= 700 : 35 = [latex]\frac { 700 }{ 35 }[/latex] : [latex]\frac { 20 }{ 1 }[/latex] = 20 : 1

Question 5
If 12 : x :: 14 : 21 then find (RBSESolutions.com) the value of x.
Solution:
12 : x :: 14 : 21

Question 6
Bhima halwai mixes 20 kg of sugar into 25 kg pulse (RBSESolutions.com) for making halwa. While Bhikha halwai mixes 15 kg. of sugar into 12 kg of pulses. Find :
(i) How kuch sugar do both the halwais mix in per kg of pulse?
(ii) The prepared halwa of which halwai is sweeter?
Solution:
(i) Bhima mixes 20 kg sugar in 25 kg pulse
∴ Quantity of sugar mixed in one kg pulse
= [latex]\frac { 20 }{ 25 }[/latex] = [latex]\frac { 4 }{ 5 }[/latex] = 800 gm
Bhikha mixes 15 kg sugar in 12 kg pulse Quantity of sugar mixed in one kg pulse
= [latex]\frac { 15 }{ 12 }[/latex] = 1.25kg = 1 kg 250 gm
(ii) Bhikha confectioner Halwa is more sweeter.

Question 7
34 labourers are required for cleaning a 10.2 km long road. Then (RBSESolutions.com) how many labourers are required for cleaning 7.5 km long road?
Solution:
10.2 km road is cleaned by 34 labourers
∴ 1 km road can be cleaned by = [latex]\frac { 34 }{ 10.2 }[/latex] labourers
∴ To clean 7.5 km road, no. of lobourors required
= [latex]\frac { 75\times 34 }{ 10.2 }[/latex] = [latex]\frac { 75\times 34 }{ 102 }[/latex] = 25 labourers

Question 8
The shadow of 7.5 mt. pole is 5 meter, then find out the (RBSESolutions.com) height of a tree standing nearby whose shadow is 10 mt. long at the same time.
Solution:
∵ 5 metre shadow then height is = 7.5 cm
∴ 1 metre shadow then height = [latex]\frac { 7.5 }{ 5 }[/latex] cm
∴ 10 metre shadow then height
= [latex]\frac { 10\times 7.5 }{ 5 }[/latex] = [latex]\frac { 75 }{ 5 }[/latex] = 15 meter
Therefore the height of tree is 15 meter.

Question 9
Ramesh covers a distance of 10 km in 15 min by his motorcycle. How (RBSESolutions.com) much time is re-quired by Ramesh to cover a distance of 26 km if the speed is same?
Solution:
∵ 10 km distance is travelled in 15 minutes
∴ 1 km distance can be travelled in = [latex]\frac { 15 }{ 10 }[/latex] minutes
∴ 26 km distance can be travelled in [latex]\frac { 26\times 15 }{ 10 }[/latex] = 39 minutes.

Question 10
3 kg pulse is required in the mid day meal of 60 students. How much (RBSESolutions.com) quantity of pulse in enough on Saturday at the time of mid day meals if 46 students were present?
Solution:
∵ 60 students pulse is required 3 kg
∴ one students required pulse = [latex]\frac { 3 }{ 60 }[/latex] kg
∴ 46 students required pulse = = [latex]\frac { 46\times 3 }{ 60 }[/latex] = [latex]\frac { 46 }{ 20 }[/latex] = 2.3 kg = 2 kg

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