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RBSE Solutions for Class 8 Maths Chapter 13 Comparison of Quantities Ex 13.4

March 23, 2019 by Fazal Leave a Comment

RBSE Solutions for Class 8 Maths Chapter 13 Comparison of Quantities Ex 13.4 is part of RBSE Solutions for Class 8 Maths. Here we have given Rajasthan Board RBSE Class 8 Maths Chapter 13 Comparison of Quantities Exercise 13.4.

Board RBSE
Textbook SIERT, Rajasthan
Class Class 8
Subject Maths
Chapter Chapter 13
Chapter Name Comparison of Quantities
Exercise Exercise 13.4
Number of Questions 11
Category RBSE Solutions

Rajasthan Board RBSE Class 8 Maths Chapter 13 Comparison of Quantities Ex 13.4

Question 1.
Vimla traveled 200 km distance by bus and she gave fare of Rs. 180. What rent she has to pay for travelling a distance of 500 km.
Solution
Let fare be Rs x.
RBSE Solutions for Class 8 Maths Chapter 13 Comparison of Quantities Ex 13.4 img-1
Here, direct relation occurred.
200 : 500 :: 180 : x
Product of(RBSESolutions.com)outer terms = Product of middle terms
⇒ 200 × x = 500 × 180
⇒ \(x=\frac { 500\times 180 }{ 200 } \)
⇒ x = 450
Hence, Rs 450 is the required fare.

RBSE Solutions

Question 2.
Shadow of a 20 metre long trees is 18 m in morning. What will be height of shadow of 120 m high tower at same time?
Solution
RBSE Solutions for Class 8 Maths Chapter 13 Comparison of Quantities Ex 13.4 img-2
Let x meter be the length of shadow.
This is direct relation.
10 : 120 :: 18 : x
Product of outer terms = Product of (RBSESolutions.com)middle terms
⇒ 10 × x = 120 × 18
⇒ \(x=\frac { 120\times 18 }{ 10 } \)
⇒ x = 216
Hence, the required shadow is 216 meter.

RBSE Solutions

Question 3.
If weight of 5 books is 2.5 kg then 30 kg will be weight of how many books?
Solution
RBSE Solutions for Class 8 Maths Chapter 13 Comparison of Quantities Ex 13.4 img-3
Let 30 kg be the weight of x books.
Here direct relation occurred.
5 : x :: 2.5 x 30
Product of outer terms = Product of middle terms
⇒ 5 × 30 = x × 2.5
⇒ \(x=\frac { 5\times 30 }{ 2.5 } \)
⇒ x = 60
Hence, 60 books having 30 kg. weight.

RBSE Solutions

Question 4.
A bus is moving with a uniform speed of 45 km per hour then what time bus will take to travel a distance of 225 km.
Solution
RBSE Solutions for Class 8 Maths Chapter 13 Comparison of Quantities Ex 13.4 img-4
Let x hours will be taken.
Here, direct relation occurred.
45 : 225 :: 1 : x
Product of outer terms = Product of(RBSESolutions.com)middle terms
⇒ 45 × x = 225 × 1
⇒ \(x=\frac { 225 }{ 45 }\)
⇒ x = 5
Hence, required time is 5 horns.

RBSE Solutions

Question 5.
Mamta can fill 30 parindahs with 15 liters of water then tell how many liters of water will be required for filling 120 such parindahs.
Solution
RBSE Solutions for Class 8 Maths Chapter 13 Comparison of Quantities Ex 13.4 img-5
Let x liter water be required.
Here, direct relation occurred.
15 : x :: 30 : 120
Product of outer(RBSESolutions.com)terms = Product of middle terms
⇒ 15 × 120 = x × 30
⇒ \(x=\frac { 15\times 120 }{ 30 } \)
⇒ x = 60
Hence, 60 liters water is required.

RBSE Solutions

Question 6.
100 liters of water can be saved by washing 5 cars with jug and buckets instead of tap. In this way how many liters of water can be saved by washing 20 such cars?
Solution
RBSE Solutions for Class 8 Maths Chapter 13 Comparison of Quantities Ex 13.4 img-6
Let x liter water can be saved.
Here, direct(RBSESolutions.com)relation occurred.
5 : 20 :: 100 : x
Product of outer terms = Product of middle terms
⇒ 5 × x = 20 × 100
⇒ \(x=\frac { 20\times 100 }{ 5 } \)
⇒ x = 400
Hence, 400 liter water can be saved.

RBSE Solutions

Question 7.
9 workers took 16 days to complete pucca boundary walls of school. If number of workers is 12 then wall can be prepared in how many days?
Solution
Let x days be required to complete the work.
RBSE Solutions for Class 8 Maths Chapter 13 Comparison of Quantities Ex 13.4 img-7
Here, inverse relation occurred.
12 : 9 :: 16 : x
Product of outer terms = Product of middle terms
⇒ 12 × x = 9 × 16
⇒ \(x=\frac { 9\times 16 }{ 12 } \)
⇒ x = 12
Hence, 12 days are(RBSESolutions.com)required to complete wall.

RBSE Solutions

Question 8.
A camp has food for 40 soldiers for 20 days. After 5 days 10 more soldiers joined then rest of the food will be sufficient for how many days?
Solution
The remaining food was enough for 40 soldiers for 20 – 5 = 25 days. Let 40 + 10 = 50 soldiers be there have enough food for x days.
RBSE Solutions for Class 8 Maths Chapter 13 Comparison of Quantities Ex 13.4 img-8
Here, inverse relation occurred.
50 : 40 :: 15 : x
Product of outer terms = Product of middle terms
⇒ 50 × x = 40 × 15
⇒ \(x=\frac { 40\times 15 }{ 50 } \)
⇒ x = 12
Hence, required(RBSESolutions.com)day are 12.

RBSE Solutions

Question 9.
Under Swachh Bharat Mission 15 volunteers clean their village in 4 days. I village needs to be cleaned in 3 days then how many workers will be required?
Solution
Let x volunteers be required.
RBSE Solutions for Class 8 Maths Chapter 13 Comparison of Quantities Ex 13.4 img-9
Here, inverse relation occurred.
3 : 4 :: 15 : x
Product of(RBSESolutions.com)outer terms = Product of middle terms
⇒ 3 × x = 4 × 15
⇒ \(x=\frac { 4\times 15 }{ 3 } \)
⇒ x = 20
Hence, the required volunteers are 20.

RBSE Solutions

Question 10.
In a school under shramdan 2 students clean for 5 hours. If same part needs to be cleaned in 3 days then how many workers will be required?
Solution
Let x students be required
RBSE Solutions for Class 8 Maths Chapter 13 Comparison of Quantities Ex 13.4 img-10
Here, inverse relation occurred,
3 : 5 :: 12 : x
Product(RBSESolutions.com)of outer terms = Product of middle terms
⇒ 3 × x = 5 × 12
⇒ \(x=\frac { 5\times 12 }{ 3 } \)
⇒ x = 20
Hence, 20 students are required.

RBSE Solutions

Question 11.
Madhu prepares food for 12 days from biogas plant by putting on 80 kilograms of cowdung. Then to prepare food for 60 days how much cow dung will be required?
Solution
Let x kg. dung is required
RBSE Solutions for Class 8 Maths Chapter 13 Comparison of Quantities Ex 13.4 img-11
Here, direct relation occurred.
12 : 60 :: 80 : x
Product of outer terms = Product(RBSESolutions.com)of middle terms
⇒ 12 × x = 60 × 80
⇒ \(x=\frac { 60\times 80 }{ 12 } \)
⇒ x = 400
Hence, the required dung is 400 kg.

RBSE Solutions

We hope the given RBSE Solutions for Class 8 Maths Chapter 13 Comparison of Quantities Ex 13.4 will help you. If you have any query regarding Rajasthan Board RBSE Class 8 Maths Chapter 13 Comparison of Quantities Exercise 13.4, drop a comment below and we will get back to you at the earliest.

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