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.

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.

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

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.

Question 3.

If weight of 5 books is 2.5 kg then 30 kg will be weight of how many books?

Solution

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.

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

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.

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

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.

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

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.

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.

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.

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.

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.

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.

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.

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

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.

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

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.

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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