RBSE Solutions for Class 7 Maths Chapter 14 Simple Equation Ex 14.2 is part of RBSE Solutions for Class 7 Maths. Here we have given Rajasthan Board RBSE Class 7 Maths Chapter 14 Simple Equation Exercise 14.2.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 7 |

Subject |
Maths |

Chapter |
Chapter 14 |

Chapter Name |
Simple Equation |

Exercise |
Ex 14.2 |

Number of Questions |
11 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 7 Maths Chapter 14 Simple Equation Ex 14.2

Question 1

By adding 12 in any number 43 is obtained. Find out (RBSESolutions.com) that number.

Solution:

Let x is the number.

According to question x + 12 = 43

⇒ x = 43 – 12 = 31

So the number is 31.

Question 2

On subtracting 5 from 4 times (RBSESolutions.com) of any number, we get 27. Find out that number.

Solution:

Let that number be x.

According to question,

4x – 5 = 27

⇒ 4 = 27 + 5 = 32

⇒ x = \(\frac { 32 }{ 4 }\) = 8

So the number is 8.

Question 3

On adding double of a number in 5 times of (RBSESolutions.com) that number, we get 42. Find out that number.

Solution:

Let the number be x

According to question,

5x + 2 = 42

⇒ 7x = 42

⇒ x = \(\frac { 42 }{ 7 }\) = 6

So, the number is 6.

Question 4

Sum of three consecutive (RBSESolutions.com) number is 27. Find out that numbers

Solution:

Let three consecutive numbers are x – 1, x andx + 1.

According to question.

x – 1 + x + x + 1 = 27

⇒ 3x + 1 – 1 = 27

⇒ 3x = 27

⇒ x = \(\frac { 27 }{ 3 }\) = 9

∴ x – 1 = 9 – 1 = 8

x = 9

and x + 1 = 9+ 1 = 10

So, three consecutive numbers are 8, 9 and 10.

Question 5

Sum of three consecutive odd (RBSESolutions.com) numbers in 39. Find out the numbers.

Solution:

Let the three odd consecutive numbers are (2 x + 1), (2x + 3) and (2x + 5).

According to question,

(2x + 1) + (2x + 3) + (2x + 5) = 39

⇒ 2x + 1 + 2x + 3 + 2x + 5 = 39

⇒ 6x + 9 = 39

⇒ 6x = 39 – 9

⇒ x = \(\frac { 30 }{ 6 }\)

⇒ x = 5

First (RBSESolutions.com) odd number = 2x +1

= 2 x 5 + 1 = 11

Second odd number= 2x +3

= 2 x 5 + 3 = 13

and third odd umber= 2x +5

= 2 x 5 + 5 = 15

So, three consecutive odd numbers are 11, 13 and 15.

Question 6

Sum of three consecutive (RBSESolutions.com) even numbers is 48. Find out the numbers.

Solution:

Let three consecutive even numbers are 2x, 2x + 2 and 2x + 4.

According to question,

2x + 2x + 2 + 2x + 4 = 48

⇒ 6 + 6 = 48

⇒ 6x = 48 – 6 = 42

⇒ x = \(\frac { 42 }{ 6 }\) = 7

First (RBSESolutions.com) even number = 2x = 2 x 7 = 14

Second even number = 2x + 2 = 2 x 7 + 2

= 14 + 2 = 16

third even number = 2x + 4 = 2 x 7 +4

= 14 + 4 = 18

So, three consecutive even numbers are 14, 16 and 18.

Question 7

Ramu’s age is 4 years more than three times of age of his son. If Ramu’s age is 37 ears, then (RBSESolutions.com) find out his son’s age.

Solution:

Let Son’s age is x

Ramu’s age = 37 years

According to question,

3x + 4 = 37

⇒ 3x = 37 – 4

⇒ 3x = 33 ⇒ x = 11

So, son’s age is 11 years.

Question 8

Age of Ishu’s father is 5 years (RBSESolutions.com) more than three times of age the Ishu. Find out Ishu’s age if her fathers’ age is 44 years.

Solution:

Let Ishu’s age is x years.

Ishu’s father age = 44 years

According to question,

3x + 5 = 44

⇒ 3x = 44 – 5 = 39

⇒ x = \(\frac { 39 }{ 3 }\) = 13

So, Ishu’s age 13 year.

Question 9

Riyaz thinks about a number is such a way that (RBSESolutions.com) if he subtracts 7 from 2 \(\frac { 1 }{ 2 }\)times of that number the result comes to be 23. Which number does Riyaz think.

Solution:

Let Riyaz think a number x.

According to question,

2\(\frac { 1 }{ 2 }\) × x – 7 = 23

⇒ \(\frac { 5 }{ 2 }\) = 23 + 7

⇒ \(\frac { 5 }{ 2 }\) x = 30

⇒ x = \(\frac { 2\times 30 }{ 5 }\) = 12

So the number thinked by Riyaz is 12.

Question 10

Age of Ramanjeet’s father is 49 years. His father’s age is 4 years more than 3 times of Ramanjeet’s age. Find out (RBSESolutions.com) Ramanjeet’s age?

Solution:

Let Ramanjeet’s age is x years.

Age of Ramanjeet’s father = 49 years

According to questions,

3x + 4 = 49

⇒ 3x = 49 – 4

⇒ 3x = 45

⇒ x = \(\frac { 45 }{ 3 }\) = 15 years

So, Ramanjeet’s age is 15 years.

Question 11

As compared to Jaipur, the (RBSESolutions.com) road accidents per month in Jodhpur are 50 less than 3 times those occur in Jaipur. Road accidents in Jaipur are 400 per month. Then how many road accidents occurred in Jodhpur?

Solution:

Number of road accidents per month in Jaipur = 400

According to question,

Let number of accidents in Jodhpur is x then,

x = 3 x 400 – 50

= 1200 – 50 = 1150

So, number of accidents occured in Jodhpur 1150 per month.

We hope the RBSE Solutions for Class 7 Maths Chapter 14 Simple Equation Ex 14.2 will help you. If you have any query regarding Rajasthan Board RBSE Class 7 Maths Chapter 14 Simple Equation Exercise 14.2, drop a comment below and we will get back to you at the earliest.

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