RBSE Solutions for Class 7 Maths Chapter 14 Simple Equation Ex 14.2

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.

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