RBSE Solutions for Class 7 Maths Chapter 14 Simple Equation Additional Questions

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

Board	RBSE
Textbook	SIERT, Rajasthan
Class	Class 7
Subject	Maths
Chapter	Chapter 14
Chapter Name	Simple Equation
Exercise	Additional Questions
Number of Questions	16
Category	RBSE Solutions

Rajasthan Board RBSE Class 7 Maths Chapter 14 Simple Equation Additional Questions

Multiple Choice Questions

Question 1
If equation x – 8 = 14, then (RBSESolutions.com) value of x will be:
(A) 22
(B) 26
(C) 20
(D) 25

Question 2
A number is 5 less than 10 times of x then that number is :
(A) 10x + 5
(B) 10x – 5
(C) 10x + 6
(D) 10x + 10

Question 3
If equation \(\frac { a }{ 2 }\) = 8, then (RBSESolutions.com) value of a will be :
(A) 16
(B) 9
(C) 10
(D) 18

Question 4
In the difference of p and 10 is 20 then equation will be :
(A) p + 10 = 10
(B) 20p + 10 = 0
(C) p – 10 = 20
(D) p + 10 = 20

Question 5
In equation 8 +10x = 2x + 24 be the (RBSESolutions.com) value of x will be :
(A) 8
(B) 6
(C) 4
(D) 2

Question 6
In equation \(\frac { x }{ 2 }\) – 2 = 4, the value of x will be :
(A) 12
(B) 13
(C) 14
(D) 25
Answer:
1. (A), 2. (B), 3. (A), 4. (C), 5. (D), 6. (A)

Fill in the Blanks
(i) In equation the value of variable which (RBSESolutions.com) satisfy the equation is known as ……….. of the equation.
(ii) In equation 2x + 8 = 10, the value of x is …………..
(iii) The numbers w’hose values are not constant is known as …………..
Solution:
(i) solution
(ii) 1,
(iii) variables

True/False
(i) If we interchange the sides of equation (RBSESolutions.com) then the value of equation will not change.
(ii) In an equation, both sides we cannot add, subtract, multiply or divide by a number one both sides simultaneously.
(iii) In equation x + 4 = 10, the value of x is 40.
(iv) x + 5 = 10 is an equation.
Solution:
(i) True
(ii) False
(iii) False
(iv) True

Very Short Answer Type Questions

Question 1
Solve the (RBSESolutions.com) equation, 10x + 20 = 40
Solution:
10x + 20 =40
(Subtract 20 from both sides)
10x + 20 – 20 = 40 – 20
⇒ 10x = 20
⇒ x = \(\frac { 20 }{ 10 }\) = 2

Question 2
Solve:
13p – 10 = 36
Solution:
13p – 10 = 36
Add 10 on both sides
13p – 10 + 10 = 36 + 10
13p = 46
⇒ p = \(\frac { 46 }{ 13 }\)

Question 3
Sum of 4 time of x and add 15 is 42. Write (RBSESolutions.com) this in form of equation.
Solution:
4x + 15 = 42

Question 4
Find the value of x in \(\frac { x }{ 4 }\) + 9 = 19
Solution:
\(\frac { x }{ 4 }\) + 9 = 19
Subtract 9 from both sides
\(\frac { x }{ 4 }\) +9 – 9 = 19 – 9
⇒ \(\frac { x }{ 4 }\) = 10 ⇒ x = 10 x 4 = 40

Short Answer Type Questions

Question 1
6(x + 9) = 15 solve it and verify (RBSESolutions.com) the answer.
Solution:
6(x + 9) = 15
(Dividing both sides by 6)

Question 2
Find out the present (RBSESolutions.com) age of Mohindar if 10years before he was of 35 years.
Solution:
Let Mohindar’s present age is x years
∴ His 10 years before age = (x – 10) year
According to question,
x – 10 = 35
⇒ x – 10 + 10 = 35 + 10
(Add 10 on both sides)
⇒ x =45
So, present age of Mohinder is 45 years.

Question 3
Sum of three consecutive (RBSESolutions.com) numbers is 81 find out the numbers.
Solution:
Let First number = x
Second number = x + 1
Third number = x + 2
Sum of three numbers = 81
∴ According to question,
x + x + 1 + x + 2 = 81
⇒ 3x + 3 = 81
⇒ 3 = 81 – 3, (shifting)
⇒ x = \(\frac { 78 }{ 3 }\) = 26
First number = 26
Second number = 26 + 1 = 27
Third number = 26 + 2 = 28

Question 4
Solve it : 4(m + 3) =18. Also (RBSESolutions.com) verify the answer.
Solution:
m + 3 = \(\frac { 18 }{ 4 }\) ⇒ m + 3 = \(\frac { 9 }{ 2 }\)
(Dividing both sides by 4)
⇒ m = \(\frac { 9 }{ 2 }\) – 3
= \(\frac { 9-6 }{ 2 }\) = \(\frac { 3 }{ 2 }\)
Verification of answer
L.H.S. = 4(m + 3) = 4m + 12
= 4 x \(\frac { 3 }{ 2 }\) + 12 (m = \(\frac { 3 }{ 2 }\) put )
= 6 + 12 = 18 = RHS
so, m = \(\frac { 3 }{ 2 }\) is correct answer.

Question 5
The cost of 5 books is ₹14 less than (RBSESolutions.com) cost of 7 books. Find the cost of one book.
Solution:
Let cost of one book is ₹x
cost of 5 books = ₹5x
cost of 7 books = ₹1x
According to question,
5x = 7x – 14
⇒ 5x – 7x = -14
⇒ -2x = -14
⇒ 2x = 14
⇒ x = 7
∴ Cost of one book ₹7.

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