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.

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

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