RBSE Solutions for Class 7 Maths Chapter 1 Integers Additional Questions is part of RBSE Solutions for Class 7 Maths. Here we have given Rajasthan Board RBSE Class 7 Maths Chapter 1 Integers Additional Questions.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 7 |

Subject |
Maths |

Chapter |
Chapter 1 |

Chapter Name |
Integers |

Exercise |
Additional Questions |

Number of Questions |
26 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 7 Maths Chapter 1 Integers Additional Questions

**Multiple Choice Questions**

Question 1

Value of (-3) x 5 will be

(A) – 15

(B) 15

(C) 30

(D) – 30

Question 2

Additive (RBSESolutions.com) inverse of 0 is

(A) 1

(B) 0

(C) 4

(D) -1

Question 3

Value of (-3) x [2 + (-1)] will be

(A) – 1

(B) – 2

(C) – 3

(D) 0

Question 4

Predecessor of (RBSESolutions.com) whole number 0 will be

(A) -1

(B) -2

(C) 1

(D) none of these

Question 5

If a is an integer then value of a + 0 will be

(A) a

(B) 0

(C) 1

(D) undefined

Question 6

What will be the (RBSESolutions.com) product of 5 negative numbers?

(A) Negative

(B) Positive

(C) 0

(D) None of these

Answers:

1. (A),

2. (B)

3. (C),

4. (D),

5. (D),

6. (A)

**Fill up the blank**

(i) 5) + (- 8) = (- 8) + (……)

(ii) -53 + …… = -53

(iii) 17 + …… = 0

(iv) [13 + (-12)] + (……) = 13 + [(- 12) + (-7)]

(v) (-4) + [15 + (-3)] = [-4 + 15] + ……

Solutions:

(i) (-5), (ii) 0, (iii) (-17), (iv) (-7), (v) (-3)

**True/False**

(i) If a positive integer is added (RBSESolutions.com) then move in right side.

(ii) If a negative integer is subtracted then move in left side.

(iii) An integers a is multiplied with 0 then product is a.

(iv) (-6) ÷ (-2) = (-2) ÷ (-6)

(v) (-1) x (-1) x (-1) x (-1) = -1

Answers:

(i) True

(ii) True

(iii) False

(iv) False

(v) False

**Very Short Answer Type Questions**

Question 1

Find the value of 8 x (6 – 3)

Solution:

8 x (6 – 3) = 8 x 3 = 24

Question 2

What is the (RBSESolutions.com) solution of [-8) + 3| ÷ (-2) +1] ?

Solution:

[(- 8) + 3] [(-2) + 1] = [ 8 + 3] ÷ [-2 + 1]

= -5 ÷ (-1) = 5

Question 3

What is the solution of (-15) x 0 x (-18)

Solution:

(-15) x 0 x (- 18) = [(-15) x 0] x ( -18)

= 0 x (- 18) = 0

Question 4

Find the (RBSESolutions.com) value of 0 ÷ (-12) .

Solution:

0 ÷ (-12) = 0

**Short Answer Type Questions**

Question 1

Verify the following

(i) 18 x 7 + (-3) = [18 x 7] + [18 x (-3)]

(ii) (-21) x [(-4) + (-6)] = [(-21) x (4)] + [(-21) x (-6)]

Solution:

(i) 18 x [7 + (-3)] = 18 x 4 = 72

and [18 x 7] + [18 x (-3)] = 126 – 54 = 72

so 18 x [7 + (-3)] = [18 x 7] + [18 x (-3)]

(ii) (-21) x [(- 4) +(6)] = (-21) x (- 4 – 6) |

= (-21) x (-10) = 210

and [(21) x(-4)] + [(-21) x (-6)] = 84 + 126 = 210

so (-21) x [(- 4) + (-6)] = [(-21) x (- 4)] + [(-21) x (-6)]

Question 2

Is for any integer a

(i) 1 ÷ a = 1?

(ii) a ÷ (-1) = -a?

Check the answer (RBSESolutions.com) of different values of a.

Solution:

(i) Let a = 5, 1 ÷ a = 1 ÷ 5 ⇒ \(\frac { 1 }{ 5 }\) ≠ 1

so, 1 ÷ a ≠ 1

(ii) Let a = 7, a ÷ (-1) = 7 ÷ (-1) = -7

so, a ÷ (-1) = -a is true.

**Long Answer Questions**

Question 1

The Temperature of Srinagar (RBSESolutions.com) on Monday was – 5°C and it decreased on Tuesday by 2°C. What will be the temperature on Tuesday? On Wednesday it increased by 4°C. What was the temperature on Wednesday?

Solution:

Temperature of Srinagar on Monday = – 5°C

It decreased 2°C on Tuesday

∴ Temperature on Tuesday = (- 5 – 2)°C = – 7°C

It increased 4°C on Wednesday

∴ Temperature on Wednesday

= (-7 + 4)°C = – 3°C.

Question 2

Sohan deposits ₹2000 in his Bank (RBSESolutions.com) account and withdraws ₹1642 on next day. If the withdrawn amount is represented by a negative number then how the amount deposited in the bank will be represented ? How much amount will be remaining after withdraws from Sohan’s account ?

Solution:

Deposited amount in Bank is represented by + sign.

Amount remaining in Sohan’s account = (+₹2000) + (-₹1642)

= ₹(2000 – 1642)

= ₹358.

Question 3

An uplift moves down words in a mine- well out the rate of 6 metre/minute. If lift starts going down wards from the height 10 m above the (RBSESolutions.com) ground, then how much time will it take to reach – 350 m?

Solution:

Difference between two heights = 10 m – (- 350 m)

= 10 m + 350 m

= 360 m

Rate of going downwards = 6 m/minute

∴ Time of reach at lowest point = \(\frac { 360 }{ 6 }\)

= 60 minute

= 1 hour.

Question 4

A cement factory gains at the rate of ₹8 per bag on white (RBSESolutions.com) cement and looses at the rate of ₹5 per bag on gray colour cement

(i) Company sells 3000 bags of white cement and 5000 bags of gray cement What is his toss or profit?

(ii) If numbers of bags of gray colour cement 6400 then how many bags of white cement should company sold so that there is no loss no profit?

Solution:

Profit on one bag of white cement = ₹8

and loss on one bag of gray cement = ₹5

(i) Profit on 3000 bags of white cement = (3000 x 8) = ₹24,000

Loss on 5000 bags of gray cement = (5000 x 5) = ₹25,000

∴ Total loss = 25,000 – 24,000 = ₹1000

(ii) Loss on 6400 bags (RBSESolutions.com) of gray cement =6400 x 5 = ₹32,000

Neither loss no profit

∴ Profit on 32000 white bags = ₹32000

∴ Number of white cement bags sold = \(\frac { 32000 }{ 8 }\)

= 4000 bags.

We hope the RBSE Solutions for Class 7 Maths Chapter 1 Integers Additional Questions will help you. If you have any query regarding Rajasthan Board RBSE Class 7 Maths Chapter 1 Integers Additional Questions, drop a comment below and we will get back to you at the earliest.

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