RBSE Solutions for Class 7 Maths Chapter 1 Integers Ex 1.1 is part of RBSE Solutions for Class 7 Maths. Here we have given Rajasthan Board RBSE Class 7 Maths Chapter 1 Integers Exercise 1.1.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 7 |

Subject |
Maths |

Chapter |
Chapter 1 |

Chapter Name |
Integers |

Exercise |
Ex 1.1 |

Number of Questions |
7 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 7 Maths Chapter 1 Integers Ex 1.1

Question 1

The temperature in Churu is measured in °C at (RBSESolutions.com) different time and represented o number line

(i) The temperature of Churu on following date from above number line :

(a) 26 January ………..

(b) 25 December ……….

(c) 25 February ……….

(d) 25 March …………

(ii) What is the difference in temperature (RBSESolutions.com) between the hottest and the coldest day?

(iii) The temperature of 26 January is how much less than the temperature of 25 February?

(iv) Can we say that sum of temperature of 25 December and 25 february is greater than the temperature of 26 jan?

Solution:

(i) Temperature of Churu on number line:

(a) -11°C on 26 January

(b) -6°C on 25 December

(c) 4°C on 25 February

(d) 14°C on 25 March

(ii) Temperature (RBSESolutions.com) of hottest day = 14°C

Temperature of coldest day = (-11°C)

Difference in temperature = 14°C – (-11°C)

= 14°C + 11°C

= 25°C

(iii) Temperature of 26 January = -11°C

Temperature of 25 February = 4°C

∴ Difference in temperature = -11°C – 4°C

= – 15°C

(iv) Temperature (RBSESolutions.com) of 25 December = -6°C

Temperature of 25 February = 4°C

∴ Sum of Temperature = -6°C + 4°C

= – 2°C

Temperature of 26 January = -11 °C,

So the sum of the temperature of 25 December and 25 February is greater than the temperature of 26 January.

Question 2

Sheela deposit ₹5000 in post office and withdraws ₹3700 after one month. If the amount withdrawn is represented in the form of negative number then how will we represent the deposited amount? What is the amount left in the account after withdrawal?

Solution:

The deposited amount (RBSESolutions.com) will be shown by a positive number.

Amount remaining after withdrawl = (+ 5000) + (- 3700 )

= ₹(5000 – 3700)

= ₹1300 (Positive)

Question 3

Solve the following

(i) (-4) + (-3)

(ii) 15 – 8 + (-9)

(iii) 400 + (-1000) + (-500)

(iv) 23 – 41 – 11

(v) -27 + (-3) + 30

Solution:

(i) (- 4) + (-3) = – 4 – 3 = -7

(ii) 15 – 8 + (-9) = 7 + (-9) = 7 – 9 = – 2

(iii) 400 + (-1000) + (-500) = 400 – 1000 – 500 = 400 – 1500 = -1100

(iv) 23 – 41 – 11 = 23 – 52 = -29

(v) -27 + (-3) + 30 = -27 -3 + 30

= – 30 + 30 = 0

Question 4

Put the appropriate sign (<, >, =) for (RBSESolutions.com) the following statements :

(i) -14 + 11 + 5 14 – 11 – 5

(ii) 30 + (-5) + (- 8) (-5) + (-8) + 30

(iii) 7+ 11 + (-5) (-7) – 11 + 5

(iv) (-14) + 11 + (-12) 14 + 11 + 12

(v) 6 +7 – 13 6 + 7 + (-13)

Solution:

(i) Here -14 + 11 + 5 = -14 + 16 = 2

and 14- 11 -5 = 14- 16 = – 2

because 2 > – 2

So, -14 + 11 + 5 14 – 11 – 5

(ii) Here 30 + (- 5) + (- 8)

= 30 – 5 – 8 = 30 – 13 = 17 and (- 5) + (- 8) + 30

= – 5 – 8 + 30 = – 13 + 30 = 17

because 17 = 17

So, 30 + (-5) + (-8) (-5) + (-8) + 30

(iii) Here 7 + 11 + (- 5)

= 7 = 11 – 5= 18 – 5 = 13

and (-7) – 11 + 5

= – 7 – 11 + 5

= -18 + 5 = -13

because 13 > -13

So, 7 +11 + (-5) (-7) -11 +5

(iv) Here (-14) +11+ (-12)

= -14 + 11 – 12

= -26 + 11 = -15

and 14 + 11 + 12 = 37

because -15 < 37

So, (-14) + 11 + (-12) 14 +11 + 12

(v) Here 6+ 7 – 13 = 13 – 13 = 0

and 6 + 7 + (-13) = 13 – 13 = 0

because 0 = 0

So, 6 +7 – 13 6 + 7 + (-13)

Question 5

Write two such (RBSESolutions.com) integers, whose :

(i) sum is (-7)

(ii) difference is 4

(iii) sum is 0

(iv) difference is – 2

Solution:

(i) Such integers are (-1) and (-6) because

(-1) + (-6) = -7

(ii) Such integers are (6) and (+ 2) because

6 + (-2) = 6- 2 = 4

(iii) Such integers may be -1 and + 1 because

(1) + (-1) = 0

(iv) Such integers may be (-3) and (-1) because

-3 – (-1) = -3 + 1 = -2

Question 6

Fill in the (RBSESolutions.com) blanks –

(i) (-3) + 5 = 5 + ….

(ii) 17 + …. = 17

(iii) …. + (-5) = 0

(iv) – 11 + [(-12) + 4] = [(-11) + (-12)] + ….

Solution:

(i) (- 3),

(ii) 0,

(iii) (+ 5),

(iv) 4.

Question 7

Examples and properties of (RBSESolutions.com) integers are given below. Match the correct property and its example.

**Example ** **Properties**

(i) (a + b) + c = a + (b + c) (a) Identity

(ii) 3 + 4 = 4 + 3 (b) Associative

(iii) (-4) + 0 = (-4) (c) Commutative

Solution:

(i) (b),

(ii) (c),

(iii) (a)

We hope the RBSE Solutions for Class 7 Maths Chapter 1 Integers Ex 1.1 will help you. If you have any query regarding Rajasthan Board RBSE Class 7 Maths Chapter 1 Integers Exercise 1.1, drop a comment below and we will get back to you at the earliest.

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