RBSE Solutions for Class 7 Maths Chapter 1 Integers Ex 1.2 is part of RBSE Solutions for Class 7 Maths. Here we have given Rajasthan Board RBSE Class 7 Maths Chapter 1 Integers Exercise 1.2.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 7 |

Subject |
Maths |

Chapter |
Chapter 1 |

Chapter Name |
Integers |

Exercise |
Ex 1.2 |

Number of Questions |
8 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 7 Maths Chapter 1 Integers Ex 1.2

Question 1

Find the product (RBSESolutions.com) of the following :

(i) (-3) x 4

(ii) (-1) x 24

(iii) (-30) x (-24)

(iv) (-214) x 0

(V) (-15) x (-7) x 6

(vi) (-5) x (-7) x (-4)

(vii) (-3) x (-2) x (-1) x (-5)

Solution:

(i) (-3) x 4 = – 12

(ii) (-1) x 24 – 24

(iii) (-30) x (- 24) = 720

(iv) (214) x 0 = 0

(v) (-15) x (7) x 6 = 105 x 6 = 630

(vi) (- 5) = (-7) x (- 4) = 35 x (- 4) = -140

(vii) (-3) x (-2) x (-1) x (5) = [(-3) x (-2)] [(-1) x (-5)] = 6 x 5 = 30

Question 2

Start with (- 1) x 5 and make (RBSESolutions.com) pattern to show that (- 1) x (- 1) = + 1

Solution:

Pattern is :

(-1) x 5 = -5 (-1) x 2 = -2

(-1) x 4 = -4 (-1) x 0 = 0

(1) x 3 = -3 (-1) x (-1) = + 1

Question 3

The rate of decreasing the temperature (RBSESolutions.com) in a refrigerator is 3°C per minute. A thing whose temperature is 25°C, is kept in refrigerator. After how much time, the temperature of thing will be – 2°C?

Solution:

Temperature of thing = 25°C

Rate of decreasing of temperature = -3°C per minute

Total decrease in temperature of thing after decreasing

from 25°C to – 2°C = (- 2) – 25

= -2 – 25

= – 27°C

Time taken to reach upto – 27°C at the rate of 3°C per minute = 27°C x \(\frac { 1 }{ { 3 }^{ 0 }C } \) per minute

= 9 minutes.

Question 4

In a game, two balls are given on selecting (RBSESolutions.com) a blue card and three balls are gained on selecting a red card. Sheetal has 27 balls with her. During game she gets 9 blue cards continuously. How much balls, she have remaining?

Solution:

Number of balls with Sheetal = 27

She gets 9 blue cards continuously

so she has to give 9 x 2 = 18 balls

∴ Number of remaining balls

= 27 – 18

= 9 balls.

So, sheetal has remaining balls = 9.

Question 5

Solve the following (RBSESolutions.com) division following :

(i) (-35) ÷ 7

(ii) 15 ÷ (-3)

(iii) – 25 ÷ (- 25)

(iv) 25 ÷ (-1)

(v) 0 ÷ (-3)

(vi) 15 ÷ [1(-2) +1]

(vii) (-6) + 3 [(-2) + 1]

Solution:

(i) (-35) ÷ 7 = -5

(ii) 15 ÷ (-3) = -5

(iii) – 25 ÷ (-25) = 1

(iv) 25 ÷ (-1) = -25

(v) 0 ÷ (3) = 0

(vi) 15 ÷ [(-2) +1] = 15 ÷ (-1) = -15

(vii) [(6) + 3] ÷ [(-2) +1]= (-3) ÷ (-1)= 3

Question 6

A shopkeeper gains ₹1 on selling (RBSESolutions.com) a pen and looses 50 paise on selling a pencil. Represent the gain and loss in terms of integers.

(i) There is a loss of ₹5 in a month. If he had sold 45 pens then find the number of pencils, sold by him in a month.

(ii) There is no profit and no loss in the second month. If he sold 70 pens and the number of pencils sold.

Solution:

Integer form, of gain of ₹1 = + 100

and integer form of loss of 50 paise = – 50

(i) Gain on (RBSESolutions.com) selling 45 pens = ₹45

= 45 x 100 = 4500 Paise

Let x pencils are sold

∴ Loss on selling x pencil = – x × 50

= – 50x Paise

According to question,

Loss on(45 Pen + x Pencil) = -₹5

= – 500 Paise

⇒ 4500 Paise – 50x Paise = -500 Paise

⇒ -50x = -500 – 4500

⇒ x = \(\frac { -5000 }{ -50 }\) = 100

∴ Number of pencils sold = 100

(ii) Gain on (RBSESolutions.com) selling 70 Pen = ₹70

= 70 x 100 = 7000 Paise

∵ There is not loss or gain

∴ Total loss = Total gain

∴ Loss on selling x pencils = 7000 Paise

Loss on 1 pencil = 50 Paise

∴ Loss on x pencils = 5Ox Paise

⇒ 50x = 7000

⇒ x = \(\frac { 7000 }{ 50 }\)

⇒ x = 140

∴Number of pencils sold = 140

Question 7

Fill up the following table (RBSESolutions.com) by multiplying the integers

Solution:

Question 8

If going up a 60 feet multi storied (RBSESolutions.com) by a lift is represented by positive integers then:

(i) How will we represented the height of flat at 60 feet above?

(ii) Represent the parking 15 feet below integer.

(iii) If lift goes upwards at the rate of 5 feet/ sec, represented as + 5 and if travels In a opposite direction, then what is the integer representing downward direction?

Solution:

(i) Height of flat at 60 feet is an integer = + 60

(ii) Position of parking 15 feet down as integer = – 15

(iii) Lift coming downwards at the rate of 5feet/ sec as an integer = – 5

We hope the RBSE Solutions for Class 7 Maths Chapter 1 Integers Ex 1.2 will help you. If you have any query regarding Rajasthan Board RBSE Class 7 Maths Chapter 1 Integers Exercise 1.2, drop a comment below and we will get back to you at the earliest.

## Leave a Reply