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.