The PDF of RD Sharma Solutions for Exercise 1.1 of Class 7 Maths Chapter 1 Integers is provided here. The questions present have been solved by BYJU’S experts in Maths, and will help students solve the problems without any difficulties. Exercise 1.1 contains ten questions with many sub-questions which deal with the multiplication of integers and their properties. Students can obtain good results in the Class 7 final exam by using RD Sharma Solutions. By practising the RD Sharma Solutions for Class 7, students will be able to grasp the concepts easily.
RD Sharma Solutions for Class 7 Maths Chapter 1 – Integers Exercise 1.1
Access answers to Maths RD Sharma Solutions For Class 7 Chapter 1 – Integers Exercise 1.1
Exercise 1.1 Page No: 1.5
1. Determine each of the following products:
(i) 12 × 7
(ii) (-15) × 8
(iii) (-25) × (-9)
(iv) 125 × (-8)
Solution:
(i) Given 12 × 7
Here we have to find the products of given numbers
12 ×7 = 84
Because the product of two integers of like signs is equal to the product of their absolute values.
(ii) Given (-15) × 8
Here we have to find the products of given numbers
(-15) ×8 = -120
Because the product of two integers of opposite signs is equal to the additive inverse of the product of their absolute values.
(iii) Given (-25) × (-9)
Here we have to find the products of given numbers
(-25) × (-9) = + (25 ×9) = +225
Because the product of two integers of opposite signs is equal to the additive inverse of the product of their absolute values.
(iv) Given 125 × (-8)
Here we have to find the products of given numbers
125 × (-8) = -1000
Because the product of two integers of opposite signs is equal to the additive inverse of the product of their absolute values.
2. Find each of the following products:
(i) 3 × (-8) × 5
(ii) 9 × (-3) × (-6)
(iii) (-2) × 36 × (-5)
(iv) (-2) × (-4) × (-6) × (-8)
Solution:
(i) Given 3 × (-8) ×5
Here we have to find the product of given number.
3 × (-8) × 5 = 3 × (-8 × 5)
=3 × -40 = -120
Since the product of two integers of opposite signs is equal to the additive inverse of the product of their absolute values.
(ii) Given 9 × (-3) × (-6)
Here we have to find the product of given number.
9 × (-3) × (-6) = 9 × (-3 × -6) [∵ the product of two integers of like signs is equal to
the product of their absolute values.]
=9 × +18 = +162
(iii) Given (-2) × 36 × (-5)
Here we have to find the product of given number.
(-2) × 36 × (-5) = (-2 × 36) × -5 [∵ the product of two integers of like signs is equal to
the product of their absolute values.]
=-72 × -5 = +360
(iv) Given (-2) × (-4) × (-6) × (-8)
Here we have to find the product of given number.
(-2) × (-4) × (-6) × (-8) = (-2 × -4) × (-6 × -8) [∵ the product of two integers of like signs is
equal to the product of their absolute values.]
=-8 × -48 = +384
3. Find the value of:
(i) 1487 × 327 + (-487) × 327
(ii) 28945 × 99 – (-28945)
Solution:
(i) Given 1487 × 327 + (-487) × 327
By using the rule of multiplication of integers, we have
1487 × 327 + (-487) × 327 = 486249 – 159249
Since the product of two integers of opposite signs is equal to the additive inverse of the product of their absolute values.
=327000
(ii) Given 28945 × 99 – (-28945)
By using the rule of multiplication of integers, we have
28945 × 99 – (-28945) = 2865555 + 28945
Since the product of two integers of like signs is equal to the product of their absolute values.
=2894500
4. Complete the following multiplication table:
Second number
| First number | x | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
| -4 | ||||||||||
| -3 | ||||||||||
| -2 | ||||||||||
| -1 | ||||||||||
| 0 | ||||||||||
| 1 | ||||||||||
| 2 | ||||||||||
| 3 | ||||||||||
| 4 |
Is the multiplication table symmetrical about the diagonal joining the upper left corner to the lower right corner?
Solution:
Second number
| First number | x | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
| -4 | 16 | 12 | 8 | 4 | 0 | -4 | -8 | -12 | -16 | |
| -3 | 12 | 9 | 6 | 3 | 0 | -3 | -6 | -9 | -12 | |
| -2 | 8 | 6 | 4 | 2 | 0 | -2 | -4 | -6 | -8 | |
| -1 | 4 | 3 | 2 | 1 | 0 | -1 | -2 | -3 | -4 | |
| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | |
| 1 | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 | |
| 2 | -8 | -6 | -4 | -2 | 0 | 2 | 4 | 6 | 8 | |
| 3 | -12 | -9 | -6 | -3 | 0 | 3 | 6 | 9 | 12 | |
| 4 | -16 | -12 | -8 | -4 | 0 | 4 | 8 | 12 | 16 |
From the table it is clear that, the table is symmetrical about the diagonal joining the upper left corner to the lower right corner.
5. Determine the integer whose product with ‘-1’ is
(i) 58
(ii) 0
(iii) -225
Solution:
(i) Given 58
Here we have to find the integer which is multiplied by -1
We get, 58 × -1 = -58
Since the product of two integers of opposite signs is equal to the additive inverse of the product of their absolute values.
(ii) Given 0
Here we have to find the integer which is multiplied by -1
We get, 0 × -1 = 0 [because anything multiplied with 0 we get 0 as their result]
(iii) Given -225
Here we have to find the integer which is multiplied by -1
We get, -225 × -1 = 225
Since the product of two integers of like signs is equal to the product of their absolute values.
RD Sharma Solutions for Class 7 Maths Chapter – 1 Integers Exercise 1.1
Exercise 1.1 of RD Sharma Solutions for Chapter 1, Integers, deals with the basic concepts related to Integers. Some of the topics focused on in Exercise 1.1 are listed below.
- Multiplication of integers
- Properties of multiplication
Students are suggested to try solving the questions from the RD Sharma book of Class 7 and then refer to these solutions to know the best way of approaching the different questions.
Comments