RD Sharma Solutions For Class 7 Chapter 1 Integers are provided here. Students can download the pdf of these solutions from the given links. Class 7 is a stage where several important topics are introduced. These crucial topics are discussed here in a particular way. The solutions are provided in accordance with the latest syllabus of CBSE which, in turn, help the students to build a strong foundation of basics and secure excellent marks in their board exams. Therefore, we at BYJU’S provide answers to all questions uniquely and briefly.
The solutions to all questions in RD Sharma books are given here in a detailed and step by step way to help the students understand in a more effective way. In this chapter, we will learn about multiplication and division of integers and various properties of these operations on integers.
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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.
Exercise 1.2 Page No: 1.8
1. Divide:
(i) 102 by 17
(ii) 85 by 5
(iii) 161 by 23
(iv) 76 by 19
(v) 17654 by 17654
(vi) (729) by (27)
(vii) 21590 by 10
(viii) 0 by 135
Solution:
(i) Given 102 by 17
We can write given question as 102 ÷ 17
102 ÷ 17 = 102/17 = 102/17 [by applying the mod]
= 102/17 = 6
(ii) Given 85 by 5
We can write given question as 85 ÷ 5
85 ÷ 5 = 85/5 = 85/5 [by applying the mod]
= 85/5 = 17
(iii) Given 161 by 23
We can write given question as 161 ÷ 23
161 ÷ 23 = 161/23 = 161/23 [by applying the mod]
= 161/23 = 7
(iv) Given 76 by 19
We can write given question as 76 ÷ 19
76 ÷ 19 = 76/19 = 76/19 [by applying the mod]
= 76/19 = 4
(v) Given 17654 by 17654
We can write given question as 17654 ÷ 17654
17654 ÷ 17654 = 17654/17654 = 17654/17654 [by applying the mod]
= 17654/17654 = 1
(vi) Given (729) by (27)
We can write given question as (729) ÷ (27)
(729) ÷ (27) = 729/27 = 729/27 [by applying the mod]
= 729/27 = 27
(vii) Given 21590 by 10
We can write given question as 21590 ÷ 10
21590 ÷ 10 = 21590/10 = 21590/10 [by applying the mod]
= 21590/10 = 2159
(viii) Given 0 by 135
We can write given question as 0 ÷ 135
0 ÷ 135 = 0 [because anything is divided by 0 we get the result as 0]
Exercise 1.3 Page No: 1.9
Find the value of
1. 36 ÷ 6 + 3
Solution:
Given 36 ÷ 6 + 3
According to BODMAS rule we have operate division first then we have to do addition
Therefore 36 ÷ 6 + 3 = 6 + 3 = 9
2. 24 + 15 ÷ 3
Solution:
Given 24 + 15 ÷ 3
According to BODMAS rule we have operate division first then we have to do addition
Therefore 24 + 15 ÷ 3 = 24 + 5 = 29
3. 120 – 20 ÷ 4
Solution:
Given 120 – 20 ÷ 4
According to BODMAS rule we have operate division first then we have to do multiplication
Therefore 120 – 20 ÷ 4 = 120 – 5 = 115
4. 32 – (3 × 5) + 4
Solution:
Given 32 – (3 × 5) + 4
According to BODMAS rule we have operate in brackets first then we have move to addition and subtraction.
Therefore 32 – (3 × 5) + 4 = 32 – 15 + 4
= 32 – 11 = 21
5. 3 – (5 – 6 ÷ 3)
Solution:
Given 3 – (5 – 6 ÷ 3)
According to BODMAS rule we have operate in brackets first then we have move to subtraction.
Therefore 3 – (5 – 6 ÷ 3) = 3 – (5 – 2)
= 3 –3 = 0
6. 21 – 12 ÷ 3 × 2
Solution:
Given 21 – 12 ÷ 3 × 2
According to BODMAS rule we have perform division first then multiplication and addition.
Therefore, 21 – 12 ÷ 3 × 2 = 21 – 4 × 2
= 21 – 8 = 13
7. 16 + 8 ÷ 4 – 2 × 3
Solution:
Given 16 + 8 ÷ 4 – 2 × 3
According to BODMAS rule we have perform division first then followed by multiplication, addition and subtraction.
Therefore, 16 + 8 ÷ 4 – 2 × 3 = 16 + 2 – 2 × 3
= 16 + 2 – 6
= 18 6
= 12
8. 28 – 5 × 6 + 2
Solution:
Given 28 – 5 × 6 + 2
According to BODMAS rule we have perform multiplication first then followed by addition and subtraction.
Therefore, 28 – 5 × 6 + 2 = 28 – 30 +2
= 28 – 28 = 0
9. (20) × (1) + (28) ÷ 7
Solution:
Given (20) × (1) + (28) ÷ 7
According to BODMAS rule we have perform division first then followed by multiplication, addition and subtraction.
Therefore, (20) × (1) + (28) ÷ 7 = (20) × (1) – 6
= 20 – 6 = 16
10. (2) + (8) ÷ (4)
Solution:
Given (2) + (8) ÷ (4)
According to BODMAS rule we have perform division first then followed by addition and subtraction.
Therefore, (2) + (8) ÷ (4) = (2) + 2
=0
11. (15) + 4 ÷ (5 – 3)
Solution:
Given (15) + 4 ÷ (5 – 3)
According to BODMAS rule we have perform division first then followed by addition and subtraction.
Therefore, (15) + 4 ÷ (5 – 3) = (15) + 4 ÷ 2
= 15 + 2
= 13
12. (40) × (1) + (28) ÷ 7
Solution:
Given (40) × (1) + (28) ÷ 7
According to BODMAS rule we have perform division first then followed by multiplication, addition and subtraction.
(40) × (1) + (28) ÷ 7 = (40) × (1) – 4
= 40 – 4
= 36
13. (3) + (8) ÷ (4) 2 × (2)
Solution:
Given (3) + (8) ÷ (4) 2 × (2)
According to BODMAS rule we have perform division first then followed by multiplication, addition and subtraction.
(3) + (8) ÷ (4) 2 × (2) = 3 + 2 2 × (2)
= 3 + 2 + 4
= 6 – 3
=3
14. (3) × (4) ÷ (2) + (1)
Solution:
Given (3) × (4) ÷ (2) + (1)
According to BODMAS rule we have perform division first then followed by multiplication, addition and subtraction.
(3) × (4) ÷ (2) + (1) = 3 × 2 1
= – 6 – 1
= 1
Exercise 1.4 Page No: 1.12
Simplify each of the following:
1. 3 – (5 – 6 ÷ 3)
Solution:
Given 3 – (5 – 6 ÷ 3)
According to removal of bracket rule firstly remove inner most bracket
We get 3 – (5 – 6 ÷ 3) = 3 – (5 – 2)
= 3 – 3
= 0
2. 25 + 14 ÷ (5 – 3)
Solution:
Given 25 + 14 ÷ (5 – 3)
According to removal of bracket rule firstly remove inner most bracket
We get 25 + 14 ÷ (5 – 3) = 25 + 14 ÷ 2
= 25 + 7
= 18
Solution:
According to removal of bracket rule first we have to remove vinculum we get
= 25 – ½ {5 + 4 – (5 – 4)}
Now by removing the innermost bracket we get
= 25 – ½ {5 + 4 – 1}
By removing the parentheses we get
= 25 – ½ (8)
Now simplifying we get
= 25 – 4
= 21
Solution:
According to removal of bracket rule first we have to remove vinculum we get
= 27 – [38 – {46 – (15 – 11)}]
Now by removing inner most bracket we get
= 27 – [38 – {46 – 4}]
By removing the parentheses we get
= 27 – [38 – 42]
Now by removing braces we get
= 27 – (4)
= 27 + 4
= 31
Solution:
By removing innermost bracket we get
= 36 – [18 – {14 – (11 ÷ 2 × 2)}]
= 36 – [18 – {14 – 11}]
Now by removing the parentheses we get
= 36 – [18 – 3]
Now remove the braces we get
= 36 – 15
= 21
6. 45 – [38 – {60 ÷ 3 – (6 – 9 ÷ 3) ÷ 3}]
Solution:
Given 45 – [38 – {60 ÷ 3 – (6 – 9 ÷ 3) ÷ 3}]
First remove the inner most brackets
= 45 – [38 – {20 – (6 – 3) ÷ 3}]
= 45 – [38 – {20 – 3 ÷ 3}]
Now remove the parentheses w get
= 45 – [38 – 19]
Now remove the braces we get
= 45 – 19
= 26
Solution:
Now first remove the vinculum we get
= 23 – [23 – {23 – (23 – 0)}]
Now remove the innermost bracket we get,
= 23 – [23 – {23 – 23}]
By removing the parentheses we get,
= 23 – [23 0]
Now we have to remove the braces and on simplifying we get,
= 23 – 23
= 0
Solution:
First we have to remove the vinculum from the given equation we get,
= 2550 – [510 – {270 – (90 – 150)}]
By removing innermost bracket we get,
= 2550 – [510 – {270 – (60)}]
= 2550 – [510 – {270 + 60}]
Now remove the parentheses we get,
= 2550 – [510 – 330]
Now we have to remove parentheses
= 2550 – 180
= 2370
Solution:
First we have to remove vinculum from the given equation,
= 4 + 1/5 [{10 × (25 – 10)} ÷ (5)]
Now remove the innermost bracket, we get
= 4 + 1/5 [{10 × 15} ÷ (5)]
Now by removing the parenthesis we get,
= 4 + 1/5 [150 ÷ 5]
By removing the braces we get,
= 4 + 1/5 (30)
On simplifying we get,
= 4 + 6
= 10
Solution:
Now we have to remove innermost bracket
= 22 – ¼ {5 – ( 48 ÷ – 16)}
= 22 – ¼ {5 – 3}
Now remove the parenthesis we get
= 22 – ¼ (8)
On simplifying we get,
= 22 + 2
= 24
Solution:
First we have to remove vinculum from the given equation then we get,
= 63 – [(3) {2 – 5}] ÷ [3 {5 + 2}]
Now remove the parenthesis from the above equation
= 63 – [(3) (7)] ÷ [3 (7)]
= 63 – [21] ÷ [21]
= 63 – 1
= 62
Solution:
First we have to remove the innermost brackets then we get,
= [29 – (2) {6 – 4}] ÷ [3 × {5 + 6}]
Now remove the parenthesis in the above equation,
= [29 + 2 (2)] ÷ [3 × 11]
Now remove all braces present in the above equation,
= 33 ÷ 33
= 1
13. Using brackets, write a mathematical expression for each of the following:
(i) Nine multiplied by the sum of two and five.
(ii) Twelve divided by the sum of one and three.
(iii) Twenty divided by the difference of seven and two.
(iv) Eight subtracted from the product of two and three.
(v) Forty divided by one more than the sum of nine and ten.
(vi) Two multiplied by one less than the difference of nineteen and six.
Solution:
(i) 9 (2 + 5)
(ii) 12 ÷ (2 + 3)
(iii) 20 ÷ (7 – 2)
(iv) 2 × 3 8
(v) 40 ÷ [1 + (9 + 10)]
(vi) 2 × [(19 6) 1]
RD Sharma solutions For Class 7 Maths Chapter 1 – Integers
Chapter 1, Integers contains four exercises. RD Sharma Solutions are given here which include the answers to all the questions present in these exercises. Let us have a look at some of the concepts that are being discussed in this chapter.
 Multiplication of integers
 Properties of multiplication
 Division of integers
 Properties of division
 Operator precedence
 Use of brackets
 Removal of brackets
Also, access RD Sharma Solutions for Class 7 Chapter 1
Chapter brief of RD Sharma Solutions for Class 7 Maths Chapter 1 – Integers
The chapter Integers primarily covers three topics such as multiplication of integers, division of integers and operator precedence. It also explains how to use and remove the brackets. Integers can be defined as a ‘’number that can be written without a fractional component”. Here the students will be thorough about the precedence of fundamental operations of addition, subtraction, multiplication and division.