RD Sharma Solutions for Class 6 Maths Chapter 5 Exercise 5.4 provide clear and detailed explanations on the topic of Negative Numbers and Integers, specifically focusing on subtraction of integers.
These solutions help students understand how to subtract positive and negative numbers, apply the rules of subtraction, and solve related problems step-by-step.
By practicing these exercises, students strengthen their grasp of integer operations, improve their calculation skills, and build confidence to tackle similar problems in exams. The solutions are designed to make learning easier and support students in mastering the concepts effectively, as per the CBSE Class 6 Maths syllabus.
RD Sharma Solutions for Class 6 Maths Chapter 5 Negative Numbers and Integers Exercise 5.4 Introduction
Exercise 5.4 focuses on subtracting integers, which means finding the difference between positive and negative numbers. This exercise helps students understand how to handle subtraction when the numbers involved can be both positive and negative.
The main idea is to learn how to change subtraction into addition by using the rule that subtracting a number is the same as adding its opposite.
For example, subtracting a negative number is the same as adding a positive number. This exercise includes various problems to practice these rules, so students get comfortable working with integers in subtraction and can solve questions correctly and confidently.
RD Sharma Solutions for Class 6 Maths Chapter 5 Exercise 5.4
Here are the detailed solutions for RD Sharma Class 6 Maths Chapter 5 Exercise 5.4 on Negative Numbers and Integers.
1. Subtract the first integer from the second in each of the following:
(i) 12, -5
(ii) – 12, 8
(iii) – 225, – 135
(iv) 1001, 101
(v) – 812, 3126
(vi) 7560, – 8
(vii) – 3978, – 4109
(viii) 0, – 1005
Solution:
(i) 12, -5
So by subtracting the first integer from the second
-5 – 12 = – 17
(ii) – 12, 8
So by subtracting the first integer from the second
8 – (-12) = 8 + 12 = 20
(iii) – 225, – 135
So by subtracting the first integer from the second
-135 – (-225) = 225 – 135 = 90
(iv) 1001, 101
So by subtracting the first integer from the second
101 – 1001 = – 900
(v) – 812, 3126
So by subtracting the first integer from the second
3126 – (-812) = 3126 + 812 = 3938
(vi) 7560, – 8
So by subtracting the first integer from the second
-8 – 7560 = – 7568
(vii) – 3978, – 4109
So by subtracting the first integer from the second
-4109 – (-3978) = – 4109 + 3978 = -131
(viii) 0, – 1005
So by subtracting the first integer from the second
-1005 – 0 = – 1005
2. Find the value of:
(i) – 27 – (- 23)
(ii) – 17 – 18 – (-35)
(iii) – 12 – (-5) – (-125) + 270
(iv) 373 + (-245) + (-373) + 145 + 3000
(v) 1 + (-475) + (-475) + (-475) + (-475) + 1900
(vi) (-1) + (-304) + 304 + 304 + (-304) + 1
Solution:
(i) – 27 – (- 23)
So we get
= – 27 + 23
On further calculation
= 23 – 27
We get
= – 4
(ii) – 17 – 18 – (-35)
So we get
= – 35 + 35
On further calculation
= 0
(iii) – 12 – (-5) – (-125) + 270
So we get
= – 12 + 5 + 125 + 270
On further calculation
= 400 – 12
We get
= 388
(iv) 373 + (-245) + (-373) + 145 + 3000
So we get
= 373 – 245 – 373 + 145 + 3000
On further calculation
= 3145 + 373 – 373 – 245
We get
= 3145 – 245
By subtraction
= 2900
(v) 1 + (-475) + (-475) + (-475) + (-475) + 1900
So we get
= 1 – 950 – 950 + 1900
On further calculation
= 1900 + 1 – 1900
We get
= 1
(vi) (-1) + (-304) + 304 + 304 + (-304) + 1
So we get
= – 1 + 1 – 304 + 304 – 304 + 304
On further calculation
= 0
3. Subtract the sum of – 5020 and 2320 from – 709.
Solution:
We know that the sum of -5020 and 2320 is
-5020 + 2320
It can be written as
= 2320 – 5020
So we get
= – 2700
Subtracting from – 709 we get
= – 709 – (-2700)
We get
= – 709 + 2700
By subtraction
= 1991
4. Subtract the sum of – 1250 and 1138 from the sum of 1136 and – 1272.
Solution:
We know that the sum of – 1250 and 1138 is
-1250 + 1138
It can be written as
= 1138 – 1250
So we get
= – 112
We know that the sum of 1136 and – 1272 is
1136 – 1272 = – 136
So we get
-136 – (-112) = – 136 + 112 = -24
5. From the sum of 233 and – 147, subtract – 284.
Solution:
We know that the sum of 233 and – 147 is
233 – 147 = 86
Subtracting – 284 we get
86 – (-284) = 86 + 284 = 370
6. The sum of two integers is 238. If one of the integers is – 122, determine the other.
Solution:
It is given that
Sum of two integers = 238
One of the integers = – 122
So the other integer = – (-122) + 238
On further calculation
Other integer = 238 + 122 = 360
7. The sum of two integers is – 223. If one of the integers is 172, find the other.
Solution:
It is given that
Sum of two integers = – 223
One of the integers = 172
So the other integer = – 223 – 172 = – 395
8. Evaluate the following:
(i) – 8 – 24 + 31 – 26 – 28 + 7 + 19 – 18 – 8 + 33
(ii) – 26 – 20 + 33 – (-33) + 21 + 24 – (-25) – 26 – 14 – 34
Solution:
(i) – 8 – 24 + 31 – 26 – 28 + 7 + 19 – 18 – 8 + 33
We get
= – 8 – 24 – 26 – 28 – 18 – 8 + 31 + 7 + 19 + 33
On further calculation
= – 32 – 26 – 28 – 26 + 38 + 19 + 33
It can be written as
= 38 – 32 – 26 – 28 + 33 – 26 + 19
So we get
= 6 – 26 – 28 + 7 + 19
By calculation
= 6 – 28 – 26 + 26
= 6 – 28
By subtraction
= – 22
(ii) – 26 – 20 + 33 – (-33) + 21 + 24 – (-25) – 26 – 14 – 34
We get
= – 46 + 33 + 33 + 21 + 24 + 25 – 26 – 14 – 34
On further calculation
= – 46 + 66 + 21 + 24 + 25 + (-74)
It can be written as
= – 46 + 66 + 70 – 74
So we get
= – 46 – 4 + 66
By calculation
= – 50 + 66
= 66 – 50
By subtraction
= 16
9. Calculate
1 – 2 + 3 – 4 + 5 – 6 + ……… + 15 – 16
Solution:
It can be written as
1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10 + 11 – 12 + 13 – 14 + 15 – 16
We get
= – 1 – 1 – 1 – 1 – 1 – 1 – 1 – 1
By calculation
= – 8
10. Calculate the sum:
5 + (-5) + 5 + (-5) + …..
(i) if the number of terms is 10.
(ii) if the number of terms is 11.
Solution:
(i) if the number of terms is 10
We get
5 + (-5) + 5 + (-5) + 5 + (-5) + 5 + (-5) + 5 + (-5)
On further calculation
= 5 – 5 + 5 – 5 + 5 – 5 + 5 – 5 + 5 – 5 = 0
(ii) if the number of terms is 11
We get
5 + (-5) + 5 + (-5) + 5 + (-5) + 5 + (-5) + 5 + (-5) + 5
On further calculation
= 5 – 5 + 5 – 5 + 5 – 5 + 5 – 5 + 5 – 5 + 5 = 5
11. Replace * by < or > in each of the following to make the statement true:
(i) (-6) + (-9) * (-6) – (-9)
(ii) (-12) – (-12) * (-12) + (-12)
(iii) (-20) – (-20) * 20 – (65)
(iv) 28 – (-10) * (-16) – (-76)
Solution:
(i) (-6) + (-9) < (-6) – (-9)
(ii) (-12) – (-12) > (-12) + (-12)
(iii) (-20) – (-20) > 20 – (65)
(iv) 28 – (-10) < (-16) – (-76)
12. If △ is an operation on integers such that a △ b = – a + b – (-2) for all integers a, b. Find the value of
(i) 4 △ 3
(ii) (-2) △ (-3)
(iii) 6 △ (-5)
(iv) (-5) △ 6
Solution:
(i) 4 △ 3
By substituting values in a △ b = – a + b – (-2)
We get
4 △ 3 = – 4 + 3 – (-2) = 1
(ii) (-2) △ (-3)
By substituting values in a △ b = – a + b – (-2)
We get
(-2) △ (-3) = – (-2) + (-3) – (-2) = 1
(iii) 6 △ (-5)
By substituting values in a △ b = – a + b – (-2)
We get
6 △ (-5) = – 6 + (-5) – (-2) = – 9
(iv) (-5) △ 6
By substituting values in a △ b = – a + b – (-2)
We get
(-5) △ 6 = – (-5) + 6 – (-2) = 13
13. If a and b are two integers such that a is the predecessor of b. Find the value of a – b.
Solution:
It is given that a is the predecessor of b
We can write it as
a + 1 = b
So we get
a – b = – 1
14. If a and b are two integers such that a is the successor of b. Find the value of a – b.
Solution:
It is given that a is the successor of b
We can write it as
a – 1 = b
So we get
a – b = 1
(i) False.
(ii) True.
(iii) True.
(iv) True.
(v) True.
(vi) False.
(vii) True.
(viii) True.
16. Fill in the blanks:
(i) – 7 + ….. = 0
(ii) 29 + ….. = 0
(iii) 132 + (-132) = ….
(iv) – 14 + ….. = 22
(v) – 1256 + ….. = – 742
(vi) ….. – 1234 = – 4539
Solution:
(i) – 7 + 7 = 0
(ii) 29 + (-29) = 0
(iii) 132 + (-132) = 0
(iv) – 14 + 36 = 22
(v) – 1256 + 514 = – 742
(vi) -3305 – 1234 = – 4539
RD Sharma Solutions for Class 6 Maths Chapter 5 Exercise 5.4 PDF Download
Students looking to strengthen their understanding of subtracting integers can refer to the detailed step-by-step solutions provided in this exercise. These solutions explain important concepts such as subtracting positive and negative integers, converting subtraction into addition, and solving related word problems with ease.
You can download the PDF of RD Sharma Solutions for Class 6 Maths Chapter 5 Exercise 5.4 from the link below and study anytime, even offline.
