RD Sharma Solutions for Class 6 Maths Exercise 5.4 PDF
RD Sharma Solutions are designed with the aim of providing better conceptual knowledge among students. Each problem contains step-wise solutions explained in an interactive manner to make the subject interesting for students to learn. The students can make use of solutions which are designed by subject experts to score well in the Class 6 exam. The solutions are prepared with the aim of helping students self-analyse their areas of weakness. RD Sharma Solutions for Class 6 Maths Chapter 5 Negative Numbers and Integers Exercise 5.4 are provided here.
RD Sharma Solutions for Class 6 Maths Chapter 5: Negative Numbers and Integers Exercise 5.4
Access answers to Maths RD Sharma Solutions for Class 6 Chapter 5: Negative Numbers and Integers Exercise 5.4
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
15. Which of the following statements are true:
(i) – 13 > – 8 – (-2)
(ii) – 4 + (-2) < 2
(iii) The negative of a negative integer is positive.
(iv) If a and b are two integers such that a > b, then a – b is always a positive integer.
(v) The difference of two integers is an integer.
(vi) Additive inverse of a negative integer is negative.
(vii) Additive inverse of a positive integer is negative.
(viii) Additive inverse of a negative integer is positive.
Solution:
(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 – Negative Numbers and Integers Exercise 5.4
RD Sharma Solutions Class 6 Maths Chapter 5 Negative Numbers and Integers Exercise 5.4 provide the students with knowledge about the subtraction of integers and various properties of subtraction.
Key Features of RD Sharma Solutions for Class 6 Maths Chapter 5: Negative Numbers and Integers Exercise 5.4
- RD Sharma Solutions help students get an overall understanding of the concepts covered in the chapter.
- The students become more familiar with the problems by solving examples and exercise-wise problems from the RD Sharma textbook.
- Good practice of problems boosts confidence among students to perform well in the exam.
- They help students answer problems more accurately, thus enhancing their problem-solving abilities.
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