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# RD Sharma Solutions -Page No.5.20, Negative Numbers And Integers, Class 6, Maths Class 6 Notes | EduRev

## RD Sharma Solutions for Class 6 Mathematics

Created by: Abhishek Kapoor

## Class 6 : RD Sharma Solutions -Page No.5.20, Negative Numbers And Integers, Class 6, Maths Class 6 Notes | EduRev

The document RD Sharma Solutions -Page No.5.20, Negative Numbers And Integers, Class 6, Maths Class 6 Notes | EduRev is a part of the Class 6 Course RD Sharma Solutions for Class 6 Mathematics.
All you need of Class 6 at this link: Class 6

#### PAGE NO 5.20:

Question 1: The successor of − 79 is
(a) − 80
(b) − 78
(c) 80
(d) 78

ANSWER: If a is an integer, then its successor is a + 1. So
Successor of − 79 = − 79 + 1
= − (79 − 1)
= − 78
Hence, the correct option is (b).

Question 2: The predecessor of − 99 is
(a) − 98
(b) − 100
(c) 98
(d) 100

If a is an integer, then its predecessor is a − 1. So
Predecessor of − 99 = − 99 − 1
= − (99 + 1)
= − 100
Hence, the correct option is (b).

Question 3: The integer 8 more than − 12 is
(a) 4
(b) − 4
(c) − 20
(d) 20

ANSWER:The integer 8 more than − 12 is (− 12) + 8.
Now
(− 12) + 8 = − (12 − 8) = − 4
Hence, the correct option is (b).

Question 4:What should be added to 18 to get − 34?
(a) 52
(b) − 52
(c) − 16
(d) 16

ANSWER:The required number is (− 34) − 18.
Now
(− 34) − 18 = − (34 + 18) = − 52
Hence, the correct option is (b).

Question 5: The additive inverse of 17 is
(a) − 17
(b) 17
(c) 117117
(d) −117-117

ANSWER: If a is an integer, then its additive inverse is − a.
So
Additive inverse of 17 = − 17
Hence, the correct option is (a).

Question 6: If an integer a is greater than 7, then |7 − a| =
(a) 7 − a
(b) a − 7
(c) 7 + a
(d) − 7 − a

ANSWER: If x is negative integer, then |x| = − x.
Here, a is greater than 7, so 7 − a is negative.
Therefore
|7 − a| = − (7 − a) = − 7 + a = a − 7
Hence, the correct option is (b).

Question 7: The additive identity element in the set of integers is
(a) 1
(b) − 1
(c) 0
(d) None of these

ANSWER: If a is an integer, then a + 0 = 0 + a = a. Here, 0 is the additive identity element in the set of integers.
Hence, the correct option is (c).

Question 8: Which of the following pairs of integers have 9 as difference?
(a) 19, 10
(b) − 19, − 10
(c) 19, − 10
(d) (a) and (b) both

ANSWER: (a) 19 − 10 = 9
(b) − 10 − (− 19) = − 10 + 19 = 19 − 10 = 9
(c) 19 − (− 10) = 19 + 10 = 29
Thus, (a) and (b) both are correct.
Hence, the correct option is (d).

Question 9: When 47 is subtracted from − 23, we get?
(a) 70
(b) 24
(c) − 24
(d) − 70

ANSWER: (− 23) − (47) = − 23 − 47
= − (23 + 47)
= − 70
Thus, when 47 is subtracted from − 23, we get − 70.
Hence, the correct option is (d).

Question 10: If ∆ is an operation on integers such that a ∆ b = a − b − 2, for all integers a, b. Then, 7 ∆ (− 4) =
(a) 11
(b) − 9
(c) 9
(d) 1

ANSWER: Here, the operation ∆ is defined as a ∆ b = a − b − 2. So
7 ∆ (− 4) = 7 − (− 4) − 2
= 7 + 4 − 2
= 11 − 2
= 9
Hence, the correct option is (c).

Question 11: The sum of two integers is 84. If one of the integers is 44, determine the other.

ANSWER: Sum of two integers = 84
One of the integers = 44
Other integer = Sum of two integers − One of the integers
= 84 − 44
= 40
Hence, the other integer is 40.

Question 12: Simplify: 9 ×× (− 16) + (− 17) ×× (− 16).

ANSWER: 9 ×× (− 16) + (− 17) ×× (− 16) = − (9 ×× 16) + (17 ×× 16)
= − 144 + 272
= 272 − 144
= 128
Hence, 9 ×× (− 16) + (− 17) ×× (− 16) = 128.

Question 13: If x = (− 23) + 22 + (− 23) + 22 + ... + (40 terms) and y = 11 + (− 10) + 11 + (− 10) + ... + (20 terms), then find y − x.

ANSWER: x = (− 23) + 22 + (− 23) + 22 + ... + (40 terms)
= (− 23) + (− 23) + ... + (20 terms) + 22 + 22 + ... + (20 terms)
= 20 ×× (− 23) + 20 ×× 22
= 20 ×× (− 23 + 22)
= 20 ×× (− 1)
= − 20
Now
y = 11 + (− 10) + 11 + (− 10) + ... + (20 terms)
= 11 + 11 + ... + (10 terms) + (− 10) + (− 10) + ... + (10 terms)
= 11 ×× 10 + (− 10) ×× 10
= 110 − 100
= 10
Therefore, y − x = 10 − (− 20) = 10 + 20 = 30.

Question 15: Calculate 1 − 2 + 3 − 4 + 5 − 6 + 7 − 8 + ... + 49 − 50.

ANSWER: 1 − 2 + 3 − 4 + 5 − 6 + 7 − 8 + ... + 49 − 50
= (1 − 2) + (3 − 4) + (5 − 6) + (7 − 8) + ... + (49 − 50)
= (− 1) + (− 1) + (− 1) + ... + 25 terms
= (− 1) ×× 25
= − 25
Hence, 1 − 2 + 3 − 4 + 5 − 6 + 7 − 8 + ... + 49 − 50 = − 25.

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