RS Aggarwal Solutions for Class 7 Maths Chapter 1–Integers
RS Aggarwal Solutions for Class 7 Maths Chapter 1–Integers

Class 7: Maths Chapter 1 solutions. Complete Class 7 Maths Chapter 1 Notes.

RS Aggarwal Solutions for Class 7 Maths Chapter 1–Integers

RS Aggarwal 7th Maths Chapter 1, Class 7 Maths Chapter 1 solutions

Ex 1A Solutions

Question 1.
Solution:
(i) 15 + (-8) =15 – 8 = 7
(ii) (-16) +9 = -7
(iii) (-7) + (-23)= -7 – 23 = -30
(iv) (-32) + 47 = -32 + 47 = 15
(v) 53 + (-26) = 53 – 26 = 27
(vi) (-48) + (-36) = -48 – 36 = -84

Question 2.
Solution:
(i) 153 and -302 = 153 + (-302) = 153 – 302 = -149
(ii) 1005 and -277 = 1005 + (-277) = 1005 – 277 = 728
(iii) -2035 and 297 = -2035 + 297 = – 1738
(iv) -489 and -324 = -489 + (-324) = -489 – 324 = -813
(v) -1000 and438 = -1000 + 438 = -562
(vi) -238 and 500 = -238 + 500 = 262

Question 3.
Solution:
Additive inverse of
(i) -83 is – (-83) = 83
(ii) 256 is -256
(iii) 0 is 0
(iv) -2001 is – (-2001) = 2001

Question 4.
Solution:
(i) 28 from – 42 = -42 – (28) = -42 – 28 = -70
(ii) – 36 from 42 = 42 – (-36) = 42 + 36 = 78
(iii) -37 from -53 = -53 – (-37) = -53 + 37= -16
(iv) -66 from -34 = -34 – (-66) = -34 + 66 = 32
(v) 318 from 0 = 0 – (318) = -318
(vi) -153 from -240= -240 – (-153) = -240 + 153 = -87
(vii) -64 from 0 = 0 – (-64) = 0 + 64 = 64
(viii) – 56 from 144 = 144 – (-56) = 144 + 56 = 200

Question 5.
Solution:
– 34 – (-1032 + 878)
= -34 – (-154) = -34 + 154 = 120

Question 6.
Solution:
38 + (-87) – 134
= (38 – 87) – 134
= -49 – 134 = -183

Question 7.
Solution:
(i) {(-13) + 27} + (-41) = (-13) + {27 + (-41)} (By Associative Law of Addition)
(ii) (-26) + {(-49) + (-83)} = {(-26) + (-49)} +(-83) (By Associative Law of Addition)
(iii) 53 + (-37) = (-37) + (53) (By Commutative Law of Addition)
(iv) (-68) + (-76) = (-76) + (-68) (By Commutative Law of Addition)
(v) (-72) + (0) = -72 (Existence of Additive identity)
(vi) -(-83) = 83
(vii) (-60) – (………) = -59 => -60 – (-1) = -59
(viii) (-31) + (……….) = -40 => -31 + (-9) = -40

Question 8.
Solution:
{-13 – (-27)} + {-25 – (-40)}
= {-13 + 27} + {-25 + 40}
= 14 + 15 = 29

Question 9.
Solution:
36 – (- 64) = 36 + 64 = 100
(-64) – 36= -64 – 36 = -100
They are not equal

Question 10.
Solution:
(a + b) + c = {-8 + (-7)} + 6 = (-8 – 7) + 6 = -15 + 6 = -9
and a + (b + c) = -8 + (-7 + 6) = -8 + (-1) = -8 – 1 = -9 Hence proved

Question 11.
Solution:
LHS = (a -b) = -9 – (-6) = -9 + 6 = -3
RHS = (b – a) = -6 – (-9) = -6 + 9 = 3
LHS ≠ RHS.
Hence (a – b) ≠ (b – a)

Question 12.
Solution:
Sum of two integers = -16
One integer = 53
Second integer = -16 – (53) = -16 – 53 = (-69)

Question 13.
Solution:
Sum of two integers = 65
One integer = -31
Second integer = 65 – (-31) = 65 + 31 = 96

Question 14.
Solution:
Difference of a and (-6) = 4
a – (-6) = 4
⇒ a + 6 = 4
⇒ a = 4 – 6
⇒ a = -2

Question 15.
Solution:
(i) We can write any two integers having opposite signs
e.g. 5, -5
Sum = 5 + (-5) = 5 – 5 = 0
(ii) The sum is a negative integer
The greater integer must be negative and smaller integer be positive
e.g. -9, 6
Sum = -9 + 6 = -3
(iii) The sum is smaller than the both integers
Both integer will be negative -4, -6
Sum = -4 + (-6) = -4 – 6 = -10
(iv) The sum is greater than the both integers
Both integers will be positive
e.g. 6, 4
(v) The sum oftwo integers is smaller than one of these integers
The greater number will be positive and smaller be negative
e.g. 6, -4
Sum = 6 + (-4) = 2

Question 16.
Solution:
(i) False: Because, all negative integers are less than zero.
(ii) False: -10 is less than -7.
(iii) Tme: Every negative integer is less than zero.
(iv) True : Sum of two negative integers is negative.
(v) False: It is not always true.

Ex 1B Solutions

Question 1.
Solution:
(i) 16 by 9 = 16 x 9 = 144
(ii) 18 by -6 = 18 x (-6) = -108
(iii) -36 by -11 = 36 x (-11) = -396
(iv) -28 by 14 = -28 x 14 = -392
(v) -53 by 18 = -53 x 18 = -954
(vi) -35 by 0 = -35 x 0 = 0
(vii) 0 by -23 = 0 x (-23) = 0
(viii) -16 by -12 = (-16) x (-12) = 192
(ix) -105 by -8 = -105 x (-8) = 840
(x) -36 by -50 = (-36) x (-50) = 1800
(xi) -28 by -1 = (-28) x (-1) = 28
(xii) 25 by -11 = 25 x -11 = -275

Question 2.
Solution:
(i) 3 x 4 x (-5) = 12 x (-5) = -60 = 60
(ii) 2 x (-5) x (-6) = (-10) x (-6) = 60
(iii) (-5) x (-8) x (-3) = 40 x (-3) = -120
(iv) (-6) x 6 x (-10) = (-36) x (-10) = 360
(v) 7 x (-8) x 3 =(-56) x 3 = -168
(vi) (-7) x (-3) x 4 = 21 x 4 = 84

Question 3.
Solution:
(i) (-4) x (-5) x (-8) x (-10) = (4 x 5) x (8 x 10)
{Number of negative integers is even}
= 20 x 80 = 1600
(ii) (-6) x (-5) x (-7) x (-2) x (-3)
Here number of negative integers is odd
= (-1) [6 x 5 x 7 x 2 x 3]
= (-1) (1260) = -1260
(iii) (-60) x (-10) x (-5) x (-1)
Here number of negative integers is even
= 60 x 10 x 5 x 1
= 3000
(iv) (-30) x (-20) x (-5)
Here number of negative integers is odd
= (-1) (30 x 20 x 5) = -1 x 3000 = -3000
(v) (-3) x (-3) x (-3) x …6 times
Here number of negative integers is even
= 3 x 3 x 3 x 3 x 3 x 3 = 729
(vi) (-5) x (-5) x (-5) x …5 times
Here number of negative integers is odd
= (-1) (5 x 5 x 5 x 5 x 5)
= (-1) (3125) = – 3125
(vii) (-1) x (-1) x (-1) x …200 times
Here number of negative integers is even
= 1 x 1 x 1 x 1 x 200 times = 1
(viii) (-1) x (-1) x (-1) x …171 times
Here number of negative integers is odd
= (-1) x (1 x 1 x 1 x ……… 171 times)
= -1 x 1 = -1

Question 4.
Solution:
Number of negative integers = 90
which is positive and 9 integers are positive
The sign of the product will be positive

Question 5.
Solution:
Number of negative integers = 103 which is negative
Product will be negative

Question 6.
Solution:
(i) (- 8) x 9 + (- 8) x 7
= (-8) {9 + 7}
= -8 x 16 = -128
(ii) 9 x (-13) + 9 x (-7)
= 9 x (-13 – 7)
= 9 x (-20) = – 180
(iii) 20 x (-16) + 20 x 14 = 20 x {-16 + 14}
= 20 x (-2)= -40
(iv) (-16) x (-15) + (-16) x (-5)
= (-16) x {-15 – 5}
= (-16) x (-20) = 320
(v) (-11) x (-15)+ (-11) x (-25)
-(-11) x {-15 – 25}
= (-11) x (-40) = -440
(vi) 10 x (-12)+ 5 x (-12)
= (-12) {10 + 5} = (-12) x 15 = -180
(vii) (-16) x (-8) + (-4) x (-8)
= (-8){-16 – 4} = (-8) x (-20) = 160
(viii) (-26) x 72 + (-26) x 28
= (-26) (72 + 28) = (-26) x 100 = -2600

Question 7.
Solution:
(i) (-6) x (………) = 6 ⇒ (-6) x (-1) = 6
(ii) (-18) x (………) = (-18) ⇒ (-18) x (1) = (-18)
(iii) (-8) x (-9) = (-9) x (……….) ⇒ (-8) x (-9) = (-9) x (-8) (By Commutative Law of Multiplication)
(iv) 7 x (-3) = (-3) x (……….) ⇒ 7 x (-3) = (-3) x (7) (By Commutative Law of Multiplication)
(v) {(-5) x 3} x (-6) = (………) x {3 x (-6)} ⇒ {(-5) x 3} x (-6) = (-5) x {3 x (-6)} (By Associative Law of Multiplication)
(vi) (-5) x (……….) = 0 ⇒ (-5) x (0) = 0 (By Property of Zero)

Question 8.
Solution:
Number of questions in a test =10
Marks awarded for every correct answer = 5
and marks deducted for every wrong answer = 2 (-2 is given)
(i) Ravi gets 4 correct and 6 incorrect answers
Total marks obtained by him = 4 x 5 – 6 x 2 = 20 – 12 = 8
(ii) Reenu gets 5 correct and 5 incorrect answers
Total marks obtained by her = 5 x 5 – 5 x 2 = 25 – 10= 15
(iii) Heena gets 2 correct and 5 incorrect answers
She gets marks = 2 x 5 – 5 x 2 = 10 – 10 = 0

Question 9.
Solution:
(i) True: As product of a positive and a negative integer is negative.
(ii) False: The product of two negative integers is positive.
(iii) True.
(iv) False: As multiplication of an integer and (-1) is negative.
(v) True as a x b = b x a.
(vi) True as (a x b) x c = a x (b x c)
(vii) False: It is not possible except integer 1.

Ex 1C Solutions

Question 1.
Solution:
(i) 65 by -13 = 65 ÷ (-13) = -5
(ii) -84 by 12 = -84 ÷ 12 = -7
(iii) -76 by 19 = -76 ÷ 19 = -4
(iv) -132 by 12 = -132 ÷ 12 = -11
(v) -150 by 25 = -150 ÷ 25 = -6
(vi) -72 by -18= -72 ÷ (-18)
(vii) -105 by -21 = -105 ÷ (-21) = 5
(viii) -36 by -1 = -36 ÷ (-1) = 36
(ix) 0 by -31 = 0 ÷ (-31) = 0
(x) -63 by 63 = -63 ÷ 63 = -1
(xi) -23 by -23 = -23 ÷ (-23)
(xii) -8 by 1 = -8 ÷ 1 = -8

Question 2.
Solution:
(i) 72 ÷ (………) = -4
⇒ 72 ÷ (-4) = -18
72 + (-18) = -4
(ii) -36 ÷ (………) = -4
⇒ -36 ÷ (-4) = 9
-36 ÷ (9) = -4
(iii) (………) ÷ (-4) = 24
⇒ -4 x 24 = -96
(-96) ÷ (-4) = 24
(iv) (……….) ÷ 25 = 0
(…….) ÷ 25 = 0 {0 ÷ a = 0}
(v) (………) ÷ (-1) = 36
⇒ 36 x (-1) = -36
(-36) ÷ (-1) = 36
(vi) (………..) + 1 = 37
⇒ (-37) x 1 = -37
(-37) ÷ 1 = -37
(vii) 39 ÷ (……….) = -1
⇒ 39 ÷ (-1) = -39
39 ÷ (-39) = -1
(viii) 1 ÷ (………) = -1
⇒ -1 ÷ 1 = -1
1 ÷ (-1) = -1
(ix) -1 + (………) = -1
-1 ÷ (1) = -1

Question 3.
Solution:
(i) True : as zero divided by non zero integer is zero.
(ii) False : as division by zero is not meaning full
(iii) False : as (-5) ÷ (-1) = 5 (product will be positive)
(iv) True : as -a ÷ 1 = -a
(v) False : as (-1) ÷ (-1) = 1
(vi) True.

Ex 1D Solutions

OBJECTIVE QUESTIONS
Mark (√) against the correct answer in each of the following.
Question 1.
Solution:
(c)
6 – (-8) = 6 + 8 = 14

Question 2.
Solution:
(b)
-9 – (-6) = -9 + 6 = -3

Question 3.
Solution:
(d)
-3 + 5 = 2

Question 4.
Solution:
(a)
-1 – (+5) = -1 – 5 = -6

Question 5.
Solution:
(a)
-2 – (4) = -2 – 4 = -6

Question 6.
Solution:
(b)
-4 – (+4) = -4 – 4 = -8

Question 7.
Solution:
(b)
-3 – (-5) = -3 + 5 = 2

Question 8.
Solution:
(c)
-3 – (-9) = -3 + 9 = 6

Question 9.
Solution:
(c)
-5 – (6) = -5 – 6 = -11

Question 10.
Solution:
(c)
-8 – (-13) = -8 + 13 = 5

Question 11.
Solution:
(a)
(-36) ÷ (-9) = 4

Question 12.
Solution:
(b)
0 ÷ (-5) = 0
(Zero divided by any integer other than zero, is zero)

Question 13.
Solution:
(c)
Division by zero is not defined

Question 14.
Solution:
(b)

Question 15.
Solution:
(b)
-3 + 9 = 6

Question 16.
Solution:
(a)
-4 – (-10) = -4 + 10 = 6

Question 17.
Solution:
(a)
Sum = 14
One integer = -8
Second = 14 – (-8) = 14 + 8 = 22

Question 18.
Solution:
(c)

Question 19.
Solution:
(b)
(-15) x 8 + (-15) x 2
= (-15) {8 + 2}
= -15 x 10 = -150

Question 20.
Solution:
(b)
(-12) x 6 – (-12) x 4 = (-12) (6 – 4) = -12 x 2 = -24

Question 21.
Solution:
(b)
(-27) x (-16)+ (-27) x (-14)
= (-27) {-16 – 14}
= (-27) x (-30)
= 810

Question 22.
Solution:
(a)
30 x (-23) + 30 x 14
= 30 x (-23 + 14)
= 30 x (-9)
= -270

Question 23.
Solution:
(c)
Sum of two integers = 93
One integer = -59
Second = 93 – (-59) = 93 + 59 = 152

Question 24.
Solution:
(b)
(?) ÷ (-18) = -5
Let x ÷ (-18) = -5
⇒ x−18 = -5
⇒ x = (-5) x (-18) = 90

RS Aggarwal Solutions for Class 7 Maths Chapter 1: Download PDF

RS Aggarwal Solutions for Class 7 Maths Chapter 1–Integers

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Investing in an R.S. Aggarwal book will never be of waste since you can use the book to prepare for various competitive exams as well. RS Aggarwal is one of the most prominent books with an endless number of problems. R.S. Aggarwal’s book very neatly explains every derivation, formula, and question in a very consolidated manner. It has tonnes of examples, practice questions, and solutions even for the NCERT questions.

He was born on January 2, 1946 in a village of Delhi. He graduated from Kirori Mal College, University of Delhi. After completing his M.Sc. in Mathematics in 1969, he joined N.A.S. College, Meerut, as a lecturer. In 1976, he was awarded a fellowship for 3 years and joined the University of Delhi for his Ph.D. Thereafter, he was promoted as a reader in N.A.S. College, Meerut. In 1999, he joined M.M.H. College, Ghaziabad, as a reader and took voluntary retirement in 2003. He has authored more than 75 titles ranging from Nursery to M. Sc. He has also written books for competitive examinations right from the clerical grade to the I.A.S. level.

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