## RS Aggarwal Class 7 Solutions Chapter 1 Integers Ex 1D

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 1 Integers Ex 1D.

Other Exercises

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
⇒ $$\frac { x }{ -18 }$$ = -5
⇒ x = (-5) x (-18) = 90

Hope given RS Aggarwal Solutions Class 7 Chapter 1 Integers Ex 1D are helpful to complete your math homework.

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## RS Aggarwal Class 7 Solutions Chapter 1 Integers Ex 1C

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 1 Integers Ex 1C.

Other Exercises

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.

Hope given RS Aggarwal Solutions Class 7 Chapter 1 Integers Ex 1C are helpful to complete your math homework.

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## RS Aggarwal Class 7 Solutions Chapter 1 Integers Ex 1B

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 1 Integers Ex 1B.

Other Exercises

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.

Hope given RS Aggarwal Solutions Class 7 Chapter 1 Integers Ex 1B are helpful to complete your math homework.

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## RS Aggarwal Class 7 Solutions Chapter 1 Integers Ex 1A

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 1 Integers Ex 1A.

Other Exercises

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:
(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.

Hope given RS Aggarwal Solutions Class 7 Chapter 1 Integers Ex 1A are helpful to complete your math homework.

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