RS Aggarwal Class 7 Solutions Chapter 1 Integers Ex 1D

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

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

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.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 7 Solutions Chapter 1 Integers Ex 1A

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:
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.

 

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

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.