RS Aggarwal Maths Solutions Class 7

RS Aggarwal Class 7 Solutions Chapter 2 Fractions Ex 2A

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2A.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2A
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2B
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2C
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions Ex 2D
• RS Aggarwal Solutions Class 7 Chapter 2 Fractions CCE Test Paper

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:
Reenu get \(\frac { 2 }{ 7 }\) of an apple while Sonal gets \(\frac { 4 }{ 5 }\) part of it

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:
For finding the required fraction, we have to subtract 7\(\frac { 3 }{ 5 }\) from 18

Question 12.
Solution:
For finding the required fraction we should subtract 7\(\frac { 4 }{ 15 }\) from 8\(\frac { 2 }{ 5 }\)

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:

Question 16.
Solution:

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RS Aggarwal Class 7 Solutions Chapter 1 Integers CCE Test Paper

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 1 Integers CCE Test Paper.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 1 Integers Ex 1A
• RS Aggarwal Solutions Class 7 Chapter 1 Integers Ex 1B
• RS Aggarwal Solutions Class 7 Chapter 1 Integers Ex 1C
• RS Aggarwal Solutions Class 7 Chapter 1 Integers Ex 1D
• RS Aggarwal Solutions Class 7 Chapter 1 Integers CCE Test Paper

Question 1.
Solution:
Let the other integer be a. Then, we have;
a + (-12) = 43
⇒ a = 43 – (-12) = 55
Hence, the other integer is 55.

Question 2.
Solution:
p – (-8) = 3
⇒ p = 3 + (-8)
⇒ p = -5
Hence, the value of p is -5.

Question 3.
Solution:
Product of (-16) and (-9) = (-16) x (-9) = 144
Now, (-132) ÷ 6 gives the quotient -22.
144 + (-22) = 122

Question 4.
Solution:
Suppose that a divides -240 to obtain 16. Then, we have:
(-240) ÷ a =16
⇒ a = (-240) ÷ 16 = -15
Hence, -15 should divide -240 to obtain 16.

Question 5.
Solution:
Let a be divided by (-7) to obtain 12. Then, we have:
a ÷ (-7) = 12
a = \(\frac { -7 }{ 12 }\)
Hence, \(\frac { -7 }{ 12 }\) should be divided by -7 to obtain 12.

Question 6.
Solution:
(i) -450
(ii) 360
(iii) -1080
(iv) -600
(v) (-5)5 = -3125
(vi) (-1)25 = -1

Question 7.
Solution:
(i) (-16) x 12 + (-16) x 8 = (-16) x (12 + 8) [Associative property]
= (-16) x 20
= -320
(ii) 25 x (-33) + 25 x (-17)
= 25 x [(-33) + (-17)] [Associative property]
= 25 x (-50)
= -1250
(iii) (-19) x (-25) + (-19) x (-15)
= (-19) x [(-25) + (-15)] [Associative property]
= (-19) x (-40)
= 760
(iv) (-47) x 68 – (-47) x 38
= (-47) x (68 – 38) [Asssociative property]
= (-47) x 30
= -1410
(v) (-105) ÷ 21
(-105) ÷ 21 = -5
(vi) (-168) ÷ (-14) = 12
(vii) 0 ÷ (-34)
= 0 (zero).
Dividing 0 by any integer gives 0.
(viii) 37 ÷ 0
Not defined.
Dividing any integer by zero is not defined.

Mark (√) against the correct answer in each of the following:
Question 8.
Solution:
(d) -8
Let the other integer be a. Then, we have:
2 + a = -6
⇒ a = -6 – 2 = -8
⇒ The other integer is -8.

Question 9.
Solution:
(b) 8
Suppose that a is subtracted from (-7). Then, we have
-7 – a = -15
a = -7 + 15 = 8
8 must be subtracted from -7 to obtain-15.

Question 10.
Solution:
(b) 108
(108) + (-18) = -6

Question 11.
Solution:
(a) 370
= (-37) x (-7) + (-37) x (-3)
= (-37) x [(-7) + (-3)] [Associative property]
= (-37) x (-10)
= 370

Question 12.
Solution:
(c) -250
=(-25) x 8 + (-25) x 2
= (-25) x (8 + 2) [Associative property]
= (-25) x 10 = -250

Question 13.
Solution:
(b) -3
(-9) – (-6)
= (-9) + 6
= -3

Question 14.
Solution:
(b) -6
-8 – (-6) = 2
Hence, -8 is -6 less than -2.

Question 15.
Solution:
(i) (-35) x -1 = 35
(ii) (-53) x (1) = -53
(iii) (-14) x (-1) = (-16) x (-14) [Commutative property]
(iv) (-21) x (0) = 0 [Property of zero]
(v) (-119) ÷ 17 = (-7)
(vi) (-247) ÷ (-19) = 13
(vii) (0) ÷ 31 = 0
(viii) (152) ÷ (-19) = -8

Question 16.
Solution:
(i) True (T)
(ii) (-8) ÷ 0 = 0
False (F). Dividing any integer by zero is not defined.
(iii) False (F).
(-1) ÷ (-1) = 1
(iv) True (T)
(v) True(T)
(vi) False (T).
68 ÷ (-17) = -4

Hope given RS Aggarwal Solutions Class 7 Chapter 1 Integers CCE Test Paper are helpful to complete your math homework.

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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

• RS Aggarwal Solutions Class 7 Chapter 1 Integers Ex 1A
• RS Aggarwal Solutions Class 7 Chapter 1 Integers Ex 1B
• RS Aggarwal Solutions Class 7 Chapter 1 Integers Ex 1C
• RS Aggarwal Solutions Class 7 Chapter 1 Integers Ex 1D
• RS Aggarwal Solutions Class 7 Chapter 1 Integers CCE Test Paper

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

• RS Aggarwal Solutions Class 7 Chapter 1 Integers Ex 1A
• RS Aggarwal Solutions Class 7 Chapter 1 Integers Ex 1B
• RS Aggarwal Solutions Class 7 Chapter 1 Integers Ex 1C
• RS Aggarwal Solutions Class 7 Chapter 1 Integers Ex 1D
• RS Aggarwal Solutions Class 7 Chapter 1 Integers CCE Test Paper

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.

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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

• RS Aggarwal Solutions Class 7 Chapter 1 Integers Ex 1A
• RS Aggarwal Solutions Class 7 Chapter 1 Integers Ex 1B
• RS Aggarwal Solutions Class 7 Chapter 1 Integers Ex 1C
• RS Aggarwal Solutions Class 7 Chapter 1 Integers Ex 1D
• RS Aggarwal Solutions Class 7 Chapter 1 Integers CCE Test Paper

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.

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