For Class 6 students the best way to score good marks in maths is by practicing more and more questions on a regular basis from RS Aggarwal Class 6 exercise questions. The exercise questions of RS Aggarwal textbook comprises greater number of questions for a certain topic. These solutions are based on helping students solve questions easily which gives an in-depth exposure of Class 6 mathematical problems which help you to score well in your examination.

All the RS Aggarwal Class 6 Solutions Chapter 4 – Integers Ex 4B (4.2) are prepared by our highly skilled subject experts keeping in mind the requirements of the Class 6 students. The topics mentioned in the Class 6 Maths CBSE syllabus are available in the series with step by step format of solution to boost the student’s knowledge and academic performance.

## Download PDF of RS Aggarwal Class 6 Solutions Chapter 4 â€“ Integers Ex 4B (4.2)

Question 1:

Subtract:

(i) 18 from -34

(ii) -15 from 25

(iii) -28 from -43

(iv) 68 from -37

(v) 219 from 0

(vi) -92 from 0

(vii) -135 from -250

(viii) -2768 from -287

(ix) 6240 from -271

(x) -3012 from 6250

Solution:

(i) -34 â€“ 18

= -52

(ii) 25 â€“ (-15)

= 25 + 15

= 40

(iii) -43 â€“ (-28)

= -43 + 28

= -15

(iv) -37 â€“ 68

= -105

(v) 0 â€“ 219

= -219

(vi) 0 â€“ (-92)

= 0 + 92

= 92

(vii) -250 â€“ (-135)

= -250 + 135

= -115

(viii) -287 â€“ (-2768)

= -287 + 2768

= 2481

(ix) -271 â€“ 6240

= -6511

(x) 6250 â€“ (-3012)

= 6250 + 3012

= 9262

Question 2:

Subtract the sum of -1050 and 813 from -23.

Solution:

Sum of -1050 and 813:

-1050 + 813

= -237

Subtracting the sum of -1050 and 813 from -23:

-23 â€“ (-237)

= -23 + 237

= 214

Question 3:

Subtract the sum of -250 and 138 from the sum of 136 and -272.

Solution:

Sum of 138 and -250:

138 + (-250)

= 138 â€“ 250

= -112

Sum of 136 and -272:

= 136 + (-272)

= 136 â€“ 272

= -136

Subtracting the sum of -250 and 138 from the sum of 136 and -272:

-136 â€“ (-112)

= -136 + 112

= -24

Question 4:

From the sum of 33 and -47, subtract -84.

Solution:

Adding 33 and -47:

33 + (-47)

= 33 â€“ 47

= -14

Subtracting -84 from -14:

-14 â€“ (-84)

= -14 + 84

= 70

Question 5:

Add -36 to the difference of -8 and -68.

Solution:

Difference of -8 and -68:

-8 â€“ (-68)

= -8 + 68

= 60

Adding -36 to 60:

-36 + 60

= 24

Question 6:

Simplify:

(i) [37 â€“ (-8)] + [11 â€“ (-30)]

(ii) [-13 â€“ (-17)] + [-22 â€“ (-40)]

Solution:

(i) [37 â€“ (-8)] + [11 â€“ (-30)]

= (37 + 8) + (11 + 30)

= 45 + 41

= 86

(ii) [-13 â€“ (-17)] + [-22 â€“ (-40)]

= (-13 + 17) + (-22 + 40)

= 4 + 18

= 22

Question 7:

Find 34 â€“ (-72) and (-72) â€“ 34. Are they equal?

Solution:

No they are not equal.

34 â€“ (-72)

= 34 + 72

= 106

(-72) â€“ 34

= -72 â€“ 34

= -106

Since 106 is not equal to -106, the two expressions are not equal.

Question 8:

The sum of two integers is -13. If one of the numbers is 170, find the other.

Solution:

Let the other integer be x.

According to question, we have:

x + 170 = -13

\(\Rightarrow\) x = -13 â€“ 170 \(\Rightarrow\) x = -183Thus, the other integer is -183.

Question 9:

The sum of two integer is 65. If one of the integers is -47, find the other.

Solution:

Let the other integer be x.

According to question, we have:

x + (-47) = 65

\(\Rightarrow\) x â€“ 47 = 65 \(\Rightarrow\) x = 65 + 47 \(\Rightarrow\) x = 112Thus, the other integer is 112.

Question 10:

Which of the following statements are true and which are false?

(i) The sum of two integer is always an integer.

(ii) The difference of two integers is always an integer.

(iii) -14 > -8 â€“ (-7)

(iv) -5 â€“ 2 > -8

(v) (-7) â€“ 3 = (-3) â€“ (-7)

Solution:

(i) True

An integer added to an integer gives an integer.

(ii) True

An integer subtracted from an integer gives an integer.

(iii) False

-8 â€“ (-7)

= -8 + 7

= -1

Since 14 is greater than 1, -1 is greater than -14.

(iv) True

-5 â€“ 2 = -7

Since 8 is greater than 7, -7 is greater than -8.

-7 > -8

(v) False

L.H.S.:

(-7) â€“ 3 = -10

R.H.S.:

(-3) â€“ (-7)

-3 + 7

= 4

Therefore, L.H.S. \(\neq\) R.H.S.

Question 11:

The point A is on a mountain which is 5700 metres above sea level and the point B is in a mine which is 39600 metres below sea level. Find the vertical distance between A and B.

Solution:

Let us consider the height above the sea level as positive and that below the sea level as negative.

Therefore, Height of point A from sea level = 5700 m

Depth of point B from sea level = -39600 m

Vertical distance between A and B = Distance of point A from sea level â€“ Distance of point B from sea level = 5700 â€“ (-39600)

= 45300 m

Question 12:

On a day in Srinagar, the temperature at 6 p.m. was 1oC but at midnight that day, it dropped to -4oC. By how many degrees Celsius did the temperature fall?

Solution:

Initial temperature of Srinagar at 6 p.m. = 1oC

Final temperature of Srinagar at midnight = -4oC

Change in temperature = Final temperature â€“ Initial temperature

= (-4 â€“ 1)oC = -5oC

So, the temperature has changed by -5oC.

The negative sign indicates that the temperature has fallen.

So, the temperature has fallen by 5oC.

Exercise 4D

Question 1:

Multiply:

(i) 15 by 9

(ii) 18 by -7

(iii) 29 by -11

(iv) -18 by 13

(v) -56 by 16

(vi) 32 by -21

(vii) -57 by 0

(viii) 0 by -31

(ix) -12 by -9

(x) -746 by -8

(xi) 118 by -7

(xii) -238 by -143

Solution:

(i) 15 by 9

= 15 x 9

= 135

(ii) 18 by -7

= -( 18 x 7 )

= -126

(iii) 29 by -11

= -( 29 x 11 )

= – 319

(iv) -18 by 13

= -( 18 x 13 )

= -234

(v) -56 by 16

= – ( 56 x 16 )

= -896

(vi) 32 by -21

= – (32 x 21)

= -672

(vii) -57 by 0

= -( 57 x 0 )

= 0

(viii) 0 by -31

= – ( 0 x 31 )

= 0

(ix) -12 by -9

= (-12 ) x ( -9)

= 108

(x) (-746) by ( -8)

= ( – 746 ) x ( – 8 )

= 5968

(xi) 118 by -7

= 118 x (-7)

= -826

(xii) -238 by -143

= (-238) x (-143)

= 34034

Question 2:

Find the products:

(i) (-2) x 3 x (-4)

(ii) 2 x (-5) x (-6)

(iii) (-8) x 3 x 5

(iv) 8 x 7 x (-10)

(v) (-3) x (-7) x (-6)

(vi) (-8) x (-3) x (-9)

Solution:

(i) (-2) x 3 x (-4)

= [(-2) x 3 ] x (-4)

= (-6) x (-4)

= 24

(ii) 2 x (-5) x (-6)

= [2 x (-5)] x (-6)

=(-10) x (-6)

= 60

(iii) (-8) x 3 x 5

= [(-8 x 3] x 5

= ( -24 ) x 5

= -120

(iv) 8 x 7 x (-10)

= [8 x 7] x (-10)

= 56 x (-10)

= -560

(v) (-3) x (-7) x (-6)

= [(-3) x (-7) ] x (-6)

= 21 x (-6)

= -126

(vi) (-8) x (-3) x (-9)

= [(-8) x (-3)] x (-9)

= 24 x (-9)

= -216

Question 3:

Use convenient groupings and find the values of

(i) 18 x (-27) x 30

(ii) (-8) x (-63) x 9

(iii) (-17) x (-23) x 41

(iv) (-51) x (-47) x (-19)

Solution:

(i) 18 x (-27) x 30

= (-27) x [18 x 30]

= (-27) x 540

= -14580

(ii) (-8) x (-63) x 9

= [(-8) x (-63)] x 9

= 504 x 9

= 4536

(iii) (-17) x (-23) x 41

= [(-17) x (-23)] x 41

= 391 x 41

= 16031

(iv) (-51) x (-47) x (-19)

= [(-51) x (-47)] x (-19)

= 2397 x (-19)

= -45543

Question 4:

Verify the following:

(i) 18 x [9 + (-7)] = 18 x 9 + 18 x (-7)

(ii) (-13) x [(-6) + (-19)] = (-13) x (-6) + (-13) x (-19)

Solution:

(i) L.H.S.

= 18 x [9 + (-7)]

= 18 x [9 -7]

= 18 x 2

= 36

R.H.S.

= 18 x 9 + 18 x (-7)

= 162 – ( 18 x 7 )

= 162 – 126

= 36

therefore, LHS = RHS

Hence, verified.

(ii) (-13) x [(-6) + (-19)] = (-13) x (-6) + (-13) x (-19)

LHS

= (-13) x [(-6) + (-19)]

= (-13) x [-6 – 19]

= (-13) x (-25)

= 325

RHS

= (-13) x (-6) + (-13) x (-19)

= 78 + 247

= 325

Therefore, LHS = RHS

Hence, verified

Question 5:

Complete the following multiplication table:

Solution

Question 6:

Which of the following statements are true and which are false?

(i) The product of a positive integer and a negative integer is negative.

(ii) The product of two negative integers is a negative integer.

(iii) The product of three negative integers is a negative integer.

(iv) Every integer when multiplied with -1 gives its multiplicative inverse.

Solution:

(i) True

(ii) False

(iii) True

(iv) False

Every integer when multiplied by (1) gives its multiplicative inverse.

Question 7:

Simplify:

(i) (-9) x 6 + (-9) x 4

(ii) 8 x (-12) + 7 x (-12)

(iii) 30 x (-22) + 30 x (14)

(iv) (-15) x (-14) + (-15) x ( -6)

(v) 43 x (-33) + (43) x (-17)

(vi) (-36) x (72) + (-36) x 28

(vii) (-27) x (-16) + (-27) x (-14)

Solution:

(i) (-9) x 6 + (-9) x 4

Using the distributive law:

(-9) x 6 + (-9) x 4

= (-9) x (6 + 9)

= (-9) x 10

= -90

(ii) 8 x (-12) + 7 x (-12)

Using the distributive law:

8 x (-12) + 7 x (-12)

= (-12) x (8 +7)

= (-12) x 15

= -180

(iii) 30 x (-22) + 30 x (14)

Using the distributive law:

30 x (-22) + 30 x (14)

= 30 x [(-22) + 14]

= 30 x (-8)

= -240

(iv) (-15) x ( -14) + (-15) x (-6)

Using the distributive law:

= (-15) x [(-14) + (-6)]

= (-15) x (-20)

= 300

(v) 43 x (-33) + 43 x (-17)

Using the distributive law:

= (43) x [-(33) + (-17)]

= (43) x [-33 – 17]

= 43 x (-50)

= -2150

(vi) (-36) x (72) + (-36) x 28

Using the distributive law:

= (-36) x (72 + 28)

= (-36) x 100

= -3600

(vii) (-27) x (-16) + (-27) x (-14)

Using the distributive law:

= (-27) x [(-16) + (-14)]

= (-27) x [-16 – 14]

= (-27) x [-30]

= 810

Exercise 4E

Question 1:

Divide:

(i) 85 by -17

(ii) -72 by 18

(iii) -80 by 16

(iv) -121 by 11

(v) 108 by -12

(vi) -161 by 23

(vii) -76 by -19

(viii) -147 by -21

(ix) -639 by -71

(x) -15625 by -125

(xi) 2067 by -1

(xii) 1765 by -1765

(xiii) 0 by -278

(xiv) 3000 by -100

Solution:

(i) = \(\frac{ -85 }{ 17 }\)

= -5

(ii) = \(\frac{ -72 }{ 18 }\)

= -4

(iii) = \(\frac{ -80 }{ 16 }\)

= -5

(iv) = \(\frac{ -121 }{ 11 }\)

= -4

(v) = \(\frac{ 108 }{ -12 }\)

= -9

(vi) = \(\frac{ -161 }{ 23 }\)

= -7

(vii) = \(\frac{ -76 }{ -19 }\)

= 4

(viii) = \(\frac{ -147 }{ -21 }\)

= 7

(ix) = \(\frac{ -639 }{ -71 }\)

= 9

(x) \(\frac{ -15625 }{ -125 }\)

= 125

(xi) \(\frac{ 2067 }{ -1 }\)

= -2067

(xii) \(\frac{ 1765 }{ -1765 }\)

= -1 x \(\frac{ 1765 }{ 1765 }\)

= -1 x 1

= -1

(xiii) = \(\frac{ 0 }{ -278 }\)

= 0

(xiv) = \(\frac{ 3000 }{ -100 }\)

= -30

Question 2:

Fill in the blanks:

(i) 80 \(\div\) ( __ ) = -5

(ii) (-84) \(\div\) ( __ ) = -7

(iii) ( __ ) \(\div\) (-5) = 25

(iv) ( __ ) \(\div\) 372 = 0

(v) ( __ ) \(\div\) 1 = -186

(vi) ( __ ) \(\div\) 17 = -2

(vii) ( __ ) \(\div\) 165 = -1

(viii) ( __ ) \(\div\) (-1) = 73

(ix) 1 \(\div\) = -1

Solution:

(i) -16

(ii) 12

(iii) -125

(iv) 0

(v) -186

(vi) -34

(vii) -165

(viii) -73

(ix) -1

Question 3:

Write (T) for True and ( F ) for false for each of the following statements:

(i) 0 \(\div\) (-6) = 0

(ii) (-8) \(\div\) 0 = 0

(iii) 15 \(\div\) (-1) = -15

(iv) (-16) \(\div\) (-4) = -4

(v) (-7) \(\div\) (-1) = 7

(vi) (-18) \(\div\) 9 = -2

(vii) 20 \(\div\) (-5) = -4

(viii) (-10) \(\div\) 1 = -10

(ix) (-1) \(\div\) (-1) = -1

Solution:

(i) True

(ii) False

This is because we cannot divide any integer by 0. If we do so , we get the quotient as infinity.

(iii) True

(iv) False

This is because the division of any two negative integers always gives a positive quotient.

(v) True

(vi) True

(vii) True

(viii) True

(ix) False

This because the division of any two negative integers always gives a positive quotient.

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- It is the best reference material for revision as it covers the entire Class 6 syllabus.
- It helps in reducing the exam stress by developing a proper understanding of the concepts.
- The RS Aggarwal Maths Solutions help students to practice different types of questions from each topic.
- It helps in clearing doubts if you face any difficulty while solving the questions mentioned in the RS Aggarwal textbook.