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Chapter 3

Integers — Exercise 3(B)

Class - 6 RS Aggarwal Mathematics Solutions



Exercise 3(B)

Question 1

Find the sum :

(i) 16 + 21

(ii) (-16) + 21

(iii) 16 + (-21)

(iv) (-25) + (-18)

(v) (-32) + (-47)

(vi) 54 + (-89)

(vii) (-38) + 45

(viii) 96 + (-103)

(ix) (-150) + (-15)

Answer

(i) 16 + 21

= 37

Hence, 16 + 21 = 37.

(ii) (-16) + 21

= -16 + 21

= 5

Hence, (-16) + 21 = 5.

(iii) 16 + (-21)

= 16 - 21

= -5

Hence, 16 + (-21) = -5.

(iv) (-25) + (-18)

= -25 - 18

= -43

Hence, (-25) + (-18) = -43.

(v) (-32) + (-47)

= -32 - 47

= -79

Hence, (-32) + (-47) = -79.

(vi) 54 + (-89)

= 54 - 89

= -35

Hence, 54 + (-89) = -35.

(vii) (-38) + 45

= -38 + 45

= 7

Hence, (-38) + 45 = 7.

(viii) 96 + (-103)

= 96 - 103

= -7

Hence, 96 + (-103) = -7.

(ix) (-150) + (-15)

= -150 - 15

= -165

Hence, (-150) + (-15) = -165.

Question 2

Write the additive inverse of :

(i) 34

(ii) -58

(iii) 0

(iv) -1

(v) 170

Answer

(i) 34

Let x be additive inverse of 34.

⇒ 34 + x = 0

⇒ x = 0 - 34

⇒ x = -34

Hence, -34 is the additive inverse of 34.

(ii) -58

Let x be additive inverse of -58.

⇒ -58 + x = 0

⇒ x = 0 - (-58)

⇒ x = 58

Hence, 58 is the additive inverse of -58.

(iii) 0

Let x be additive inverse of 0.

⇒ 0 + x = 0

⇒ x = 0 - 0

⇒ x = 0

Hence, 0 is the additive inverse of 0.

(iv) -1

Let x be additive inverse of -1.

⇒ -1 + x = 0

⇒ x = 0 - (-1)

⇒ x = 1

Hence, 1 is the additive inverse of -1.

(v) 170

Let x be additive inverse of 170.

⇒ 170 + x = 0

⇒ x = 0 - 170

⇒ x = -170

Hence, -170 is the additive inverse of 170.

Question 3

Write the successor of :

(i) 101

(ii) -47

(iii) -1

(iv) -80

(v) -301

Answer

(i) 101

Successor of 101 = 101 + 1 = 102

Hence, 102 is the successor of 101.

(ii) -47

Successor of -47 = (-47) + 1 = -46

Hence, -46 is the successor of -47.

(iii) -1

Successor of -1 = (-1) + 1 = 0

Hence, 0 is the successor of -1.

(iv) -80

Successor of -80 = (-80) + 1 = -79

Hence, -79 is the successor of -80.

(v) -301

Successor of -301 = (-301) + 1 = -300

Hence, -300 is the successor of -301.

Question 4

Write the predecessor of :

(i) 40

(ii) -32

(iii) -70

(iv) -91

(v) 0

Answer

(i) 40

Predecessor of 40 = 40 - 1 = 39

Hence, 39 is the predecessor of 40.

(ii) -32

Predecessor of -32 = -32 - 1 = -33

Hence, -33 is the predecessor of -32.

(iii) -70

Predecessor of -70 = -70 - 1 = -71

Hence, -71 is the predecessor of -70.

(iv) -91

Predecessor of -91 = -91 - 1 = -92

Hence, -92 is the predecessor of -91.

(v) 0

Predecessor of 0 = 0 - 1 = -1

Hence, -1 is the predecessor of 0.

Question 5

Find the difference :

(i) (15) - (21)

(ii) (-29) - (9)

(iii) 70 - (-8)

(iv) 24 - (-24)

(v) (-36) - (64)

(vi) 0 - (-20)

(vii) (-63) - (-7)

(viii) (-80) - (-20)

(ix) (-12) - (-71)

Answer

(i) (15) - (21)

= 15 - 21

= -6

Hence, (15) - (21) = -6.

(ii) (-29) - (9)

= -29 - 9

= -38

Hence, (-29) - (9) = -38.

(iii) 70 - (-8)

= 70 + 8

= 78

Hence, 70 - (-8) = 78.

(iv) 24 - (-24)

= 24 + 24

= 48

Hence, 24 - (-24) = 48.

(v) (-36) - (64)

= -36 - 64

= -100

Hence, (-36) - (64) = -100.

(vi) 0 - (-20)

= 0 + 20

= 20

Hence, 0 - (-20) = 20.

(vii) (-63) - (-7)

= -63 + 7

= -56

Hence, (-63) - (-7) = -56.

(viii) (-80) - (-20)

= -80 + 20

= -60

Hence, (-80) - (-20) = -60.

(ix) (-12) - (-71)

= -12 + 71

= 59

Hence, (-12) - (-71) = 59.

Question 6

Subtract :

(i) -180 from 180

(ii) 75 from -75

(iii) -630 from -70

(iv) -95 from 0

(v) -90 from -1

(vi) -16 from -25

Answer

(i) -180 from 180

= 180 - (-180)

= 180 + 180

= 360.

Hence, final result = 360.

(ii) 75 from -75

= -75 - 75

= -150

Hence, final result = -150.

(iii) -630 from -70

= -70 - (-630)

= -70 + 630

= 560

Hence, final result = 560.

(iv) -95 from 0

= 0 - (-95)

= 0 + 95

= 95

Hence, final result = 95.

(v) -90 from -1

= -1 - (-90)

= -1 + 90

= 89

Hence, final result = 89.

(vi) -16 from -25

= -25 - (-16)

= -25 + 16

= -9

Hence, final result = -9.

Question 7

Fill in the blanks :

(i) (+23) + (...............) = 0

(ii) (-13) + (...............) = + 20

(iii) (-14) + (...............) = -34

(iv) (-20) + (...............) = -8

(v) (-30) - (...............) = -14

(vi) (...............) - (-25) = 2

Answer

(i) (+23) + (...............) = 0

Let x be the number in blanks,

⇒ 23 + x = 0

⇒ x = 0 - 23

⇒ x = -23

(+23) + (-23) = 0

(ii) (-13) + (...............) = + 20

Let x be the number in blanks,

⇒ -13 + x = 20

⇒ x = 20 - (-13)

⇒ x = 20 + 13

⇒ x = 33

(-13) + (33) = + 20

(iii) (-14) + (...............) = -34

Let x be the number in blanks,

⇒ -14 + x = -34

⇒ x = -34 + 14

⇒ x = -20

(-14) + (-20) = -34

(iv) (-20) + (...............) = -8

Let x be the number in blanks,

⇒ -20 + x = -8

⇒ x = -8 + 20

⇒ x = 12

(-20) + (12) = -8

(v) (-30) - (...............) = -14

Let x be the number in blanks,

⇒ -30 - x = -14

⇒ -x = -14 + 30

⇒ -x = 16

⇒ x = -16

(-30) - (-16) = -14

(vi) (...............) - (-25) = 2

Let x be the number in blanks,

⇒ x + 25 = 2

⇒ x = 2 - 25

⇒ x = -23

(-23) - (-25) = 2

Question 8

Use number line to find :

(i) 5 + 4

(ii) 4 + (-6)

(iii) (-4) + 8

(iv) (-5) + 3

(v) (-3) + (-5)

(vi) (-6) + (-3)

(vii) 5 - (-2)

(viii) (-4) - 5

(ix) 4 - (-4)

Answer

(i) 5 + 4 = 9

Starting from 0 on the number line, move 5 steps to the right and from there move 4 steps to the right to reach 9 as shown below.

Starting from 0 on the number line, move 5 steps to the right and from there move 4 steps to the right to reach 9 as shown below. Integers, R.S. Aggarwal Mathematics Solutions ICSE Class 6.

(ii) 4 + (-6) = 4 - 6 = -2

Starting from 0 on the number line, move 4 steps to the right and from there move 6 steps to the left to reach -2 as shown below.

Starting from 0 on the number line, move 4 steps to the right and from there move 6 steps to the left to reach -2 as shown below. Integers, R.S. Aggarwal Mathematics Solutions ICSE Class 6.

(iii) (-4) + 8 = -4 + 8 = 4

Starting from 0 on the number line, move 4 steps to the left and from there move 8 steps to the right to reach 4 as shown below.

Starting from 0 on the number line, move 4 steps to the left and from there move 8 steps to the right to reach 4 as shown below. Integers, R.S. Aggarwal Mathematics Solutions ICSE Class 6.

(iv) (-5) + 3 = -2

Starting from 0 on the number line, move 5 steps to the left and from there move 3 steps to the right to reach -2 as shown below.

Starting from 0 on the number line, move 5 steps to the left and from there move 3 steps to the right to reach -2 as shown below. Integers, R.S. Aggarwal Mathematics Solutions ICSE Class 6.

(v) (-3) + (-5) = -3 - 5 = -8

Starting from 0 on the number line, move 3 steps to the left and from there move 5 steps to the left to reach -8 as shown below.

Starting from 0 on the number line, move 3 steps to the left and from there move 5 steps to the left to reach -8 as shown below. Integers, R.S. Aggarwal Mathematics Solutions ICSE Class 6.

(vi) (-6) + (-3) = -6 - 3 = -9

Starting from 0 on the number line, move 6 steps to the left and from there move 3 steps to the left to reach -9 as shown below.

What fractions do the shaded parts in each of the following figures represent? Integers, R.S. Aggarwal Mathematics Solutions ICSE Class 6.

(vii) 5 - (-2) = 5 + 2 = 7

Starting from 0 on the number line, move 5 steps to the right and from there move 2 steps to the right to reach 7 as shown below.

Starting from 0 on the number line, move 5 steps to the right and from there move 2 steps to the right to reach 7 as shown below. Integers, R.S. Aggarwal Mathematics Solutions ICSE Class 6.

(viii) (-4) - 5 = -4 - 5 = -9

Starting from 0 on the number line, move 4 steps to the left and from there move 5 steps to the left to reach -9 as shown below.

Starting from 0 on the number line, move 4 steps to the left and from there move 5 steps to the left to reach -9 as shown below. Integers, R.S. Aggarwal Mathematics Solutions ICSE Class 6.

(ix) 4 - (-4) = 4 + 4 = 8

Starting from 0 on the number line, move 4 steps to the right and from there move 4 steps to the right to reach 8 as shown below.

Starting from 0 on the number line, move 4 steps to the right and from there move 4 steps to the right to reach 8 as shown below. Integers, R.S. Aggarwal Mathematics Solutions ICSE Class 6.

Question 9

State whether the given statement is true or false :

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

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

(iii) The absolute value of every integer is a positive integer.

(iv) 0 is the smallest positive integer.

(v) Every whole number is an integer.

(vi) The sum of two integers can never be zero.

Answer

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

True

Explanation:

The set of integers is closed under addition. This means that when you add any two integers, the result will always be another integer. For example, 5 + (−3) = 2, and 2 is an integer.

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

True

Explanation:

The set of integers is also closed under subtraction. When you subtract one integer from another, the result is always an integer. For example, 3−8=−5, and -5 is an integer.

(iii) The absolute value of every integer is a positive integer.

True

Explanation:

The absolute value of an integer is its distance from zero, which is always non-negative.

(iv) 0 is the smallest positive integer.

False

Explanation:

The set of positive integers starts with 1. Zero is an integer that is neither positive nor negative. The smallest positive integer is 1.

(v) Every whole number is an integer.

True

Explanation:

The set of whole numbers is {0, 1, 2, 3, ...}. The set of integers is {..., -2, -1, 0, 1, 2, ...}. All the numbers in the set of whole numbers are also included in the set of integers.

(vi) The sum of two integers can never be zero.

False

Explanation:

The sum of an integer and its additive inverse (its opposite) is always zero. For example, the sum of 5 and -5 is 0.

Question 10

What should be added to 15 to get (-15)?

Answer

Let the number be x.

∴ 15 + x = -15

⇒ x = -15 - 15

⇒ x = -30

Hence, -30 should be added to 15 to get -15.

Question 11

What should be subtracted from (-3) to get +18?

Answer

Let the number be x.

∴ -3 - x = 18

⇒ -x = 18 + 3

⇒ -x = 21

⇒ x = -21

Hence, -21 should be subtracted from (-3) to get +18.

Question 12

The sum of two integers is -23. If one of them is 12, find the other.

Answer

Given, sum of two integers = -23

One of the numbers = 12

Let x be the number.

∴ 12 + x = -23

⇒ x = -23 - 12

⇒ x = -35

Hence, the other number = -35.

Question 13

While playing children's cards, Amit lost 70 points in the first game, 50 in the second game and 35 in the third game. He gained 60 in the fourth game and 80 in the fifth game. What was his net loss or gain?

Answer

Given,

Amit lost in the first game = -70 points

Amit lost in the second game = -50 points

Amit lost in the third game = -35 points

Amit gained in the fourth game = +60 points

Amit gained in the fifth game = +80 points

Net loss or gain = (-70) + (-50) + (-35) + 60 + 80

= -70 - 50 - 35 + 60 + 80

= -120 - 35 + 60 + 80

= -155 + 60 + 80

= -95 + 80

= -15

Hence, Amit had a net loss of 15 points.

Question 14

On one day on a hill, the temperature at 8 p.m. was 2°C but at mid-night that day, it fell down to -3°C. By how many degrees did the temperature fall?

Answer

Initial temperature: 2° C

Final temperature: −3° C

Difference = Initial temperature − Final temperature

= 2° C − (−3° C)

= 2° C + 3° C

= 5° C

Hence, the temperature fell by 5° C.

Question 15

A car travelled east of Delhi by 100 km and then to the west of it by 130 km. How far from Delhi was the car finally?

Answer

Given, travelled east: +100 km

Travelled west: −130 km

Final position = (+ 100) km + (− 130) km

= 100 − 130

= −30 km

The negative sign indicates that the final position is to the west of Delhi. The distance from Delhi is the absolute value of this number.

Distance = ∣−30∣ km

Distance = 30 km

Hence, the car was finally 30 km to the west of Delhi.

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