## RS Aggarwal Class 6 Math Fourth Chapter Integers Exercise 4B Solution

## EXERCISE 4B

**(1) Use the number line and add the following integers:**

(i) 9 + (-6)

= 9 – 6

= 3

(ii) (- 3)+ 7

= 7 – 3

= 4

(iii) 8 + (-8)

= 8 – 8

= 0

(iv) (-1) + (-3)

= -1 – 3

= – 4

(v) (-4) + (-7)

= – 4 – 7

= – 11

(vi) (-2) + (-8)

= – 2 – 8

= – 10

(vii) 3 + (-2) + (-4)

= 3 – 2 – 4

= 3 – 6

= – 3

(viii) (-1) + (-2) +(-3)

= -1 -2 -3

= -3 -3

= -6

(ix) 5 + (-2) + (-6)

= 5 -2 -6

= 5 -8

= -3

**(2) Fill in the blanks:**

(i) (-3) +(-9)

= -3 -9

= -12

(ii) (-7) + (-8)

= -7 -8

= -15

(iii) (-9) + 16

= -9 +16

= 5

(iv) (-13) + 25

= -13 +25

= 12

(v) 8 + (-17)

= 8 -17

= -9

(vi) 2 + (-12)

= 2 -12

= -10

**(3) Add:**

**(4) Add:**

**(5) Find the sum of**

(i) 137 and -354

= 137 + (-354)

= 137 -354

= – 217

(ii) 1001 and -13

= 1001 + (-13)

= 1001 -13

= 988

(iii) – 3057 and 199

= -3057 + 199

= – 2858

(iv) -36 and 1027

= -36 + 1027

= 991

(v) -389 and -1032

= -389 + (-1032)

= – 389 – 1032

= – 1421

(vi) -36 and 100

= -36 + 100

= 64

(vii) 3002 and -888

= 3002 + (-888)

= 2114

(viii) -18, +25 and -37

= -18 +25 +(-37)

= 25 -18 -37

= 25 – 55

= – 30

(ix) -312, 39 and 192

= -312 + 39 + 192

= -312 + 231

= – 81

(x) -51, -203, 36 and -28

= -51 +(-203) + 36+ (-28)

= -51 -203 – 28 +36

= – 282 + 36

= -246

(6) Find the additive inverse of

(i) -57

= 57

(ii) 183

= – 183

(iii) 0

= 0

(iv) – 1001

= 1001

(v) 2054

= – 2054

**(7) Write the successor of each one of the following:**

(i) 201

= 201 +1

= 202

(ii) 70

= 70 + 1

=71

(iii) – 5

= -5 + 1

= – 4

(iv) – 99

= -99 + 1

= – 98

(v) – 500

= -500 + 1

= – 499

**(8) Write the predecessor of each one of the following:**

(i) 120

= 120 – 1

= 119

(ii) 79

= 79 – 1

= 78

(iii) -8

= – 8 – 1

= – 9

(iv) – 141

= -141 – 1

= – 142

(v) – 300

= -300 – 1

= – 301

**(9) Simplify:**

(i) (-7) + (-9) + 12 + (-16)

= -7 -9 +12 -16

=[ -(7 + 9+ 16)] + 12

= (- 32) +12

= -32 +12

= – 20

(ii) 37 + (-23) + (-65) + 9 +(-12)

= 37 -23 -65 +9-12

= (37+ 9) – (23+ 65+ 12)

= 46 – 100

= – 54

(iii) (-145) +79 + (-265)+ (-41) + 2

= – 145 + 79 – 265 – 41 +2

= – (145+ 265+41) + (79+ 2)

= – 451 + 81

= – 370

(iv) 1056 + (-798) +(-38) + 44+ (-1)

= 1056 – 798 – 38 + 44 – 1

= (1056+ 44)- (798+38+1)

= 1100 – 837

= 263

**(10) A car traveled 60 km to the north of Patna and then 90 km to the south from there. How far from Patna was the car finally?**

= 60 – 90

= – 30

Ans: The car finally travelled30 km south.

**(11) A man bought some pencils for Rs 30 and some pens for Rs 90. The next day, he again bought some pencils for Rs 25. Then, he sold all the pencils for Rs 20 and the pens for Rs 70. What was his net gain or loss?**

Solution: 30 + 90+ 25 – (20+ 70)

= 145 – 90

= 55

Ans: The man loss 55 rupees.

**(12) For each of the following statement write (T) for true and (F) for false:**

(i) The sum of two negative integers is always a negative integers.

Ans: T

(ii) The sum of a negative integer and a positive integers is always a negative integer.

Ans: F

(iii) The sum of an integer and its negative is zero.

Ans: T

(iv) The sum of three different integers can never be zero.

Ans: F

(v)

-5<-3

Ans: F

(vi)

8-5=8+-5

Ans: F

(13) Find an integer a such that

Thanks