RS Aggarwal 2021-2022 for Class 6 Maths Chapter 4 - Integer
RS Aggarwal Class 6 Math Solution Chapter 4 Integer Exercise 4B is available here. RS Aggarwal Class 6 Math Solutions are solved by expert teachers in step by step, which help the students to understand easily. RS Aggarwal textbooks are responsible for a strong foundation in Maths. These textbook solutions help students in exams as well as their daily homework routine. The solutions included are easy to understand, and each step in the solution is described to match the students’ understanding.
RS Aggarwal Class 6 Math Solution Chapter 4 Whole Number
Exercise 4B
1: Use the number line and add, the following integers:
(i) 9 + (– 6)
(ii) (– 3) + 7
(iii) 8 + ( – 8)
(iv) ( – 1 ) + ( – 3)
(v) ( – 4 ) + ( – 7)
(vi) ( – 2) + ( – 8)
(vii) 3 + ( – 2) + ( – 4)
(viii) (– 1) + (– 2) + ( – 3)
(ix) 5 + ( –2) + ( – 6)
(iii) On the number line, we start from 0 and move 8 steps to the right to reach point A.
(iv) On the number line, we start from 0 and move 1 step to the left to reach point A.
(v) On the number line, we start from 0 and move 4 steps to the left to reach point A.
∴ (5 + ( –2) + ( – 6)= – 3
2: Fill in the blanks:
(i) (– 3) + (– 9) =
(ii) (– 7) + (– 8) =
(iii) (– 9) + 16 =
(iv) ( –13 ) + 25 =
(v) 8 + ( –17) =
(vi) 2 + (– 12) =
Solution:
(i) (−3) + (−9) = −3 − 9 = −12 (ii) (−7) + (−8) = −7 − 8 = −15 (iii) (−9) + 16 = −9 + 16 = 7 (iv) (−13) + 25 = −13 + 25 = 12 (v) 8 + (−17) = 8 − 17 = −9 (v) 2 + (−12) = 2 − 12 = −10
4: Add :
(i) − 2 0 6 (ii) + 1 7 8
Solution:
(i) − 2 0 6 (ii) + 1 7 8
5: Find the sum of
(i) 137 and – 354
(ii) 1001 and – 13
(iii) – 3057 and 199
(iv) – 36 and 1027
(v) – 389 and – 1032
(vi) – 36 and 100
(vii) 3002 and – 888
(viii) – 18, + 25 and – 37
(ix) – 312, 39 and 192
(x) – 51, – 203, 36 and – 28
Solution:
= 1001 – 13
(vi) – 36 + 100
= – 36 + 100
= 64
(vii) 3002 and – 888
= 3002 + (– 888)
= 3002 – 888
= 2114
(viii) – 18, + 25 + (– 37)
= – 18 + 25 – 37
= – 55 + 25
= – 30
(ix) – 312 + 39 + 192
= – 312 + 131
= – 81
(x) – 51 + (– 203) + 36 + (– 28)
= – 51 – 203 + 36 – 28
= – 282 + 36
= – 246
6: Find the additive inverse of
(i) – 57
(ii) 183
(iii) 0
(iv) – 1001
(v) 2054
Solution:
(i) The additive inverse of – 57 is 57
(ii) The additive inverse of 183 is –183
(iii) The additive inverse of 0 is 0
(iv) The additive inverse of – 1001 is 1001
(v) The additive inverse of 2054 is – 2054
7: Write the successor of each one of the following:
(i) 201
(ii) 70
(iii) – 5
(iv) – 99
(v) – 500
Solution:
(i) Successor of 201 = 201 + 1 = 202
(ii) Successor of 70 = 70 + 1 = 71
(iii) Successor of – 5 = – 5 + 1 = – 4
(iv) Successor of – 99 = – 99 + 1 = – 98
(v) Successor of – 500 = – 500 + 1 = – 499
8: Write the predecessor of each one of the following:
(i) 120
(ii) 79
(iii) – 8
(iv) – 141
(v) – 300
Solution:
(i) Predecessor of 120 = 120 – 1 = 119
(ii) Predecessor of 79 = 79 – 1 = 78
(iii) Predecessor of – 8 = – 8 – 1 = – 9
(iv) Predecessor of – 141 = – 141 – 1 = – 142
(v) Predecessor of – 300 = – 300 – 1 = – 301
9: Simplify:
(i) ( – 7) + ( – 9) + 12 + ( – 16)
(ii) 37 + ( – 23) + ( – 65) + 9 + ( – 12)
(iii) ( – 145) + 79 + ( – 265) + ( – 41) + 2
(iv) 1056 + ( – 798) + (– 38) + 44 + (– 1)
Solution:
(i) (−7) + (−9) + 12 + (−16)
= − 7 − 9 + 12 − 16
= 12 − (7 + 9 + 16)
= 12 − 32
= −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
= 79 +2 − ( 145 + 265 + 41)
= 81 − 451
= −370
(iv) 1056 + (−798) + (−38) + 44 + (−1)
= 1056 − 798 − 38 + 44 − 1
= 1056 + 44 − (798 + 38 + 1)
= 1100 − 837
= 263
10: A car travelled 60 km to the North of Patna and then 90 km to the South from there. How far from Patha was the car finally?
Solution:
Let the direction of north be positive and the direction of south be negative.
Then, Distance travelled to the north of Patna = 60 km Distance travelled to the south of Patna = − 90 km
Total distance travelled by the car = 60 + (− 90)
= − 30 km
The car was 30 km south of Patna.
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:
Total amount spent on pencils = Rs. 30 + Rs. 25
= Rs. 55
Total amount spent on pens (C.P) = Rs. 90
Total cost price of pencils and pens (C.P) = Rs. 55 + Rs. 90 = Rs. 145
Total sale price of pencils and pens (S.P)= Rs 20 + Rs. 70
= Rs. 90
Loss = cost price – selling price
= Rs. 145 – Rs. 90
= Rs.55 Hence, his net loss was Rs 55.
12: For each of the following statements write (T) for true and (F) for false:
(i) The sum of two negative integers is always a negative integer.
(ii) The sum of a negative integer and a positive integer is always a negative integer.
(iii) The sum of an integer and its negative is zero.
(iv) The sum of three different integers can never be zero.
(v) | –5 | < | – 3 |
(vi) | 8 – 5 | = | 8 | + | – 5 |
Solution:
(i) True
(ii) False: As if a positive integer is greater then it will be positive.
(iii) True: As ( – a + a = 0 ).
(iv) False: As the sum of three integers can be zero or non-zero
(v) False: As | – 5 | = 5 and | – 3 | = 3 and 5 > 3.
(vi) False: | 8 – 5 | = | 3 | = 3 and | 8 | + | – 5 | = 8 + 5 = 13.
13: Find an integer a such that
(i) a + 6 = 0
(ii) 5 + a = 0
(iii) a + ( – 4) = 0
(iv) – 8 + a = 0
Solution:
(i) a + 6 = 0 ⇒ a = 0 − 6 ⇒ a = − 6 (ii) 5 + a = 0 ⇒ a = 0 − 5
⇒ a = − 5 (iii) a + (−4) = 0 ⇒ a = 0 − (−4) ⇒ a = 4 (iv) −8 + a = 0 ⇒ a = 0 + 8 ⇒ a = 8
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