EVENTS CONVENT HIGH SCHOOL
24/08/2021 CLASS-6 SESSION2021-22(SLOT-1)
Maths
Chapter-2
WHOLE NUMBERS
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Question 1.Write the next three natural numbers after 10999.
Solution:
The next three natural numbers after 10999 are 11000, 11001 and 11002.
Question 2.Write three whole numbers occurring just before 10001.
Solution:
10001 – 1 = 10000
10000 – 1 = 9999
9999 – 1 = 9998
Hence, three whole numbers just before 10001 are 10000, 9999 and 9998.
Question 3.Which is the smallest whole number?
Solution:
0 is the smallest whole number.
Question 4.How many whole numbers are there between 32 and 53?
Solution:
The whole numbers between 32 and 53 are 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52.
Question 5.Write the successor of:
(a) 2440701
(b) 100199
(c) 1099999
(d) 2345670
Solution:
(a) Successor of 244070 is 244070 + 1 = 244071
Hence, successor of 244070 is 244071.
(b) Successor of 100199 is 100199 + 1 = 100200
Hence, successor of 100199 is 100200.
(c) Successor of 1099999 is 1099999 + 1 = 1100000
Hence, successor of 1099999 is 1100000.
(d) Successor of 2345670 is 2345670 + 1 = 2345671
Hence, successor of 2345670 is 2345671
Question 6.Write the predecessor of:
(a) 94
(b) 10000
(c) 208090
(d) 7654321
Solution:
(a) Predecessor of 94 is 94 – 1 = 93
Hence, predecessor of 94 is 93.
(b) Predecessor of 1000 is 10000 – 1 = 9999
Hence, predecessor of 10000 is 9999.
(c) Predecessor of 208090 is 208090 -1 = 208089
Hence, predecessor of 208090 is 208089.
(d) Predecessor of 7654321 is 7654321 – 1 = 7654320
Hence, predecessor of 7654321 is 7654320.
Exercise 2.2
Question 1.Find the sum by suitable arrangement:
(a) 837 + 208 + 363
(b) 1962 + 453,+ 1538 + 647
Solution:
(a) 837 + 208 + 363 = (837 + 363) + 208
= 1200 + 208 [Using associative property]
= 1408
(b) 1962 + 453 + 1538 + 647
= (1962 + 1538) + (453 + 647)
= 3500 + 1100 = 4600
Question 2.Find the product by suitable arrangement:
(а) 2 x 1768 x 50
(b) 4 x 166 x 25
(c) 8 x 291 x 125
(d) 625 x 279 x 16
(e) 285 x 5 x 60
(f) 125 x 40 x 8 x 25
Solution:
(a) 2 x 1768 x 50 = (2 x 50) x 1768 = 176800
(b) 4 x 166 x 25 = 166 x (25 x 4) = 166 x 100 = 16600
(c) 8 x 291 x 125 = (8 x 125) x 291 = 1000 x 291 = 291000
(d) 625 x 279 x 16 = (625 x 16) x 279 = 10000 x 279 = 2790000
(e) 285 x 5 x 60 = 285 x (5 x 60) = 285 x 300 = (300 – 15)x 300 = 300 x 300 – 15 x 300 = 90000 – 4500 = 85500
(f) 125 x 40 x 8 x 25 = (125 x 8) x (40 x 25) = 1000 X 1000 = 1000000
Question 3.Find the value of the following:
(а) 297 x 17 + 297 x 3
(б) 54279 x 92 + 8 x 54279
(c) 81265 x 169 – 81265 x 69
(d) 3845 x 5 x 782 + 769 x 25 x 218
Solution:
(a) 297 x 17 x 297 x 3 = 297 x (17 + 3)
= 297 x 20 = 297 x 2 x 10
= 594 x 10 = 5940
(b) 54279 x 92 + 8 x 54279 = 54279 x (92 + 8)
= 54279 x 100 = 5427900
(c) 81265 x 169 – 81265 x 69
= 81265 x (169 – 69)
= 81265 x 100 = 8126500
(d) 3845 x 5 x 782 + 769 x 25 x 218 = 3845 x 5 x 782 + 769 x 5 x 5 x 218
= 3845 x 5 x 782 + (769 x 5) x 5 x 218
= 3845 x 5 x 782 + 3845 x 5 x 218
= 3845 x 5 x 782 + 3845 x 5 x 218
= 3845 x 5 x (782 + 218)
= 3845 x 5 x 1000
= 19225 x 1000
= 19225000
