ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 2 Whole Numbers Check Your Progress
Question 1.
Write next three consecutive whole numbers of the number 9998.
Solution:
The next three consecutive whole number of 9998 are:
9998 + 1 = 9999
9999 + 1 = 10000
10000 + 1 = 10001
∴ Numbers are = 9999, 10000, 10001
Question 2.
Write three consecutive whole numbers occurring just before 567890.
Solution:
The three consecutive whole numbers just before 567890 are:
567890 – 1 = 567889 – 1 = 567888 – 1
= 567887
∴ These are : 567889, 567888, 567887
Question 3.
Find the product of the successor and the predecessor of the smallest number of 3-digits.
Solution:
Smallest number of 3-digits = 100
Successor = 100 + 1
Predecessor =100 – 1
∴ Product = 100 + 1 × 100 – 1
= 101 × 99 = 9999
Question 4.
Find the number of whole numbers between the smallest and the greatest numbers of 2-digits.
Solution:
Smallest number of 2-digits = 10
Greatest number of 2-digits = 99
Numbers between 10 and 99
11, 12, …………, 98
= 98 – 10 = 88
Question 5.
Find the following sum by suitable arrangements:
(i) 678 + 1319 + 322 + 5681
(ii) 777 + 546 + 1463 + 223 + 537
Solution:
(i) 678 + 1319 + 322 + 5681
= (678 + 322) + (5681 + 1319)
= 1000 + 7000 = 8000
(ii) 777 + 546 + 1463 + 223 + 537
= (777 + 223) + (1463 + 537) + 546
= 1000 + 2000 + 546 = 3546
Question 6.
Determine the following products by suitable arrangements:
(i) 625 × 437 × 16
(ii) 309 × 25 × 7 × 8
Solution:
(i) 625 × 437 × 16
= 437 × (625 × 16)
= 437 × 10000 = 4370000
(ii) 309 × 25 × 7 × 8
= (309 × 7) × (25 × 8)
= 2163 × 200 = 432600
Question 7.
Find the value of the following by using suitable properties:
(i) 236 × 414 + 236 × 563 + 236 × 23
(ii) 370 × 1587 – 37 × 10 × 587
Solution:
(i) 236 × 414 + 236 × 563 + 236 × 23
= 236 × (414 + 563 + 23)
= 236 × (1000) = 236000
(ii) 370 × 1587 – 37 × 10 × 587
= 37 × 10(1587 – 587)
= 370 × 1000 = 370000
Question 8.
Divide 6528 by 29 and check the result by division algorithm.
Solution:
6528 ÷ 29
∴ Quotient = 225
and Remainder = 3
Question 9.
Find the greatest 4-digit number which is exactly divisible by 357.
Solution:
Largest 4 digit number is 9999
Dividing 9999 by 357, we get
Remainder = 3
Subtracting 3 from 9999, 9999 – 3 = 9996,
we get the required number divisible by 357.
So 9996
Question 10.
Find the smallest 5-digit number which is exactly divisible by 279.
Solution:
Smallest 5-digit number is 10000
Dividing it by 279, we get remainder = 235
To make the smallest 5-digit number exactly divisible by 279, we have to add 279 – 235 = 44 in 10000
∴ 10000 + 44 = 10044.