## 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.

## Leave a Reply