Test: Operations on Large Numbers- 1 - Class 5 MCQ

d) 4 x 100000000 + 5 x 10000000 + 7 x 1000000 + 6 x 100000 + 9 x 10000 + 0 x 1000 + 2 x 100 + 1 x 10 + 3 x 1

Corrected Solution:

The number 45,76,90,213 has nine digits in the places from hundred millions down to ones. Multiply each digit by its place value:

4 x 100000000 + 5 x 10000000 + 7 x 1000000 + 6 x 100000 + 9 x 10000 + 0 x 1000 + 2 x 100 + 1 x 10 + 3 x 1 = 45,76,90,213.

The numbers in ascending order are:
89,72,340; 3,04,63,500; 8,36,04,295; 40,34,51,276; 99,00,32,941.

The smallest digit is 0, but we cannot start with 0. The next smallest digit is 2. So, the smallest 7-digit number is 2045689.

The greatest 7-digit number is formed by using the largest digit available, which is 9, and repeating it for all seven positions since repetition is allowed. This results in:

9999999

  14,39,64,709
+ 13,07,59,074
= 27,47,23,783
So, D is the correct option.

Option A: 99,43,460 corresponds exactly to the calculated difference when considering standard comma placement for large numbers (i.e., separating every three digits from the right).

In the Indian numbering system, 1 lakh equals 100,000. Therefore, 10 lakh is calculated as:

• 10 × 100,000 = 1,000,000

In the International system, 1,000,000 is referred to as 1 million. Thus, 10 lakh corresponds to 1 million in the International system.

The numeral in the Indian system is 69,04,41,627. In the International system, it is written as 690,441,627.