# NCERT Class 6 Solutions: Whole Numbers (Chapter 2) Exercise 2.2–Part 1

• Previous

Q-1 Find the sum by suitable rearrangement:

Solution:

The idea is to pair the numbers so that their least significant (right-most) digits add to either 5 or better still to 10. For example when asked to add 11 + 12 + 19. First do 11 + 19 = 30 (notice how the total has 0 in right-most digit). Now it is easy to 30 + 12 = 42.

• Rearrange the sum,

Q-2 Find the product by suitable arrangement

Solution:

The idea is to pair the numbers so that a number ending in 5 is multiplied with an even number. This would produce 0 in the least significant. For example when asked to calculate . First do = 180 (notice how the total has 0 in right-most digit). Now it is easy to do = 1260 (we could ignore the zero, then multiply and then add back the zero).

1. Note here we choose to multiply 4 with 25 (even though 166 was also even), this was because we got 100 makes it very easy to multiply further.

2. Here 8 is even.

3. Here 16 is even.

4. Here 60 is the even number.

5. We know that product of 8 and 125 is 1000 and so we make a pair out of them. We also know that 4 multiplied with 25 would be 100 so product of 40 and 25 would also be 1000.

Q-3 Find the value of the following:

Solution:

You can distribute multiplication over addition and subtraction:

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