Operations On Whole Numbers

Operations on whole numbers are discussed here:

The four basic operations on whole numbers are addition; subtraction; multiplication and division. We will learn about the basic operations in more detailed explanations along with the examples.

Worked-out problems related to Operations on whole numbers

1. Solve using rearrangement: 

(a) 784 + 127 + 216

Solution: 

784 + 127 + 216

= (784 + 216) + 127

= 1000 + 127 

= 1127 

(b) 25 × 8 × 125 × 4 

Solution: 

25 × 8 × 125 × 4 

= (125 × 8) × (25 × 4) 

= 1000 × 100 

= 100000

2. Find the value using distributive property. 

(a) 2651 × 62 + 2651 × 38 

Solution: 

2651 × 62 + 2651 × 38 

Property: a × b + a × c = a × (b + c) 

= 2651 × (62 + 38) 

= 2651 × 100 

= 265100 

(b) 347 × 163 - 347 × 63 

Solution: 

347 × 163 - 347 × 63 

Property: a × b - a × c = a × (b - c) 

= 347 × (163 - 63) 

= 347 × 100 

= 34700 

(c) 128 × 99 + 128 

Solution: 

128 × 99 + 128 

Property: a × b - a × c = a × (b + c) 

= 128 × 99 + 128 

= 128 × (99 + 1) 

= 12800 

3. Find the product using distributive property: 

(a) 237 × 103 

Solution: 

237 × 103 

237 × (100 + 3) 

Property: a × (b + c) = a × b + a × c 

Therefore, 237 × (100 + 3) 

= 237 × 100 + 237 × 3 

= 23700 + 711

= 24411 

(b) 510 × 99 

Solution:

510 × 99

510 × (100 - 1)

Property: a × (b - c) = a × b - a × c

Therefore, 510 × (100 - 1)

= 510 × 100 - 510 × 1

= 51000 - 510

= 50490

4. Verify the following:

(a) 537 + 265 = 265 + 537

Solution:

537 + 265 =265 + 537

L.H.S. = 537 + 265 = 802

R.H.S. = 265 + 537 = 802

Property: a + b =b + a

Therefore, L.H.S. = R.H.S.

Hence, verified.

(b) 25 × (36 × 50) = (25 × 36) × 50

Solution:

25 × (36 × 50) = (25 × 36) × 50

L.H.S.= 25 × (36 × 50) = 25 × 1800 = 45000

R.H.S. = (25 × 36) × 50 = 900 × 50 = 45000

Property: a × (a × c) = (a × b) × c

Therefore, L.H.S. = R.H.S.

Hence, verified.

5. Find the least number that must be subtracted from 1000 so that 45 divides the difference exactly.

Solution:

Divide 1000 by 45.

Now 1000 - 10 = 990

Therefore, 10 should be subtracted from 1000 so that difference 990 is divisible by 45.

6. Find the least number that should be added to 1000 so that 65 divides the sum exactly.

Solution:

Divide 1000 by 65.

Now finding the difference between the divisor and remainder, we get

65 - 25 = 40

Therefore, 40 must be added to 1000 so that the sum 1040 is exactly divisible by 65.

7. Find the number which when divided by 15 gives 7 as the quotient and 3 as the remainder.

Solution:

Dividend = divisor × quotient + remainder

= 15 × 7 + 3

= 105 + 3 = 108

Therefore, the required number is 108

● Operations On Whole Numbers

● Addition Of Whole Numbers.

● Subtraction Of Whole Numbers.

● Multiplication Of Whole Numbers.

● Properties Of Multiplication.

● Division Of Whole Numbers.

● Properties Of Division.

5th Grade Math Problems 

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