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
● Subtraction Of Whole Numbers.
● Multiplication Of Whole Numbers.
● Properties Of Multiplication.
5th Grade Math Problems
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