Using your knowledge of Rounding Numbers, you can develop a "feel" for
the numbers involved in a calculation, and estimate the general range in which the answer
lies, before actually carrying out the calculation itself. This skill obviously enables
to reduce errors in your work.
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Addition
Example
Add
£ 3.90, £ 4.26, £ 13.29
Roughly speaking, this would be equal to
4 + 4 + 13 = 21
Compare this with the true answer of £ 21.45
So you can see that using this procedure (having a "feel" for the numbers, estimating
the answer beforehand), we can at least eliminate totally wrong answers.
If you had placed the decimal point in the wrong place in the one of the numbers, this
would be picked up.
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Multiplication
Example
22.1 x 5.2
Roughly speaking, this would be equal to
22 x 5 = 110
Compare this with the correct value of 114.92
Example
27.18 x 28.39
Roughly speaking, this would be equal to
30 x 30 = 900
Compare this with the correct value of 771.6402, which is not too close
to the estimation, but near enough to be confident that it could
indeed be the correct answer.
If you are confident with decimal numbers, it might be possible to approximate
to an appropriate non-whole number (i.e. a number with figures to the right of the decimal
point.
| Original Problem | Approximated by
|
| 10.89 x 2.53 | 11 x 2.5 = 27.5
|
1.456 x 16.3 | 1.5 x 16 = 24
|
| 7.79 x 16.12 | 7.75 x 16 = 124
|
The above procedure, although useful, is limited to use in a small number of cases.
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Division
Example
123 / 5.8
can be approximated by
123 / 6
which is approximately
20 or 21
So we know the result of 123 / 5.8
will be close to 20 or 21 (the real figure will be slightly higher). Therefore we know the
approximate value of the answer before we have started, and this will help to exclude any
totally-incorrect calculations.
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Fractions
Sometimes, a "difficult" fraction can be approximated by an "easier" fraction.
For example, a fraction like
17 / 168
could be estimated as
1 / 10
Sometimes a decimal can be approximated by an fraction, making estimation easier (you might
need to consult the fractions and/or decimal modules before reading this).
Examples
| Decimal | Approximated by
|
| 0.57 | 1/2 (0.50)
|
0.26 | 1/4 (0.25)
|
| 0.129 | 1/8 (0.125)
|
Examples of Calculation
| Original Problem | Approximated by
|
| 24.6 x 0.56 | 25 x (1/2) = 12.5
|
42.3 x 0.12 | 42 x (1/10) = 4.2
|
| 51.21 x 0.027 | 50 x (1/40) = 1.25
|
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Expressions involving a mixture of Operations
Example
Even something as 'complicated' as
( 48.3 x 34.29 )
¸
( 9.73 x
4.63 )
can be approximated by
( 50 x 35 ) / ( 10 x 5 ) = 35
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More complicated procedures
Example
What about
7.23 x 0.032
If you are fully happy with the decimal system, then
- Using a knowledge of the decimal system, you could recognize that multiplication by
0.01 shifts
the decimal point in 7.23 by two places to the left, i.e. 0.0723, approximately 0.07.
- Since 0.032 is approximately 3 times 0.1, multiply 0.07 by 3 to give
approximately 0.21.
-
(Compare this with the true value of 0.23136)
0.093 x 0.059
-
Recognize that multiplying 0.093 by 0.01 will shift the decimal point two places to the
left, producing 0.00093
- Since 0.059 is approximately 6 times the value of 0.01, multiply 0.0009 (0.00093 from above, but
rounded for simplicity)
by 6, to produce 0.0054
- (Compare this with the real value of 0.005487)
- You obviously need to have a good understanding of the decimal system to use this method.
Sometimes it is easier to use the methods outlined above, under 'Fractions'.
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Past Test Questions
Chris buys 4 CDs at £13.99 each. He gives the cashier £60.
Which method would give Chris the closest estimate to let him check his change?
- A 60 - 13 x 4
- B 60 - 4 - 13
- C 60 - 4 + 14
- D 60 - 4 x 14
Cathy buys 13 T-Shirts that cost £ 9.89 each.
She estimates the total cost in four ways. Each way gives a different result. Which
of the estimates below is closest to the total cost ?
- A £ 130.00
- B £ 128.70
- C £ 127.60
- D £ 123.50
Jess has to quickly work out an estimate of
784 / (39.2 × 0.47)
in her head.
Show a method which she could use.
Use it to give an estimate for the value of the calculation.
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