4.01b - Estimating calculations

4.01b - Estimating calculations

An estimate can expose a misplaced decimal point before you trust a calculator answer. You will choose manageable approximations, estimate expressions with several operations and roots, and explain what an estimate can tell you.

Choosing helpful approximations

Could three kilograms of fruit costing about two pounds per kilogram really cost sixty pounds? A quick estimate tells you the scale before you do the exact calculation.

Suppose the mass is 2.9 kg2.9\,\mathrm{kg} and the price is £1.961.96 per kilogram. Replacing these by 33 and 22 gives about £66. The exact cost will be near that, so £6060 deserves another look.

Estimate

An estimate is an approximate value found using simpler numbers or information. The symbol ≈\approx means “is approximately equal to”. An estimate is useful for judging size; it is not automatically an upper or lower bound.

Use the accuracy specified in the question. If it asks you to round each number to one significant figure, do that before calculating. Otherwise choose nearby numbers that make the calculation useful and manageable. For example, 198÷49198\div49 is easily estimated by 200÷50200\div50, while rounding 0.0490.049 to zero would destroy the division.

Worked example 1: a price estimate

Question. By rounding each number to one significant figure, estimate 2.9×1.962.9\times1.96.

Plan. Round both factors, then multiply.

2.9≈3one significant figure1.96≈2one significant figure2.9×1.96≈3×2=6multiply the approximations\begin{aligned} 2.9&\approx3 &&\text{one significant figure}\\ 1.96&\approx2 &&\text{one significant figure}\\ 2.9\times1.96&\approx3\times2=6 &&\text{multiply the approximations} \end{aligned}

Answer. An estimate is 66.

Check. Both positive factors were increased, so the exact product is below 66. It is still above 2.5×1.9=4.752.5\times1.9=4.75, consistent with this scale.

Change only the second factor's size. Keeping its decimal place value matters much more than keeping many digits.

Worked example 2: a small factor

Question. By rounding each number to one significant figure, estimate 2.9×0.1962.9\times0.196.

Plan. Keep the first significant digit of the small factor, then multiply.

2.9≈3round the first factor0.196≈0.2the first significant digit is tenths2.9×0.196≈3×0.2=0.6three lots of two tenths\begin{aligned} 2.9&\approx3 &&\text{round the first factor}\\ 0.196&\approx0.2 &&\text{the first significant digit is tenths}\\ 2.9\times0.196&\approx3\times0.2=0.6 &&\text{three lots of two tenths} \end{aligned}

Answer. An estimate is 0.60.6.

Check. The second factor is one tenth of the previous one, so the new product and its estimate should also be one tenth as large.

An approximation sign belongs where you change a value. Once you have chosen the rounded numbers, the arithmetic on those numbers can use equals signs.

Watch out

Do not write 2.9×0.196=0.62.9\times0.196=0.6. The original product is not exactly 0.60.6. Write 2.9×0.196≈3×0.2=0.62.9\times0.196\approx3\times0.2=0.6.

Keep the small factor's scale in the next estimate.

Estimating products and quotients

For an expression with several operations, keep its structure while replacing its inputs.

A fraction bar groups the whole numerator and the whole denominator. In a+bc\frac{a+b}{c}, add the numerator before dividing; in abcd\frac{ab}{cd}, both denominator factors divide the numerator. Round inputs before calculating, rather than trying to round an exact answer you have not found.

Worked example 3: a product over a divisor

Question. By rounding each number to one significant figure, estimate 5.8×390.21\frac{5.8\times39}{0.21}.

Plan. Round the three inputs and simplify the resulting quotient.

5.8×390.21≈6×400.2round each input=240×5dividing by a fifth multiplies by 5=1200finish the arithmetic\begin{aligned} \frac{5.8\times39}{0.21}&\approx\frac{6\times40}{0.2} &&\text{round each input}\\ &=240\times5 &&\text{dividing by a fifth multiplies by 5}\\ &=1200 &&\text{finish the arithmetic} \end{aligned}

Answer. An estimate is 12001200.

Check. A numerator near 240240 divided by a positive number below 11 must increase. An answer such as 4848 would have the wrong scale.

Now change only the numerator operation from multiplication to addition. The rounded numbers are the same, but the calculation is quite different.

Worked example 4: a sum over a divisor

Question. By rounding each number to one significant figure, estimate 5.8+390.21\frac{5.8+39}{0.21}.

Plan. Round, add the whole numerator, then divide.

5.8+390.21≈6+400.2preserve the addition=460.2evaluate the numerator=230multiply 46 by 5\begin{aligned} \frac{5.8+39}{0.21}&\approx\frac{6+40}{0.2} &&\text{preserve the addition}\\ &=\frac{46}{0.2} &&\text{evaluate the numerator}\\ &=230 &&\text{multiply 46 by 5} \end{aligned}

Answer. An estimate is 230230.

Check. 230×0.2=46230\times0.2=46, which reconstructs the rounded numerator.

The digits alone do not identify the method: the operation signs and grouping do.

Watch out

6+400.2\frac{6+40}{0.2} is not 6+400.26+\frac{40}{0.2}. The fraction bar makes both terms part of the numerator, so the correct value is 230230, not 206206.

Use the complete numerator before dividing.

Estimating roots and checking scale

For a square root, choose a nearby square number or simplify the expression inside the root to a familiar square.

Since 72=497^2=49, 48\sqrt{48} is a little below 77. The square-root sign means the non-negative square root: 49=7\sqrt{49}=7, not ±7\pm7. For non-negative inputs, increasing the input increases its square root. This lets you check which side of your estimate the true answer lies on.

Worked example 5: a root of a quotient

Question. Estimate 3.90.041\sqrt{\frac{3.9}{0.041}} without a calculator.

Plan. Use nearby inputs that make the value inside the root a perfect square.

3.90.041≈40.04use manageable nearby values=100evaluate inside the root=10take the non-negative root\begin{aligned} \sqrt{\frac{3.9}{0.041}}&\approx\sqrt{\frac{4}{0.04}} &&\text{use manageable nearby values}\\ &=\sqrt{100} &&\text{evaluate inside the root}\\ &=10 &&\text{take the non-negative root} \end{aligned}

Answer. An estimate is 1010.

Check. The exact numerator is smaller than 44 and its denominator larger than 0.040.04. The exact quotient is below 100100, so its root is below 1010.

Change only the denominator's scale by a factor of one hundred. The square root changes by a factor of ten, a useful connection between place value and powers.

Worked example 6: checking a decimal-point error

Question. A calculator answer to 3.90.00041\sqrt{\frac{3.9}{0.00041}} has been copied as 9.7539.753. Use an estimate to decide whether this is plausible.

Plan. Estimate the expression inside the root and compare the scales.

3.90.00041≈40.0004use nearby values=10000divide before taking the root=100since 1002=10000\begin{aligned} \sqrt{\frac{3.9}{0.00041}}&\approx\sqrt{\frac{4}{0.0004}} &&\text{use nearby values}\\ &=\sqrt{10000} &&\text{divide before taking the root}\\ &=100 &&\text{since }100^2=10000 \end{aligned}

Answer. 9.7539.753 is implausible: the answer should be near 100100.

Check. 9.75329.753^2 is below 102=10010^2=100, whereas the original quotient exceeds 3÷0.001=30003\div0.001=3000. The copied value cannot be its square root.

A plausible scale does not prove every digit is correct. Estimation is a filter: it catches a cost near £6060 when the opening fruit calculation should be near £66, but it cannot distinguish every pair of nearby answers.

Key idea

A good estimate keeps the operations and the size of the answer visible. For roots, estimate inside the root first, then take the root.

Use an estimate to assess a reported answer, rather than to reproduce it.

Quick Recap

  • Choose useful nearby numbers, or use the accuracy specified.
  • Keep the operations and grouping of the original expression.
  • Use ≈\approx when replacing a value by an approximation.
  • Estimate inside a square root, then take the root.

The fruit should cost about £66. An estimate can reveal a tenfold error before a detailed calculation.