C1.9; E1.9 - Estimation and sensible accuracy

C1.9; E1.9 - Estimation and sensible accuracy

A calculator reports that 48.6 kilograms at 19.70perkilogramwouldcostabout19.70 per kilogram would cost about 96. Does the scale look right? An estimate can expose an entry error before you trust the digits. This lesson develops the calculation estimates, scale checks and sensible final accuracy required by C1.9 and E1.9.

Estimating calculations

An estimate replaces awkward inputs with nearby useful values, then calculates with those values. For example, 19.6×3.819.6\times3.8 is close to 20×420\times4, so its value should be around 80.

Estimate

An estimate is an approximate result obtained from simpler values. Use ≈\approx, rather than ==, when you replace an exact value with an approximation.

If a question says to round each input to one significant figure, follow that instruction and show the rounded expression. Keep the original operations and brackets. Estimating is a method of calculation, rather than an instruction to calculate exactly and round the final answer.

Worked example 1: a product estimate

Question. By writing each number correct to one significant figure, estimate 19.6×3.819.6\times3.8.

Plan. Round the inputs first, then multiply mentally.

19.6≈203.8≈419.6×3.8≈20×4=80\begin{aligned} 19.6&\approx20\\ 3.8&\approx4\\ 19.6\times3.8&\approx20\times4=80 \end{aligned}

Answer. An estimate is 8080.

Check. Both positive inputs were rounded up, so 80 should be above the exact product. Indeed, 19.6<2019.6<20 and 3.8<43.8<4 confirm this without calculating the original product.

Now add a denominator. The rounding decisions stay the same, but the fraction bar requires the whole denominator to be evaluated before division.

Worked example 2: a fraction with a denominator sum

Question. By writing each number correct to one significant figure, estimate

19.6×3.80.49+0.31.\frac{19.6\times3.8}{0.49+0.31}.

Plan. Round all four inputs, keeping the numerator product and denominator sum grouped.

19.6≈20,3.8≈40.49≈0.5,0.31≈0.319.6×3.80.49+0.31≈20×40.5+0.3=800.8=100\begin{aligned} 19.6&\approx20,\quad3.8\approx4\\ 0.49&\approx0.5,\quad0.31\approx0.3\\ \frac{19.6\times3.8}{0.49+0.31} &\approx\frac{20\times4}{0.5+0.3}\\ &=\frac{80}{0.8}=100 \end{aligned}

Answer. An estimate is 100100.

Check. The rounded denominator is below 1, so dividing 80 by it must increase 80. Also, 100×0.8=80100\times0.8=80.

Without a prescribed rounding rule, choose nearby values that make the arithmetic useful. An estimate need not be the closest possible approximation, but it must preserve enough scale to answer the question.

Watch out

Do not change a sum into a product while simplifying. Here the denominator is 0.5+0.3=0.80.5+0.3=0.8, not 0.5×0.30.5\times0.3. Write the rounded expression before evaluating it.

Show all the rounded inputs in your own estimate.

Checking scale

An estimate tells you whether a reported answer has a plausible size. It is particularly good at exposing an omitted zero, a misplaced decimal point, an incorrect sign or division entered as multiplication. A close estimate supports an answer's scale; it does not prove every digit is correct.

Worked example 3: finding a factor-of-ten error

Question. A calculator calculation of 48.6×19.748.6\times19.7 is reported as 95.742. Use an estimate to assess the result, then find the correct value.

Plan. Estimate the product, compare its size with the report, then enter the calculation again.

48.6×19.7≈50×20=100048.6×19.7=957.42correct entry\begin{aligned} 48.6\times19.7&\approx50\times20=1000\\ 48.6\times19.7&=957.42 &&\text{correct entry} \end{aligned}

95.742 is near 100, rather than 1000. It is ten times too small.

Answer. The reported result is implausible; the correct product is 957.42957.42.

Check. Independently, 48.6(20−0.3)=972−14.58=957.4248.6(20-0.3)=972-14.58=957.42. Both positive inputs are slightly below the rounded ones, so the exact product should be just below 1000.

Keep the same inputs and change multiplication to division. The estimate must now become a quotient too.

Worked example 4: a division result

Question. A calculator calculation of 48.6÷19.748.6\div19.7 is reported as 24.6701. Use an estimate to assess the result, then calculate the quotient to three significant figures.

Plan. Compare with 50÷2050\div20, then re-enter the division.

48.6÷19.7≈50÷20=2.548.6÷19.7=2.467005…≈2.47three significant figures\begin{aligned} 48.6\div19.7&\approx50\div20=2.5\\ 48.6\div19.7&=2.467005\ldots\\ &\approx2.47 &&\text{three significant figures} \end{aligned}

Answer. 24.6701 is implausible; the quotient is 2.472.47 to three significant figures.

Check. 19.7×2=39.419.7\times2=39.4 and 19.7×3=59.119.7\times3=59.1, so the quotient must lie between 2 and 3. The reported value fails this independent bound.

For the opening question, 48.6×19.748.6\times19.7 should be near 1000, so a value near $96 has the wrong scale. The exact product confirms the diagnosis.

Key idea

A plausible estimate is a warning system. If an answer fails it, check the entry and units. If an answer passes it, still check the operation and any requested accuracy.

Use a scale argument rather than trusting the number of displayed digits.

Contextual rounding

Final accuracy depends on what an answer represents. A payment usually needs cents, a count of vehicles must be a whole number that meets the need, and a measured length should not pretend to be more precise than the task allows. Keep the full working value until you make that final decision.

Key idea

Follow any stated accuracy first. On calculator papers, otherwise give non-exact numerical answers to three significant figures, or degree angles to one decimal place. Preserve an exact answer unless the question asks for rounding; later topics may request exact answers in π\pi or surd form.

An exact result such as 12.48 is not automatically shortened to three significant figures. For money, the written $12.48 is also the exact amount in cents.

Worked example 5: a price per litre

Question. A container holds 3 L3\,\mathrm{L} and costs $23.47. Calculate the price per litre, giving your answer to the nearest cent.

Plan. Divide the total cost by the number of litres, then round the final price.

price per litre=23.47÷3=7.823333…≈7.82nearest cent\begin{aligned} \text{price per litre}&=23.47\div3\\ &=7.823333\ldots\\ &\approx7.82 &&\text{nearest cent} \end{aligned}

Answer. $7.82 per litre.

Check. 3×7.82=23.463\times7.82=23.46, only one cent below the original total; the difference comes from rounding the unit price.

A whole-number decision changes the rule. Ordinary rounding can leave some people without transport.

Worked example 6: enough vehicles

Question. 74 people need transport. Each minibus holds 12 people. Find the minimum number of minibuses needed.

Plan. Divide by capacity, then choose the next whole number that accommodates everyone.

74÷12=6.1666…6×12=72two people would be left7×12=84enough places\begin{aligned} 74\div12&=6.1666\ldots\\ 6\times12&=72 &&\text{two people would be left}\\ 7\times12&=84 &&\text{enough places} \end{aligned}

Answer. 7 minibuses.

Check. Six are insufficient and seven are sufficient, so seven is the minimum.

For a different whole-number context, round the other way: 26 m26\,\mathrm{m} of cord makes only 14 complete 1.8 m1.8\,\mathrm{m} pieces, since 26÷1.8=14.444…26\div1.8=14.444\ldots. Fifteen pieces would need 27 m27\,\mathrm{m}. Read what is being counted.

Worked example 7: showing a rounded measurement

Question. A 7.36 m7.36\,\mathrm{m} ribbon is cut into five equal lengths. Show that each length is 1.47 m1.47\,\mathrm{m}, correct to three significant figures.

Plan. Calculate the length with more figures than the stated answer, then show the rounding.

length=7.365=1.472 m1.472 m≈1.47 mfourth significant digit is 2\begin{aligned} \text{length}&=\frac{7.36}{5}=1.472\,\mathrm{m}\\ 1.472\,\mathrm{m}&\approx1.47\,\mathrm{m} &&\text{fourth significant digit is 2} \end{aligned}

Answer. Each length is 1.47 m1.47\,\mathrm{m} to three significant figures.

Check. Before rounding, 5×1.472=7.365\times1.472=7.36. The extra digit demonstrates why the stated rounded value is correct.

When a value feeds into a later calculation, use its unrounded stored value. For instance, 10÷3=3.333…10\div3=3.333\ldots, and tripling that exact quotient gives 10; tripling a prematurely rounded 3.33 gives 9.99. Write enough working to show your method, especially for “Show that” tasks.

Watch out

Round only after the calculation is finished. A request to show a result correct to three significant figures needs a value with at least one more significant figure before the final rounding; writing only the given answer does not demonstrate it.

Decide the needed whole number from the situation.

Quick Recap

  • Estimate by simplifying inputs while preserving operations and grouping.
  • Use an estimate to detect implausible scale, then check the actual calculation.
  • Keep full working precision and round only the final required value.
  • Money, capacity and complete-piece contexts can require different final rounding decisions.

The opening cost should be around 1000,soareportedvaluearound1000, so a reported value around 96 deserves an immediate entry check.