C1.9; E1.9 - Rounding and estimation

C1.9; E1.9 - Rounding and estimation

A rounded answer should keep the information the problem needs. You will round to decimal places, significant figures and named place values, make quick estimates, and decide whether a practical answer needs rounding up or down. This lesson is for Core and Extended.

Decimal places and significant figures

A calculation says that 142142 pupils need 5.9166…5.9166\ldots buses with 2424 seats each. Would reporting 5.925.92 buses solve the problem? Before choosing a practical answer, distinguish two ways to round a number.

Decimal places and significant figures

Decimal places count digits after the decimal point. For example, 2.3072.307 has three decimal places.

Significant figures count from the first nonzero digit, including later zeros. Thus 0.004070.00407 has three significant figures: 4,0,74,0,7. The zeros before 44 locate its size; they do not count. The written value 2.302.30 has three significant figures and two decimal places.

To round, locate the last digit you will keep, then inspect the next digit. If the next digit is 00 to 44, keep the retained digit unchanged. If it is 55 to 99, increase the retained digit by one, carrying if necessary. Remove later decimal digits; replace later whole-number places with zeros so the place value stays correct. For negative values, apply this rule to the magnitude and restore the minus sign.

For example, 56845684 to the nearest thousand is 60006000: the hundreds digit is 66, so round the thousands digit up. Do not write 66, because that loses the thousands place.

[DIAGRAM: asset_name: Rounding to the nearest thousand; asset_slug: rounding_thousand; caption: 5684 is above the midpoint 5500, so it is nearer to 6000 than 5000.; description: An accurately scaled number line from 5000 to 6000, with the midpoint 5500 marked and a point at 5684. The two halves have pale blue and pale sage shading. A labelled arrow links 5684 to its nearest thousand 6000.]
5684 is above the midpoint 5500, so it is nearer to 6000 than 5000.

At an exact halfway value, this rule rounds the magnitude up: 55005500 to the nearest thousand is 60006000.

Worked example 1: rounding to decimal places

Question. Write 0.047860.04786 correct to 33 decimal places.

Plan. Keep through the thousandths place and inspect the next digit.

0.04786:keep 0.047three places after the pointnext digit=8round the 7 up0.04786≈0.048discard later decimal digits\begin{aligned} 0.04786&:\quad \text{keep }0.047 && \text{three places after the point}\\ \text{next digit}&=8 && \text{round the }7\text{ up}\\ 0.04786&\approx0.048 && \text{discard later decimal digits} \end{aligned}

Answer. 0.0480.048.

Check. 0.047860.04786 is 0.000140.00014 from 0.0480.048, but 0.000860.00086 from 0.0470.047, so 0.0480.048 is nearer.

Keep the number and the requested count 33, but change decimal places to significant figures. Counting now starts at the first nonzero digit.

Worked example 2: rounding to significant figures

Question. Write 0.047860.04786 correct to 33 significant figures.

Plan. Count 4,7,84,7,8 as the first three significant digits and inspect the following 66.

0.04786:keep 0.0478first three significant digitsnext digit=6round the 8 up0.04786≈0.0479retain three significant figures\begin{aligned} 0.04786&:\quad \text{keep }0.0478 && \text{first three significant digits}\\ \text{next digit}&=6 && \text{round the }8\text{ up}\\ 0.04786&\approx0.0479 && \text{retain three significant figures} \end{aligned}

Answer. 0.04790.0479.

Check. The rounding step is 0.00010.0001. The value is 0.000040.00004 from 0.04790.0479, nearer than 0.04780.0478.

Watch out

0.047860.04786 to 33 significant figures is not 0.0480.048: that has only two significant figures. Also, keep a final zero when it communicates requested accuracy: 4.1964.196 to 22 decimal places is 4.204.20.

Count from the first nonzero digit, then inspect the next one.

Estimating calculations

An estimate replaces awkward numbers with nearby convenient values before calculating. It gives a quick idea of the size of an answer. If told to round each number to one significant figure, do exactly that; otherwise, choose values that make the arithmetic manageable and stay close to the originals.

The approximation symbol ≈\approx means “is approximately equal to”. Rounding inputs changes the calculation, so an estimated answer is not an exact equality. Estimation is useful before using a calculator: it can reveal a misplaced decimal point.

Worked example 3: estimating a product divided by a number

Question. By writing each number correct to 11 significant figure, estimate

58.4×0.2194.87\frac{58.4\times0.219}{4.87}

Plan. Round each input separately, then simplify the easier calculation.

58.4≈60,0.219≈0.2,4.87≈5one significant figure each58.4×0.2194.87≈60×0.25replace the inputs=125=2.4calculate the estimate\begin{aligned} 58.4&\approx60,\quad0.219\approx0.2,\quad4.87\approx5 && \text{one significant figure each}\\ \frac{58.4\times0.219}{4.87} &\approx\frac{60\times0.2}{5} && \text{replace the inputs}\\ &=\frac{12}{5}=2.4 && \text{calculate the estimate} \end{aligned}

Answer. An estimate is 2.42.4.

Check. The numerator is about 1212, and five groups of about 2.42.4 make 1212. An answer such as 240240 would have the wrong scale.

Change only the denominator from 4.874.87 to 0.4870.487. It is ten times smaller, so the estimated quotient should be ten times larger.

Worked example 4: a smaller denominator

Question. By writing each number correct to 11 significant figure, estimate

58.4×0.2190.487\frac{58.4\times0.219}{0.487}

Plan. Use the same rounded numerator, but now round the denominator to 0.50.5.

58.4≈60,0.219≈0.2,0.487≈0.5one significant figure each58.4×0.2190.487≈60×0.20.5preserve the denominator’s scale=12÷0.5=24dividing by a half doubles\begin{aligned} 58.4&\approx60,\quad0.219\approx0.2,\quad0.487\approx0.5 && \text{one significant figure each}\\ \frac{58.4\times0.219}{0.487} &\approx\frac{60\times0.2}{0.5} && \text{preserve the denominator's scale}\\ &=12\div0.5=24 && \text{dividing by a half doubles} \end{aligned}

Answer. An estimate is 2424.

Check. 2424 is ten times the previous estimate 2.42.4, matching the factor-of-ten change in the denominator.

An estimate need not lie above the exact answer or below it: different inputs may round in different directions. It checks plausibility, rather than proving a precise limit.

Watch out

Rounding 0.4870.487 to 11 significant figure gives 0.50.5, not 00. Significant figures start at its first nonzero digit. Do not round every number to the nearest whole number.

Make the inputs easy before calculating.

Choosing sensible accuracy

The required accuracy depends on what the answer represents. A measurement might sensibly be 5.92 m5.92\,\mathrm{m}; a number of buses must be a whole number. Sometimes you must round up to provide enough, while sometimes you must round down because only complete groups count.

When an exam question specifies accuracy, follow it. On Cambridge calculator papers, non-exact numerical answers are normally given to 33 significant figures, and angles in degrees to 11 decimal place, unless a different accuracy is requested. An exact answer such as 12.7512.75 does not need shortening merely because it has more than three significant figures. Keep the full unrounded value during a chain of calculations, then round the final answer.

For a money amount, use the currency's normal smallest unit when appropriate: a calculated price of 7.87.8 dollars is written \7.80$. For a count, ask whether a remaining fraction requires an extra whole item or cannot form a complete group.

Worked example 5: enough capacity

Question. There are 142142 pupils. Each bus has 2424 pupil seats. Work out the smallest number of buses needed.

Plan. Divide to find the number of busloads, then allow a whole bus for any pupils left over.

142÷24=5+2224five full buses leave 225×24=120<142five buses are insufficient6×24=144≥142six buses have enough seats\begin{aligned} 142\div24&=5+\frac{22}{24} && \text{five full buses leave }22\\ 5\times24&=120<142 && \text{five buses are insufficient}\\ 6\times24&=144\ge142 && \text{six buses have enough seats} \end{aligned}

Answer. 66 buses.

Check. Six buses leave 144−142=2144-142=2 spare seats. Five buses would leave 2222 pupils without seats.

Keep the same division, but change what a remainder means. This time a partial group cannot be counted.

Worked example 6: complete groups only

Question. A club has 142142 counters. Each complete game set needs 2424 counters. Work out the greatest number of complete sets it can make.

Plan. Count only whole groups of 2424; leftover counters do not make another complete set.

142÷24=5+2224five groups and a remainder5×24=120counters used for five sets142−120=22<24not enough for a sixth set\begin{aligned} 142\div24&=5+\frac{22}{24} && \text{five groups and a remainder}\\ 5\times24&=120 && \text{counters used for five sets}\\ 142-120&=22<24 && \text{not enough for a sixth set} \end{aligned}

Answer. 55 complete sets.

Check. Six sets would need 6×24=1446\times24=144 counters, two more than the club has.

Watch out

“Round every practical answer to the nearest whole number” is unsafe. Enough buses requires rounding up; complete sets requires rounding down. Decide from the situation before rounding.

The opening 5.9166…5.9166\ldots busloads therefore means 66 whole buses, not 5.925.92 buses.

Quick Recap

  • Decimal places count from the decimal point; significant figures count from the first nonzero digit.
  • Inspect the next digit to round, keeping place-value zeros and any zero needed to show the requested accuracy.
  • Estimate by replacing inputs with convenient nearby values before calculating.
  • Keep full working accuracy, then give a final answer that fits both the instruction and the context.

The opening journey needs 66 buses: a fraction of a busload still needs a whole bus.