FN14–FN15, HN14–HN15 - Rounding and estimation

FN14–FN15, HN14–HN15 - Rounding and estimation

Learn to round numbers without losing their place value, and to estimate calculations before trusting a calculator display. FN15 and HN15 cover decimal places and significant figures; FN14 and HN14 use those ideas to judge whether an answer has a sensible size.

Decimal places

A calculator gives 100.7100.7 for 49.7×6.12÷0.30249.7\times6.12\div0.302. Could that be right? We will build a quick check, starting with what rounding changes and what it keeps.

Rounding replaces a number by a nearby value at a chosen place. For example, 73647364 lies between the neighbouring hundreds 73007300 and 74007400. It is 6464 above 73007300 but only 3636 below 74007400, so it rounds to 74007400 to the nearest hundred. Keep the zeros: 7474 would be a different size.

Decimal places

The number of decimal places is the number of digits written after the decimal point. To round to two decimal places, choose the nearest hundredth: for example, 4.2764.276 becomes 4.284.28.

Locate the last place you must keep. The next digit decides which neighbouring value is nearer: 00 to 44 leaves the retained digit unchanged; 55 to 99 increases it by one. For the positive numbers here, an exact halfway value goes to the larger neighbour. Delete later decimal digits, but keep any zeros needed to show the requested decimal places.

Worked example 1: rounding down to a hundredth

Question. Write 6.2946.294 correct to two decimal places.

Plan. Keep the hundredths and use the thousandths digit to choose the nearer hundredth.

6.294−6.29=0.004distance from the lower hundredth6.30−6.294=0.006distance from the upper hundredth\begin{aligned} 6.294-6.29&=0.004 &&\text{distance from the lower hundredth}\\ 6.30-6.294&=0.006 &&\text{distance from the upper hundredth} \end{aligned}

The next digit is 44, so the lower neighbour is nearer.

Answer. 6.296.29.

Check. 6.2946.294 is below the midpoint 6.2956.295, agreeing with the decision.

Change only the final digit. The number is now exactly halfway, so the tie rule matters.

Worked example 2: halfway and a carry

Question. Write 6.2956.295 correct to two decimal places.

Plan. Compare the neighbouring hundredths 6.296.29 and 6.306.30, using the halfway rule.

6.295−6.29=0.0056.30−6.295=0.005\begin{aligned} 6.295-6.29&=0.005\\ 6.30-6.295&=0.005 \end{aligned}

The distances are equal. Choose 6.306.30, the larger neighbour. Increasing the 99 in the hundredths makes a carry into the tenths.

Answer. 6.306.30 to two decimal places.

Check. The value is between 6.296.29 and 6.306.30, exactly at their midpoint. The final zero records the requested hundredths accuracy.

For a negative number, still choose the nearest value: −4.283-4.283 is nearer −4.28-4.28 than −4.29-4.29. A halfway negative value needs a stated tie convention; the positive halfway rule above is all that the measurement examples in this lesson need.

Round directly from the original value. For example, 2.3492.349 goes straight to 2.32.3 to one decimal place. Going first to 2.352.35, then to 2.42.4, gives the wrong result.

Watch out

Writing 6.36.3 instead of 6.306.30 keeps the numerical value but hides the requested two decimal places. When the question specifies decimal places, show them, including a final zero if needed.

Use the next digit once, working from the original number.

Significant figures

Decimal places count from the decimal point; significant figures count from the first nonzero digit. This lets you describe the accuracy of very small and very large numbers in the same way.

Significant figures

The significant figures of a written nonzero number start at its first nonzero digit. Count every following digit that is retained, including zeros between digits and zeros at the end of a decimal.

In 0.004500.00450, the leading zeros locate the digits but do not count as significant figures. The digits 4,5,04,5,0 give three significant figures. In 4.0504.050, all four digits count: the internal zero and the final decimal zero both belong to the stated precision.

Written numberSignificant figuresWhy
0.00620.006222Start at the 66, not the decimal point.
0.006200.0062033The final zero records an extra retained place.
6.0206.02044Zeros after the first nonzero digit count.

To round, find the last significant figure required and inspect the next digit. After the decision, preserve the number's size. For example, 52 68452\,684 to two significant figures is 53 00053\,000, not 5353. Whole-number trailing zeros alone can leave the intended precision unclear, so write “53 00053\,000 (2 significant figures)” if it needs stating explicitly.

Worked example 3: ignoring leading zeros

Question. Write 0.0040560.004056 correct to three significant figures.

Plan. Count from the 44, keeping 4,0,54,0,5, then inspect the next digit.

The third significant figure is the 55 in the hundred-thousandths place. The next digit is 66, so increase that 55 to 66.

0.004056≈0.004060.004056\approx0.00406

Answer. 0.004060.00406.

Check. The answer has three significant figures, 4,0,64,0,6, and differs from the original by only 0.0000040.000004.

The next number changes one of the retained digits to 99. Rounding now makes a carry, so a final zero becomes essential.

Worked example 4: keeping a significant zero

Question. Write 0.0040960.004096 correct to three significant figures.

Plan. Keep 4,0,94,0,9, then round using the following 66.

0.004096≈0.00410round to the nearest 0.00001\begin{aligned} 0.004096&\approx0.00410 &&\text{round to the nearest }0.00001 \end{aligned}

Increasing the 99 carries into the preceding place. The final zero shows that the third significant figure is still retained.

Answer. 0.004100.00410.

Check. Its significant figures are 4,1,04,1,0. The original is 0.0000040.000004 below 0.004100.00410, but 0.0000060.000006 above 0.004090.00409, so the answer is the nearer neighbour.

The two ways of reporting accuracy can give different answers. To two decimal places, 0.0040960.004096 is 0.000.00; to two significant figures, it is 0.00410.0041. Always read which is requested.

Watch out

0.004100.00410 and 0.00410.0041 have the same value, but the first explicitly shows three significant figures and the second shows two. Leading zeros locate the first significant digit; they do not use up the requested count.

Check both the starting digit and the final written precision.

Estimating calculations

An estimate is a deliberately simpler calculation that gives the approximate size of an answer. It can catch a misplaced decimal point, a wrong operation or a calculator entry error.

Estimate

An estimate is an approximate value found using sensible simplifications. Write ≈\approx when rounded inputs make the result approximate, rather than claim exact equality.

Rounding each input to one significant figure is often a useful start, but it is not a rule to round every number to the nearest whole number. A denominator such as 0.3020.302 should become 0.30.3, not zero. If an instruction specifies the rounding to use, follow it; otherwise choose nearby values that make the arithmetic manageable while keeping its scale.

Worked example 5: a product estimate

Question. Estimate 49.7×6.1249.7\times6.12 by rounding each number to one significant figure.

Plan. Replace each factor by a nearby simple number, then multiply.

49.7≈50one significant figure6.12≈6one significant figure49.7×6.12≈50×6=300\begin{aligned} 49.7&\approx50 &&\text{one significant figure}\\ 6.12&\approx6 &&\text{one significant figure}\\ 49.7\times6.12&\approx50\times6=300 \end{aligned}

Answer. An estimate is 300300.

Check. Five lots of about 6060 also gives about 300300. A result near 3030 or 30003000 would have the wrong scale.

Now keep that product but divide by a number close to 0.30.3. Dividing by a positive number below one makes the result larger.

Worked example 6: checking a calculator answer

Question. A calculator entry gives 100.7100.7 for 49.7×6.12÷0.30249.7\times6.12\div0.302. Use an estimate to decide whether this is plausible.

Plan. Round each input to one significant figure and keep the same operations.

49.7×6.120.302≈50×60.3=30003multiply numerator and denominator by 10=1000\begin{aligned} \frac{49.7\times6.12}{0.302} &\approx\frac{50\times6}{0.3}\\ &=\frac{3000}{3} &&\text{multiply numerator and denominator by }10\\ &=1000 \end{aligned}

Answer. 100.7100.7 is not plausible: it is about one tenth of the estimate. Check the entry, including the decimal in the divisor. A correct calculation gives about 1007.171007.17.

Check. The numerator is about 300300. Multiplying the claimed 100.7100.7 by about 0.30.3 gives only about 3030, so it cannot recover the numerator.

This answers the opening question: the result should be near 10001000. An estimate detects a large error; it does not prove every digit of a nearby calculator answer is right. Keep full calculator values in an exact calculation and round the final answer as requested. Rounding 49.749.7 and 6.126.12 is appropriate for an estimate, but would throw away information if an accurate product were required.

Exam tip

Show the rounded calculation with its operations. A list such as “50,6,0.350,6,0.3” does not show how you used those numbers. Retain brackets, fraction bars and operation order when you estimate.

Use a fresh set of values to judge the size of a display.

Quick Recap

  • For decimal places, count from the decimal point; for significant figures, start at the first nonzero digit.
  • Use the next digit to round once, directly from the original number.
  • Keep zeros that preserve place value or show the requested precision.
  • Estimate with simple nearby values and the original operations, then compare the size of the result.

The opening calculator answer fails the check: the calculation is near 10001000, not 100100.