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Approximation and Estimation

Secondary 1 Mathematics · G3 focus · Study guide

Round with purpose, distinguish decimal places from significant figures, estimate calculations and interpret rounded measurements.

Concept → worked examples → practice13 min read · practice linked below
9.996 is closer to 10.00 than to 9.99The line runs from 9.990 to 10.000. The midpoint is 9.995. The number 9.996 is just right of the midpoint, at six tenths of the interval, so rounds to 10.00.Round to the nearest hundredth9.9969.999.99510.00Midpoint: round to 10.00 from here onwards.

An approximate answer should say how precise it is

Rounding replaces a number with a nearby value that is easier to use. Estimation uses convenient values to judge roughly how large an answer should be. The symbol ≈\approx means “is approximately equal to”; reserve == for an exact equality.

The required accuracy tells you which digit to keep last. For the usual school rule for positive numbers, inspect the next digit: 0–4 leaves the kept digit unchanged; 5–9 increases it by one. Discard later digits, keeping any zeros needed for place value or the stated accuracy. For a negative number, round its magnitude using the same rule and then restore the negative sign: −2.76≈−2.8-2.76\approx-2.8 to 1 decimal place.

Decimal places and significant figures answer different questions

Decimal places (d.p.) count digits to the right of the decimal point. Significant figures (s.f.) count from the first non-zero digit. For 0.00405060.0040506, the leading zeros locate the number but are not significant; the zero between 4 and 5 is significant. A final zero after a decimal can show accuracy: 2.50 has three significant figures and two decimal places.

Round to decimal places and ordinary place values

Worked example 1: a carry crosses the decimal point

Round 9.996 to 2 decimal places, and 4736 to the nearest hundred.

  1. In 9.996, keep through the hundredths digit: 9.999.99. The next digit is 6, so add one hundredth.
  2. The two 9s cause a carry: 9.99+0.01=10.009.99+0.01=10.00. Hence 9.996≈10.00 (2 d.p.).\boxed{9.996\approx10.00\text{ (2 d.p.)}}.
  3. For 4736, the hundreds digit is 7 and the next digit is 3. Keep 7 and replace the tens and units by zeros: 4736≈4700 (nearest hundred).\boxed{4736\approx4700\text{ (nearest hundred)}}.

Check: 9.996 is 0.004 away from 10.00 but 0.006 away from 9.99. Writing 10 alone gives the same value but does not communicate the requested two decimal places.

9.996 is closer to 10.00 than to 9.99The line runs from 9.990 to 10.000. The midpoint is 9.995. The number 9.996 is just right of the midpoint, at six tenths of the interval, so rounds to 10.00.Round to the nearest hundredth9.9969.999.99510.00Midpoint: round to 10.00 from here onwards.
The chosen place value is 0.01. Its halfway point is 0.005 from either end.

Count significant figures from the first non-zero digit

Worked example 2: small numbers and significant zeros

Round 0.0040506 to 3 significant figures, and 70496 to 3 significant figures.

  1. The first three significant digits of 0.0040506 are 4, 0 and 5. The next digit is 0, so do not increase the 5: 0.0040506≈0.00405 (3 s.f.).\boxed{0.0040506\approx0.00405\text{ (3 s.f.)}}.
  2. The first three significant digits of 70496 are 7, 0 and 4. The next digit is 9, so increase the hundreds digit from 4 to 5: 70496≈70500 (3 s.f.).\boxed{70496\approx70500\text{ (3 s.f.)}}.

Check the scale: 0.00405 is close to four thousandths; 0.405 would be 100 times too large. In 70500, the last two zeros hold the tens and units places. State “3 s.f.” to make the accuracy clear.

Explore the rounding boundary

Choose a positive number and an accuracy. Compare the number with the midpoint between the two neighbouring rounding values.

9.996 rounds to 10.00 to 2 decimal places. The rounding unit is 0.01; the midpoint is 9.995.

Nearest rounding values and their midpoint9.996 lies between 9.99 and 10.00, above the midpoint 9.995. It rounds to 10.00.Number: 9.9969.9910.009.995The middle tick marks halfway.
The diagram enlarges one rounding interval, so the scale changes when the accuracy changes.

Estimate before trusting a calculator

Replace awkward numbers with convenient nearby numbers, often to one significant figure. Write the approximate expression so your estimate can be checked. A quick estimate is a reasonableness check, not a replacement for a more accurate calculation when one is required. Follow any rounding instructions in the question.

Worked example 3: estimate a quotient

Without a calculator, estimate 49.6×19.80.51\frac{49.6\times19.8}{0.51}.

  1. Use 49.6≈5049.6\approx50, 19.8≈2019.8\approx20 and 0.51≈0.50.51\approx0.5.
  2. Calculate with these convenient numbers: 49.6×19.80.51≈50×200.5=10000.5=2000.\frac{49.6\times19.8}{0.51}\approx\frac{50\times20}{0.5}=\frac{1000}{0.5}=\boxed{2000}.

Dividing by 0.5 doubles a number; it does not halve it. The exact value is a little below 2000 because we increased both numerator factors and decreased the positive denominator. A calculator result near 200 or 20000 would signal an entry error.

Worked example 4: retain accuracy until the final line

Evaluate 4.8221.37+0.63\frac{4.82^2}{1.37+0.63}, giving the answer to 3 significant figures.

  1. Calculate the full denominator: 1.37+0.63=21.37+0.63=2.
  2. Keep the exact square: 4.822=23.23244.82^2=23.2324.
  3. Divide, then round: 23.23242=11.6162≈11.6 (3 s.f.).\frac{23.2324}{2}=11.6162\approx\boxed{11.6\text{ (3 s.f.)}}.

Check: using 4.82≈54.82\approx5 gives 25÷2=12.525\div2=12.5, the same approximate size. Rounding 4.82 to 5 before squaring would change the required answer; use the estimate as a separate check.

What values could have rounded to this measurement?

A rounded positive measurement represents a range. If the rounding unit is uu, move half a unit below and above the rounded value. Under the usual half-up rule, the lower endpoint is included and the upper endpoint is excluded. The upper endpoint would round to the next value.

Worked example 5: a measurement interval and whole-number endpoints

(a) A length is 8.4 cm correct to 1 decimal place. Find its possible range. (b) A whole-number count is 2840 when rounded to the nearest ten. Find the smallest and largest possible counts.

  1. For the length, the rounding unit is 0.1 cm. Half of this is 0.05 cm, so 8.35≤L<8.45 cm.\boxed{8.35\le L<8.45\text{ cm}}.
  2. 8.35 rounds to 8.4 and is included. 8.45 rounds to 8.5 and is excluded. There is no greatest possible real length below 8.45; 8.45 is the upper bound.
  3. For the count, half of ten is 5, giving 2835≤N<28452835\le N<2845. Because NN is a whole number, the smallest count is 2835\boxed{2835} and the largest is 2844\boxed{2844}.

Check: 2834 rounds to 2830, while 2845 rounds to 2850. These are just outside the valid integer range.

For a measurement rounded to the nearest unit, its distance from the reported value is at most half that unit. Counts known exactly, such as “there are 12 students”, do not automatically carry a rounding interval.

Common mistakes and next steps

  • Counting leading zeros as significant: start at the first non-zero digit.
  • Dropping a required zero: 7.2 to 3 significant figures is written 7.20.
  • Rounding repeatedly: round the original number once. For example, 2.449 to 1 d.p. is 2.4, not 2.5 from first rounding it to 2.45.
  • Using equality for an estimate: write ≈\approx when you change a value by rounding.
  • Including the upper endpoint: 8.45 would round up to 8.5, so it is excluded from the range that rounds to 8.4.

Use the approximation and estimation practice bank to work on rounding, calculator expressions and ranges of possible values.