This note covers estimating answers, rounding to decimal places and significant figures, upper and lower bounds, ratio, direct and inverse proportion, and money calculations such as exchange rates and interest. These skills turn raw numbers into sensible, checkable answers in real situations such as budgets, recipes and measurements. Bounds and compound interest are Extended topics.
Key ideas
Rounding. To round to a given number of decimal places (d.p.), look at the next digit: 5 or more rounds up, 4 or less leaves the digit unchanged. To round to significant figures (s.f.), start counting at the first non-zero digit. Leading zeros are not significant, but zeros between or after non-zero digits can be.
Estimation. Round every number to 1 significant figure, then calculate. The estimate is a quick check that the exact answer is sensible. It is also useful if a calculator slips: if your exact answer is far from the estimate, look for a mistake.
Bounds (Extended). A measurement given to some accuracy stands for a range of true values. A length of 12.4 cm to the nearest 0.1 cm lies from 12.35 cm (lower bound) up to, but not including, 12.45 cm (upper bound). The general idea is to go half a unit of accuracy down and half a unit up. In calculations:
- Sum: add the upper bounds for the upper bound, and the lower bounds for the lower bound.
- Difference \(a-b\): the largest value is (upper \(a\)) \(-\) (lower \(b\)), and the smallest is (lower \(a\)) \(-\) (upper \(b\)).
- Product: use upper times upper for the maximum, and lower times lower for the minimum.
- Quotient \(a\div b\): the largest value is (upper \(a\)) \(\div\) (lower \(b\)), and the smallest is (lower \(a\)) \(\div\) (upper \(b\)).
To quote a result to a stated accuracy, find both bounds. If they round to the same value at that accuracy, that value is correct.
Ratio. A ratio compares quantities, and it can be simplified by dividing every part by a common factor. To share an amount in a ratio, add the parts to find the total number of parts, find the value of one part, then multiply.
Proportion. If two quantities are in direct proportion, one doubles when the other doubles, and the ratio between them is constant. If they are in inverse proportion, one halves when the other doubles, and their product is constant. A common approach is to find the value of one unit first, then scale.
Money. To convert currencies, multiply by the rate when going to the currency the rate is quoted in, and divide when going back. Simple interest is calculated on the original amount only. Compound interest (Extended) is calculated on the growing balance, so the multiplier is raised to the power of the number of years.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| Decimal places | Digits after the decimal point. Check the digit after the last one you keep. |
| Significant figures | Digits counted from the first non-zero digit. |
| Lower bound, upper bound | The smallest and largest values that round to the given value. Subtract and add half the accuracy. |
| Share of a ratio \(a:b\) | \(\dfrac{a}{a+b}\) of the total and \(\dfrac{b}{a+b}\) of the total. |
| Direct proportion | \(y=kx\); the ratio \(y\div x\) stays the same. |
| Inverse proportion | \(y=\dfrac{k}{x}\); the product \(xy\) stays the same. |
| Simple interest | \(I=\dfrac{PRT}{100}\), with principal \(P\), rate \(R\)% per year and time \(T\) years. |
| Compound interest | \(A=P\left(1+\dfrac{R}{100}\right)^{n}\) after \(n\) years (Extended). |
| Exchange rate | A multiplier between currencies, for example €1 = £0.86. |
Worked examples
Example 1 (a) Estimate \(\dfrac{49.2\times3.08}{0.52}\) to 1 s.f. in each number. (b) (Extended) A sprinter runs 100 m, measured to the nearest metre, in 12.4 s, measured to the nearest 0.1 s. Find the greatest and least possible average speed, and give the speed to a suitable accuracy.
(a) Round to 1 s.f.: \(49.2\approx50\), \(3.08\approx3\), \(0.52\approx0.5\). Then \(\dfrac{50\times3}{0.5}=\dfrac{150}{0.5}=300\). The calculator gives about 291, which is close, so it is sensible.
(b) Distance: \(99.5\le d\lt100.5\). Time: \(12.35\le t\lt12.45\).
Speed is distance divided by time. The greatest speed uses the largest distance and the smallest time: \(\dfrac{100.5}{12.35}=8.14\) m/s (3 s.f.).
The least speed uses the smallest distance and the largest time: \(\dfrac{99.5}{12.45}=7.99\) m/s (3 s.f.).
Both bounds round to 8 to 1 s.f., so the speed can be quoted as 8 m/s to 1 s.f. (Quoting 8.1 or 8.0 would not be justified, because the bounds disagree at 2 s.f.)
Answer: (a) 300; (b) greatest 8.14 m/s, least 7.99 m/s, so 8 m/s to 1 s.f.
Example 2 (a) A prize fund of £840 is shared between three teams in the ratio 3 : 4 : 5. How much does each team receive? (b) Six workers take 10 days to build a stage. How long would four workers take, working at the same rate?
(a) Total parts: \(3+4+5=12\). One part: \(840\div12=70\).
The teams receive \(3\times70=210\), \(4\times70=280\) and \(5\times70=350\) pounds. Check: \(210+280+350=840\).
(b) More workers means fewer days, so this is inverse proportion. Total work: \(6\times10=60\) worker-days.
With four workers: \(60\div4=15\) days.
Answer: (a) £210, £280, £350; (b) 15 days
Example 3 (a) A student exchanges €240 at a rate of €1 = £0.86. How many pounds does she get? (b) (Extended) She also invests £2500 at 3.2% per year compound interest for 5 years. Find the final value and the interest earned, and compare it with simple interest at the same rate.
(a) Multiply by the rate: \(240\times0.86=206.40\). She receives £206.40.
(b) The yearly multiplier is \(1+\dfrac{3.2}{100}=1.032\). After 5 years: \(2500\times1.032^{5}\).
Use the power key on your calculator: \(1.032^5=1.17057\ldots\), so the value is £2926.43 to the nearest penny.
Interest earned: \(2926.43-2500=426.43\), which is £426.43.
Simple interest: \(\dfrac{2500\times3.2\times5}{100}=400\), which is £400. Compound interest is greater by £26.43 because interest is earned on earlier interest.
Answer: (a) £206.40; (b) £2926.43 (interest £426.43, which is £26.43 more than simple interest)
Common errors
- Rounding too early. Keep full calculator values in the middle of a calculation and round only the final answer.
- Taking the upper bound of everything in a quotient. To make \(a\div b\) as large as possible, use the upper bound of \(a\) and the lower bound of \(b\).
- Writing the upper bound of 12.4 to the nearest 0.1 as 12.5 or 12.4999. The bound is 12.45, because you add half of 0.1.
- Counting leading zeros as significant: 0.00380 has 3 significant figures, not 5.
- Forgetting to check a ratio share. The parts should always add back to the original amount.
Exam tip: When an answer asks "to a suitable degree of accuracy" after bounds, write both bounds to more figures than needed, then state the one value they share at the stated accuracy.
Check yourself
- Round 0.0070468 to 3 significant figures and to 2 decimal places.
Answer
The first significant figure is 7. Three s.f.: 0.00705 (the next digit is 6, so 4 rounds up to 5). Two d.p.: 0.01 (the third decimal place is 7, so round up). - A rectangle has sides 8 cm and 5 cm, each to the nearest cm. Find the least possible perimeter.
Answer
Lower bounds are 7.5 and 4.5. Perimeter \(=2(7.5+4.5)=24\) cm. - Divide 56 sweets between Ali and Bea in the ratio 5 : 3.
Answer
Total parts 8, one part \(=56\div8=7\). Ali gets 35 and Bea gets 21. - (Extended) A car is worth £12000 and loses 15% of its value each year. Find its value after 3 years.
Answer
The multiplier is 0.85. Value \(=12000\times0.85^3=12000\times0.614125=7369.5\), so £7369.50.