This note covers ratio, proportion and degree of accuracy: simplifying and sharing in a ratio, comparing value for money, rounding to decimal places and significant figures, estimating, and upper and lower bounds. These skills are used in money, science, scale drawings and every calculation where an answer has to be reported sensibly.
Key ideas
Ratio. A ratio compares quantities in the same units. To simplify, divide every part by a common factor, as with a fraction. If the quantities have different units, convert first. For example, 40 cm to 2 m becomes 40 : 200, which simplifies to 1 : 5.
A ratio can be written as \(1:n\) by dividing by the first part, so \(4:10\) becomes \(1:2.5\).
Sharing in a ratio. Add the parts to get the total number of parts. Divide the total amount by this number to find one part. Then multiply by each share. Check that your shares add up to the original amount.
Proportion. Two quantities are in direct proportion when one is a fixed multiple of the other, so doubling one doubles the other. They are in inverse proportion when one increases as the other decreases, so doubling one halves the other. The unitary method (find the value of one unit first) solves most simple problems. For best-buy questions, work out the price per unit, or the amount per pound, and compare.
Accuracy.
- To round to a given number of decimal places (d.p.), look at the digit after the last one you keep. If it is 5 or more, round up.
- Significant figures (s.f.) count from the first non-zero digit. In 0.00407 the significant figures are 4, 0 and 7, so it has 3 s.f.
- To estimate, round every number to 1 significant figure and then calculate.
- A value given to a stated accuracy lies in a range. A length of 12 cm to the nearest cm lies in the interval \(11.5\le l\lt12.5\). The lower bound is 11.5 and the upper bound is 12.5. This range is called an error interval.
- Truncation cuts digits off with no rounding. A length of 12 cm truncated to the nearest cm lies in \(12\le l\lt13\).
Higher tier: when you combine measurements, use bounds to find the largest and smallest possible results. For \(a+b\) and \(a\times b\), use upper with upper for the maximum and lower with lower for the minimum. For \(a-b\) and \(a\div b\), the maximum uses the upper bound of \(a\) with the lower bound of \(b\), and the minimum uses the lower bound of \(a\) with the upper bound of \(b\).
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(a:b\) | Ratio of \(a\) to \(b\). Both parts must have the same units. |
| One part \(=\dfrac{\text{total}}{\text{sum of parts}}\) | Sharing an amount in a ratio. |
| Direct proportion | Ratio of the two quantities stays constant. |
| Inverse proportion | The product of the two quantities stays constant. |
| Decimal places | Digits after the decimal point. |
| Significant figures | Digits counted from the first non-zero digit. |
| Error interval | \(\text{lower bound}\le x\lt\text{upper bound}\) for a rounded value. |
| Bounds of a rounded value | Half of the rounding unit below and above the stated value. |
Worked examples
Example 1 Three friends, Asha, Ben and Chloe, invest in a small business in the ratio \(3:5:4\). The profit is £840. Share the profit in the same ratio.
Total parts: \(3+5+4=12\).
One part: \(840\div12=70\), so one part is £70.
Asha: \(3\times70=210\). Ben: \(5\times70=350\). Chloe: \(4\times70=280\).
Check: \(210+350+280=840\). This matches the total.
Answer: Asha £210, Ben £350, Chloe £280.
Example 2 A shop sells rice in two bags. A 450 g bag costs £1.35 and a 700 g bag costs £2.03. Which bag is better value?
Work out the cost per 100 g for each bag.
Small bag: \(1.35\div4.5=0.30\), so £0.30 per 100 g.
Large bag: \(2.03\div7=0.29\), so £0.29 per 100 g.
The large bag costs less for the same mass.
Answer: the 700 g bag is better value (29p per 100 g compared with 30p per 100 g).
Example 3 (Higher) The sides of a rectangle are measured as 12.4 cm and 8.5 cm, each to 1 decimal place. Find the upper and lower bounds of the area, and give the area to a suitable degree of accuracy.
Each value to 1 d.p. can be 0.05 either side. So \(12.35\le l\lt12.45\) and \(8.45\le w\lt8.55\).
Upper bound of the area: \(12.45\times8.55=106.4475\).
Lower bound of the area: \(12.35\times8.45=104.3575\).
Round both to 2 s.f.: 110 and 100. They are different, so 2 s.f. is not justified.
Round both to 1 s.f.: both give 100.
Answer: upper bound \(106.4475\) cm\(^2\), lower bound \(104.3575\) cm\(^2\). The area is 100 cm\(^2\) to 1 s.f.
Common errors
- Sharing in a ratio by dividing the total by the number of people instead of the total number of parts.
- Comparing ratios with different units, such as 50 cm : 2 m, without converting first.
- Counting zeros as significant when they are only place holders, for example treating 0.0305 as having 5 s.f. (it has 3).
- Rounding in the middle of a calculation. Keep full calculator values and round only the final answer.
- Writing the upper bound as 12.49 or 12.4999. Use 12.5 and an inequality \(x\lt12.5\) that is strict.
Exam tip: for best-buy questions, state the comparison clearly, for example "per 100 g", and write a short sentence naming the better value. Numbers alone do not answer the question.
Check yourself
- The ratio of boys to girls in a club is \(5:7\). There are 18 more girls than boys. How many members are there?
Answer
The difference is \(7-5=2\) parts, which equals 18, so 1 part is 9. Total parts \(=12\), so members \(=12\times9=108\). - Write 0.0040762 to 3 significant figures and to 3 decimal places.
Answer
3 s.f.: the digits are 4, 0, 7 and the next digit is 6, so round up to 0.00408. 3 d.p.: 0.004. - A mass is 63 kg to the nearest kg. Write the error interval.
Answer
\(62.5\le m\lt63.5\). - (Higher) A car travels 150 m, to the nearest 10 m, in 12 s, to the nearest second. Find the greatest possible average speed.
Answer
Greatest speed uses the upper bound of distance and the lower bound of time: \(155\div11.5=13.478\ldots\), so 13.5 m/s to 3 s.f.