Ratio and proportion, Year 10

24; 920

HintFind 3% of the starting amount once, repeat it for every year, and remember what was there to begin with.

The why3% of £800 is 0.03 × 800 = £24. Simple interest pays that same £24 every year, because it is always worked out on the original £800. So 5 years earn 5 × 24 = £120 and the account then holds £920.

3

HintShare the interest equally between the years first, then compare one year's worth with the amount invested.

The why£60 over 4 years is £15 a year. 15 ÷ 500 = 0.03, which is 3%. Simple interest is the same every year, so the total can be split evenly.

VAT (value added tax)

HintLook near the bottom of a shop receipt, where it is often shown as a percentage.

The whyVAT is a percentage put on top of the price before tax. At a rate of 20%, an item priced at £50 before VAT costs 50 × 1.2 = £60.

40

HintFind the money gained, then compare it with what the shop paid, not with what the customer paid.

The whyProfit = 49 − 35 = £14. As a percentage of the cost price: 14 ÷ 35 = 0.4 = 40%.

£3500

HintOnly part of her pay is taxed: take off the tax-free slice before you find the percentage.

The whyTaxable income = 30 000 − 12 500 = £17 500. 20% of 17 500 = £3500.

£195

HintOne of the two words means money going out of the account and the other means money coming in.

The whyA debit takes money out and a credit pays money in: 240 − 75 + 30 = £195.

30 000 N/m²

HintAsk how many square centimetres cover one square metre: it is not 100.

The why1 m = 100 cm, so 1 m² = 100 × 100 = 10 000 cm². Each of those square centimetres carries 3 N, so one square metre carries 3 × 10 000 = 30 000 N.

force; area

HintRead the unit aloud as "newtons per square metre" and ask what each part measures.

The whyA compound unit is a formula in disguise. Newtons measure force and square metres measure area, and "per" means divide, so pressure = force ÷ area. In the same way g/cm³ shows that density = mass ÷ volume.

2700 kg/m³

HintConvert the mass part and the volume part separately; a cubic metre holds a million cubic centimetres.

The why1 m³ = 100³ = 1 000 000 cm³, so one cubic metre of the metal has a mass of 2.7 × 1 000 000 = 2 700 000 g, which is 2700 kg. Overall, changing g/cm³ to kg/m³ multiplies by 1000.

1200

HintDividing by a number smaller than 1 makes the result bigger than what you started with.

The why600 ÷ 0.5 = 1200. A whole square metre under the same load would carry 1200 N; the crate covers only half a square metre, so its 600 N is concentrated on that half.

Block B (3 g/cm³, against 2.7 g/cm³)

HintWork out the mass packed into one cubic centimetre of each.

The whyDensity = mass ÷ volume. A: 540 ÷ 200 = 2.7 g/cm³. B: 480 ÷ 160 = 3 g/cm³. B has less mass in total, but more in each cubic centimetre.

5 g/cm³

HintWrite an expression for how much space the solid takes up, then divide.

The whyThe volume of the cube is x × x × x = x³ cm³. Density = mass ÷ volume = 5x³ ÷ x³ = 5 g/cm³. The x³ cancels, so the density is the same whatever the size of the cube.

y = 5x

HintStart from y = kx and use the pair of values to find the missing multiplier.

The whyDirect proportion means y = kx. Putting in the values: 20 = k × 4, so k = 5.

3

HintIn this kind of relationship the product of the two quantities never changes.

The whyInverse proportion means y = k ÷ x, so k = xy = 4 × 6 = 24. When x = 8, y = 24 ÷ 8 = 3. Doubling x halves y.

2; 50

HintSquare the x-value before you use it, both times.

The why18 = k × 3² = 9k, so k = 2 and the equation is y = 2x². When x = 5, y = 2 × 25 = 50.

It becomes a quarter of its old value

HintWrite y = k ÷ x² and replace x by 2x; look at the denominator.

The whyy = k ÷ x². Doubling x makes x² four times as large, and dividing by a number four times as large gives a result one quarter the size.

The constant of proportionality

HintIt never changes, and its name says which kind of relationship it belongs to.

The whyEvery proportion statement becomes an equation by bringing in this one number. Find it from a known pair of values, then use the equation for any other value.