Ratio

Count: 3 squares are shaded and $8 - 3 = 5$ are not.

Shaded : unshaded $= 3 : 5$

Red : blue $= 24 : 40$

The highest common factor of 24 and 40 is 8.

$24 \div 8 = 3$ and $40 \div 8 = 5$, so the ratio is $3 : 5$.

(a) The February cross is at 23.

(b) January: 10 houses. June: 56 houses.

Ratio January : June $= 10 : 56 = 5 : 28$

Cows : sheep $= 24 : 36$

Divide both by the HCF, 12: $2 : 3$

Red : yellow $= 90 : 150$.

Divide both by 30 (their HCF): $3 : 5$.

(a) $100 - 73 = 27\%$

(b) Total parts $= 2 + 5 = 7$, so adults are $\frac{2}{7}$.

(c) Friday: $0.7 \times 500 = 350$ children

Saturday: $720 \div 8 \times 5 = 450$ children

$450 > 350$, so no, Kasim is wrong.

Total parts $= 2 + 16 + 7 = 25$

Green is 7 of these parts, so the fraction is $\frac{7}{25}$.

1 part is 35 beads, and there are $1 + 4 = 5$ parts.

$35 \times 5 = 175$

Let Amol have 1 sweet. Gemma has $6 \times 1 = 6$. Harry has $6 \div 2 = 3$.

Ratio: $1 : 6 : 3$

(a) There are $2 + 3 = 5$ parts, and land is 2 of them.

$\frac{2}{5} \times 100 = 40\%$

(b) Water is $100 - 20 = 80\%$.

Land : water $= 20 : 80 = 1 : 4$

(a) There are $3 + 4 = 7$ parts and 3 are red, so the fraction is $\frac{3}{7}$.

(b) Divide both sides by 12: $12 \div 12 = 1$, $30 \div 12 = 2.5$.

So the ratio is $1 : 2.5$.

(a) 8 litres is $8 \div 2 = 4$ lots of 2 litres.

Oranges needed: $4 \times 30 = 120$

Boxes: $120 \div 24 = 5$

(b) $1260 : 280$. Divide both by 140: $9 : 2$

(a) Shop A: $30 \div 4 = 7.5$, so 8 packs: $8 \times £1.60 = £12.80$

Shop B: $30 \div 6 = 5$ packs: $5 \times £2.70 = £13.50$

Shop A is cheaper.

(b) Shop A: 7 packs + 2 singles $= £11.20 + £0.80 = £12$. Shop B is still £13.50, so no change: shop A.

20 litre offer: 2 bags = 40 litres for £3.50, so $3.50 \div 40 = £0.0875$ per litre.

40 litre offer: 3 bags = 120 litres for £9, so $9 \div 120 = £0.075$ per litre.

The 40 litre bags are cheaper per litre, so they are better value.

$19.2\text{ m} = 1920\text{ cm}$

Model length: $1920 \div 24 = 80\text{ cm}$

Divide both sides by 2.25:

$4.5 \div 2.25 = 2$ and $2.25 \div 2.25 = 1$

So the ratio is $2 : 1$

Total parts: $6 + 9 + 10 = 25$

Ben gets 9 of the 25 parts, so $\frac{9}{25}$.

Let Ryan's score be 1.

Azmol $= 4 \times 1 = 4$, Kim $= 4 \div 2 = 2$.

Azmol : Ryan : Kim $= 4 : 1 : 2$

$450 : 15$

Divide both sides by 15: $30 : 1$

Red : black $= 28 : 84$

The HCF of 28 and 84 is 28, so divide both by 28:

$28 : 84 = 1 : 3$

Group 1: $36 : 48$. Divide both by 12 to get $3 : 4$.

Group 2: $\frac{3}{7}$ are boys, so $\frac{4}{7}$ are girls. Boys : girls $= 3 : 4$.

Both are $3 : 4$, so Ann is correct.

One pencil: $£1.80 \div 12 = £0.15$

One ratio part: $£0.15 \div 3 = £0.05$

One pen: $7 \times £0.05 = £0.35$

Five pens: $5 \times £0.35 = £1.75$

Batches of 16 each ingredient allows:

Butter $500 \div 175 = 2.86$, sugar $300 \div 75 = 4$, flour $625 \div 250 = 2.5$

Flour runs out first: $2.5 \times 16 = 40$ biscuits

Alan: $100 - 32 - 40 = 28\%$, and $0.28 \times 400 = 112$

Beryl: $1 - \frac{3}{10} - \frac{1}{10} = \frac{6}{10}$, and $\frac{6}{10} \times 500 = 300$

Charlie: $150 \div 3 = 50$, so medium $= 4 \times 50 = 200$

Total: $112 + 300 + 200 = 612$

(a) Orange juice limits it: $300 \div 5 = 60$ ml per part.

Lemonade needed $= 2 \times 60 = 120$ ml (he has 200 ml, so enough).

Total $= 7 \times 60 = 420$ ml.

(b) No: he still only needs 120 ml of lemonade, and 160 ml is enough.

Nuts: $\frac{1}{4}$, no nuts: $1 - \frac{1}{4} = \frac{3}{4}$

Ratio $\frac{1}{4} : \frac{3}{4} = 1 : 3$

(a) Alan's £2.50 is $\frac{1}{8}$, so the total is $2.50 \times 8 = £20$.

Bispah gets $\frac{1}{2} \times 20 = £10$.

(b) Chan gets $20 - 2.50 - 10 = £7.50$.

Alan : Chan $= 2.5 : 7.5 = 1 : 3$.

(a) $30 \div 6 = 5$ red counters.

(b) For 1 : 2 with 30 yellow, there must be $30 \div 2 = 15$ red.

$15 - 5 = 10$ red counters were added.

(a) Fruit is 40%, so oats are $100 - 40 = 60\%$.

$40 : 60$, divide both by 40: $1 : 1.5$

(b) $1 + 3 = 4$ parts, 3 of them bran, so $\frac{3}{4}$.

(a)(i) Double $1 : 3$ to $2 : 6$ so $b$ is 6 in both. $a : b : c = 2 : 6 : 5$

(a)(ii) Total parts $2 + 6 + 5 = 13$, so $a$ is $\frac{2}{13}$.

(b) $p = 5n = 5 \times 2m = 10m$, so $m : p = 1 : 10$

Cars: $160 \div 10 = 16$, $16 \times 3 = 48$

Electric: $48 \div 8 = 6$

Diesel: $25\%$ of 48 $= 12$

Petrol: $48 - 6 - 12 = 30$

From L to M: $x$ goes up $4 - (-3) = 7$, $y$ goes up $9 - 1 = 8$. This is 2 parts.

One part: $x$ up 3.5, $y$ up 4.

MN is 3 parts: $x$ up 10.5, $y$ up 12.

$N = (4 + 10.5,\ 9 + 12) = (14.5, 21)$

Let Rick have $n$. Selma has $2n$, Tony has $2n - 6$.

$n + 2n + 2n - 6 = 54 \Rightarrow 5n = 60 \Rightarrow n = 12$

Tony has $24 - 6 = 18$

$12 : 18 = 1 : 1.5$, so $p = 1.5$

$\frac{a}{b} = \frac{2}{5}$ gives $a : b = 2 : 5$, and $\frac{b}{c} = \frac{3}{4}$ gives $b : c = 3 : 4$.

Make b the same in both (15): $a : b = 6 : 15$ and $b : c = 15 : 20$.

So $a : b : c = 6 : 15 : 20$.

Let Rytis get 1 part. Linda gets $3 \times 1 = 3$ parts. Adam gets double Linda: 6 parts.

Total: $1 + 3 + 6 = 10$ parts.

Linda gets $\frac{3}{10}$.

One part: $3000 \div 5 = 600$ tins.

Small boxes: $2 \times 600 = 1200$ tins, $1200 \div 6 = 200$ boxes.

Large boxes: $3 \times 600 = 1800$ tins, $1800 \div 20 = 90$ boxes.

$\frac{90}{290} \times 100 = 31.0\%$, which is more than 30%, so Carlo is not correct.

Women: $60 \div 2 = 30$

Men: $30 \div 3 = 10$

Children: $60 - 30 - 10 = 20$

Children : men $= 20 : 10 = 2 : 1$, so $n = 2$

In the sample, $\frac{5}{30} = \frac{1}{6}$ of the counters are yellow.

So 12 yellow counters is about $\frac{1}{6}$ of the bag.

$p \approx 12 \times 6 = 72$

Flats : bungalows $= 8 : 5$, and $50 \div 5 = 10$, so flats $= 8 \times 10 = 80$

Houses : flats $= 7 : 4$, and $80 \div 4 = 20$, so houses $= 7 \times 20 = 140$

Make the flats match: houses : flats $= 7 : 4 = 14 : 8$, so houses : flats : bungalows $= 14 : 8 : 5$.

5 parts $= 50$, so 1 part $= 10$.

Houses $= 14 \times 10 = 140$

Price with VAT: $8500 \times 1.2 = £10\,200$

Monthly payments: $12 \times 531.25 = £6375$

Deposit: $10\,200 - 6375 = £3825$

Ratio: $3825 : 6375$. Divide both by 1275: $3 : 5$

Empty jars: $\frac{3}{8} \times 400 = 150$

S : M $= 3 : 4$ and M : L $= 1 : 2 = 4 : 8$, so S : M : L $= 3 : 4 : 8$

Small empty: $\frac{3}{15} \times 150 = 30$

Percentage: $\frac{30}{400} \times 100 = 7.5\%$

Red : yellow $= 3 : 5$, so yellow $= \frac{5}{3}x$.

Bag A $= x + \frac{5}{3}x = \frac{8}{3}x$

Bag B $= \frac12 \times \frac{8}{3}x = \frac{4}{3}x$

Total $= \frac{8}{3}x + \frac{4}{3}x = \frac{12}{3}x = 4x$

Let 1 kg of carrots cost $5k$ and 1 kg of tomatoes cost $9k$.

$7(5k) + 5(9k) = 480$, so $35k + 45k = 80k = 480$ and $k = 6$

Carrots: $5 \times 6 = 30$p. Tomatoes: $9 \times 6 = 54$p

$\frac{3}{7} = \frac{9}{21}$, so the ratio has 21 parts in total.

$9 + 4 + x = 21$

$x = 8$

Fran and Gary share 5 parts; Olivia and Jessica share 10 parts.

Olivia and Jessica have half as many, so Fran and Gary's total must be 20 parts: multiply $2 : 3$ by 4 to get $8 : 12$.

Olivia and Jessica: $9 : 1$ (total 10, which is half of 20).

$w : x : y : z = 8 : 12 : 9 : 1$

Let the pay be $6x$ and $5x$. After the rise: $\frac{6x + 1.5}{5x + 1.5} = \frac{13}{11}$

$11(6x + 1.5) = 13(5x + 1.5)$

$66x + 16.5 = 65x + 19.5$, so $x = 3$

Marta: $6 \times 3 = £18$, Khalid: $5 \times 3 = £15$

$A : B = 2 : 3$ and $C : D = 3 : 4$ (75% = $\frac{3}{4}$)

Let $A = 2k$, $B = 3k$, $C = 3m$, $D = 4m$. Then $5k = 3 \times 7m$, so $5k = 21m$: take $k = 21$, $m = 5$

$A : B : C : D = 42 : 63 : 15 : 20$

Check: $105 = 3 \times 35$

Soup $= \frac{2}{5}$, prawns $= \frac{3}{5}$

Soup then curry: $\frac{2}{5} \times \frac{3}{8} = \frac{3}{20}$

Prawns then curry: $\frac{3}{5} \times \frac{5}{6} = \frac{1}{2}$

Curry total $= \frac{3}{20} + \frac{10}{20} = \frac{13}{20}$

$2a : 5c = 6 : 25 \Rightarrow 50a = 30c \Rightarrow a = \frac{3c}{5}$

$4b : 7c = 20 : 21 \Rightarrow 84b = 140c \Rightarrow b = \frac{5c}{3}$

Take $c = 15$: $a = 9$, $b = 25$.

$a + b : b + c = 34 : 40 = 17 : 20$

$11(2x^2 + y^2) = 43(4x^2 - y^2)$

$22x^2 + 11y^2 = 172x^2 - 43y^2$

$54y^2 = 150x^2$, so $\dfrac{x^2}{y^2} = \dfrac{54}{150} = \dfrac{9}{25}$

$x$ and $y$ are positive, so $\dfrac{x}{y} = \dfrac{3}{5}$ and $x : y = 3 : 5$

Black cards are $\frac{3}{8}$ of all cards; triangles are $\frac{7}{9}$ of all cards.

Use a total of 72 (LCM of 8 and 9): black $= 27$, triangles $= 56$

So black as a fraction of triangles $= \frac{27}{56}$

(a) $5c - c = 4d - d$, so $4c = 3d$, giving $c : d = 3 : 4$

(b) $6x^2 - 7xy - 20y^2 = 0$

$(2x - 5y)(3x + 4y) = 0$

Since $x, y > 0$, $3x + 4y \ne 0$, so $2x = 5y$ and $x : y = 5 : 2$

(a) Small : large $= 4 : 7$ has 11 parts; red : yellow $= 3 : 5$ has 8 parts. The total must divide by both 11 and 8, and LCM(11, 8) $= 88$.

(b) With 88 cubes: small $= 4 \times 8 = 32$, yellow $= 5 \times 11 = 55$.

All small cubes are yellow, so large yellow $= 55 - 32 = 23$.

$\frac{p-5}{q-5} = 5$ gives $p = 5q - 20$

$\frac{p+20}{q+20} = \frac{5}{2}$ gives $2p + 40 = 5q + 100$

Substitute: $10q - 40 + 40 = 5q + 100$, so $q = 20$ and $p = 80$.

$p : q = 80 : 20 = 4 : 1$