Direct and inverse proportion

60 biscuits is $60 \div 20 = 3$ times as many.

Flour: $250 \times 3 = 750$ g

25 scones is $25 \div 10 = 2.5$ times the recipe.

$40 \times 2.5 = 100$ g of sugar

1 kg costs $54 \div 3 = £18$

2 kg cost $2 \times 18 = £36$

One pen: $2.38 \div 2 = £1.19$, so 20 pens: $20 \times 1.19 = £23.80$

One folder: $5.60 \div 5 = £1.12$, so 20 folders: $20 \times 1.12 = £22.40$

Total: $23.80 + 22.40 = £46.20$

$£46.20 < £50$, so yes, Ben has enough (with £3.80 left).

Scale factor: $30 \div 20 = 1.5$

Butter: $120 \times 1.5 = 180$ g

Flour: $200 \times 1.5 = 300$ g

Sugar: $50 \times 1.5 = 75$ g

60 cookies is $60 \div 20 = 3$ times the recipe.

Peanut butter needed: $3 \times 250 = 750$ g

$750 < 900$, so yes, he has enough (150 g spare).

(a) Packs: $180 \div 12 = 15$. Cost: $15 \times £3 = £45$

(b) Total volume: $24 \times 330 = 7920$ ml

Cost of 1 ml: $700 \div 7920 = 0.0884$p, so 100 ml costs $8.84$p

To the nearest penny: 9p

One box: $600 \div 120 = 5$p

One bag: $1000 \div 270 = 3.7037...$p

Difference: $5 - 3.7037... = 1.296...$, which is $1.3$p to 1 d.p.

Banana: $\frac{60}{100} \times 84 = 50.4$

Yogurt: $\frac{150}{100} \times 87 = 130.5$

Total: $50.4 + 130.5 = 180.9$ calories

Scale factor: $72 \div 16 = 4.5$

Butter: $240 \times 4.5 = 1080$ g

Flour: $350 \times 4.5 = 1575$ g

Sugar: $100 \times 4.5 = 450$ g

Mincemeat: $280 \times 4.5 = 1260$ g

The formula is $H = \dfrac{72}{n}$.

Monday: $72 \div 6 = 12$ hours

Tuesday: $72 \div 9 = 8$ hours

Difference $= 12 - 8 = 4$ hours

Cost per battery:

4-pack: $180 \div 4 = 45$p

8-pack: $320 \div 8 = 40$p

12-pack: $600 \div 12 = 50$p

The pack of 8 is best value.

Total work $= 30 \times 4 = 120$ machine-hours.

In 6 hours: $120 \div 6 = 20$ machines.

Total work: $5 \times 14 = 70$ pump-hours.

With 4 pumps: $70 \div 4 = 17.5$ hours

Offer 1: pay $2 \times 75 = 150$p for 3 bottles, which is 6 pints. $150 \div 6 = 25$p per pint.

Offer 2: pay $128 + 64 = 192$p for 2 bottles, which is 8 pints. $192 \div 8 = 24$p per pint.

The 4-pint offer is better value.

One pipe would take $4 \times 12 = 48$ hours to fill the lake.

Six pipes take $48 \div 6 = 8$ hours for the whole lake.

A quarter of the lake: $8 \div 4 = 2$ hours.

Carrots: $1.74 \div 3 = £0.58$ per kg, so 2 kg cost $£1.16$

Onions: $2.36 - 1.16 = £1.20$ for 2.5 kg, so $1.20 \div 2.5 = £0.48$ per kg

4 kg: $4 \times 0.48 = £1.92$, which is less than £2, so yes.

Scale factor: $25 \div 10 = 2.5$

Butter $80 \times 2.5 = 200$ g, flour $350 \times 2.5 = 875$ g, sugar $30 \times 2.5 = 75$ g, eggs $2 \times 2.5 = 5$

Compare: butter $200 - 100 = 100$ g more, sugar $75 - 50 = 25$ g more, eggs $5 - 4 = 1$ more. Flour is fine (875 g < 1 kg).

Carrots: $£1.80 \div 3 = £0.60$ per kg

2 kg carrots cost $£1.20$, so 5 kg potatoes cost $£3.45 - £1.20 = £2.25$

Potatoes: $£2.25 \div 5 = £0.45$ per kg

Total: $4 \times £0.60 + 2 \times £0.45 = £2.40 + £0.90 = £3.30$

(a) Batches each ingredient allows (12 biscuits per batch):

Butter: $500 \div 125 = 4$. Flour: $700 \div 200 = 3.5$. Sugar: $250 \div 50 = 5$

Flour runs out first: $3.5 \times 12 = 42$ biscuits.

(b) No. Flour is the limit, so extra sugar makes no difference.

(a) Total work $= 12 \times 5 = 60$ person-hours. $60 \div 15 = 4$ hours.

(b) If the people work at different rates, it could take longer (if some are slower) or less time (if some are faster).

(a) $60 \div 15 = 4$, so she needs $4 \times 50 = 200$ g of sugar.

Flour: $3 \times 200 = 600$ g

(b) Butter: $2 \times 200 = 400$ g

$400 \div 250 = 1.6$, so she must buy 2 packs.

Machine A: 6 parts per 10 min is $6 \times 6 = 36$ per hour.

$36 \times 12 = 432$

Machine B: 13 parts per 15 min is $13 \times 4 = 52$ per hour.

$52 \times 10 = 520$

Total: $432 + 520 = 952$

Batches each ingredient allows:

Butter $900 \div 225 = 4$; sugar $1000 \div 110 \approx 9.1$; flour $1000 \div 275 \approx 3.6$; chocolate $225 \div 75 = 3$

Chocolate runs out first, after 3 batches.

$3 \times 30 = 90$ biscuits

Total work $= 6 \times 14.5 = 87$ worker-days.

With 13 workers: $87 \div 13 = 6.69...$ days.

$6.69 < 7$, so 13 workers are enough.

(a) 3 kg covers $5 \times 9 = 45$ m², so 1 kg covers $15$ m².

New lawn: $10 \times 14 = 140$ m² needs $140 \div 15 = 9.33$ kg.

She has $5 \times 2 = 10$ kg, so yes.

(b) Now she has $4 \times 2 + 1 = 9$ kg, less than 9.33 kg, so she does not have enough.

Rearrange: $n = \frac{720}{d}$

Ali: $720 \div 40 = 18$ workers

Hayley: $720 \div 30 = 24$ workers

$24 - 18 = 6$ more

Rearrange: $n = \frac{720}{d}$

Ali: $720 \div 40 = 18$ workers

Hayley: $720 \div 30 = 24$ workers

Difference: $24 - 18 = 6$ more workers.

Total work: $6 \times 9 = 54$ machine-days

First 3 days: $3 + 4 + 5 = 12$ machine-days

Left: $54 - 12 = 42$, done by 6 machines in $42 \div 6 = 7$ days

Total: $3 + 7 = 10$ days

(a) One tap alone would take $120 \times 5 = 600$ minutes.

Three taps: $600 \div 3 = 200$ minutes.

(b) Assume each tap gives water at the same rate, and that rate stays constant.

(a) $f = \dfrac{k}{d^2}$. With $f = 3.5, d = 8$: $k = 3.5 \times 64 = 224$, so $f = \dfrac{224}{d^2}$.

(b) $10 = \dfrac{224}{d^2} \Rightarrow d^2 = 22.4 \Rightarrow d = \sqrt{22.4} = 4.73$ (3 s.f.)

$y = kx$, so $24 = k \times 1.5$ and $k = 16$

When $x = 5$: $y = 16 \times 5 = 80$

$p \propto \frac{1}{t}$ so $p \times t$ is constant: $1 \times 100 = 100$

$t = 25$: $p = \frac{100}{25} = 4$

$p = 5$: $t = \frac{100}{5} = 20$

$t = 2$: $p = \frac{100}{2} = 50$

(a) $y = kx$: a straight line starting at the origin and going up to the right.

(b) $y = \frac{k}{x}$: a curve that is high near the $y$-axis, falls as $x$ grows and gets closer to the $x$-axis, never touching either axis.

$y \propto x$ is a straight line through the origin: B

$y \propto x^2$ curves upwards more and more steeply: D

$y \propto \sqrt{x}$ rises quickly then levels off: A

$y \propto \frac{1}{x}$ falls towards the axes: C

(a) In direct proportion $y \div x$ is constant. Here $0 \div 2 = 0$ but $4 \div 4 = 1$, so Nadia is wrong.

(b)(i) $w = k\sqrt{t}$: $140 = k \times 8$, so $k = 17.5$

$w = 17.5 \times \sqrt{7.84} = 17.5 \times 2.8 = 49$

(ii) Start at O and rise, curving so it gets less steep.

$y \propto x$ is a straight line through the origin: graph A.

$y \propto x^2$ curves upwards more and more steeply: graph D.

$y \propto \frac{1}{x}$ falls as $x$ grows and never reaches the axes: graph B.

$y \propto \sqrt{x}$ rises steeply at first and then flattens: graph C.

$h = \frac{k}{p}$: $10 = \frac{k}{6}$, so $k = 60$

$p = K\sqrt{t}$: $6 = K \times 12$, so $K = \frac{1}{2}$

$h = \frac{60}{\frac{1}{2}\sqrt{t}} = \frac{120}{\sqrt{t}}$

$y = \dfrac{k}{x^2}$, so $8 = \dfrac{k}{2.5^2}$ and $k = 8 \times 6.25 = 50$

When $y = \frac{8}{9}$: $x^2 = 50 \div \frac{8}{9} = 56.25$

$x = \pm 7.5$, so the negative value is $-7.5$

$y = \frac{k}{x^3}$

$44 = \frac{k}{a^3}$, so $k = 44a^3$

When $x = 2a$: $y = \frac{44a^3}{(2a)^3} = \frac{44a^3}{8a^3} = 5.5$

Direct proportion is a straight line through the origin: D.

Inverse proportion $y = \frac{k}{x}$ has two curves in opposite quadrants: A.

$y = kx^2$ is a U-shaped parabola through the origin: B.

$y = \frac{k}{x^2}$ is always positive, so both branches are above the $x$-axis: C.

$T = k\sqrt{L}$

New length $= 1.4L$, so new $T = k\sqrt{1.4L} = \sqrt{1.4} \times k\sqrt{L}$

$\sqrt{1.4} = 1.1832$

Increase $= 18.3\%$

$y = kx^2$: $3 = k \times 0.5^2$, so $k = 12$.

$x = \frac{j}{w}$: $2 = \frac{j}{0.2}$, so $j = 0.4$.

When $w = 2$: $x = \frac{0.4}{2} = 0.2$

$y = 12 \times 0.2^2 = 0.48$

$y = k\sqrt{t}$: $15 = k \times 3$, so $k = 5$

$t = \frac{K}{x^3}$: $8 = \frac{K}{8}$, so $K = 64$

$y = 5\sqrt{\frac{64}{x^3}} = \frac{5 \times 8}{x^{3/2}} = \frac{40}{x^{3/2}}$

$x = ky^2$ and $y = cz^3$, so $x = k(cz^3)^2 = mz^6$

Substitute $z = 2$, $x = 32$: $32 = m \times 64$, so $m = \frac{1}{2}$

$x = \frac{1}{2}z^6$

$x = k\sqrt{y}$

New $y$ is $1.44y$, so new $x = k\sqrt{1.44y} = 1.2k\sqrt{y} = 1.2x$

So $x$ rises by $20\%$.

$y = \frac{K}{d^2}$: $4 = \frac{K}{100}$, so $K = 400$

$d = kx^2$: $24 = k \times 4$, so $k = 6$ and $d = 6x^2$

$y = \frac{400}{(6x^2)^2} = \frac{400}{36x^4} = \frac{100}{9x^4}$