R Ratio, proportion and rates of change — revision question pack

16 specification points · notes, questions, answers and worked methods

Checked against Edexcel 1MA1 section R. Review basis: the qualification registry sourced from the Pearson Edexcel Level 1/Level 2 GCSE (9-1) in Mathematics (1MA1) specification; registry verification recorded 9 July 2026.

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R1 · Change freely between related standard units (time, length, area, volume/capacity, mass) and compound units (speed, rates of pay, prices, density, pressure) in numerical and algebraic contexts

Explanation

Unit conversion means multiplying or dividing by a fixed conversion factor. Convert quantities to compatible units before substituting into a formula, and convert the numerator and denominator of a compound unit separately.
Length factors must be raised to the correct power: because $1\,\text{m}=100\,\text{cm}$ , then $1\,\text{m}^2=100^2\,\text{cm}^2$ and $1\,\text{m}^3=100^3\,\text{cm}^3$ .
In algebraic contexts the same factor applies to the whole expression.
A sensible estimate can expose a conversion in the wrong direction.
Examiners expect the conversion step to be visible, especially when the final unit is compound.

Worked example

A block is $2.4\,\text{m}$ long, $35\,\text{cm}$ wide and $80\,\text{mm}$ high. Its mass is $26.88\,\text{kg}$ . Find its density in $\text{kg}/\text{m}^3$ .

1.Convert to metres: $35\,\text{cm}=0.35\,\text{m}$ and $80\,\text{mm}=0.08\,\text{m}$ .
2.Volume $=2.4\times0.35\times0.08=0.0672\,\text{m}^3$ .
3.Density $=\dfrac{26.88}{0.0672}=400\,\text{kg}/\text{m}^3$ .

Answer: $400\,\text{kg}/\text{m}^3$ .

Common mistakes

Don't use the length factor $100$ for an area or volume conversion instead of $100^2$ or $100^3$ .
Don't substitute centimetres and metres into the same formula without converting first.
Don't convert only the numerator of a compound unit such as kilometres per hour.

Exam tip

For a multi-mark conversion, write the compatible units before the formula so the method mark is unambiguous.

Tier 1 · Easy

1
A charity walk is $3.6\,\text{km}$ long. Change this distance to metres.
(1)
(Total for Question 1 is 1 mark)
2
Change $2.7\,\text{m}^2$ to square centimetres.
(1)
(Total for Question 2 is 1 mark)

Tier 2 · Standard

1
A pump moves water at $540$ litres per minute. Change this rate to cubic metres per second.
(3)
(Total for Question 1 is 3 marks)
2
A rectangle is $3x$ metres long and $40\,\text{cm}$ wide. Express its area in square metres in terms of $x$ .
(2)
(Total for Question 2 is 2 marks)
3
A dairy uses $22.32\,\text{m}^3$ of milk in $31$ days. Work out the average amount used in litres per hour.
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1
A solid block measures $2.4\,\text{m}$ by $35\,\text{cm}$ by $80\,\text{mm}$ and has mass $26.88\,\text{kg}$ . Work out its density in $\text{kg}/\text{m}^3$ .
(4)
(Total for Question 1 is 4 marks)
2
A machine prints $480$ labels in $6$ minutes, and each label has area $35\,\text{cm}^2$ . Work out the area of labels printed in one hour, in square metres.
(4)
(Total for Question 2 is 4 marks)
3
A uniform metal sheet has area $0.84\,\text{m}^2$ and mass $6.72\,\text{kg}$ . Work out the mass, in grams, of a piece with area $350\,\text{cm}^2$ .
(4)
(Total for Question 3 is 4 marks)
4
A roll of material has mass $9.6\,\text{kg}$ . Its mass per unit area is $320\,\text{g}/\text{m}^2$ and its width is $75\,\text{cm}$ . Work out the length of the roll in metres.
(4)
(Total for Question 4 is 4 marks)
5
A pipe carries water at $420\,\text{cm}^3$ per second from 08:35 until 10:20. Calculate the volume carried, in cubic metres.
(4)
(Total for Question 5 is 4 marks)

R2 · Use scale factors, scale diagrams and maps

Explanation

A scale written $1:n$ compares a diagram length with the corresponding real length in the same units. Multiply a diagram length by $n$ to obtain the real length, or divide the real length by $n$ to obtain the diagram length.
Convert units only after establishing which direction the scale factor acts.
For areas, square the linear scale factor; for volumes, cube it.
On a map, measure between the stated points accurately and use any given scale bar or numerical scale.
Examiners award method for applying the scale in the correct direction and then converting units correctly.

A scale of

1:n

multiplies every diagram length by

n

to give the corresponding real length.

Worked example

A map has scale $1:25\,000$ . A route measures $6.4\,\text{cm}$ on the map. Find the real distance in kilometres.

1.Real distance $=6.4\times25\,000=160\,000\,\text{cm}$ .
2. $160\,000\,\text{cm}=1600\,\text{m}$ .
3. $1600\,\text{m}=1.6\,\text{km}$ .

Answer: $1.6\,\text{km}$ .

Common mistakes

Don't divide by the scale factor when converting a diagram length to a real length.
Don't mix centimetres and metres inside the ratio before making the units consistent.
Don't use $n$ rather than $n^2$ when a scale-diagram question asks for area.

Exam tip

Write what $1\,\text{cm}$ represents before scaling, because this makes the direction of the conversion clear.

Tier 1 · Easy

1
A ranger's sketch uses scale $1:25\,000$ . A footpath trace measures $6.4\,\text{cm}$ . Find the real distance in kilometres.
(2)
(Total for Question 1 is 2 marks)
2
A shape is enlarged so that a side of length $8\,\text{cm}$ becomes $14\,\text{cm}$ . Work out the scale factor.
(1)
(Total for Question 2 is 1 mark)

Tier 2 · Standard

1
A model theatre uses scale $1:40$ . A real doorway is $2.04\,\text{m}$ high. Work out the model doorway height in centimetres.
(2)
(Total for Question 1 is 2 marks)
2
A real wall is $18\,\text{m}$ long and is drawn as $7.2\,\text{cm}$ . Write the scale of the drawing in the form $1:n$ .
(2)
(Total for Question 2 is 2 marks)
3
A floor plan uses scale $1 : 75$ . A rectangular storage bay measures $6.4\,\text{cm}$ by $3.6\,\text{cm}$ on the plan. Work out its real perimeter in metres.
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1
A site plan has scale $1:2500$ . A garden occupies $96\,\text{cm}^2$ on the plan. Calculate its real area in hectares. Use $1$ hectare $=10\,000\,\text{m}^2$ .
(4)
(Total for Question 1 is 4 marks)
2
A road is $9.6\,\text{cm}$ long on a map with scale $1:12\,500$ . Work out its length on a second map with scale $1:30\,000$ .
(3)
(Total for Question 2 is 3 marks)
3
A rectangular terrace measures $9.6\,\text{cm}$ by $6.4\,\text{cm}$ on a plan with scale $1 : 125$ . The terrace will be covered with square paving slabs of side $40\,\text{cm}$ . Work out the number of slabs needed. Assume that no slabs are wasted.
(5)
(Total for Question 3 is 5 marks)
4
A model storage tank is made to scale $1 : 40$ . The model has volume $54\,\text{cm}^3$ . Work out the volume of the real tank in litres.
(4)
(Total for Question 4 is 4 marks)
5
A lake covers $28.8\,\text{cm}^2$ on an aerial photograph with scale $1 : 8000$ . Work out the area it would cover on a plan with scale $1 : 12\,000$ .
(4)
(Total for Question 5 is 4 marks)

R3 · Express one quantity as a fraction of another, where the fraction is less than 1 or greater than 1

Explanation

To express quantity $A$ as a fraction of quantity $B$ , form $\dfrac{A}{B}$ and simplify.
The order of the wording fixes the order of the fraction: the quantity before “as a fraction of” is the numerator.
Quantities must first be written in the same units, so that the units cancel.
The result may be less than $1$ , equal to $1$ , or greater than $1$ ; an improper fraction is valid when the first quantity is larger.
Examiners expect a simplified exact fraction unless the question asks for another form.

Worked example

Express $0.84\,\text{m}^2$ as a fraction of $2400\,\text{cm}^2$ . Give the fraction in its simplest form.

1.Convert the area: $0.84\,\text{m}^2=0.84\times10\,000=8400\,\text{cm}^2$ .
2.Form the fraction in the stated order: $\dfrac{8400}{2400}$ .
3.Simplify: $\dfrac{8400}{2400}=\dfrac{84}{24}=\dfrac{7}{2}$ .

Answer: $\dfrac{7}{2}$ .

Common mistakes

Don't reverse the fraction and write the second quantity over the first.
Don't form a fraction before converting the two quantities to the same units.
Don't reject an answer greater than $1$ even though the first quantity is larger.

Exam tip

Underline the quantity named first and place it in the numerator before simplifying.

Tier 1 · Easy

1
Express $18$ as a fraction of $30$ . Give the fraction in its simplest form.
(2)
(Total for Question 1 is 2 marks)
2
Express $45$ pence as a fraction of £1.20. Give your answer in its simplest form.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1
Express $84$ minutes as a fraction of $35$ minutes. Simplify your answer.
(2)
(Total for Question 1 is 2 marks)
2
A park has $18$ oak trees, $12$ birch trees and $15$ beech trees. What fraction of these trees are birch? Give your answer in its simplest form.
(2)
(Total for Question 2 is 2 marks)
3
A runner travels $1.8\,\text{km}$ uphill and then $750\,\text{m}$ downhill. Express the total distance as a fraction of the uphill distance. Give your answer in its simplest form.
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1
A bottle contains $1.2$ litres of juice. Sam pours $350\,\text{ml}$ of the juice into a jug. Express the amount left in the bottle as a fraction of the amount poured into the jug. Give your answer in its simplest form.
(3)
(Total for Question 1 is 3 marks)
2
Two parcels have masses $1.8\,\text{kg}$ and $650\,\text{g}$ , then $240\,\text{g}$ is removed from the first parcel. Express the first parcel's new mass as a fraction of the new total mass.
(3)
(Total for Question 2 is 3 marks)
3
A floor has area $7.2\,\text{m}^2$ . Tiles cover $18\,000\,\text{cm}^2$ and a mat covers $1.35\,\text{m}^2$ . Express the uncovered area as a fraction of the covered area. Give your answer in its simplest form.
(4)
(Total for Question 3 is 4 marks)
4
Ribbon A is $2.4\,\text{m}$ long and ribbon B is $90\,\text{cm}$ long. A $35\,\text{cm}$ piece is cut from ribbon A and joined to ribbon B. Express the new length of ribbon B as a fraction of the new length of ribbon A. Give the fraction in its simplest form.
(4)
(Total for Question 4 is 4 marks)
5
Tank A has capacity $1.8\,\text{m}^3$ and is $65\%$ full. Tank B contains $420$ litres. Express the amount in tank A as a fraction of the total amount in both tanks. Give the fraction in its simplest form.
(4)
(Total for Question 5 is 4 marks)

R4 · Use ratio notation, including reduction to simplest form

Explanation

Ratio notation compares quantities in a fixed order, for example $a:b$ . Before simplifying, express every quantity in the same units.
An equivalent ratio is produced by multiplying or dividing every part by the same non-zero value.
For an integer ratio, divide all parts by their highest common factor; when decimals occur, first scale them to integers if helpful.
Ratios can have more than two parts, and a shared part can be matched using a common multiple.
Examiners require the stated order to be preserved and usually expect the simplest integer form.

Worked example

Given $x:y=4:7$ and $y:z=6:5$ , find $x:y:z$ in its simplest integer form.

1.Match the shared $y$ using $\operatorname{lcm}(7,6)=42$ .
2.Scale $4:7$ by $6$ to obtain $x:y=24:42$ .
3.Scale $6:5$ by $7$ to obtain $y:z=42:35$ , so $x:y:z=24:42:35$ .

Answer: $24:42:35$ .

Common mistakes

Don't change the order of the quantities while simplifying the ratio.
Don't divide only one part rather than every part by the same value.
Don't join two ratios without first making their shared part equal.

Exam tip

For linked ratios, show the common value of the repeated quantity to secure the method mark.

Tier 1 · Easy

1
Write the ratio $42:63$ in its simplest form.
(1)
(Total for Question 1 is 1 mark)
2
Write $0.6:1.5$ as a ratio in its simplest form.
(1)
(Total for Question 2 is 1 mark)

Tier 2 · Standard

1
Write $1.8\,\text{m}:75\,\text{cm}$ as a ratio in its simplest form.
(2)
(Total for Question 1 is 2 marks)
2
Write $1.2\,\text{kg}:450\,\text{g}:750\,\text{g}$ as a ratio in its simplest form.
(2)
(Total for Question 2 is 2 marks)
3
A class contains $18$ boys and $30$ girls. On one day, $\dfrac{1}{3}$ of the boys and $\dfrac{1}{5}$ of the girls are absent. Write the ratio of boys present to girls present in its simplest form.
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1
Write $\dfrac{2}{3}:1.5:\dfrac{5}{6}$ as a ratio in its simplest integer form.
(3)
(Total for Question 1 is 3 marks)
2
The ratio $p:q:r$ is $4:7:9$ . Write $(p+2q):(2r-p)$ in its simplest form.
(3)
(Total for Question 2 is 3 marks)
3
Write $2\,\text{m}/\text{s} : 150\,\text{cm}/\text{s} : 0.09\,\text{km}/\text{min}$ as a ratio in its simplest integer form.
(4)
(Total for Question 3 is 4 marks)
4
Write $32\% : 0.44 : \dfrac{3}{8}$ as a ratio of whole numbers in its simplest form.
(3)
(Total for Question 4 is 3 marks)
5
The ratios $a:b=3:5$ and $b:c=10:7$ . Work out $(a+c):(b-a)$ in its simplest form.
(4)
(Total for Question 5 is 4 marks)

R5 · Divide a quantity in a given part:part or part:whole ratio; express division into two parts as a ratio; apply ratio to real problems (conversion, comparison, scaling, mixing, concentrations)

Explanation

In a part:part ratio $a:b$ , the whole contains $a+b$ equal shares.
Divide the total by the sum of the ratio parts to find one share, then multiply by each part.
A part:whole statement needs different reading: if one part is $\dfrac{3}{8}$ of the whole, the other part is $\dfrac{5}{8}$ , giving ratio $3:5$ .
In mixing or concentration problems, track the amount of the relevant ingredient rather than simply averaging percentages.
Examiners expect both final parts to be labelled and to add back to the original total.

Worked example

A technician mixes a $20\%$ solution with a $50\%$ solution to make $18$ litres of a $40\%$ solution. Find the volume of each starting solution.

1.Let $x$ litres be the $50\%$ solution, so $18-x$ litres is the $20\%$ solution.
2.Equate solute amounts: $0.50x+0.20(18-x)=0.40\times18$ .
3. $0.30x=3.6$ , so $x=12$ and $18-x=6$ .

Answer: $12$ litres of the $50\%$ solution and $6$ litres of the $20\%$ solution.

Common mistakes

Don't divide the total by one ratio part instead of by the sum of all parts.
Don't treat a part:whole ratio as though both numbers were separate parts.
Don't average two concentrations without accounting for the volumes mixed.

Exam tip

Label each share or variable, then check that the parts sum to the stated whole.

Tier 1 · Easy

1
Divide £84 in the ratio $3:4$ .
(2)
(Total for Question 1 is 2 marks)
2
Two lengths are in the ratio $3:7$ and differ by $28\,\text{cm}$ . Work out the two lengths.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1
A box holds $240$ counters. Red counters make up $\frac{3}{8}$ of the whole. Find the number of red and non-red counters, and state their ratio.
(3)
(Total for Question 1 is 3 marks)
2
Red, blue and white paint are mixed in the ratio $2:5:3$ , and $1.8$ litres of white paint is used. Work out the total volume of paint.
(3)
(Total for Question 2 is 3 marks)
3
A theatre sells $288$ tickets. The ratio of adult tickets to child tickets is $7 : 5$ . Adult tickets cost £11 and child tickets cost £7. Work out the total money received.
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1
Aisha and Ben share some money in the ratio $3:5$ . Aisha gives £20 of her share to Ben. Their shares are now in the ratio $1:2$ . Work out the total amount of money they shared.
(5)
(Total for Question 1 is 5 marks)
2
A drink is mixed using concentrate and water in the ratio $3:11$ . Mia has $1.8$ litres of concentrate and $7.2$ litres of water. Work out the greatest volume of drink she can make.
(3)
(Total for Question 2 is 3 marks)
3
A $14$ -litre fruit drink contains mango juice and apple juice in the ratio $2 : 5$ . A different $12$ -litre drink contains mango juice and apple juice in the ratio $1 : 3$ . The two drinks are combined. Work out the ratio of mango juice to apple juice in the combined drink.
(4)
(Total for Question 3 is 4 marks)
4
A fund is divided between A, B and C. A receives $\dfrac{2}{5}$ of the whole fund. The rest is divided between B and C in the ratio $3 : 4$ . C receives £840. Work out the value of the whole fund.
(4)
(Total for Question 4 is 4 marks)
5
A container holds $12$ litres of a drink that is $30\%$ concentrate. Water is added so that the new drink is $18\%$ concentrate. Work out the volume of water added.
(4)
(Total for Question 5 is 4 marks)

R6 · Express a multiplicative relationship between two quantities as a ratio or a fraction

Explanation

A multiplicative relationship compares quantities by division rather than subtraction. If $A=kB$ , then $\dfrac{A}{B}=k$ , so $A:B=k:1$ and $A$ is the fraction $k$ of $B$ .
Convert decimal multipliers to fractions when an integer ratio is required; for example, $1.75=\dfrac{7}{4}$ gives $A:B=7:4$ .
For a chain of relationships, substitute one equation into another or match the shared ratio part.
Check that a multiplier greater than $1$ corresponds to the larger quantity.
Examiners expect the direction of the comparison to agree with the wording, because reversing it produces the reciprocal.

Worked example

Quantities $P$ , $Q$ and $R$ satisfy $P=\dfrac{5}{3}Q$ and $Q=\dfrac{3}{4}R$ . Express $P:R$ in simplest form.

1.Substitute $Q=\dfrac{3}{4}R$ into $P=\dfrac{5}{3}Q$ .
2. $P=\dfrac{5}{3}\times\dfrac{3}{4}R=\dfrac{5}{4}R$ .
3.Therefore $P:R=\dfrac{5}{4}:1=5:4$ .

Answer: $P:R=5:4$ .

Common mistakes

Don't subtract the quantities and report an additive difference instead of a multiplier.
Don't reverse the comparison and give the reciprocal ratio.
Don't leave a ratio containing fractions when a simplest integer ratio is required.

Exam tip

Translate “ $A$ is $k$ times $B$ ” into $A=kB$ before converting it to a ratio.

Tier 1 · Easy

1
Quantity $A$ is $18$ and quantity $B$ is $30$ . Express $A$ as a fraction of $B$ and write $A:B$ in simplest form.
(2)
(Total for Question 1 is 2 marks)
2
Quantity $A$ is $\dfrac{3}{8}$ of quantity $B$ . Express $A:B$ in its simplest form.
(1)
(Total for Question 2 is 1 mark)

Tier 2 · Standard

1
The mass of parcel $P$ is $1.75$ times the mass of parcel $Q$ . Express $P:Q$ as an integer ratio and express $P$ as a fraction of $Q$ .
(2)
(Total for Question 1 is 2 marks)
2
Quantity $P$ is $40\%$ greater than quantity $Q$ . Express $P:Q$ in its simplest form and write $Q$ as a fraction of $P$ .
(2)
(Total for Question 2 is 2 marks)
3
The ratio $M : N$ is $7 : 12$ . Express $N-M$ as a fraction of $M$ , and write $M : (N-M)$ in its simplest form.
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1
Quantities $P$ , $Q$ and $R$ satisfy $P=\frac{5}{3}Q$ and $Q=\frac{3}{4}R$ . Express $P:R$ in simplest form and write $P$ as a fraction of $R$ .
(3)
(Total for Question 1 is 3 marks)
2
Quantity $A$ is $35\%$ greater than quantity $B$ . Express the difference $A-B$ as a fraction of $A$ , and write $(A-B):A$ in its simplest form.
(3)
(Total for Question 2 is 3 marks)
3
Quantity $A$ is $20\%$ less than quantity $B$ . Quantity $C$ is $50\%$ greater than quantity $A$ . Express $C : B$ in its simplest form and write $B$ as a fraction of $C$ .
(4)
(Total for Question 3 is 4 marks)
4
Quantity A is $1.6$ times quantity B. The difference between A and B is $42$ . Express this difference as a fraction of the total of A and B. Give the fraction in its simplest form.
(4)
(Total for Question 4 is 4 marks)
5
The price of item P is $\dfrac{5}{6}$ of the price of item Q. The price of item Q is $30\%$ less than the price of item R. Express the price of P as a fraction of the price of R, and write P:R in its simplest form.
(4)
(Total for Question 5 is 4 marks)

R7 · Understand and use proportion as equality of ratios

Explanation

Proportion means that two corresponding ratios are equal. Keep the quantities in the same order on both sides, then solve by scaling, finding a unit value, or cross-multiplying.
If $\dfrac{a}{b}=\dfrac{c}{d}$ with non-zero denominators, then $ad=bc$ .
A unitary method is often clearest in context: find the amount for one unit and scale to the required number.
Decimal answers involving money must be interpreted in pounds and pence.
Examiners award method for a correct proportional relationship even if a later arithmetic slip occurs, so show the equation or unit value.

Worked example

Eight identical notebooks cost £11.20. Use proportion to find the cost of $14$ notebooks.

1.Find the cost of one notebook: $11.20\div8=1.40$ .
2.Scale to $14$ notebooks: $14\times1.40=19.60$ .
3.State the money answer using two decimal places.

Answer: £19.60.

Common mistakes

Don't place corresponding quantities in different orders in the two ratios.
Don't add the same amount instead of multiplying by the same scale factor.
Don't write £19.6 without interpreting the final zero as pence in a money context.

Exam tip

Show either the equal-ratios equation or the value of one unit before giving the scaled answer.

Tier 1 · Easy

1
Solve the proportion $\frac{x}{15}=\frac{6}{10}$ .
(2)
(Total for Question 1 is 2 marks)
2
Which ratio is equal to $5:8$ ? A $10:18$ B $15:24$ C $20:30$ . Write down the letter of the correct ratio.
(1)
(Total for Question 2 is 1 mark)

Tier 2 · Standard

1
A recipe uses $180$ g of flour for every $120$ g of sugar. Leo uses $315$ g of flour and $200$ g of sugar. Has Leo used flour and sugar in the correct proportion? Show your working.
(3)
(Total for Question 1 is 3 marks)
2
A photograph has width to height ratio $7:4$ . Work out its width when its height is $18\,\text{cm}$ .
(2)
(Total for Question 2 is 2 marks)
3
Nine identical bolts have a total mass of $234\,\text{g}$ . Work out the total mass of $14$ of these bolts.
(2)
(Total for Question 3 is 2 marks)

Tier 3 · Hard

1
Solve $\frac{2x+3}{18}=\frac{x+6}{12}$ and verify that the two ratios are equal.
(4)
(Total for Question 1 is 4 marks)
2
A printer produces $126$ pages in $3.5$ minutes at a constant rate. Work out how long it takes to produce $270$ pages.
(3)
(Total for Question 2 is 3 marks)
3
A fruit drink contains syrup and water in the ratio $3 : 8$ . After $1.5$ litres of water evaporates, the ratio is $3 : 7$ . Work out the original volume of the drink.
(4)
(Total for Question 3 is 4 marks)
4
A photograph is $24\,\text{cm}$ wide and $16\,\text{cm}$ high. A strip of width $x\,\text{cm}$ is cut from both the left edge and the right edge. The new width and height are in the ratio $4 : 3$ . Work out $x$ .
(4)
(Total for Question 4 is 4 marks)
5
A recipe uses $400\,\text{g}$ of toasted oats to make $14$ snack bars. During toasting, oats lose $4\%$ of their mass. Work out the greatest number of complete snack bars that can be made from $1.875\,\text{kg}$ of untoasted oats.
(4)
(Total for Question 5 is 4 marks)

R8 · Relate ratios to fractions and to linear functions

Explanation

For parts in ratio $a:b$ , the whole has $a+b$ equal parts, so their fractions of the whole are $\dfrac{a}{a+b}$ and $\dfrac{b}{a+b}$ . A constant ratio between variables can also define a linear function.
If $y:x=m:1$ , then $\dfrac{y}{x}=m$ and $y=mx$ .
Its graph is a straight line through the origin, with gradient $m$ .
A line with a non-zero intercept does not represent a constant ratio.
Examiners may ask you to move between ratio, fraction, equation and graph, so always state the link explicitly.

Worked example

Concentrate and water are mixed in the ratio $2:7$ . Let $c$ be the concentrate volume and $V$ the total volume. Express $V$ as a function of $c$ , then find both volumes when $V=54$ .

1.Concentrate is $\dfrac{2}{2+7}=\dfrac{2}{9}$ of the total, so $c=\dfrac{2}{9}V$ .
2.Rearrange to $V=\dfrac{9}{2}c$ .
3.When $V=54$ , $c=\dfrac{2}{9}\times54=12$ and water $=54-12=42$ .

Answer: $V=\dfrac{9}{2}c$ ; $12$ litres concentrate and $42$ litres water.

Common mistakes

Don't use $\dfrac{a}{b}$ instead of $\dfrac{a}{a+b}$ for a fraction of the whole.
Don't write a constant-ratio graph with a non-zero intercept.
Don't treat the total as one ratio part rather than the sum of the parts.

Exam tip

When linking ratio to a function, verify that the equation gives $y=0$ when $x=0$ .

Tier 1 · Easy

1
The ratio of cats to dogs at a shelter is $3:5$ . What fraction of the animals are cats?
(2)
(Total for Question 1 is 2 marks)
2
The equation $y=4x$ describes a constant ratio. Write $y:x$ in its simplest form.
(1)
(Total for Question 2 is 1 mark)

Tier 2 · Standard

1
Quantities $y$ and $x$ are always in the ratio $5:2$ . Write $y$ as a linear function of $x$ , then find $y$ when $x=14$ .
(3)
(Total for Question 1 is 3 marks)
2
A rectangle's length and width are in the ratio $5:3$ . Let $P$ be its perimeter and $w$ its width. Express $P$ as a linear function of $w$ .
(3)
(Total for Question 2 is 3 marks)
3
In a choir, the numbers of sopranos, altos and tenors are in the ratio $5 : 3 : 2$ . Let $t$ be the number of tenors and $N$ the total number of singers. Express $N$ as a linear function of $t$ . Then find $N$ when $t=18$ .
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1
A mixture contains concentrate and water in the ratio $2:7$ . Let $c$ litres be the concentrate and $V$ litres be the total mixture. Express $V$ as a linear function of $c$ , then find both component volumes when $V=54$ .
(4)
(Total for Question 1 is 4 marks)
2
The pairs $(4,14)$ , $(7,24.5)$ and $(10,35)$ follow a constant-ratio linear function connecting $x$ and $y$ . Work out $x$ when $y=56$ .
(3)
(Total for Question 2 is 3 marks)
3
Blue and red beads are used in the ratio $4 : 7$ . Each blue bead has mass $3\,\text{g}$ and each red bead has mass $5\,\text{g}$ . Let $b$ be the number of blue beads and $M$ grams be the total mass. Express $M$ as a linear function of $b$ . Work out the number of red beads when $M=282$ .
(4)
(Total for Question 3 is 4 marks)
4
Three quantities are in the ratio $2 : 3 : 7$ . Let $d$ be the difference between the largest quantity and the sum of the other two, and let $T$ be the total of all three quantities. Express $T$ as a linear function of $d$ . Work out $T$ when $d=54$ .
(4)
(Total for Question 4 is 4 marks)
5
A straight line connecting $x$ and $y$ passes through $(4,11)$ and $(10,23)$ . Work out the equation of the line. Does the relationship give a constant ratio $y:x$ ? You must show all your working.
(4)
(Total for Question 5 is 4 marks)

R9 · Define percentage as 'number of parts per hundred'; interpret percentages and percentage changes as fractions/decimals, multiplicatively; percentages > 100%; percentage change and simple interest

Explanation

A percentage is a number of parts per hundred, so $p\%=\dfrac{p}{100}$ . Percentage change is multiplicative: an increase of $r\%$ uses multiplier $1+\dfrac{r}{100}$ and a decrease uses $1-\dfrac{r}{100}$ .
Percentages above $100\%$ correspond to multipliers greater than $1$ .
To reverse a change, divide by the multiplier; do not apply the opposite percentage to the changed value.
Simple interest is calculated each year from the original principal, so the yearly interest is constant.
Examiners expect the original amount to be the denominator when calculating percentage change.

Worked example

After a $12\%$ decrease, a machine is worth £704. Find its value before the decrease.

1.A $12\%$ decrease leaves $100\%-12\%=88\%$ .
2.Write $0.88\times\text{original}=704$ .
3.Original $=704\div0.88=800$ .

Answer: £800.

Common mistakes

Don't add $12\%$ to the reduced value when reversing a $12\%$ decrease.
Don't use the new value rather than the original value as the denominator for percentage change.
Don't calculate simple interest from a growing balance as though it were compound interest.

Exam tip

For reverse percentage, write the multiplier equation and divide by the multiplier.

Tier 1 · Easy

1
Work out $35\%$ of $240$ .
(2)
(Total for Question 1 is 2 marks)
2
Work out $135\%$ of $80$ .
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1
After a $12\%$ decrease, a machine is valued at £704. Work out its value before the decrease.
(3)
(Total for Question 1 is 3 marks)
2
The price of an item increases from £72 to £81. Work out the percentage increase.
(3)
(Total for Question 2 is 3 marks)
3
A workshop makes $840$ parts in one month. This is $140\%$ of the number made in the previous month. Work out the number of parts made in the previous month.
(2)
(Total for Question 3 is 2 marks)

Tier 3 · Hard

1
A saver deposits £2500 in an account paying $3.6\%$ simple interest each year. After $5$ years, a fee equal to $2\%$ of the final balance is charged. Calculate the amount left after the fee.
(5)
(Total for Question 1 is 5 marks)
2
An investment earns £378 in simple interest over $3$ years at a rate of $4.5\%$ per year. Work out the original amount invested.
(3)
(Total for Question 2 is 3 marks)
3
Nadia invests £1800 at $4\%$ simple interest per year. She invests another amount at $2.5\%$ simple interest per year. After $3$ years, the total interest from the two investments is £396. Work out the second amount invested.
(4)
(Total for Question 3 is 4 marks)
4
A quantity is increased by $20\%$ and then decreased by $p\%$ . Its final value is $108\%$ of its original value. Work out $p$ .
(4)
(Total for Question 4 is 4 marks)
5
A tank is initially $80\%$ full. Then $15\%$ of the water in it is used. After $102$ litres are added, the tank is $92\%$ full. Work out the capacity of the tank.
(4)
(Total for Question 5 is 4 marks)

R10 · Solve problems involving direct and inverse proportion, including graphical and algebraic representations

Explanation

For direct proportion, $y\propto x$ means $y=kx$ .
The ratio $\dfrac{y}{x}$ is constant and the graph is a straight line through the origin.
For inverse proportion, $y\propto\dfrac{1}{x}$ means $y=\dfrac{k}{x}$ , so the product $xy$ is constant and the graph is a decreasing curve for positive values.
Find $k$ from a known pair, write the equation, then substitute the required value.
Examiners expect the equation with the constant of proportionality, not only a numerical scaling argument, when the command is “find a formula”.

Direct proportion gives a straight line through the origin; inverse proportion gives a decreasing reciprocal curve for positive values.

Worked example

$x$ and $y$ are inversely proportional. When $x=6$ , $y=12$ . Find $y$ after $x$ increases by $25\%$ .

1.Use $xy=k$ : $k=6\times12=72$ .
2.New $x=6\times1.25=7.5$ .
3.New $y=\dfrac{72}{7.5}=9.6$ .

Answer: $y=9.6$ .

Common mistakes

Don't write $y=kx$ for an inverse-proportion relationship.
Don't draw a direct-proportion line that does not pass through the origin.
Don't find the constant correctly and fail to use it in a complete equation.

Exam tip

Write $y=kx$ or $y=\dfrac{k}{x}$ before substituting values; this is the key method step.

Tier 1 · Easy

1
A direct variation links $x$ and $y$ . The pair $x=6$ , $y=18$ is known. Determine $y$ at $x=10$ .
(3)
(Total for Question 1 is 3 marks)
2
A $y=11x$ B $y=\dfrac{11}{x}$ C $y=x+11$ . Write down the letter of the equation that shows $y$ is directly proportional to $x$ .
(1)
(Total for Question 2 is 1 mark)

Tier 2 · Standard

1
Variables $x$ and $y$ vary inversely. One recorded pair is $x=3$ , $y=14$ . Determine $y$ at $x=7$ .
(3)
(Total for Question 1 is 3 marks)
2
A direct-proportion graph has gradient $2.8$ . Work out the $x$ -coordinate of the point on the graph where $y=35$ .
(2)
(Total for Question 2 is 2 marks)
3
A fixed amount of feed lasts $18$ goats for $28$ days. Each goat eats the same amount each day. Work out how many days the feed would last $24$ goats.
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1
$x$ and $y$ are inversely proportional. Initially $x=6$ and $y=12$ . The value of $x$ is increased by $25\%$ . Find the new value of $y$ and the percentage decrease in $y$ .
(5)
(Total for Question 1 is 5 marks)
2
The pairs $(2,30)$ , $(3,20)$ , $(5,12)$ and $(8,8)$ are meant to show inverse proportion. Work out the corrected value of $y$ when $x=8$ .
(3)
(Total for Question 2 is 3 marks)
3
The mass of a cable is directly proportional to its length. A $3.6\,\text{m}$ length has mass $1.26\,\text{kg}$ . A drum holding some of this cable has total mass $8.4\,\text{kg}$ , and the empty drum has mass $1.4\,\text{kg}$ . Work out the length of cable on the drum.
(4)
(Total for Question 3 is 4 marks)
4
Six identical scanners can process one batch of documents in $14$ minutes. After all six scanners have worked for $4$ minutes, two scanners stop. The remaining scanners continue at the same rate. Work out the total time taken to process the batch.
(5)
(Total for Question 4 is 5 marks)
5
The volume $V$ litres delivered by a pump is directly proportional to the running time $t$ minutes. The pump delivers $270$ litres in $7.5$ minutes. Write an equation connecting $V$ and $t$ . A tank of capacity $750$ litres already contains $138$ litres. Work out how long the pump must run to fill the tank.
(4)
(Total for Question 5 is 4 marks)

R11 · Use compound units such as speed, rates of pay, unit pricing, density and pressure

Explanation

A compound unit combines quantities, usually through division. Use $\text{speed}=\dfrac{\text{distance}}{\text{time}}$ , $\text{density}=\dfrac{\text{mass}}{\text{volume}}$ and $\text{pressure}=\dfrac{\text{force}}{\text{area}}$ .
Rates of pay and unit prices are totals divided by the relevant time or number of items.
Make the units compatible before substituting, and rearrange the formula if the unknown is in the numerator or denominator.
The units provide a useful check on the operation and show which quantity should be divided by which.
Examiners expect both the numerical value and the correct compound unit, such as $\text{kg}/\text{m}^3$ .

Worked example

A solid has mass $18.9\,\text{kg}$ and volume $0.0075\,\text{m}^3$ . Find its density.

1.Select $\text{density}=\dfrac{\text{mass}}{\text{volume}}$ .
2.Substitute: $\text{density}=\dfrac{18.9}{0.0075}$ .
3.Evaluate and attach the units: $2520\,\text{kg}/\text{m}^3$ .

Answer: $2520\,\text{kg}/\text{m}^3$ .

Common mistakes

Don't divide volume by mass when calculating density.
Don't substitute minutes into a formula when the requested speed is per hour.
Don't give a numerical answer without the required compound unit.

Exam tip

Write the compound-unit formula first, because a correct formula can earn a method mark despite arithmetic error.

Tier 1 · Easy

1
A shift lasting $7$ hours pays £52.50. Work out the hourly rate of pay.
(2)
(Total for Question 1 is 2 marks)
2
A machine packs $54$ cartons in $6$ minutes. Work out the rate in cartons per minute.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1
A coach travels $156\,\text{km}$ in $2$ hours $24$ minutes. Calculate its average speed in $\text{km}/\text{h}$ .
(3)
(Total for Question 1 is 3 marks)
2
A material has mass $5.46\,\text{kg}$ and density $780\,\text{kg}/\text{m}^3$ . Work out its volume.
(3)
(Total for Question 2 is 3 marks)
3
A $750\,\text{g}$ pack of cereal costs £3.30. A $1.2\,\text{kg}$ pack costs £5.16. Work out which pack is better value and by how much per kilogram.
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1
A force of $3.6\,\text{kN}$ acts uniformly on a rectangular pad measuring $24\,\text{cm}$ by $15\,\text{cm}$ . Calculate the pressure in pascals, where $1\,\text{Pa}=1\,\text{N}/\text{m}^2$ .
(4)
(Total for Question 1 is 4 marks)
2
A printer uses $0.45$ litres of ink to print $18\,000$ pages, and the ink costs £28 per litre. Work out the ink cost per $1000$ pages.
(4)
(Total for Question 2 is 4 marks)
3
A technician works $40$ hours in one week. The first $36$ hours are paid at £12.40 per hour and the remaining hours are paid at one and a half times this rate. Work out the technician's average rate of pay for the week.
(4)
(Total for Question 3 is 4 marks)
4
A driver travels $72\,\text{km}$ at $48\,\text{km}/\text{h}$ , stops for $18$ minutes, then travels $96\,\text{km}$ at $64\,\text{km}/\text{h}$ . Work out the average speed for the whole journey, including the stop. Round your answer to the nearest $0.1\,\text{km}/\text{h}$ .
(5)
(Total for Question 4 is 5 marks)
5
A cuboid has dimensions $0.5\,\text{m}$ by $0.25\,\text{m}$ by $0.08\,\text{m}$ and density $2400\,\text{kg}/\text{m}^3$ . It rests on its $0.5\,\text{m}$ by $0.25\,\text{m}$ face. Work out the pressure it exerts on the floor. Use $10\,\text{N}/\text{kg}$ for gravitational field strength.
(5)
(Total for Question 5 is 5 marks)

R12 · Compare lengths, areas and volumes using ratio notation; make links to similarity (including trigonometric ratios) and scale factors

Explanation

Similar shapes have equal corresponding angles and proportional corresponding lengths. If the linear scale factor from one shape to another is $k$ , the length ratio is $k$ , the area ratio is $k^2$ , and the volume ratio is $k^3$ .
Match corresponding measurements and keep the comparison order consistent.
To recover a length factor from an area ratio take a square root; from a volume ratio take a cube root.
Trigonometric ratios remain constant in similar right-angled triangles.
Examiners expect you to identify whether the measurements are lengths, areas or volumes before applying the scale factor.

Corresponding lengths in similar shapes scale by

k

, so their areas scale by

k^2

Worked example

Two similar solids have smaller-to-larger volume ratio $125:216$ . The smaller surface area is $275\,\text{cm}^2$ . Find the larger surface area.

1. $125:216=5^3:6^3$ , so the length ratio is $5:6$ .
2.The surface-area ratio is $5^2:6^2=25:36$ .
3.Larger area $=275\times\dfrac{36}{25}=396\,\text{cm}^2$ .

Answer: $396\,\text{cm}^2$ .

Common mistakes

Don't use the linear scale factor directly for an area or volume.
Don't pair non-corresponding sides when forming the scale factor.
Don't take a square root when recovering a length factor from a volume ratio.

Exam tip

Annotate the scale factor as $k$ , $k^2$ or $k^3$ before calculating.

Tier 1 · Easy

1
Two similar shapes have corresponding sides of $6\,\text{cm}$ and $15\,\text{cm}$ . Write the smaller-to-larger length ratio in simplest form.
(1)
(Total for Question 1 is 1 mark)
2
Two similar posters have lengths in the ratio $2 : 5$ . Write down the ratio of their areas.
(1)
(Total for Question 2 is 1 mark)

Tier 2 · Standard

1
The corresponding length ratio of two similar tiles is $3:7$ . The smaller tile has area $54\,\text{cm}^2$ . Find the area of the larger tile.
(3)
(Total for Question 1 is 3 marks)
2
Two similar right-angled triangles have the same acute angle. In the smaller triangle, the side opposite this angle is $4\,\text{cm}$ and the hypotenuse is $5\,\text{cm}$ . Work out the opposite side in the larger triangle when its hypotenuse is $17.5\,\text{cm}$ .
(3)
(Total for Question 2 is 3 marks)
3
Two similar shapes have areas $96\,\text{cm}^2$ and $150\,\text{cm}^2$ . A side of the smaller shape is $12\,\text{cm}$ . Work out the length of the corresponding side of the larger shape.
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1
Two similar solids have smaller-to-larger volume ratio $125:216$ . The smaller solid has surface area $275\,\text{cm}^2$ . Work out the larger surface area.
(4)
(Total for Question 1 is 4 marks)
2
A rectangular photograph measures $8\,\text{cm}$ by $12\,\text{cm}$ and is enlarged so that its perimeter is $75\,\text{cm}$ . Work out the area of the enlarged photograph.
(4)
(Total for Question 2 is 4 marks)
3
Two similar rectangles have corresponding lengths in the ratio $3 : 5$ . The two corresponding longer sides differ by $14\,\text{cm}$ . The smaller rectangle has area $189\,\text{cm}^2$ . Work out the area of the larger rectangle.
(4)
(Total for Question 3 is 4 marks)
4
Two similar solids have smaller-to-larger surface-area ratio $81 : 144$ . The larger solid has volume $512\,\text{cm}^3$ . Work out the volume of the smaller solid.
(4)
(Total for Question 4 is 4 marks)
5
Two similar containers have volumes in the ratio $64 : 125$ . Their corresponding heights differ by $9\,\text{cm}$ . Work out the sum of their heights.
(4)
(Total for Question 5 is 4 marks)

R13 · Understand that X is inversely proportional to Y is equivalent to X is proportional to 1/Y; construct and interpret equations that describe direct and inverse proportion

Explanation

If $X$ is inversely proportional to $Y$ , then $X\propto\dfrac{1}{Y}$ and $X=\dfrac{k}{Y}$ for a constant $k$ ; equivalently, $XY=k$ . Foundation questions can require interpreting equations that describe direct and inverse proportion.
Higher tier: construct equations involving a power, such as $y=kx^2$ or $y=\dfrac{k}{x^n}$ .
Substitute a known pair to find $k$ , write the complete equation, then use it to find an unknown.
Check any stated restrictions, such as a positive length, before choosing a root.
Examiners usually award separate method marks for the correct proportional form and for finding the constant.

Worked example

Higher tier: For positive $b$ , $a$ varies inversely as $b^2$ . Given $a=20$ when $b=3$ , find $b$ when $a=7.2$ .

1.Write $a=\dfrac{k}{b^2}$ .
2.Use the first pair: $k=ab^2=20\times3^2=180$ .
3. $7.2=\dfrac{180}{b^2}$ , so $b^2=25$ and the positive value is $b=5$ .

Answer: $b=5$ .

Common mistakes

Don't write $X=kY$ when the relationship is inverse proportion.
Don't use $y=kx$ when the stated proportional quantity is $x^2$ .
Don't find $k$ but never write or use the complete proportional equation.

Exam tip

Translate the wording into a formula containing $k$ before substituting any values.

Tier 1 · Easy

1
Variables $p$ and $q$ vary inversely, with recorded values $p=12$ and $q=5$ . Write their connecting equation.
(2)
(Total for Question 1 is 2 marks)
2
A $xy=14$ B $y=14x$ C $y=x-14$ . Write down the letter of the equation that shows $y$ is inversely proportional to $x$ .
(1)
(Total for Question 2 is 1 mark)

Tier 2 · Standard

1
The relationship between $y$ and $x^2$ is direct proportion. Given $x=3$ when $y=45$ , form the equation and evaluate $y$ at $x=4$ .
(4)
(Total for Question 1 is 4 marks)
2
The variables $p$ and $q$ satisfy $p=\dfrac{48}{q}$ . When $q$ increases from $3$ to $12$ , work out the fraction of its original value that $p$ becomes.
(3)
(Total for Question 2 is 3 marks)
3
The equation $p=\dfrac{72}{q}$ connects two inversely proportional variables. State whether $p=9$ , $q=8$ satisfies this equation. Work out $q$ when $p=12$ .
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1
For positive $b$ , the variable $a$ varies inversely with $b^2$ . Given $a=20$ at $b=3$ , determine $b$ when $a=7.2$ .
(4)
(Total for Question 1 is 4 marks)
2
A rectangular field has length $x$ metres and width $y$ metres. The width is inversely proportional to the length. When $x=18$ , $y=14$ . Find a formula for $y$ in terms of $x$ . Then work out the perimeter of the field when $x=21$ .
(4)
(Total for Question 2 is 4 marks)
3
Higher tier: $u$ is directly proportional to $v^3$ . When $v=2$ , $u=28$ . Write down an equation connecting $u$ and $v$ . Work out the positive value of $v$ when $u=756$ .
(4)
(Total for Question 3 is 4 marks)
4
Higher tier: For positive $x$ , $y$ is inversely proportional to $x^3$ . When $x=2$ , $y=16$ . Write an equation connecting $x$ and $y$ . Work out $x$ when $y=2$ .
(4)
(Total for Question 4 is 4 marks)
5
Higher tier: $A$ is directly proportional to $b^2$ . When $b=3$ , $A=54$ . The value of $b$ is decreased by $20\%$ . Work out the percentage decrease in $A$ .
(4)
(Total for Question 5 is 4 marks)

R14 · Interpret the gradient of a straight line graph as a rate of change; recognise and interpret graphs that illustrate direct and inverse proportion

Explanation

The gradient of a straight line is $\dfrac{\text{change in }y}{\text{change in }x}$ , so it represents a rate of change with units taken from the axes.
Choose two well-separated points on the line, not merely nearby grid intersections, and calculate rise over run.
In context, state what the rate means: on a distance-time graph it is speed; on a cost-time graph it is cost per unit time.
A direct-proportion graph is straight through the origin, while an inverse-proportion graph has constant product $xy$ .
Examiners expect both the gradient and its contextual interpretation.

The gradient of a straight line is rise divided by run, with units of vertical-axis units per horizontal-axis unit.

Worked example

A distance-time line passes through $(2,10)$ and $(7,35)$ , with time in seconds and distance in metres. Find and interpret its gradient.

1.Change in distance $=35-10=25\,\text{m}$ .
2.Change in time $=7-2=5\,\text{s}$ .
3.Gradient $=25\div5=5\,\text{m}/\text{s}$ , which is the speed.

Answer: The object travels at $5\,\text{m}/\text{s}$ .

Common mistakes

Don't calculate run divided by rise instead of rise divided by run.
Don't use points that are not both on the straight line.
Don't give a bare gradient without its units or contextual meaning.

Exam tip

For “interpret the gradient”, give a value, compound unit and sentence explaining the rate.

Tier 1 · Easy

1
A straight distance-time graph passes through $(2,10)$ and $(7,35)$ , where time is in seconds and distance in metres. Find and interpret its gradient.
(3)
(Total for Question 1 is 3 marks)
2
A graph shows the mass of grain processed, in kilograms, against time in minutes. The straight line passes through the origin and $(6,27)$ . Work out its gradient and give the units.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1
A phone-call cost graph is modelled by $C=18+0.12m$ , where $C$ is cost in pounds and $m$ is time in minutes. Interpret the gradient and calculate the cost of a $35$ -minute call.
(3)
(Total for Question 1 is 3 marks)
2
Line A passes through $(2,7)$ and $(8,25)$ , while line B has gradient $2.5$ . Work out which line has the greater rate of change.
(3)
(Total for Question 2 is 3 marks)
3
A straight-line graph shows the mass of ice left in a cooler. It passes through $(3,870)$ and $(11,710)$ , where time is in minutes and mass is in grams. Work out the rate at which the ice is melting and the mass of ice at time $0$ .
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1
An inverse-proportion model contains the points $(2,18)$ , $(3,12)$ and $(6,6)$ . Check that all three coordinates are consistent with the model, write its equation, and find $y$ when $x=9$ .
(4)
(Total for Question 1 is 4 marks)
2
A straight line passes through $(3,14)$ and $(7,30)$ . Work out its gradient and give a reason why the line does not represent direct proportion.
(3)
(Total for Question 2 is 3 marks)
3
A graph shows the volume $V$ litres left in a tank after $t$ minutes. Two straight-line sections join the points $(0,150)$ , $(5,120)$ and $(13,56)$ . Work out during which time interval the tank is draining faster, and by how many litres per minute.
(4)
(Total for Question 3 is 4 marks)
4
Graph A is straight; it passes through $(0,0)$ and through $(5,35)$ . Graph B is an inverse-proportion curve through $(4,30)$ . Work out the sum of the two $y$ -values when $x=10$ .
(4)
(Total for Question 4 is 4 marks)
5
A straight-line graph shows energy used, in kilowatt-hours, against running time, in hours. The line passes through $(1.5,4.8)$ and $(6.5,16.8)$ . Work out and interpret the gradient. At this constant rate, work out the running time needed for the energy used to increase by $30$ kilowatt-hours.
(4)
(Total for Question 5 is 4 marks)

R15 · Interpret the gradient at a point on a curve as the instantaneous rate of change; apply average and instantaneous rates of change (gradients of chords and tangents) (not calculus) [Higher only]

Explanation

For a curve, the gradient changes. The gradient of the chord joining two curve points gives the average rate of change over that interval.
The gradient of a tangent at one point estimates the instantaneous rate of change there; GCSE questions use a drawn tangent, not calculus. Choose two well-separated, readable points on the chord or tangent and calculate $\dfrac{\Delta y}{\Delta x}$ .
The chosen tangent points need not lie on the original curve.
Include compound units and compare rates using their values.
Examiners allow a sensible range when answers depend on drawing and reading a tangent.

A chord estimates average rate over an interval; a tangent estimates instantaneous rate at one point.

Worked example

A tangent to a curve at $x=7$ passes through $(4,11)$ and $(10,32)$ . Estimate the instantaneous rate of change.

1.Use two clear points on the tangent.
2.Change in $y=32-11=21$ and change in $x=10-4=6$ .
3.Tangent gradient $=\dfrac{21}{6}=3.5$ .

Answer: $3.5$ units of $y$ per unit of $x$ .

Common mistakes

Don't use two points on the curve instead of two points on the drawn tangent.
Don't call a chord gradient the instantaneous rate of change.
Don't read points too close together, magnifying graph-reading error.

Exam tip

Draw a large tangent triangle and show $\dfrac{\Delta y}{\Delta x}$ ; a sensible estimate range is normally accepted.

Tier 1 · Easy

1
A curve passes through $(2,5)$ and $(8,23)$ . Calculate the average rate of change of $y$ with respect to $x$ between these points.
(2)
(Total for Question 1 is 2 marks)
2
Write down whether a chord or a tangent is used to estimate the instantaneous rate of change at a point on a curve.
(1)
(Total for Question 2 is 1 mark)

Tier 2 · Standard

1
A tangent to a curve at $x=7$ passes through the grid points $(4,11)$ and $(10,32)$ . Estimate the instantaneous rate of change at $x=7$ .
(3)
(Total for Question 1 is 3 marks)
2
A tangent to a temperature-time curve passes through $(1,46)$ and $(7,19)$ , where time is in minutes and temperature is in degrees Celsius. Estimate the instantaneous rate of change of temperature.
(3)
(Total for Question 2 is 3 marks)
3
A curve shows the fuel remaining in a vehicle, in litres, after travelling $d$ kilometres. The tangent at $d=170$ passes through $(80,52)$ and $(260,43)$ . Estimate the instantaneous rate at which fuel is being used, in litres per $100\,\text{km}$ .
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1
A curve shows water volume $V$ litres after $t$ minutes and passes through $(2,46)$ and $(8,118)$ . The tangent at $t=5$ passes through $(4,70)$ and $(7,112)$ . Find the average rate from $t=2$ to $t=8$ , estimate the instantaneous rate at $t=5$ , and compare them.
(5)
(Total for Question 1 is 5 marks)
2
A graph shows distance $d$ metres travelled after time $t$ seconds. The tangent at $t=4$ passes through $(2,15)$ and $(8,51)$ , while the tangent at $t=10$ passes through $(8,55)$ and $(14,73)$ . Compare the two instantaneous speeds.
(4)
(Total for Question 2 is 4 marks)
3
A distance-time curve has a tangent at $t=45$ seconds passing through $(20,140)$ and $(70,390)$ , where distance is in metres. Estimate the instantaneous speed at $t=45$ . If this speed is maintained, work out the time needed to travel a further $1.2\,\text{km}$ .
(4)
(Total for Question 3 is 4 marks)
4
At a chosen point on a distance-time curve, the drawn tangent goes through $(8,1.9)$ and $(23,2.5)$ , where time is in minutes and distance is in kilometres. Estimate the instantaneous speed there, in kilometres per hour.
(4)
(Total for Question 4 is 4 marks)
5
A graph shows the volume of liquid in a vertical tank against time. A tangent to the curve passes through $(12,920)$ and $(27,560)$ , where time is in minutes and volume is in litres. The tank has constant horizontal cross-sectional area $0.8\,\text{m}^2$ . Estimate the instantaneous rate at which the liquid level is falling, in centimetres per minute.
(5)
(Total for Question 5 is 5 marks)

R16 · Set up, solve and interpret the answers in growth and decay problems, including compound interest and work with general iterative processes

Explanation

Repeated percentage growth by $r\%$ uses multiplier $1+\dfrac{r}{100}$ each period; repeated decay uses $1-\dfrac{r}{100}$ . After $n$ equal periods, $\text{final}=\text{initial}\times(\text{multiplier})^n$ .
Compound interest is repeated growth because each period acts on the latest balance.
Higher tier: an iterative process defines each new term from the previous term, so calculate successive values in order and identify the first one satisfying the condition.
Keep full calculator precision until the requested final rounding.
Examiners expect the multiplier, exponent or iteration trail, followed by an interpretation in context.

Worked example

Higher tier: A cooling model is $T_{n+1}=0.65T_n+12$ with $T_0=80$ . Find the first $n$ for which $T_n<40$ .

1. $T_1=64$ , $T_2=53.6$ and $T_3=46.84$ .
2. $T_4=42.446$ and $T_5=39.5899$ .
3. $T_4$ is not below $40$ but $T_5$ is, so the first value is $n=5$ .

Answer: $n=5$ , with $T_5=39.6$ to one decimal place.

Common mistakes

Don't calculate repeated percentage change as simple change from the original amount.
Don't use $1.15$ for a $15\%$ decay instead of $0.85$ .
Don't round each intermediate iteration and change the first term meeting the condition.

Exam tip

For “first time” questions, show the last value that fails and the next value that meets the condition.

Tier 1 · Easy

1
£600 is invested at $4\%$ compound interest per year. Work out the value after $2$ years.
(2)
(Total for Question 1 is 2 marks)
2
Write down the multiplier for an increase of $6.5\%$ .
(1)
(Total for Question 2 is 1 mark)

Tier 2 · Standard

1
A culture initially contains $960$ cells and decreases by $15\%$ each hour. Calculate the expected number after $3$ hours.
(3)
(Total for Question 1 is 3 marks)
2
An account pays $5\%$ compound interest each year, and the interest earned in the second year is £31.50. Work out the amount originally invested.
(3)
(Total for Question 2 is 3 marks)
3
A warehouse starts with $6400$ filters. It dispatches $15\%$ of the current stock in the first week and $10\%$ of the remaining stock in the second week. Work out the number of filters left.
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1
A forestry model starts with $4096$ young trees. Model S increases the number each year by $12.5\%$ of the original number. Model C increases the previous year's number by $12.5\%$ each year. Work out how many more young trees Model C predicts than Model S after $3$ years.
(3)
(Total for Question 1 is 3 marks)
2
£ $3000$ is invested. The value grows by $4\%$ in year one, then by $1.5\%$ in each of the following two years. Work out the overall percentage increase across the three years, giving the percentage correct to $1$ decimal place.
(4)
(Total for Question 2 is 4 marks)
3
A ball is dropped from a height of $10\,\text{m}$ . After each impact, its greatest rebound height is $72\%$ of the preceding greatest height. Work out the total vertical distance travelled from release until the ball hits the floor after its third rebound. Give the distance to the nearest centimetre.
(5)
(Total for Question 3 is 5 marks)
4
Higher tier: A tank initially contains $500$ litres. At the start of each week, $80$ litres are added. During that week, $20\%$ of the resulting amount is used. Let $S_n$ be the amount left after $n$ weeks. Write an iterative formula for $S_{n+1}$ in terms of $S_n$ , and work out the amount left after $4$ weeks.
(5)
(Total for Question 4 is 5 marks)

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

R1 · Change freely between related standard units (time, length, area, volume/capacity, mass) and compound units (speed, rates of pay, prices, density, pressure) in numerical and algebraic contexts

Tier 1 · Easy

Mark scheme for R1 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$3600\,\text{m}$	1	$3.6\times1000=3600$ , so the distance is $3600\,\text{m}$ .
2	$27\,000\,\text{cm}^2$	1	$1\,\text{m}^2=10\,000\,\text{cm}^2$ , so $2.7\times10\,000=27\,000\,\text{cm}^2$ .

Tier 2 · Standard

Mark scheme for R1 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$0.009\,\text{m}^3/\text{s}$	3	$540$ litres is $0.540\,\text{m}^3$ . Dividing by $60$ converts per minute to per second: $0.540/60=0.009\,\text{m}^3/\text{s}$ .
2	$1.2x\,\text{m}^2$	2	$40\,\text{cm}=0.4\,\text{m}$ . The area is $3x\times0.4=1.2x\,\text{m}^2$ .
3	$30$ litres per hour	3	$22.32\,\text{m}^3=22\,320$ litres and $31$ days is $31\times24=744$ hours. The average amount used is $22\,320\div744=30$ litres per hour.

Tier 3 · Hard

Mark scheme for R1 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	$400\,\text{kg}/\text{m}^3$	4	Convert the lengths to metres: $35\,\text{cm}=0.35\,\text{m}$ and $80\,\text{mm}=0.08\,\text{m}$ . The volume is $2.4\times0.35\times0.08=0.0672\,\text{m}^3$ , so the density is $26.88/0.0672=400\,\text{kg}/\text{m}^3$ .
2	$16.8\,\text{m}^2$	4	The machine prints $480\div6=80$ labels per minute, so it prints $80\times60=4800$ labels per hour. Their area is $4800\times35=168\,000\,\text{cm}^2=16.8\,\text{m}^2$ .
3	$280\,\text{g}$	4	$0.84\,\text{m}^2=8400\,\text{cm}^2$ , so the piece is $350\div8400=\dfrac{1}{24}$ of the sheet. Its mass is $6.72\div24=0.28\,\text{kg}=280\,\text{g}$ .
4	$40\,\text{m}$	4	$320\,\text{g}/\text{m}^2=0.32\,\text{kg}/\text{m}^2$ , so the area is $9.6\div0.32=30\,\text{m}^2$ . The width is $0.75\,\text{m}$ , giving length $30\div0.75=40\,\text{m}$ .
5	$2.646\,\text{m}^3$	4	The time is $105\times60=6300$ seconds. The volume is $420\times6300=2\,646\,000\,\text{cm}^3$ . Since $1\,\text{m}^3=1\,000\,000\,\text{cm}^3$ , the volume is $2.646\,\text{m}^3$ .

R2 · Use scale factors, scale diagrams and maps

Tier 1 · Easy

Mark scheme for R2 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$1.6\,\text{km}$	2	$6.4\times25\,000=160\,000\,\text{cm}=1600\,\text{m}=1.6\,\text{km}$ .
2	$1.75$	1	The scale factor is $14\div8=1.75$ .

Tier 2 · Standard

Mark scheme for R2 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$5.1\,\text{cm}$	2	The real height is $204\,\text{cm}$ . Divide by $40$ : $204/40=5.1\,\text{cm}$ .
2	$1 : 250$	2	$18\,\text{m}=1800\,\text{cm}$ . The scale is $7.2:1800=1:250$ .
3	$15\,\text{m}$	3	The real dimensions are $6.4\times75=480\,\text{cm}=4.8\,\text{m}$ and $3.6\times75=270\,\text{cm}=2.7\,\text{m}$ . The perimeter is $2(4.8+2.7)=15\,\text{m}$ .

Tier 3 · Hard

Mark scheme for R2 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	$6$ hectares	4	At this scale, $1\,\text{cm}$ represents $25\,\text{m}$ , so $1\,\text{cm}^2$ represents $25^2=625\,\text{m}^2$ . The real area is $96\times625=60\,000\,\text{m}^2=6$ hectares.
2	$4\,\text{cm}$	3	The real road length is $9.6\times12\,500=120\,000\,\text{cm}$ . On the second map its length is $120\,000\div30\,000=4\,\text{cm}$ .
3	$600$ slabs	5	The real dimensions are $9.6\times125=1200\,\text{cm}=12\,\text{m}$ and $6.4\times125=800\,\text{cm}=8\,\text{m}$ , so the terrace area is $96\,\text{m}^2$ . Each slab has area $0.4^2=0.16\,\text{m}^2$ . The number needed is $96\div0.16=600$ .
4	$3456$ litres	4	The volume scale factor is $40^3=64\,000$ . The real volume is $54\times64\,000=3\,456\,000\,\text{cm}^3$ . Since $1000\,\text{cm}^3=1$ litre, this is $3456$ litres.
5	$12.8\,\text{cm}^2$	4	Lengths on the second plan are $8000\div12\,000=\dfrac{2}{3}$ of those on the photograph. Areas are therefore multiplied by $\left(\dfrac{2}{3}\right)^2=\dfrac{4}{9}$ . The area is $28.8\times\dfrac{4}{9}=12.8\,\text{cm}^2$ .

R3 · Express one quantity as a fraction of another, where the fraction is less than 1 or greater than 1

Tier 1 · Easy

Mark scheme for R3 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$\frac{3}{5}$	2	$\frac{18}{30}=\frac{3}{5}$ after dividing the numerator and denominator by $6$ .
2	$\dfrac{3}{8}$	2	£1.20 is $120$ pence, so the fraction is $\dfrac{45}{120}=\dfrac{3}{8}$ .

Tier 2 · Standard

Mark scheme for R3 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$\frac{12}{5}$	2	The units already match, so form $\frac{84}{35}$ . Dividing both parts by $7$ gives $\frac{12}{5}$ .
2	$\dfrac{4}{15}$	2	There are $18+12+15=45$ trees in total. The required fraction is $\dfrac{12}{45}=\dfrac{4}{15}$ .
3	$\dfrac{17}{12}$	3	$1.8\,\text{km}=1800\,\text{m}$ , so the total distance is $1800+750=2550\,\text{m}$ . The required fraction is $\dfrac{2550}{1800}=\dfrac{17}{12}$ .

Tier 3 · Hard

Mark scheme for R3 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	$\dfrac{17}{7}$	3	Convert $1.2$ litres to $1200$ ml, so $1200-350=850$ ml remains. The required fraction is $850/350=17/7$ .
2	$\dfrac{12}{17}$	3	The first new mass is $1800-240=1560\,\text{g}$ . The new total is $1560+650=2210\,\text{g}$ , so the fraction is $\dfrac{1560}{2210}=\dfrac{12}{17}$ .
3	$\dfrac{9}{7}$	4	$18\,000\,\text{cm}^2=1.8\,\text{m}^2$ , so the covered area is $1.8+1.35=3.15\,\text{m}^2$ . The uncovered area is $7.2-3.15=4.05\,\text{m}^2$ . The required fraction is $\dfrac{4.05}{3.15}=\dfrac{9}{7}$ .
4	$\dfrac{25}{41}$	4	Ribbon A starts at $240\,\text{cm}$ . The new lengths are $240-35=205\,\text{cm}$ and $90+35=125\,\text{cm}$ . The required fraction is $\dfrac{125}{205}=\dfrac{25}{41}$ .
5	$\dfrac{39}{53}$	4	$1.8\,\text{m}^3=1800$ litres, so tank A contains $0.65\times1800=1170$ litres. The total is $1170+420=1590$ litres. The fraction is $\dfrac{1170}{1590}=\dfrac{39}{53}$ .

R4 · Use ratio notation, including reduction to simplest form

Tier 1 · Easy

Mark scheme for R4 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$2:3$	1	The highest common factor of $42$ and $63$ is $21$ . Dividing both parts by $21$ gives $2:3$ .
2	$2:5$	1	Multiply both parts by $10$ to get $6:15$ , then divide both parts by $3$ to get $2:5$ .

Tier 2 · Standard

Mark scheme for R4 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$12:5$	2	Convert $1.8\,\text{m}$ to $180\,\text{cm}$ . Then $180:75=12:5$ after dividing both parts by $15$ .
2	$8:3:5$	2	Convert $1.2\,\text{kg}$ to $1200\,\text{g}$ . Then $1200:450:750=8:3:5$ after dividing every part by $150$ .
3	$1 : 2$	3	$18\times\dfrac{2}{3}=12$ boys and $30\times\dfrac{4}{5}=24$ girls are present. The ratio $12 : 24$ simplifies to $1 : 2$ .

Tier 3 · Hard

Mark scheme for R4 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	$4:9:5$	3	Write $1.5=3/2$ . Multiplying every part by $6$ clears all the denominators: $6(2/3):6(3/2):6(5/6)=4:9:5$ .
2	$9:7$	3	Write $p=4k$ , $q=7k$ and $r=9k$ . Then $p+2q=4k+14k=18k$ and $2r-p=18k-4k=14k$ . The ratio $18k:14k$ simplifies to $9:7$ .
3	$4 : 3 : 3$	4	In metres per minute, the rates are $2\times60=120$ , $1.5\times60=90$ and $0.09\times1000=90$ . The ratio $120 : 90 : 90$ simplifies to $4 : 3 : 3$ .
4	$64 : 88 : 75$	3	$32\%=\dfrac{8}{25}$ and $0.44=\dfrac{11}{25}$ . Multiplying all three parts by $200$ gives $64 : 88 : 75$ , which has no common factor.
5	$13 : 4$	4	Match the shared quantity $b$ : $a:b=6:10$ and $b:c=10:7$ , so take $a=6k$ , $b=10k$ and $c=7k$ . Then $a+c=13k$ and $b-a=4k$ , giving $13 : 4$ .

R5 · Divide a quantity in a given part:part or part:whole ratio; express division into two parts as a ratio; apply ratio to real problems (conversion, comparison, scaling, mixing, concentrations)

Tier 1 · Easy

Mark scheme for R5 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	£36 and £48	2	There are $3+4=7$ shares, so one share is £ $84/7=£12$ . The two amounts are $3\times£12=£36$ and $4\times£12=£48$ .
2	$21\,\text{cm}$ and $49\,\text{cm}$	2	The difference of $7-3=4$ shares represents $28\,\text{cm}$ , so one share is $7\,\text{cm}$ . The lengths are $3\times7=21\,\text{cm}$ and $7\times7=49\,\text{cm}$ .

Tier 2 · Standard

Mark scheme for R5 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$90$ red counters $150$ non-red counters Ratio $3:5$	3	Red counters: $\frac{3}{8}\times240=90$ . The remainder is $240-90=150$ . Thus red:non-red is $90:150=3:5$ .
2	$6$ litres	3	The $3$ white shares equal $1.8$ litres, so one share is $1.8\div3=0.6$ litres. There are $2+5+3=10$ shares, giving $10\times0.6=6$ litres in total.
3	£2688	4	There are $7+5=12$ shares, so one share is $288\div12=24$ tickets. There are $168$ adult tickets and $120$ child tickets. The total received is $168\times£11+120\times£7=£1848+£840=£2688$ .

Tier 3 · Hard

Mark scheme for R5 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	£480	5	Let the original shares be $3k$ pounds and $5k$ pounds. After the transfer, $2(3k-20)=5k+20$ , so $k=60$ . The original total was $8k=8\times60=£480$ .
2	$8.4$ litres	3	$1.8$ litres of concentrate needs $1.8\times\dfrac{11}{3}=6.6$ litres of water. Mia has enough water, so the concentrate is used up. The greatest volume is $1.8+6.6=8.4$ litres.
3	$7 : 19$	4	The first drink contains $14\times\dfrac{2}{7}=4$ litres of mango and $10$ litres of apple. The second contains $12\times\dfrac{1}{4}=3$ litres of mango and $9$ litres of apple. The combined amounts are $7$ litres and $19$ litres, giving ratio $7 : 19$ .
4	£2450	4	B and C share $\dfrac{3}{5}$ of the fund. C receives $\dfrac{4}{7}$ of this remainder, so C receives $\dfrac{4}{7}\times\dfrac{3}{5}=\dfrac{12}{35}$ of the whole fund. The whole fund is £ $840\times\dfrac{35}{12}=£2450$ .
5	$8$ litres	4	The amount of concentrate stays at $0.30\times12=3.6$ litres. If the final volume is $V$ litres, $0.18V=3.6$ , so $V=20$ . The volume of water added is $20-12=8$ litres.

R6 · Express a multiplicative relationship between two quantities as a ratio or a fraction

Tier 1 · Easy

Mark scheme for R6 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$A=\frac{3}{5}B$ $A:B=3:5$	2	$A/B=18/30=3/5$ , so $A=\frac{3}{5}B$ and the corresponding ratio is $3:5$ .
2	$A:B=3:8$	1	$A=\dfrac{3}{8}B$ , so the corresponding ratio is $A:B=3:8$ .

Tier 2 · Standard

Mark scheme for R6 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$P:Q=7:4$ $P=\frac{7}{4}Q$	2	$1.75=\frac{7}{4}$ . Therefore $P=\frac{7}{4}Q$ , which gives $P:Q=7:4$ .
2	$P:Q=7:5$ $Q=\dfrac{5}{7}P$	2	$P=1.4Q=\dfrac{7}{5}Q$ , so $P:Q=7:5$ . Reversing the comparison gives $Q=\dfrac{5}{7}P$ .
3	$\dfrac{5}{7}$ $M : (N-M)=7 : 5$	3	Let $M=7k$ and $N=12k$ . Then $N-M=5k$ , so $\dfrac{N-M}{M}=\dfrac{5k}{7k}=\dfrac{5}{7}$ and $M : (N-M)=7 : 5$ .

Tier 3 · Hard

Mark scheme for R6 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	$P:R=5:4$ $P=\frac{5}{4}R$	3	Substitute $Q=\frac{3}{4}R$ into the first relationship: $P=\frac{5}{3}\times\frac{3}{4}R=\frac{5}{4}R$ . Therefore $P:R=5:4$ .
2	$\dfrac{7}{27}$ $(A-B):A=7:27$	3	Taking $B$ as $100$ gives $A=135$ , so the difference is $35$ . Therefore $\dfrac{A-B}{A}=\dfrac{35}{135}=\dfrac{7}{27}$ , and $(A-B):A=7:27$ .
3	$C : B=6 : 5$ $B=\dfrac{5}{6}C$	4	$A=0.8B$ . Then $C=1.5A=1.5\times0.8B=1.2B=\dfrac{6}{5}B$ . Therefore $C : B=6 : 5$ and, reversing the comparison, $B=\dfrac{5}{6}C$ .
4	$\dfrac{3}{13}$	4	Let B be $x$ , so A is $1.6x$ . Their difference is $0.6x=42$ , giving $x=70$ . Therefore B is $70$ , A is $112$ , and the total is $182$ . The required fraction is $\dfrac{42}{182}=\dfrac{3}{13}$ .
5	P is $\dfrac{7}{12}$ of R $P : R = 7 : 12$	4	Q is $0.7=\dfrac{7}{10}$ of R. Therefore P is $\dfrac{5}{6}\times\dfrac{7}{10}=\dfrac{35}{60}=\dfrac{7}{12}$ of R, so $\text{P}:\text{R}=7:12$ .

R7 · Understand and use proportion as equality of ratios

Tier 1 · Easy

Mark scheme for R7 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$x=9$	2	$x=15\times\frac{6}{10}=9$ .
2	B	1	$15:24$ simplifies to $5:8$ after dividing both parts by $3$ , so the correct letter is B.

Tier 2 · Standard

Mark scheme for R7 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	No, because $180:120=3:2$ but $315:200=63:40$ .	3	Simplify the recipe ratio to $180:120=3:2$ . Leo's ratio is $315:200=63:40$ , which is not equal to $3:2$ , so the ingredients are not in the correct proportion.
2	$31.5\,\text{cm}$	2	The scale factor from $4$ to $18$ is $18\div4=4.5$ . The width is $7\times4.5=31.5\,\text{cm}$ .
3	$364\,\text{g}$	2	One bolt has mass $234\div9=26\,\text{g}$ . Therefore $14$ bolts have mass $14\times26=364\,\text{g}$ .

Tier 3 · Hard

Mark scheme for R7 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	$x=12$ Both ratios equal $\frac{3}{2}$	4	Cross-multiply: $12(2x+3)=18(x+6)$ . This gives $24x+36=18x+108$ , so $6x=72$ and $x=12$ . Substitution gives $27/18=18/12=3/2$ .
2	$7.5$ minutes	3	The printer produces $126\div3.5=36$ pages per minute. The time for $270$ pages is $270\div36=7.5$ minutes.
3	$16.5$ litres	4	Let the original amounts be $3k$ and $8k$ litres. Equality of the new ratios gives $\dfrac{3k}{8k-1.5}=\dfrac{3}{7}$ . Hence $21k=24k-4.5$ , so $k=1.5$ . The original volume was $11k=16.5$ litres.
4	$x=\dfrac{4}{3}\,\text{cm}$	4	The new width is $24-2x$ . Equality of the ratios gives $\dfrac{24-2x}{16}=\dfrac{4}{3}$ . Hence $72-6x=64$ , so $6x=8$ and $x=\dfrac{4}{3}\,\text{cm}$ .
5	$63$ complete snack bars	4	$1.875\,\text{kg}=1875\,\text{g}$ . After toasting, $1875\times0.96=1800\,\text{g}$ remains. Since $1800\div400=4.5$ recipe batches, the number of bars is $4.5\times14=63$ .

R8 · Relate ratios to fractions and to linear functions

Tier 1 · Easy

Mark scheme for R8 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$\frac{3}{8}$	2	There are $3+5=8$ equal parts altogether, of which $3$ are cats. The fraction is $\frac{3}{8}$ .
2	$4:1$	1	$y=4x$ means that $y$ is four times $x$ , so $y:x=4:1$ .

Tier 2 · Standard

Mark scheme for R8 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$y=\frac{5}{2}x$ $y=35$	3	$y:x=5:2$ means $y/x=5/2$ , so $y=\frac{5}{2}x$ . At $x=14$ , $y=\frac{5}{2}\times14=35$ .
2	$P=\dfrac{16}{3}w$	3	Since length:width is $5:3$ , the length is $\dfrac{5}{3}w$ . Therefore $P=2\left(\dfrac{5}{3}w+w\right)=2\times\dfrac{8}{3}w=\dfrac{16}{3}w$ .
3	$N=5t$ $N=90$	3	The tenors are $2$ of the $10$ ratio parts, so $t=2k$ and $N=10k=5t$ . When $t=18$ , $N=5\times18=90$ .

Tier 3 · Hard

Mark scheme for R8 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	$V=\frac{9}{2}c$ $12$ litres concentrate $42$ litres water	4	The total has $2+7=9$ parts, so $c$ is $2/9$ of $V$ . Hence $V=\frac{9}{2}c$ . If $V=54$ , then $c=\frac{2}{9}\times54=12$ , leaving $54-12=42$ litres of water.
2	$x=16$	3	$14\div4=24.5\div7=35\div10=3.5$ , so the constant-ratio function is $y=3.5x$ . When $y=56$ , $3.5x=56$ , giving $x=16$ .
3	$M=\dfrac{47}{4}b$ $42$ red beads	4	There are $\dfrac{7}{4}b$ red beads, so $M=3b+5\left(\dfrac{7}{4}b\right)=\dfrac{47}{4}b$ . When $M=282$ , $b=282\times\dfrac{4}{47}=24$ . The number of red beads is $\dfrac{7}{4}\times24=42$ .
4	$T=6d$ $T=324$	4	Write the quantities as $2k$ , $3k$ and $7k$ . Then $d=7k-(2k+3k)=2k$ , while $T=12k$ . Therefore $T=6d$ , and when $d=54$ , $T=6\times54=324$ .
5	$y=2x+3$ No, the ratio $y:x$ is not constant.	4	The gradient is $\dfrac{23-11}{10-4}=2$ . Using $(4,11)$ gives $11=2\times4+c$ , so $c=3$ and $y=2x+3$ . Also $11:4$ is not equal to $23:10$ , so the ratio $y:x$ is not constant; equivalently, the line does not pass through the origin.

R9 · Define percentage as 'number of parts per hundred'; interpret percentages and percentage changes as fractions/decimals, multiplicatively; percentages > 100%; percentage change and simple interest

Tier 1 · Easy

Mark scheme for R9 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$84$	2	$35\%=0.35$ , so $0.35\times240=84$ .
2	$108$	2	$135\%=1.35$ , so $1.35\times80=108$ .

Tier 2 · Standard

Mark scheme for R9 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	£800	3	A $12\%$ decrease leaves $88\%=0.88$ of the original value. Divide by the multiplier: £ $704/0.88=£800$ .
2	$12.5\%$	3	The increase is £ $81-£72=£9$ . As a fraction of the original price this is $\dfrac{9}{72}=0.125$ , so the percentage increase is $12.5\%$ .
3	$600$ parts	2	$140\%=1.4$ , so $1.4\times\text{previous number}=840$ . The previous number was $840\div1.4=600$ .

Tier 3 · Hard

Mark scheme for R9 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	£2891	5	The yearly simple interest is $0.036\times£2500=£90$ . Over $5$ years this is £450, giving £2950. The fee is $0.02\times£2950=£59$ , so £ $2950-£59=£2891$ remains.
2	£2800	3	Over $3$ years the simple interest is $3\times4.5\%=13.5\%$ of the original amount. Therefore the original amount is £ $378\div0.135=£2800$ .
3	£2400	4	The first investment earns $1800\times0.04\times3=£216$ . The second investment therefore earns £ $396-£216=£180$ . Over $3$ years its simple-interest rate is $3\times2.5\%=7.5\%$ , so the second amount is £ $180\div0.075=£2400$ .
4	$p=10$	4	The multipliers satisfy $1.20\left(1-\dfrac{p}{100}\right)=1.08$ . Hence $1-\dfrac{p}{100}=1.08\div1.20=0.9$ , so $p=10$ .
5	$425$ litres	4	Let the capacity be $C$ litres. After $15\%$ is used, $0.8C\times0.85=0.68C$ litres remain. Therefore $0.68C+102=0.92C$ , so $0.24C=102$ and $C=425$ litres.

R10 · Solve problems involving direct and inverse proportion, including graphical and algebraic representations

Tier 1 · Easy

Mark scheme for R10 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$y=30$	3	$y=kx$ . Using $18=6k$ gives $k=3$ , so when $x=10$ , $y=3\times10=30$ .
2	A	1	A direct-proportion equation has the form $y=kx$ , so the correct letter is A.

Tier 2 · Standard

Mark scheme for R10 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$y=6$	3	For inverse proportion, $xy=k$ . Here $k=3\times14=42$ , so at $x=7$ , $y=42/7=6$ .
2	$x=12.5$	2	The graph has equation $y=2.8x$ . Therefore $x=35\div2.8=12.5$ .
3	$21$ days	3	The number of goats and the number of days are inversely proportional, so the constant product is $18\times28=504$ . For $24$ goats, the feed lasts $504\div24=21$ days.

Tier 3 · Hard

Mark scheme for R10 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	New $y=9.6$ $20\%$ decrease	5	The constant product is $xy=6\times12=72$ . Increasing $x$ by $25\%$ gives $x=6\times1.25=7.5$ , so $y=72/7.5=9.6$ . The decrease is $12-9.6=2.4$ , and $2.4/12\times100=20\%$ .
2	$y=7.5$	3	For the first three pairs, $xy=2\times30=3\times20=5\times12=60$ . The constant product is $60$ , so when $x=8$ , $y=60\div8=7.5$ .
3	$20\,\text{m}$	4	The mass per metre is $1.26\div3.6=0.35\,\text{kg}/\text{m}$ . The cable on the drum has mass $8.4-1.4=7\,\text{kg}$ . Its length is $7\div0.35=20\,\text{m}$ .
4	$19$ minutes	5	The complete batch requires $6\times14=84$ scanner-minutes. In the first $4$ minutes, $6\times4=24$ scanner-minutes of work are completed, leaving $60$ scanner-minutes. Four scanners need $60\div4=15$ more minutes, so the total time is $4+15=19$ minutes.
5	$V=36t$ $17$ minutes	4	Write $V=kt$ . Then $k=270\div7.5=36$ , so $V=36t$ . The tank needs $750-138=612$ litres. Therefore $36t=612$ , giving $t=17$ minutes.

R11 · Use compound units such as speed, rates of pay, unit pricing, density and pressure

Tier 1 · Easy

Mark scheme for R11 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	£7.50 per hour	2	Divide total pay by time: £ $52.50/7=£7.50$ per hour.
2	$9$ cartons per minute	2	Divide the number of cartons by the time: $54\div6=9$ cartons per minute.

Tier 2 · Standard

Mark scheme for R11 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$65\,\text{km}/\text{h}$	3	$24$ minutes is $24/60=0.4$ hours, so the time is $2.4$ hours. Average speed is $156/2.4=65\,\text{km}/\text{h}$ .
2	$0.007\,\text{m}^3$	3	From $\text{density}=\dfrac{\text{mass}}{\text{volume}}$ , volume $=\dfrac{\text{mass}}{\text{density}}$ . Therefore the volume is $5.46\div780=0.007\,\text{m}^3$ .
3	The $1.2\,\text{kg}$ pack, by £0.10 per kilogram	3	The smaller pack costs £ $3.30\div0.75=£4.40$ per kilogram. The larger pack costs £ $5.16\div1.2=£4.30$ per kilogram. The $1.2\,\text{kg}$ pack is cheaper by £ $4.40-£4.30=£0.10$ per kilogram.

Tier 3 · Hard

Mark scheme for R11 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	$100\,000\,\text{Pa}$	4	Convert the force to $3600\,\text{N}$ and the dimensions to $0.24\,\text{m}$ and $0.15\,\text{m}$ . The area is $0.24\times0.15=0.036\,\text{m}^2$ , so the pressure is $3600/0.036=100\,000\,\text{Pa}$ .
2	£0.70 per $1000$ pages	4	The ink used per $1000$ pages is $0.45\div18=0.025$ litres. The cost is $0.025\times£28=£0.70$ per $1000$ pages.
3	£13.02 per hour	4	The overtime rate is $1.5\times£12.40=£18.60$ per hour. The total pay is $36\times£12.40+4\times£18.60=£446.40+£74.40=£520.80$ . The average rate is £ $520.80\div40=£13.02$ per hour.
4	$50.9\,\text{km}/\text{h}$	5	The travel times are $72\div48=1.5$ hours and $96\div64=1.5$ hours. The stop is $18\div60=0.3$ hours, so the total time is $3.3$ hours and the distance is $168\,\text{km}$ . The average speed is $168\div3.3=50.909\ldots$ , which rounds to $50.9\,\text{km}/\text{h}$ . This is $0.0409\ldots$ from the upper rounding boundary $50.95$ , a margin of more than $0.03\,\text{km}/\text{h}$ .
5	$1920\,\text{Pa}$	5	The volume is $0.5\times0.25\times0.08=0.01\,\text{m}^3$ , so the mass is $2400\times0.01=24\,\text{kg}$ . Its weight is $24\times10=240\,\text{N}$ . The contact area is $0.5\times0.25=0.125\,\text{m}^2$ , so the pressure is $240\div0.125=1920\,\text{Pa}$ .

R12 · Compare lengths, areas and volumes using ratio notation; make links to similarity (including trigonometric ratios) and scale factors

Tier 1 · Easy

Mark scheme for R12 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$2:5$	1	$6:15$ simplifies by dividing both parts by $3$ , giving $2:5$ .
2	$4 : 25$	1	For similar shapes, areas are in the ratio of the squares of corresponding lengths: $2^2 : 5^2 = 4 : 25$ .

Tier 2 · Standard

Mark scheme for R12 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$294\,\text{cm}^2$	3	The area ratio is $3^2:7^2=9:49$ . Therefore the larger area is $54\times\frac{49}{9}=294\,\text{cm}^2$ .
2	$14\,\text{cm}$	3	For the same acute angle, the ratio $\dfrac{\text{opposite}}{\text{hypotenuse}}$ is constant. Therefore the larger opposite side is $17.5\times\dfrac{4}{5}=14\,\text{cm}$ .
3	$15\,\text{cm}$	3	The area ratio is $96 : 150=16 : 25$ . Therefore the corresponding length ratio is $4 : 5$ . The larger side is $12\times\dfrac{5}{4}=15\,\text{cm}$ .

Tier 3 · Hard

Mark scheme for R12 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	$396\,\text{cm}^2$	4	Since $125:216=5^3:6^3$ , the length ratio is $5:6$ . The surface-area ratio is therefore $25:36$ . The larger area is $275\times\frac{36}{25}=396\,\text{cm}^2$ .
2	$337.5\,\text{cm}^2$	4	The original perimeter is $2(8+12)=40\,\text{cm}$ , so the length scale factor is $75\div40=1.875$ . The enlarged dimensions are $8\times1.875=15\,\text{cm}$ and $12\times1.875=22.5\,\text{cm}$ . Its area is $15\times22.5=337.5\,\text{cm}^2$ .
3	$525\,\text{cm}^2$	4	The difference of $5-3=2$ ratio parts is $14\,\text{cm}$ , so the corresponding longer sides are $21\,\text{cm}$ and $35\,\text{cm}$ . The area scale factor is $\left(\dfrac{5}{3}\right)^2=\dfrac{25}{9}$ . The larger area is $189\times\dfrac{25}{9}=525\,\text{cm}^2$ .
4	$216\,\text{cm}^3$	4	The surface-area ratio $81:144$ simplifies to $9:16=3^2:4^2$ , so the length ratio is $3:4$ . The volume ratio is $3^3:4^3=27:64$ . The smaller volume is $512\times\dfrac{27}{64}=216\,\text{cm}^3$ .
5	$81\,\text{cm}$	4	Since $64:125=4^3:5^3$ , the corresponding height ratio is $4:5$ . The difference of one ratio part is $9\,\text{cm}$ , so the heights are $36\,\text{cm}$ and $45\,\text{cm}$ . Their sum is $81\,\text{cm}$ .

R13 · Understand that X is inversely proportional to Y is equivalent to X is proportional to 1/Y; construct and interpret equations that describe direct and inverse proportion

Tier 1 · Easy

Mark scheme for R13 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$p=\frac{60}{q}$	2	Write $p=k/q$ . Using the given pair, $k=pq=12\times5=60$ , so $p=\frac{60}{q}$ .
2	A	1	Inverse proportion has constant product $xy=k$ , so the correct letter is A.

Tier 2 · Standard

Mark scheme for R13 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$y=5x^2$ $y=80$	4	Write $y=kx^2$ . Then $45=k\times3^2$ , so $k=5$ and $y=5x^2$ . At $x=4$ , $y=5\times4^2=80$ .
2	$\dfrac{1}{4}$	3	Originally $p=48\div3=16$ . After the change, $p=48\div12=4$ . The new value as a fraction of the original is $\dfrac{4}{16}=\dfrac{1}{4}$ .
3	Yes, because $9\times8=72$ . $q=6$	3	For the proposed pair, $pq=9\times8=72$ , so it satisfies the equation. When $p=12$ , $12=\dfrac{72}{q}$ , so $12q=72$ and $q=6$ .

Tier 3 · Hard

Mark scheme for R13 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	$b=5$	4	Use $a=k/b^2$ . The first pair gives $k=ab^2=20\times9=180$ . Then $7.2=180/b^2$ , so $b^2=25$ . The requested positive value is $b=5$ .
2	$y=\dfrac{252}{x}$ $66\,\text{m}$	4	Write $y=\dfrac{k}{x}$ . Since $k=18\times14=252$ , the equation is $y=\dfrac{252}{x}$ . When $x=21$ , $y=252\div21=12$ , so the perimeter is $2(21+12)=66\,\text{m}$ .
3	$u=3.5v^3$ $v=6$	4	Write $u=kv^3$ . Since $28=k\times2^3$ , $k=3.5$ and $u=3.5v^3$ . When $u=756$ , $v^3=756\div3.5=216$ . The positive value is $v=6$ .
4	$y=\dfrac{128}{x^3}$ $x=4$	4	Write $y=\dfrac{k}{x^3}$ . Using $x=2$ and $y=16$ gives $k=16\times2^3=128$ , so $y=\dfrac{128}{x^3}$ . When $y=2$ , $2=\dfrac{128}{x^3}$ , hence $x^3=64$ and the positive value is $x=4$ .
5	$36\%$ decrease	4	The equation has the form $A=kb^2$ . A $20\%$ decrease multiplies $b$ by $0.8$ , so it multiplies $A$ by $0.8^2=0.64$ . The new value is $64\%$ of the old value, giving a $100\%-64\%=36\%$ decrease. (Using the given pair, $k=54\div9=6$ gives the same result.)

R14 · Interpret the gradient of a straight line graph as a rate of change; recognise and interpret graphs that illustrate direct and inverse proportion

Tier 1 · Easy

Mark scheme for R14 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	Gradient $=5\,\text{m}/\text{s}$ The object travels at $5\,\text{m}/\text{s}$ .	3	Gradient $=(35-10)/(7-2)=25/5=5\,\text{m}/\text{s}$ . On a distance-time graph this is the speed.
2	$4.5\,\text{kg}$ per minute	2	Using $(0,0)$ and $(6,27)$ , the gradient is $\dfrac{27-0}{6-0}=4.5\,\text{kg}$ per minute. This is the rate at which grain is processed.

Tier 2 · Standard

Mark scheme for R14 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	The cost increases by £0.12 per minute. £22.20	3	The coefficient of $m$ is the gradient, so the rate is £0.12 per minute. At $m=35$ , $C=18+0.12\times35=18+4.20=£22.20$ .
2	Line A Line A has gradient $3$ , which is greater than $2.5$ .	3	The gradient of line A is $\dfrac{25-7}{8-2}=\dfrac{18}{6}=3$ . Since $3>2.5$ , line A has the greater rate of change.
3	$20$ grams per minute $930$ grams at time $0$	3	The gradient is $\dfrac{710-870}{11-3}=\dfrac{-160}{8}=-20$ grams per minute. The negative sign shows the mass is decreasing at $20$ grams per minute. Using $(3,870)$ , the mass at time $0$ is $870+3\times20=930$ grams.

Tier 3 · Hard

Mark scheme for R14 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	$xy=36$ for all three points $y=\frac{36}{x}$ $y=4$ when $x=9$	4	The products are $2\times18=36$ , $3\times12=36$ and $6\times6=36$ , so all three points are consistent with the stated inverse-proportion model. Thus $y=36/x$ , and at $x=9$ , $y=36/9=4$ .
2	Gradient $=4$ It has equation $y=4x+2$ , so it does not pass through the origin.	3	The gradient is $\dfrac{30-14}{7-3}=4$ . Using $(3,14)$ gives $14=4\times3+c$ , so $c=2$ . The line has equation $y=4x+2$ and does not pass through the origin, so it is not direct proportion.
3	From $5$ to $13$ minutes, by $2$ litres per minute	4	From $0$ to $5$ minutes, the gradient is $\dfrac{120-150}{5-0}=-6$ litres per minute. From $5$ to $13$ minutes, it is $\dfrac{56-120}{13-5}=-8$ litres per minute. The second interval has the greater draining rate, by $8-6=2$ litres per minute.
4	$82$	4	For graph A, $y=7x$ , so at $x=10$ , $y=70$ . For graph B, $xy=4\times30=120$ , so at $x=10$ , $y=12$ . The sum is $70+12=82$ .
5	$2.4$ kilowatt-hours per hour $12.5$ hours	4	The gradient is $\dfrac{16.8-4.8}{6.5-1.5}=\dfrac{12}{5}=2.4$ kilowatt-hours per hour, so energy use increases by $2.4$ kilowatt-hours for each hour of running. An increase of $30$ kilowatt-hours takes $30\div2.4=12.5$ hours.

R15 · Interpret the gradient at a point on a curve as the instantaneous rate of change; apply average and instantaneous rates of change (gradients of chords and tangents) (not calculus) [Higher only]

Tier 1 · Easy

Mark scheme for R15 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$3$ units of $y$ per unit of $x$	2	The chord gradient is $(23-5)/(8-2)=18/6=3$ .
2	A tangent	1	The gradient of a tangent at the point estimates the instantaneous rate of change.

Tier 2 · Standard

Mark scheme for R15 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$3.5$ units of $y$ per unit of $x$	3	Use two points on the tangent: $(32-11)/(10-4)=21/6=3.5$ . This tangent gradient estimates the instantaneous rate at $x=7$ .
2	$-4.5\,{}^\circ\text{C}$ per minute	3	The tangent gradient is $\dfrac{19-46}{7-1}=\dfrac{-27}{6}=-4.5$ . The instantaneous rate of change is $-4.5\,{}^\circ\text{C}$ per minute.
3	$5$ litres per $100\,\text{km}$	3	The tangent gradient is $\dfrac{43-52}{260-80}=\dfrac{-9}{180}=-0.05$ litre per kilometre. The negative sign shows fuel is being used. Its rate of use is $0.05\times100=5$ litres per $100\,\text{km}$ .

Tier 3 · Hard

Mark scheme for R15 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	Average rate $=12$ litres per minute Instantaneous rate $=14$ litres per minute The instantaneous rate is $2$ litres per minute greater.	5	The chord gradient is $(118-46)/(8-2)=72/6=12$ litres per minute. The tangent gradient is $(112-70)/(7-4)=42/3=14$ litres per minute. Therefore the instantaneous rate is $14-12=2$ litres per minute greater.
2	Speed at $t=4$ is $6\,\text{m}/\text{s}$ Speed at $t=10$ is $3\,\text{m}/\text{s}$ The speed at $t=10$ is half the speed at $t=4$ .	4	At $t=4$ , the tangent gradient is $\dfrac{51-15}{8-2}=\dfrac{36}{6}=6\,\text{m}/\text{s}$ . At $t=10$ , it is $\dfrac{73-55}{14-8}=\dfrac{18}{6}=3\,\text{m}/\text{s}$ . Therefore the later instantaneous speed is half the earlier speed.
3	$5\,\text{m}/\text{s}$ $240$ seconds (or $4$ minutes)	4	The tangent gradient is $\dfrac{390-140}{70-20}=\dfrac{250}{50}=5\,\text{m}/\text{s}$ . A further $1.2\,\text{km}$ is $1200\,\text{m}$ , so the time is $1200\div5=240$ seconds, which is $4$ minutes.
4	$2.4\,\text{km}/\text{h}$	4	The tangent gradient is $\dfrac{2.5-1.9}{23-8}=\dfrac{0.6}{15}=0.04$ kilometres per minute. Multiplying by $60$ gives an instantaneous speed of $2.4\,\text{km}/\text{h}$ .
5	$3\,\text{cm}$ per minute	5	The tangent gradient is $\dfrac{560-920}{27-12}=-24$ litres per minute, so volume is falling at $24$ litres per minute. This is $0.024\,\text{m}^3$ per minute. Using $\text{change in height}=\dfrac{\text{change in volume}}{\text{area}}$ gives $0.024\div0.8=0.03\,\text{m}$ per minute, which is $3\,\text{cm}$ per minute.

R16 · Set up, solve and interpret the answers in growth and decay problems, including compound interest and work with general iterative processes

Tier 1 · Easy

Mark scheme for R16 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	£648.96	2	Use multiplier $1.04$ twice: £ $600\times1.04^2=£648.96$ .
2	$1.065$	1	An increase of $6.5\%$ uses multiplier $1+0.065=1.065$ .

Tier 2 · Standard

Mark scheme for R16 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	About $590$ cells Unrounded model value $=589.56$	3	The decay multiplier is $0.85$ . After $3$ hours the model gives $960\times0.85^3=589.56$ , which is about $590$ whole cells.
2	£600	3	The balance before the second-year interest was added is £ $31.50\div0.05=£630$ . This is $105\%$ of the original amount, so the original investment was £ $630\div1.05=£600$ .
3	$4896$ filters	3	After the first week, $6400\times0.85=5440$ filters remain. After the second week, $5440\times0.90=4896$ filters remain.

Tier 3 · Hard

Mark scheme for R16 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	$200$ more young trees	3	Model S adds $4096\times0.125=512$ trees each year, so after $3$ years it predicts $4096+3\times512=5632$ . Model C predicts $4096\times1.125^3=5832$ . The difference is $5832-5632=200$ young trees.
2	$7.1\%$	4	The overall multiplier is $1.04\times1.015^2=1.04\times1.030225=1.071434\ldots$ , so the value is multiplied by $1.0714\ldots$ overall. The overall increase is $7.1434\ldots\%$ , which is $7.1\%$ correct to $1$ decimal place. (Equivalently, £ $3000$ grows to £ $3214.30$ , an increase of £ $214.30$ , and $214.30\div3000=7.14\ldots\%$ .)
3	$42.23\,\text{m}$	5	The first three rebound heights are $10\times0.72=7.2$ , $7.2\times0.72=5.184$ and $5.184\times0.72=3.73248$ metres. Each rebound height is travelled upwards and downwards, so the total distance is $10+2(7.2+5.184+3.73248)=42.23296\,\text{m}$ . This rounds to $42.23\,\text{m}$ .
4	$S_{n+1}=0.8(S_n+80)$ , with $S_0=500$ $393.728$ litres	5	Adding $80$ litres and then keeping $80\%$ gives $S_{n+1}=0.8(S_n+80)$ . Starting with $S_0=500$ , the values are $S_1=464$ , $S_2=435.2$ , $S_3=412.16$ and $S_4=393.728$ . Therefore $393.728$ litres remain after $4$ weeks.