R Ratio, proportion and rates of change — printable coverage pack

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

Standard-unit conversions multiply or divide by a fixed factor, such as $1\,\text{km}=1000\,\text{m}$ and $1\,\text{litre}=10^{-3}\,\text{m}^3$ .
Convert every measurement to compatible units before using a formula; for a rate, convert the numerator and denominator separately.
Area and volume factors are powered: since $1\,\text{m}=100\,\text{cm}$ , $1\,\text{m}^2=10^4\,\text{cm}^2$ and $1\,\text{m}^3=10^6\,\text{cm}^3$ .
A common error is to use a length conversion unchanged for an area or volume conversion.

Tier 1 · Easy

1. A charity walk is $3.6\,\text{km}$ long. Change this distance to metres.[1 mark]
Answer
- $3600\,\text{m}$
Method: $3.6\times1000=3600$ , so the distance is $3600\,\text{m}$ .

Tier 2 · Standard

1. A pump moves water at $540$ litres per minute. Change this rate to cubic metres per second.[3 marks]
Answer
- $0.009\,\text{m}^3/\text{s}$
Method: $540$ litres is $0.540\,\text{m}^3$ . Dividing by $60$ converts per minute to per second: $0.540/60=0.009\,\text{m}^3/\text{s}$ .

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 marks]
Answer
- $400\,\text{kg}/\text{m}^3$
Method: 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 $26.88/0.0672=400\,\text{kg}/\text{m}^3$ .

R2 · Use scale factors, scale diagrams and maps

A scale $1:n$ means one unit on the diagram represents $n$ of the same units in reality.
Measure or calculate the diagram length, apply the scale factor, and then convert to the unit requested.
For example, at $1:25\,000$ , $1\,\text{cm}$ represents $25\,000\,\text{cm}=250\,\text{m}$ .
A common error is to mix centimetres and metres before applying the scale factor.

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 marks]
Answer
- $1.6\,\text{km}$
Method: $6.4\times25\,000=160\,000\,\text{cm}=1600\,\text{m}=1.6\,\text{km}$ .

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 marks]
Answer
- $5.1\,\text{cm}$
Method: The real height is $204\,\text{cm}$ . Divide by $40$ : $204/40=5.1\,\text{cm}$ .

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 marks]
Answer
- $6$ hectares
Method: 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.

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

To express quantity $A$ as a fraction of quantity $B$ , write $\frac{A}{B}$ and simplify.
First convert the quantities to the same units; the units then cancel in the fraction.
The fraction is greater than $1$ when the first quantity is larger than the quantity it is compared with.
A common error is to reverse the numerator and denominator because the phrase 'of another' is misread.

Tier 1 · Easy

1. Express $18$ as a fraction of $30$ . Give the fraction in its simplest form.[2 marks]
Answer
- $\frac{3}{5}$
Method: $\frac{18}{30}=\frac{3}{5}$ after dividing the numerator and denominator by $6$ .

Tier 2 · Standard

1. Express $84$ minutes as a fraction of $35$ minutes. Simplify your answer.[2 marks]
Answer
- $\frac{12}{5}$
Method: The units already match, so form $\frac{84}{35}$ . Dividing both parts by $7$ gives $\frac{12}{5}$ .

Tier 3 · Hard

1. Express $0.84\,\text{m}^2$ as a fraction of $2400\,\text{cm}^2$ . Give the fraction in its simplest form.[3 marks]
Answer
- $\frac{7}{2}$
Method: $0.84\,\text{m}^2=8400\,\text{cm}^2$ . Therefore the required fraction is $\frac{8400}{2400}=\frac{84}{24}=\frac{7}{2}$ .

R4 · Use ratio notation, including reduction to simplest form

A ratio compares quantities in a stated order, using notation such as $a:b$ .
Convert all quantities to the same units, then divide every part by their highest common factor.
Equivalent ratios are made by multiplying or dividing every part by the same non-zero number.
A common error is to simplify only one part or to change the order of the quantities.

Tier 1 · Easy

1. Write the ratio $42:63$ in its simplest form.[1 mark]
Answer
- $2:3$
Method: The highest common factor of $42$ and $63$ is $21$ . Dividing both parts by $21$ gives $2:3$ .

Tier 2 · Standard

1. Write $1.8\,\text{m}:75\,\text{cm}$ as a ratio in its simplest form.[2 marks]
Answer
- $12:5$
Method: Convert $1.8\,\text{m}$ to $180\,\text{cm}$ . Then $180:75=12:5$ after dividing both parts by $15$ .

Tier 3 · Hard

1. Given $x:y=4:7$ and $y:z=6:5$ , find $x:y:z$ in its simplest integer form.[3 marks]
Answer
- $24:42:35$
Method: Use a common value for $y$ . The lowest common multiple of $7$ and $6$ is $42$ . Multiply $4:7$ by $6$ to get $24:42$ , and multiply $6:5$ by $7$ to get $42:35$ . Hence $x:y:z=24:42:35$ .

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)

For a part:part ratio $a:b$ , the whole contains $a+b$ equal shares.
Find one share by dividing the total by the sum of the parts, then multiply by each stated part.
A part:whole fraction such as $\frac{3}{8}$ gives the complementary part $\frac{5}{8}$ , so the two parts are in ratio $3:5$ .
A common error is to divide a total by one ratio number instead of by the sum of all parts.

Tier 1 · Easy

1. Divide £84 in the ratio $3:4$ .[2 marks]
Answer
- £36 and £48
Method: There are $3+4=7$ shares, so one share is £ $84/7=£12$ . The two amounts are $3\times£12=£36$ and $4\times£12=£48$ .

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 marks]
Answer
- $90$ red counters
- $150$ non-red counters
- Ratio $3:5$
Method: Red counters: $\frac{3}{8}\times240=90$ . The remainder is $240-90=150$ . Thus red:non-red is $90:150=3:5$ .

Tier 3 · Hard

1. 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.[5 marks]
Answer
- $6$ litres of the $20\%$ solution
- $12$ litres of the $50\%$ solution
Method: Let $x$ be the litres of $50\%$ solution, so $18-x$ litres has strength $20\%$ . Equating the amount of solute gives $0.50x+0.20(18-x)=0.40\times18$ . Hence $0.30x=3.6$ , so $x=12$ and $18-x=6$ .

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

A multiplicative comparison states how many times one quantity is another, not their additive difference.
If $A=kB$ , then $A:B=k:1$ and $A$ is the fraction $k$ of $B$ .
For integer ratio notation, rewrite a fractional multiplier such as $\frac{5}{3}$ as the ratio $5:3$ .
A common error is to subtract the quantities and report the difference as though it were 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 marks]
Answer
- $A=\frac{3}{5}B$
- $A:B=3:5$
Method: $A/B=18/30=3/5$ , so $A=\frac{3}{5}B$ and the corresponding ratio is $3:5$ .

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 marks]
Answer
- $P:Q=7:4$
- $P=\frac{7}{4}Q$
Method: $1.75=\frac{7}{4}$ . Therefore $P=\frac{7}{4}Q$ , which gives $P:Q=7:4$ .

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 marks]
Answer
- $P:R=5:4$
- $P=\frac{5}{4}R$
Method: Substitute $Q=\frac{3}{4}R$ into the first relationship: $P=\frac{5}{3}\times\frac{3}{4}R=\frac{5}{4}R$ . Therefore $P:R=5:4$ .

R7 · Understand and use proportion as equality of ratios

Two quantities are in proportion when corresponding ratios are equal.
Write an equation between matching ratios, then use scaling or cross-multiplication to find an unknown.
For example, $\frac{a}{b}=\frac{c}{d}$ implies $ad=bc$ when $b$ and $d$ are non-zero.
A common error is to compare quantities in different orders on the two sides of the equation.

Tier 1 · Easy

1. Solve the proportion $\frac{x}{15}=\frac{6}{10}$ .[2 marks]
Answer
- $x=9$
Method: $x=15\times\frac{6}{10}=9$ .

Tier 2 · Standard

1. Eight identical notebooks cost £11.20. Use proportion to find the cost of $14$ notebooks.[3 marks]
Answer
- £19.60
Method: One notebook costs £ $11.20/8=£1.40$ . Therefore $14$ notebooks cost $14\times£1.40=£19.60$ .

Tier 3 · Hard

1. Solve $\frac{2x+3}{18}=\frac{x+6}{12}$ and verify that the two ratios are equal.[4 marks]
Answer
- $x=12$
- Both ratios equal $\frac{3}{2}$
Method: Cross-multiply: $12(2x+3)=18(x+6)$ . This gives $24x+36=18x+108$ , so $6x=72$ and $x=12$ . Substitution gives $27/18=18/12=3/2$ .

R8 · Relate ratios to fractions and to linear functions

If two parts are in ratio $a:b$ , their fractions of the whole are $\frac{a}{a+b}$ and $\frac{b}{a+b}$ .
A constant ratio $y:x=m:1$ gives the linear relationship $y=mx$ .
The graph of a constant ratio is a straight line through the origin, with gradient equal to the multiplier.
A common error is to use $\frac{a}{b}$ for the fraction of the whole instead of $\frac{a}{a+b}$ .

Tier 1 · Easy

1. The ratio of cats to dogs at a shelter is $3:5$ . What fraction of the animals are cats?[2 marks]
Answer
- $\frac{3}{8}$
Method: There are $3+5=8$ equal parts altogether, of which $3$ are cats. The fraction is $\frac{3}{8}$ .

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 marks]
Answer
- $y=\frac{5}{2}x$
- $y=35$
Method: $y:x=5:2$ means $y/x=5/2$ , so $y=\frac{5}{2}x$ . At $x=14$ , $y=\frac{5}{2}\times14=35$ .

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 marks]
Answer
- $V=\frac{9}{2}c$
- $12$ litres concentrate
- $42$ litres water
Method: 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.

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

A percentage is a number of parts per hundred, so $p\%=\frac{p}{100}$ .
Use a multiplier: an increase of $r\%$ multiplies by $1+\frac{r}{100}$ , while a decrease multiplies by $1-\frac{r}{100}$ .
Percentages above $100\%$ have multipliers above $1$ ; simple interest is calculated repeatedly from the original principal only.
A common error in reverse percentage problems is to undo a decrease by adding the same percentage to the reduced value.

Tier 1 · Easy

1. Work out $35\%$ of $240$ .[2 marks]
Answer
- $84$
Method: $35\%=0.35$ , so $0.35\times240=84$ .

Tier 2 · Standard

1. After a $12\%$ decrease, a machine is valued at £704. Work out its value before the decrease.[3 marks]
Answer
- £800
Method: A $12\%$ decrease leaves $88\%=0.88$ of the original value. Divide by the multiplier: £ $704/0.88=£800$ .

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 marks]
Answer
- £2891
Method: 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.

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

In direct proportion, both quantities change by the same factor and the graph is a straight line through the origin.
In inverse proportion, multiplying one quantity by a factor divides the other by that factor, so their product is constant.
Find the constant from a known pair, write the relationship, and then substitute the required value.
A common error is to treat an inverse relationship as linear and add or subtract the same amount.

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 marks]
Answer
- $y=30$
Method: $y=kx$ . Using $18=6k$ gives $k=3$ , so when $x=10$ , $y=3\times10=30$ .

Tier 2 · Standard

1. Variables $x$ and $y$ vary inversely. One recorded pair is $x=3$ , $y=14$ . Determine $y$ at $x=7$ .[3 marks]
Answer
- $y=6$
Method: For inverse proportion, $xy=k$ . Here $k=3\times14=42$ , so at $x=7$ , $y=42/7=6$ .

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 marks]
Answer
- New $y=9.6$
- $20\%$ decrease
Method: 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 $2.4/12\times100=20\%$ .

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

A compound unit combines two or more units, such as $\text{km}/\text{h}$ , £ per hour, $\text{kg}/\text{m}^3$ or $\text{N}/\text{m}^2$ .
Use $\text{speed}=\frac{\text{distance}}{\text{time}}$ , $\text{density}=\frac{\text{mass}}{\text{volume}}$ and $\text{pressure}=\frac{\text{force}}{\text{area}}$ with compatible units.
Unit pricing and rates of pay are found by dividing the total cost or pay by the relevant number of units or hours.
A common error is to substitute minutes into a calculation whose required rate is per hour.

Tier 1 · Easy

1. A shift lasting $7$ hours pays £52.50. Work out the hourly rate of pay.[2 marks]
Answer
- £7.50 per hour
Method: Divide total pay by time: £ $52.50/7=£7.50$ per hour.

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 marks]
Answer
- $65\,\text{km}/\text{h}$
Method: $24$ minutes is $24/60=0.4$ hours, so the time is $2.4$ hours. Average speed is $156/2.4=65\,\text{km}/\text{h}$ .

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 marks]
Answer
- $100\,000\,\text{Pa}$
Method: 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 $3600/0.036=100\,000\,\text{Pa}$ .

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

For similar shapes with length scale factor $k$ , corresponding lengths are in ratio $k$ , areas in ratio $k^2$ and volumes in ratio $k^3$ .
Find the linear scale factor first, then square or cube it to compare areas or volumes.
Corresponding angles and trigonometric ratios are unchanged in similar right-angled triangles, while all corresponding lengths share one scale factor.
A common error is to apply the linear scale factor directly to an area or a volume.

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 mark]
Answer
- $2:5$
Method: $6:15$ simplifies by dividing both parts by $3$ , giving $2:5$ .

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 marks]
Answer
- $294\,\text{cm}^2$
Method: The area ratio is $3^2:7^2=9:49$ . Therefore the larger area is $54\times\frac{49}{9}=294\,\text{cm}^2$ .

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 marks]
Answer
- $396\,\text{cm}^2$
Method: 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 $275\times\frac{36}{25}=396\,\text{cm}^2$ .

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

If $X$ is inversely proportional to $Y$ , then $X\propto\frac{1}{Y}$ and $X=\frac{k}{Y}$ for a constant $k$ .
Use one known pair to calculate $k$ , then write the complete equation before finding another value.
Direct proportion may involve powers, such as $y=kx^2$ , while inverse proportion may be written $y=\frac{k}{x^n}$ .
A common error is to write $X=kY$ for inverse proportion or to invert the wrong variable.

Tier 1 · Easy

1. Variables $p$ and $q$ vary inversely, with recorded values $p=12$ and $q=5$ . Write their connecting equation.[2 marks]
Answer
- $p=\frac{60}{q}$
Method: Write $p=k/q$ . Using the given pair, $k=pq=12\times5=60$ , so $p=\frac{60}{q}$ .

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 marks]
Answer
- $y=5x^2$
- $y=80$
Method: Write $y=kx^2$ . Then $45=k\times3^2$ , so $k=5$ and $y=5x^2$ . At $x=4$ , $y=5\times4^2=80$ .

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 marks]
Answer
- $b=5$
Method: 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 $b=5$ .

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

The gradient $\frac{\text{change in }y}{\text{change in }x}$ is a rate of change, with units formed from the vertical and horizontal axes.
Choose two well-separated points on a straight line, calculate the changes, and interpret the result in context.
A direct-proportion graph is a straight line through the origin; an inverse-proportion graph has constant product $xy$ and approaches the axes.
A common error is to calculate run over rise or to omit the rate's units and meaning.

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 marks]
Answer
- Gradient $=5\,\text{m}/\text{s}$
- The object travels at $5\,\text{m}/\text{s}$ .
Method: Gradient $=(35-10)/(7-2)=25/5=5\,\text{m}/\text{s}$ . On a distance-time graph this is the speed.

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 marks]
Answer
- The cost increases by £0.12 per minute.
- £22.20
Method: The coefficient of $m$ is the gradient, so the rate is £0.12 per minute. At $m=35$ , $C=18+0.12\times35=18+4.20=£22.20$ .

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 marks]
Answer
- $xy=36$ for all three points
- $y=\frac{36}{x}$
- $y=4$ when $x=9$
Method: 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 $x=9$ , $y=36/9=4$ .

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]

The gradient of a chord between two points on a curve gives the average rate of change over that interval.
The gradient of the tangent at one point estimates the instantaneous rate of change there.
Use two clear points on the chord or tangent and calculate $\frac{\Delta y}{\Delta x}$ , including the compound units.
A common error is to use two points on the curve when the question requires two points on the drawn tangent.

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 marks]
Answer
- $3$ units of $y$ per unit of $x$
Method: The chord gradient is $(23-5)/(8-2)=18/6=3$ .

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 marks]
Answer
- $3.5$ units of $y$ per unit of $x$
Method: Use two points on the tangent: $(32-11)/(10-4)=21/6=3.5$ . This tangent gradient estimates the instantaneous rate at $x=7$ .

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 marks]
Answer
- Average rate $=12$ litres per minute
- Instantaneous rate $=14$ litres per minute
- The instantaneous rate is $2$ litres per minute greater.
Method: 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.

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

Repeated growth by $r\%$ uses multiplier $1+\frac{r}{100}$ each period; repeated decay uses $1-\frac{r}{100}$ .
For $n$ equal periods, use $\text{final}=\text{initial}\times(\text{multiplier})^n$ and keep full calculator precision until the end.
An iterative process defines each new value from the previous one, so substitute successively and interpret the first term meeting the stated condition.
A common error is to calculate compound change as simple change or to round every intermediate iteration.

Tier 1 · Easy

1. £600 is invested at $4\%$ compound interest per year. Work out the value after $2$ years.[2 marks]
Answer
- £648.96
Method: Use multiplier $1.04$ twice: £ $600\times1.04^2=£648.96$ .

Tier 2 · Standard

1. A culture initially contains $960$ cells and decreases by $15\%$ each hour. Calculate the expected number after $3$ hours.[3 marks]
Answer
- About $590$ cells
- Unrounded model value $=589.56$
Method: The decay multiplier is $0.85$ . After $3$ hours the model gives $960\times0.85^3=589.56$ , which is about $590$ whole cells.

Tier 3 · Hard

1. A cooling model uses $T_{n+1}=0.65T_n+12$ with $T_0=80$ . Find the first value of $n$ for which $T_n<40$ , and give that temperature to one decimal place.[5 marks]
Answer
- $n=5$
- $T_5=39.6$ to one decimal place
Method: Iterate without premature rounding: $T_1=64$ , $T_2=53.6$ , $T_3=46.84$ , $T_4=42.446$ and $T_5=39.5899$ . The first value below $40$ occurs at $n=5$ , with temperature $39.6$ to one decimal place.