Edexcel GCSE Maths coverage

Ratio, proportion and rates of change

Section R
16 spec leafs

Notes and three levels of exam-style practice for each registered specification leaf in this section.

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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

  • Standard-unit conversions multiply or divide by a fixed factor, such as 1km=1000m1\,\text{km}=1000\,\text{m} and 1litre=103m31\,\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 1m=100cm1\,\text{m}=100\,\text{cm}, 1m2=104cm21\,\text{m}^2=10^4\,\text{cm}^2 and 1m3=106cm31\,\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 mark
ORIGINAL

A charity walk is 3.6km3.6\,\text{km} long. Change this distance to metres.

Tier 2 · Standard

3 marks
ORIGINAL

A pump moves water at 540540 litres per minute. Change this rate to cubic metres per second.

Tier 3 · Hard

4 marks
ORIGINAL

A solid block measures 2.4m2.4\,\text{m} by 35cm35\,\text{cm} by 80mm80\,\text{mm} and has mass 26.88kg26.88\,\text{kg}. Work out its density in kg/m3\text{kg}/\text{m}^3.

R2

Use scale factors, scale diagrams and maps

  • A scale 1:n1:n means one unit on the diagram represents nn 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 1:250001:25\,000, 1cm1\,\text{cm} represents 25000cm=250m25\,000\,\text{cm}=250\,\text{m}.
  • A common error is to mix centimetres and metres before applying the scale factor.

Tier 1 · Easy

2 marks
ORIGINAL

A ranger's sketch uses scale 1:250001:25\,000. A footpath trace measures 6.4cm6.4\,\text{cm}. Find the real distance in kilometres.

Tier 2 · Standard

2 marks
ORIGINAL

A model theatre uses scale 1:401:40. A real doorway is 2.04m2.04\,\text{m} high. Work out the model doorway height in centimetres.

Tier 3 · Hard

4 marks
ORIGINAL

A site plan has scale 1:25001:2500. A garden occupies 96cm296\,\text{cm}^2 on the plan. Calculate its real area in hectares. Use 11 hectare =10000m2=10\,000\,\text{m}^2.

R3

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

  • To express quantity AA as a fraction of quantity BB, write AB\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 11 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

2 marks
ORIGINAL

Express 1818 as a fraction of 3030. Give the fraction in its simplest form.

Tier 2 · Standard

2 marks
ORIGINAL

Express 8484 minutes as a fraction of 3535 minutes. Simplify your answer.

Tier 3 · Hard

3 marks
ORIGINAL

Express 0.84m20.84\,\text{m}^2 as a fraction of 2400cm22400\,\text{cm}^2. Give the fraction in its simplest form.

R4

Use ratio notation, including reduction to simplest form

  • A ratio compares quantities in a stated order, using notation such as a:ba: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 mark
ORIGINAL

Write the ratio 42:6342:63 in its simplest form.

Tier 2 · Standard

2 marks
ORIGINAL

Write 1.8m:75cm1.8\,\text{m}:75\,\text{cm} as a ratio in its simplest form.

Tier 3 · Hard

3 marks
ORIGINAL

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

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:ba:b, the whole contains a+ba+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 38\frac{3}{8} gives the complementary part 58\frac{5}{8}, so the two parts are in ratio 3:53:5.
  • A common error is to divide a total by one ratio number instead of by the sum of all parts.

Tier 1 · Easy

2 marks
ORIGINAL

Divide £84 in the ratio 3:43:4.

Tier 2 · Standard

3 marks
ORIGINAL

A box holds 240240 counters. Red counters make up 38\frac{3}{8} of the whole. Find the number of red and non-red counters, and state their ratio.

Tier 3 · Hard

5 marks
ORIGINAL

A technician mixes a 20%20\% solution with a 50%50\% solution to make 1818 litres of a 40%40\% solution. Find the volume of each starting solution.

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=kBA=kB, then A:B=k:1A:B=k:1 and AA is the fraction kk of BB.
  • For integer ratio notation, rewrite a fractional multiplier such as 53\frac{5}{3} as the ratio 5:35:3.
  • A common error is to subtract the quantities and report the difference as though it were a ratio.

Tier 1 · Easy

2 marks
ORIGINAL

Quantity AA is 1818 and quantity BB is 3030. Express AA as a fraction of BB and write A:BA:B in simplest form.

Tier 2 · Standard

2 marks
ORIGINAL

The mass of parcel PP is 1.751.75 times the mass of parcel QQ. Express P:QP:Q as an integer ratio and express PP as a fraction of QQ.

Tier 3 · Hard

3 marks
ORIGINAL

Quantities PP, QQ and RR satisfy P=53QP=\frac{5}{3}Q and Q=34RQ=\frac{3}{4}R. Express P:RP:R in simplest form and write PP as a fraction of RR.

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, ab=cd\frac{a}{b}=\frac{c}{d} implies ad=bcad=bc when bb and dd are non-zero.
  • A common error is to compare quantities in different orders on the two sides of the equation.

Tier 1 · Easy

2 marks
ORIGINAL

Solve the proportion x15=610\frac{x}{15}=\frac{6}{10}.

Tier 2 · Standard

3 marks
ORIGINAL

Eight identical notebooks cost £11.20. Use proportion to find the cost of 1414 notebooks.

Tier 3 · Hard

4 marks
ORIGINAL

Solve 2x+318=x+612\frac{2x+3}{18}=\frac{x+6}{12} and verify that the two ratios are equal.

R8

Relate ratios to fractions and to linear functions

  • If two parts are in ratio a:ba:b, their fractions of the whole are aa+b\frac{a}{a+b} and ba+b\frac{b}{a+b}.
  • A constant ratio y:x=m:1y:x=m:1 gives the linear relationship y=mxy=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 ab\frac{a}{b} for the fraction of the whole instead of aa+b\frac{a}{a+b}.

Tier 1 · Easy

2 marks
ORIGINAL

The ratio of cats to dogs at a shelter is 3:53:5. What fraction of the animals are cats?

Tier 2 · Standard

3 marks
ORIGINAL

Quantities yy and xx are always in the ratio 5:25:2. Write yy as a linear function of xx, then find yy when x=14x=14.

Tier 3 · Hard

4 marks
ORIGINAL

A mixture contains concentrate and water in the ratio 2:72:7. Let cc litres be the concentrate and VV litres be the total mixture. Express VV as a linear function of cc, then find both component volumes when V=54V=54.

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%=p100p\%=\frac{p}{100}.
  • Use a multiplier: an increase of r%r\% multiplies by 1+r1001+\frac{r}{100}, while a decrease multiplies by 1r1001-\frac{r}{100}.
  • Percentages above 100%100\% have multipliers above 11; 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

2 marks
ORIGINAL

Work out 35%35\% of 240240.

Tier 2 · Standard

3 marks
ORIGINAL

After a 12%12\% decrease, a machine is valued at £704. Work out its value before the decrease.

Tier 3 · Hard

5 marks
ORIGINAL

A saver deposits £2500 in an account paying 3.6%3.6\% simple interest each year. After 55 years, a fee equal to 2%2\% of the final balance is charged. Calculate the amount left after the fee.

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

3 marks
ORIGINAL

A direct variation links xx and yy. The pair x=6x=6, y=18y=18 is known. Determine yy at x=10x=10.

Tier 2 · Standard

3 marks
ORIGINAL

Variables xx and yy vary inversely. One recorded pair is x=3x=3, y=14y=14. Determine yy at x=7x=7.

Tier 3 · Hard

5 marks
ORIGINAL

xx and yy are inversely proportional. Initially x=6x=6 and y=12y=12. The value of xx is increased by 25%25\%. Find the new value of yy and the percentage decrease in yy.

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 km/h\text{km}/\text{h}, £ per hour, kg/m3\text{kg}/\text{m}^3 or N/m2\text{N}/\text{m}^2.
  • Use speed=distancetime\text{speed}=\frac{\text{distance}}{\text{time}}, density=massvolume\text{density}=\frac{\text{mass}}{\text{volume}} and pressure=forcearea\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

2 marks
ORIGINAL

A shift lasting 77 hours pays £52.50. Work out the hourly rate of pay.

Tier 2 · Standard

3 marks
ORIGINAL

A coach travels 156km156\,\text{km} in 22 hours 2424 minutes. Calculate its average speed in km/h\text{km}/\text{h}.

Tier 3 · Hard

4 marks
ORIGINAL

A force of 3.6kN3.6\,\text{kN} acts uniformly on a rectangular pad measuring 24cm24\,\text{cm} by 15cm15\,\text{cm}. Calculate the pressure in pascals, where 1Pa=1N/m21\,\text{Pa}=1\,\text{N}/\text{m}^2.

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 kk, corresponding lengths are in ratio kk, areas in ratio k2k^2 and volumes in ratio k3k^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 mark
ORIGINAL

Two similar shapes have corresponding sides of 6cm6\,\text{cm} and 15cm15\,\text{cm}. Write the smaller-to-larger length ratio in simplest form.

Tier 2 · Standard

3 marks
ORIGINAL

The corresponding length ratio of two similar tiles is 3:73:7. The smaller tile has area 54cm254\,\text{cm}^2. Find the area of the larger tile.

Tier 3 · Hard

4 marks
ORIGINAL

Two similar solids have smaller-to-larger volume ratio 125:216125:216. The smaller solid has surface area 275cm2275\,\text{cm}^2. Work out the larger surface area.

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 XX is inversely proportional to YY, then X1YX\propto\frac{1}{Y} and X=kYX=\frac{k}{Y} for a constant kk.
  • Use one known pair to calculate kk, then write the complete equation before finding another value.
  • Direct proportion may involve powers, such as y=kx2y=kx^2, while inverse proportion may be written y=kxny=\frac{k}{x^n}.
  • A common error is to write X=kYX=kY for inverse proportion or to invert the wrong variable.

Tier 1 · Easy

2 marks
ORIGINAL

Variables pp and qq vary inversely, with recorded values p=12p=12 and q=5q=5. Write their connecting equation.

Tier 2 · Standard

4 marks
ORIGINAL

The relationship between yy and x2x^2 is direct proportion. Given x=3x=3 when y=45y=45, form the equation and evaluate yy at x=4x=4.

Tier 3 · Hard

4 marks
ORIGINAL

For positive bb, the variable aa varies inversely with b2b^2. Given a=20a=20 at b=3b=3, determine bb when a=7.2a=7.2.

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 change in ychange in x\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 xyxy 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

3 marks
ORIGINAL

A straight distance-time graph passes through (2,10)(2,10) and (7,35)(7,35), where time is in seconds and distance in metres. Find and interpret its gradient.

Tier 2 · Standard

3 marks
ORIGINAL

A phone-call cost graph is modelled by C=18+0.12mC=18+0.12m, where CC is cost in pounds and mm is time in minutes. Interpret the gradient and calculate the cost of a 3535-minute call.

Tier 3 · Hard

4 marks
ORIGINAL

An inverse-proportion model contains the points (2,18)(2,18), (3,12)(3,12) and (6,6)(6,6). Check that all three coordinates are consistent with the model, write its equation, and find yy when x=9x=9.

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 ΔyΔx\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

2 marks
ORIGINAL

A curve passes through (2,5)(2,5) and (8,23)(8,23). Calculate the average rate of change of yy with respect to xx between these points.

Tier 2 · Standard

3 marks
ORIGINAL

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

Tier 3 · Hard

5 marks
ORIGINAL

A curve shows water volume VV litres after tt minutes and passes through (2,46)(2,46) and (8,118)(8,118). The tangent at t=5t=5 passes through (4,70)(4,70) and (7,112)(7,112). Find the average rate from t=2t=2 to t=8t=8, estimate the instantaneous rate at t=5t=5, and compare them.

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%r\% uses multiplier 1+r1001+\frac{r}{100} each period; repeated decay uses 1r1001-\frac{r}{100}.
  • For nn equal periods, use final=initial×(multiplier)n\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

2 marks
ORIGINAL

£600 is invested at 4%4\% compound interest per year. Work out the value after 22 years.

Tier 2 · Standard

3 marks
ORIGINAL

A culture initially contains 960960 cells and decreases by 15%15\% each hour. Calculate the expected number after 33 hours.

Tier 3 · Hard

5 marks
ORIGINAL

A cooling model uses Tn+1=0.65Tn+12T_{n+1}=0.65T_n+12 with T0=80T_0=80. Find the first value of nn for which Tn<40T_n<40, and give that temperature to one decimal place.