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Edexcel IGCSE Maths 4MA1 · Concept clusters

Edexcel IGCSE Maths concept clusters

Original, past-paper-style synoptic questions. Each one is built around a single setup and needs ideas from several subchapters at once, so you practise connecting topics the way long exam questions expect. 24 questions · 462 marks · two per subchapter or merged group · every part with a worked solution · All courses

Topic

Subchapter

Level

Skill

Topic 1 · Numbers and the number system

● home subchapter · ○ linked subchapter · last column: links to other topics

Setup	1.1	1.2	1.3	1.4	1.5	Other
A concert tour: ticket prices, VAT and revenue shares	●		○			3.2
Gears on a road bike: ratios, proportion and accuracy			●			4.3 3.2
Signals from a deep-space probe: standard form and travel times	○	●	○			3.2
A-series paper: surds, powers and the ratio root 2		○			●	4.2 3.1
Streaming habits: a three-set Venn diagram survey	○			●		6.2 6.1
A health clinic screening: two-set Venn diagram with algebra				●		2.2 6.4 6.1

1.1 Integers, fractions, decimals and percentages · 1.3 Ratio, proportion and degree of accuracymerged group

+ Add to test✓ In your testCC-EIM-11-13-118 marks~20 min

A concert tour: ticket prices, VAT and revenue shares

1.1 Integers, fractions, decimals and percentages1.3 Ratio, proportion and degree of accuracy3.2 Graphs, gradients and real-life graphsFoundation

A band plays one concert at an arena with 12 000 seats. Every seat is either a standard seat or a VIP seat. \(\frac{1}{10}\) of the seats are VIP seats.

All ticket prices include VAT at 20%.

The graph in Figure 1 shows the total number of standard tickets sold by the end of each week after tickets went on sale. All the standard tickets were sold by the end of week 5.

Figure 1: the total number of standard tickets sold, recorded at the end of each week

(a)

A standard ticket costs £57.60, including VAT.

Work out the price of a standard ticket before VAT is added.

ROUTINENotes

Show solution

\(57.60 \div 1.2\) or \(\frac{57.60}{120}\times100\)	M1	Method may be seen in stages: 1% \(= 0.48\).
£48	A1	Accept £48.00. \(57.60\times0.8 = 46.08\) scores M0 A0.

Answer: £48

(2)

(b)

Work out the number of standard seats.

ROUTINENotes

Show solution

\(12\,000\times\frac{1}{10}\) \((=1200)\) or \(12\,000\times\frac{9}{10}\)	M1
10 800	A1

Answer: 10 800

(2)

(c)

Use the graph to find the fraction of the standard seats that had been sold by the end of week 2.

Give your fraction in its simplest form.

GRAPHNotes

Show solution

7200 read from the graph	B1
\(\frac{7200}{10\,800}\)	M1	Allow ECF from their answer to (b) as the denominator.
\(\frac{2}{3}\)	A1	Correct simplest form of their fraction for ECF; 0.666… or 66.7% alone scores A0.

Answer: \(\frac{2}{3}\)

(3)

(d)

Show that the number of standard tickets sold during week 5 was 50% more than the number sold during week 4.

PROOFNotes

Show solution

\(9600-8800 = 800\) and \(10\,800-9600 = 1200\)	M1	Both weekly sales found from the graph.
\(\frac{1200-800}{800}\times100\) or \(\frac{1200}{800} = 1.5\) or \(800\times1.5 = 1200\)	M1	Dependent on the first M1.
50% with full working shown	A1	cso. A conclusion such as \(1.5 = 1 + 0.5\), so 50% more, is needed for the multiplier method.

Answer: 800 → 1200, an increase of 50%

(3)

(e)(i)

Write down what the gradient of each line segment in Figure 1 represents.

INTERPRETNotes

Show solution

the number of standard tickets sold per week (in that week)

Accept the rate of ticket sales. Do not accept the total number of tickets sold.

(1)

(e)(ii)

Kai says,

“The graph shows that more standard tickets were sold in week 5 than in any other week.”

Is Kai correct? Give a reason for your answer.

INTERPRETNotes

Show solution

No, with a reason, e.g. 4800 tickets were sold in week 1 but only 1200 in week 5, or the line is steepest in week 1

Accept a comparison with week 2 (2400) or week 3 (1600) instead of week 1.

Answer: No: 4800 were sold in week 1 (the steepest segment)

(1)

(f)

A VIP ticket costs £138, including VAT. Every seat for the concert is sold.

The money from all the ticket sales, not including VAT, is shared between the band, the promoter and the venue in the ratio \(5:3:2\).

Work out the amount of money the band receives.

MULTINotes

Show solution

\(10\,800\times48\) \((=518\,400)\) or \(10\,800\times57.60\) \((=622\,080)\)	M1	Allow ECF from (a) and (b).
\(1200\times115\) \((=138\,000)\) or \(1200\times138\) \((=165\,600)\)	M1	VIP takings; \(138\div1.2 = 115\).
\(656\,400\div10\times5\) or \(787\,680\div1.2\div10\times5\)	M1	Total without VAT divided by 10 and multiplied by 5.
£328 200	A1	cao

Answer: £328 200

(4)

(g)

For the next concert the ticket money is the same, but the ratio changes to \(9:7:4\) for the band, the promoter and the venue.

Work out how much less money the band receives than in part (f).

MULTINotes

Show solution

\(656\,400\div20\times9\) \((=295\,380)\) or \(656\,400\times\left(\frac{5}{10}-\frac{9}{20}\right)\)	M1	Allow ECF from their total in (f).
£32 820	A1	Accept £32 820 from a correct follow-through of their (f).

Answer: £32 820

(2)

Show full solution

Working	Mark	Notes
Part (a)
\(57.60 \div 1.2\) or \(\frac{57.60}{120}\times100\)	M1	Method may be seen in stages: 1% \(= 0.48\).
£48	A1	Accept £48.00. \(57.60\times0.8 = 46.08\) scores M0 A0.
Part (b)
\(12\,000\times\frac{1}{10}\) \((=1200)\) or \(12\,000\times\frac{9}{10}\)	M1
10 800	A1
Part (c)
7200 read from the graph	B1
\(\frac{7200}{10\,800}\)	M1	Allow ECF from their answer to (b) as the denominator.
\(\frac{2}{3}\)	A1	Correct simplest form of their fraction for ECF; 0.666… or 66.7% alone scores A0.
Part (d)
\(9600-8800 = 800\) and \(10\,800-9600 = 1200\)	M1	Both weekly sales found from the graph.
\(\frac{1200-800}{800}\times100\) or \(\frac{1200}{800} = 1.5\) or \(800\times1.5 = 1200\)	M1	Dependent on the first M1.
50% with full working shown	A1	cso. A conclusion such as \(1.5 = 1 + 0.5\), so 50% more, is needed for the multiplier method.
Part (e)(i)
the number of standard tickets sold per week (in that week)	B1	Accept the rate of ticket sales. Do not accept the total number of tickets sold.
Part (e)(ii)
No, with a reason, e.g. 4800 tickets were sold in week 1 but only 1200 in week 5, or the line is steepest in week 1	B1	Accept a comparison with week 2 (2400) or week 3 (1600) instead of week 1.
Part (f)
\(10\,800\times48\) \((=518\,400)\) or \(10\,800\times57.60\) \((=622\,080)\)	M1	Allow ECF from (a) and (b).
\(1200\times115\) \((=138\,000)\) or \(1200\times138\) \((=165\,600)\)	M1	VIP takings; \(138\div1.2 = 115\).
\(656\,400\div10\times5\) or \(787\,680\div1.2\div10\times5\)	M1	Total without VAT divided by 10 and multiplied by 5.
£328 200	A1	cao
Part (g)
\(656\,400\div20\times9\) \((=295\,380)\) or \(656\,400\times\left(\frac{5}{10}-\frac{9}{20}\right)\)	M1	Allow ECF from their total in (f).
£32 820	A1	Accept £32 820 from a correct follow-through of their (f).

Answers: (a) £48 · (b) 10 800 · (c) \(\frac{2}{3}\) · (d) 800 → 1200, an increase of 50% · (e)(ii) No: 4800 were sold in week 1 (the steepest segment) · (f) £328 200 · (g) £32 820

Common error (a) Taking 20% off £57.60 gives £46.08, which is wrong: the 20% was added to the price before VAT, so divide by the multiplier 1.2.

Exam tip (e)(ii) On a graph of a running total, the number sold in one week is the rise of that segment, not the height of the point at the end of the week.

More 1.1 questions in the question bank

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Gears on a road bike: ratios, proportion and accuracy

1.3 Ratio, proportion and degree of accuracy4.3 Area, surface area and volume3.2 Graphs, gradients and real-life graphsFoundation

Mia rides a road bike. The pedals turn a chainring with 52 teeth. A chain joins the chainring to a rear sprocket with 16 teeth, which turns the rear wheel. The diameter of the rear wheel is 680 mm, correct to the nearest millimetre.

The gear ratio is the ratio of the number of teeth on the chainring to the number of teeth on the rear sprocket. In one complete turn of the pedals, the rear wheel turns \(52 \div 16\) times.

The cadence is the number of complete turns of the pedals per minute.

Figure 1: the rear wheel and gears of the bike (not to scale)

(a)(i)

Write the gear ratio \(52:16\) in its simplest form.

ROUTINENotes

Show solution

\(13:4\)

Accept \(3.25:1\).

Answer: \(13:4\)

(1)

(a)(ii)

Mia turns the pedals 20 times.

Work out the number of times the rear wheel turns.

ROUTINENotes

Show solution

Answer: 65

(1)

(b)

Work out the distance the bike travels in one complete turn of the pedals.

Give your answer in metres, correct to 3 significant figures.

MULTINotes

Show solution

\(\pi\times680\) \((=2136.2\ldots\text{ mm})\) or \(\pi\times0.68\) \((=2.136\ldots\text{ m})\)	M1	Circumference of the wheel. \(\pi\times340^2\) scores M0.
\(\pi\times680\times\frac{52}{16}\) \((=6942.9\ldots)\)	M1	Their circumference multiplied by 3.25.
6.94 m	A1	Accept 6.94 only (correct to 3 s.f., in metres).

Answer: 6.94 m

(3)

(c)

Mia pedals with a cadence of 80 turns per minute.

Show that her speed is 33 km/h, correct to the nearest km/h.

PROOFNotes

Show solution

\(80\times6.94\) \((=555.2\text{ m per minute})\)	M1	Allow \(80\times6942.9\ldots\) mm; ECF from (b).
their \(555.2\times60\div1000\)	M1	Correct conversion of their distance per minute to km/h.
33.3 (km/h), so 33 km/h to the nearest km/h	A1	cso. Must see a value such as 33.3 before rounding to 33.

Answer: 33.3 km/h, which is 33 km/h to the nearest km/h

(3)

(d)

Mia's speed is directly proportional to her cadence. Figure 2 shows the graph of speed against cadence when the 16-tooth sprocket is used.

Use the graph to find the gradient of the line, and give a practical interpretation of this gradient.

Figure 2: speed of the bike against cadence when the 16-tooth sprocket is used

GRAPHNotes

Show solution

a correct gradient triangle or two points read, e.g. \(\frac{50}{120}\) or \(\frac{25}{60}\)	M1
0.42	A1	Accept answers from 0.41 to 0.42 inclusive, or \(\frac{5}{12}\).
the speed increases by about 0.42 km/h for each extra pedal turn per minute	B1	Interpretation must refer to speed and cadence (with ‘per pedal turn per minute’ or equivalent).

Answer: 0.42 (km/h per pedal turn per minute)

(3)

(e)(i)

For a hill, Mia changes to a rear sprocket with 32 teeth. The chainring and the wheel are not changed.

Work out Mia's speed in km/h when her cadence is 96 turns per minute.

MULTINotes

Show solution

\(\pi\times0.68\times\frac{52}{32}\) \((=3.471\ldots\text{ m})\)	M1	Distance per pedal turn. Or half of their (b), since the 32-tooth sprocket has twice as many teeth as the 16-tooth sprocket.
their \(3.471\ldots\times96\times60\div1000\)	M1
20 (km/h)	A1	Accept 19.9 to 20.0. Using 6.94 ÷ 2 gives 19.987…, which is acceptable.

Answer: 20 km/h

(3)

(e)(ii)

On Figure 2, draw the graph of speed against cadence for the 32-tooth sprocket, for cadences from 0 to 120 turns per minute.

GRAPHNotes

Show solution

a straight line from \((0, 0)\) through \((96, 20)\) to about \((120, 25)\)

Ruled line through the origin and their point from (e)(i) (ECF).

Answer: Straight line through \((0, 0)\) and \((96, 20)\)

(1)

(f)

On the hill Mia rides at a speed of 14 km/h. She likes to keep her cadence between 60 and 90 turns per minute.

Use your graphs to decide which sprocket, 16-tooth or 32-tooth, Mia should use on the hill. Give a reason for your answer.

INTERPRETNotes

Show solution

reads a cadence at 14 km/h from either line: about 34 (16-tooth) or about 67 (32-tooth)	M1	Accept 32 to 36 and 65 to 70, or ECF from their line in (e)(ii).
32-tooth, because its cadence (about 67) is between 60 and 90, while the 16-tooth cadence (about 34) is below 60	A1	Both cadences must be seen or clearly used.

Answer: 32-tooth sprocket: cadence about 67, inside 60–90 (16-tooth: about 34)

(2)

(g)

The diameter of the wheel, 680 mm, is correct to the nearest millimetre.

By considering bounds, decide whether your answer to (b) for the 16-tooth sprocket is correct to 3 significant figures. You must show your working.

Higher onlyINTERPRETNotes

Show solution

679.5 or 680.5 seen	B1
\(\pi\times679.5\times3.25\) \((=6937.8\ldots)\) or \(\pi\times680.5\times3.25\) \((=6948.0\ldots)\)	M1	One bound of the distance per pedal turn, in mm or m.
both 6.937(8…) m and 6.948(0…) m, and the conclusion that the lower bound rounds to 6.94 but the upper bound rounds to 6.95, so the answer is not certain to be correct to 3 s.f.	A1	Accept values in mm. The conclusion must follow from both bounds; "yes" scores A0.

Answer: Lower bound 6.938 m, upper bound 6.948 m; these round to 6.94 and 6.95, so no

(3)

Show full solution

Working	Mark	Notes
Part (a)(i)
\(13:4\)	B1	Accept \(3.25:1\).
Part (a)(ii)
65	B1
Part (b)
\(\pi\times680\) \((=2136.2\ldots\text{ mm})\) or \(\pi\times0.68\) \((=2.136\ldots\text{ m})\)	M1	Circumference of the wheel. \(\pi\times340^2\) scores M0.
\(\pi\times680\times\frac{52}{16}\) \((=6942.9\ldots)\)	M1	Their circumference multiplied by 3.25.
6.94 m	A1	Accept 6.94 only (correct to 3 s.f., in metres).
Part (c)
\(80\times6.94\) \((=555.2\text{ m per minute})\)	M1	Allow \(80\times6942.9\ldots\) mm; ECF from (b).
their \(555.2\times60\div1000\)	M1	Correct conversion of their distance per minute to km/h.
33.3 (km/h), so 33 km/h to the nearest km/h	A1	cso. Must see a value such as 33.3 before rounding to 33.
Part (d)
a correct gradient triangle or two points read, e.g. \(\frac{50}{120}\) or \(\frac{25}{60}\)	M1
0.42	A1	Accept answers from 0.41 to 0.42 inclusive, or \(\frac{5}{12}\).
the speed increases by about 0.42 km/h for each extra pedal turn per minute	B1	Interpretation must refer to speed and cadence (with ‘per pedal turn per minute’ or equivalent).
Part (e)(i)
\(\pi\times0.68\times\frac{52}{32}\) \((=3.471\ldots\text{ m})\)	M1	Distance per pedal turn. Or half of their (b), since the 32-tooth sprocket has twice as many teeth as the 16-tooth sprocket.
their \(3.471\ldots\times96\times60\div1000\)	M1
20 (km/h)	A1	Accept 19.9 to 20.0. Using 6.94 ÷ 2 gives 19.987…, which is acceptable.
Part (e)(ii)
a straight line from \((0, 0)\) through \((96, 20)\) to about \((120, 25)\)	B1	Ruled line through the origin and their point from (e)(i) (ECF).
Part (f)
reads a cadence at 14 km/h from either line: about 34 (16-tooth) or about 67 (32-tooth)	M1	Accept 32 to 36 and 65 to 70, or ECF from their line in (e)(ii).
32-tooth, because its cadence (about 67) is between 60 and 90, while the 16-tooth cadence (about 34) is below 60	A1	Both cadences must be seen or clearly used.
Part (g)
679.5 or 680.5 seen	B1
\(\pi\times679.5\times3.25\) \((=6937.8\ldots)\) or \(\pi\times680.5\times3.25\) \((=6948.0\ldots)\)	M1	One bound of the distance per pedal turn, in mm or m.
both 6.937(8…) m and 6.948(0…) m, and the conclusion that the lower bound rounds to 6.94 but the upper bound rounds to 6.95, so the answer is not certain to be correct to 3 s.f.	A1	Accept values in mm. The conclusion must follow from both bounds; "yes" scores A0.

Answers: (a)(i) \(13:4\) · (a)(ii) 65 · (b) 6.94 m · (c) 33.3 km/h, which is 33 km/h to the nearest km/h · (d) 0.42 (km/h per pedal turn per minute) · (e)(i) 20 km/h · (e)(ii) Straight line through \((0, 0)\) and \((96, 20)\) · (f) 32-tooth sprocket: cadence about 67, inside 60–90 (16-tooth: about 34) · (g) Lower bound 6.938 m, upper bound 6.948 m; these round to 6.94 and 6.95, so no

Common error (b) Using \(\pi r^2\) gives an area, not a distance. One turn of the wheel moves the bike forward one circumference, \(\pi d\).

More 1.3 questions in the question bank

1.2 Powers, roots and standard form · 1.5 Surdsmerged group

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Signals from a deep-space probe: standard form and travel times

1.2 Powers, roots and standard form1.3 Ratio, proportion and degree of accuracy1.1 Integers, fractions, decimals and percentages3.2 Graphs, gradients and real-life graphsFoundation

A space probe is travelling away from the Sun in a straight line at a constant speed.

Figure 1 shows the distance, \(d\times10^{9}\) km, of the probe from the Sun, \(t\) years after 1 January 2030.

Figure 1: the distance, d × 10⁹ km, of the probe from the Sun, t years after 1 January 2030

(a)(i)

The planet Neptune is about 4 500 000 000 km from the Sun.

Write 4 500 000 000 in standard form.

ROUTINENotes

Show solution

\(4.5\times10^{9}\)

Do not accept \(45\times10^{8}\).

Answer: \(4.5\times10^{9}\)

(1)

(a)(ii)

Use Figure 1 to find the date on which the probe was 4.5 × 10⁹ km from the Sun.

GRAPHNotes

Show solution

1 January 2031

Accept \(t=1\) clearly linked to the start of 2031. “2031” alone is enough.

Answer: 1 January 2031

(1)

(b)

Take one year to be 365 days.

Show that the speed of the probe is about 16 km/s.

PROOFNotes

Show solution

gradient \(=\frac{8\times10^{9}-4\times10^{9}}{8-0}\) or any correct pair of points from the line	M1	Change in \(d\) divided by change in \(t\), using the scale \(\times10^{9}\).
\(5\times10^{8}\) km per year	A1	Accept 500 000 000 km per year.
\(\frac{5\times10^{8}}{365\times24\times60\times60}\) or \(\frac{5\times10^{8}}{3.1536\times10^{7}}\)	M1	Dividing their distance per year by the number of seconds in a year. Allow ECF from an incorrect gradient for this mark only.
\(15.85\ldots\) which rounds to 16 km/s	A1	cso. Must see a value such as 15.8 or 15.85 before stating 16.

Answer: \(15.85\ldots\approx16\) km/s

(4)

(c)

Radio signals travel at \(3.0\times10^{5}\) km/s.

On 1 July 2035 the probe was \(6.75\times10^{9}\) km from the Earth. A signal was sent from the Earth to the probe.

Work out the time the signal took to reach the probe. Give your answer in hours and minutes.

MULTINotes

Show solution

\(\frac{6.75\times10^{9}}{3.0\times10^{5}}\) \((=22\,500\) s\()\)	M1	Time \(=\) distance \(\div\) speed.
\(22\,500\div3600\) \((=6.25\) hours\()\) or \(22\,500\div60\) \((=375\) minutes\()\)	M1	Converting their time in seconds.
6 hours 15 minutes	A1	6.25 hours alone or 6 hours 25 minutes scores A0.

Answer: 6 hours 15 minutes

(3)

(d)(i)

Use Figure 1 to write down the distance of the probe from the Sun on 1 January 2036. Give your answer in standard form.

GRAPHNotes

Show solution

\(7.0\times10^{9}\) km

Accept \(7\times10^{9}\).

Answer: \(7.0\times10^{9}\) km

(1)

(d)(ii)

The distance from the Earth to the Sun is \(1.5\times10^{8}\) km.

Write the ratio

distance from the Earth to the Sun : distance from the probe to the Sun on 1 January 2036

in the form \(1:n\). Give \(n\) correct to 3 significant figures.

ROUTINENotes

Show solution

\(\frac{7.0\times10^{9}}{1.5\times10^{8}}\) or \(1.5\times10^{8}:7.0\times10^{9}\) with both parts divided by \(1.5\times10^{8}\)	M1	Allow ECF from (d)(i).
\(1:46.7\)	A1	Accept more accurate values such as \(1:46.67\). \(1:47\) scores A0.

Answer: \(1:46.7\)

(2)

(e)

As the Earth moves around the Sun, the distance from the Earth to the probe can be up to \(1.5\times10^{8}\) km more or less than the distance from the Sun to the probe.

The mission team works out signal times using the distance from the Sun to the probe instead of the distance from the Earth to the probe.

Work out \(1.5\times10^{8}\) km as a percentage of the distance from the probe to the Sun on 1 January 2036, and comment on whether the team's method is sensible.

INTERPRETNotes

Show solution

\(\frac{1.5\times10^{8}}{7.0\times10^{9}}\times100\) or \(\frac{1}{46.7}\times100\)	M1	Allow ECF from (d)(i) or (d)(ii).
2.14(%)	A1	Accept 2.1 or 2.142…
Sensible, with a reason, e.g. the error in the distance (and so in the time) is only about 2%	B1	Accept a statement that the time would be wrong by at most about 2% or by only a few minutes. Dependent on a percentage less than 10%.

Answer: 2.14%; sensible, as the error is only about 2%

(3)

(f)

Later in the mission, the probe is \(1.08\times10^{10}\) km from the Earth.

Work out the percentage increase in the time a signal takes to reach the probe, compared with your answer to part (c).

MULTINotes

Show solution

\(\frac{1.08\times10^{10}}{6.75\times10^{9}}\) \((=1.6)\) or \(\frac{1.08\times10^{10}}{3.0\times10^{5}}\) \((=36\,000\) s or 10 hours\()\)	M1	Either the scale factor for the distance or the new time.
\((1.6-1)\times100\) or \(\frac{10-6.25}{6.25}\times100\) or \(\frac{36\,000-22\,500}{22\,500}\times100\)	M1	Allow ECF from (c) for their time.
60(%)	A1	cao

Answer: 60%

(3)

(g)

The time, \(T\) years, that an object takes to travel once around the Sun is given by

\(T=a^{\frac32}\)

where \(a\) is the average distance of the object from the Sun, measured in astronomical units (AU).

The probe flies past a dwarf planet that takes 216 years to travel once around the Sun.

Find the average distance of the dwarf planet from the Sun, in AU.

Higher onlyALGEBRANotes

Show solution

\(a=216^{\frac23}\) or \(a^{3}=216^{2}\) or \(\left(\sqrt[3]{216}\right)^{2}\)	M1	Correct inverse operation on the index.
36	A1	cao

Answer: 36 AU

(2)

Show full solution

Working	Mark	Notes
Part (a)(i)
\(4.5\times10^{9}\)	B1	Do not accept \(45\times10^{8}\).
Part (a)(ii)
1 January 2031	B1	Accept \(t=1\) clearly linked to the start of 2031. “2031” alone is enough.
Part (b)
gradient \(=\frac{8\times10^{9}-4\times10^{9}}{8-0}\) or any correct pair of points from the line	M1	Change in \(d\) divided by change in \(t\), using the scale \(\times10^{9}\).
\(5\times10^{8}\) km per year	A1	Accept 500 000 000 km per year.
\(\frac{5\times10^{8}}{365\times24\times60\times60}\) or \(\frac{5\times10^{8}}{3.1536\times10^{7}}\)	M1	Dividing their distance per year by the number of seconds in a year. Allow ECF from an incorrect gradient for this mark only.
\(15.85\ldots\) which rounds to 16 km/s	A1	cso. Must see a value such as 15.8 or 15.85 before stating 16.
Part (c)
\(\frac{6.75\times10^{9}}{3.0\times10^{5}}\) \((=22\,500\) s\()\)	M1	Time \(=\) distance \(\div\) speed.
\(22\,500\div3600\) \((=6.25\) hours\()\) or \(22\,500\div60\) \((=375\) minutes\()\)	M1	Converting their time in seconds.
6 hours 15 minutes	A1	6.25 hours alone or 6 hours 25 minutes scores A0.
Part (d)(i)
\(7.0\times10^{9}\) km	B1	Accept \(7\times10^{9}\).
Part (d)(ii)
\(\frac{7.0\times10^{9}}{1.5\times10^{8}}\) or \(1.5\times10^{8}:7.0\times10^{9}\) with both parts divided by \(1.5\times10^{8}\)	M1	Allow ECF from (d)(i).
\(1:46.7\)	A1	Accept more accurate values such as \(1:46.67\). \(1:47\) scores A0.
Part (e)
\(\frac{1.5\times10^{8}}{7.0\times10^{9}}\times100\) or \(\frac{1}{46.7}\times100\)	M1	Allow ECF from (d)(i) or (d)(ii).
2.14(%)	A1	Accept 2.1 or 2.142…
Sensible, with a reason, e.g. the error in the distance (and so in the time) is only about 2%	B1	Accept a statement that the time would be wrong by at most about 2% or by only a few minutes. Dependent on a percentage less than 10%.
Part (f)
\(\frac{1.08\times10^{10}}{6.75\times10^{9}}\) \((=1.6)\) or \(\frac{1.08\times10^{10}}{3.0\times10^{5}}\) \((=36\,000\) s or 10 hours\()\)	M1	Either the scale factor for the distance or the new time.
\((1.6-1)\times100\) or \(\frac{10-6.25}{6.25}\times100\) or \(\frac{36\,000-22\,500}{22\,500}\times100\)	M1	Allow ECF from (c) for their time.
60(%)	A1	cao
Part (g)
\(a=216^{\frac23}\) or \(a^{3}=216^{2}\) or \(\left(\sqrt[3]{216}\right)^{2}\)	M1	Correct inverse operation on the index.
36	A1	cao

Answers: (a)(i) \(4.5\times10^{9}\) · (a)(ii) 1 January 2031 · (b) \(15.85\ldots\approx16\) km/s · (c) 6 hours 15 minutes · (d)(i) \(7.0\times10^{9}\) km · (d)(ii) \(1:46.7\) · (e) 2.14%; sensible, as the error is only about 2% · (f) 60% · (g) 36 AU

Common error (b) The vertical scale is in units of \(10^{9}\) km, so a rise of 4 on the graph is \(4\times10^{9}\) km, not 4 km. Write the units next to the gradient before converting to km/s.

Exam tip (c) 0.25 of an hour is \(0.25\times60=15\) minutes, not 25 minutes. Convert the decimal part of the hours separately.

More 1.2 questions in the question bank

+ Add to test✓ In your testCC-EIM-12-15-220 marks~20 min

A-series paper: surds, powers and the ratio root 2

1.5 Surds1.2 Powers, roots and standard form4.2 Pythagoras' theorem and trigonometry3.1 Sequences and nth termsHigher

Paper sizes A0, A1, A2, A3, … are all rectangles with sides in the ratio \(1:\sqrt2\).

A sheet of each size is made by cutting a sheet of the size before it exactly in half, as shown in Figure 1.

Figure 1: a sheet of A0 paper and the smaller sheets it can be cut into

(a)

A sheet of A0 paper has an area of 1 m². Its short side is \(x\) metres and its long side is \(x\sqrt2\) metres.

Show that \(x^{2}=\dfrac{\sqrt2}{2}\).

PROOFNotes

Show solution

\(x\times x\sqrt2=1\), so \(x^{2}=\frac{1}{\sqrt2}\)	M1	Area equation formed and rearranged for \(x^{2}\).
\(\frac{1}{\sqrt2}\times\frac{\sqrt2}{\sqrt2}=\frac{\sqrt2}{2}\)	A1	cso. The rationalising step must be seen.

Answer: \(x^{2}=\dfrac{1}{\sqrt2}=\dfrac{\sqrt2}{2}\)

(2)

(b)

Write \(x\) in the form \(2^{k}\), where \(k\) is a fraction.

Hence find the short side of a sheet of A0 paper, in millimetres, correct to the nearest millimetre.

ALGEBRANotes

Show solution

\(\frac{\sqrt2}{2}=2^{\frac12}\times2^{-1}=2^{-\frac12}\) or \(\frac{1}{\sqrt2}=2^{-\frac12}\)	M1	\(x^{2}\) written as a single power of 2.
\(x=2^{-\frac14}\)	A1	Accept \(k=-\frac14\) or \(-0.25\).
841 (mm)	B1	Accept 841 from any correct method, e.g. \(\sqrt{0.7071\ldots}=0.8408\ldots\) m.

Answer: \(x=2^{-\frac14}\); 841 mm

(3)

(c)

A sheet of A-series paper has short side \(s\) and long side \(s\sqrt2\). It is cut in half, parallel to its short side, to make two equal sheets of the next size.

Show that each new sheet also has sides in the ratio \(1:\sqrt2\).

PROOFNotes

Show solution

new sides are \(s\) and \(\frac{s\sqrt2}{2}\)	M1	Halving the long side; may be seen on a sketch.
\(\frac{s}{\frac{s\sqrt2}{2}}=\frac{2}{\sqrt2}\) or \(\frac{s\sqrt2}{2}:s=1:\frac{2}{\sqrt2}\)	M1	Long side divided by short side, or ratio scaled to the form \(1:n\).
\(\frac{2}{\sqrt2}=\frac{2\sqrt2}{2}=\sqrt2\), so the ratio is \(1:\sqrt2\)	A1	cso. Accept \(\frac{2}{\sqrt2}=\sqrt2\) shown by \(\sqrt2\times\sqrt2=2\). Decimal checks only score M1 M0 A0.

Answer: \(\dfrac{2}{\sqrt2}=\sqrt2\), so \(1:\sqrt2\)

(3)

(d)

A designer makes a greetings card with sides in the ratio \(1:\sqrt2\). The long side of the card is 12 cm.

Find the exact length of the short side. Give your answer in the form \(a\sqrt2\), where \(a\) is an integer.

ROUTINENotes

Show solution

\(\frac{12}{\sqrt2}\)	M1
\(6\sqrt2\) cm	A1	Decimal answer 8.49 scores M1 A0.

Answer: \(6\sqrt2\) cm

(2)

(e)

Find the exact length of a diagonal of the card. Give your answer as a surd in its simplest form.

MULTINotes

Show solution

\((6\sqrt2)^{2}+12^{2}\) \((=72+144)\)	M1	Pythagoras with their short side from (d).
\(\sqrt{216}\)	M1	Square root of their total.
\(6\sqrt6\) cm	A1	cao. 14.7 scores M1 M1 A0.

Answer: \(6\sqrt6\) cm

(3)

(f)

The long sides of sheets of size A0, A1, A2, A3, … form a geometric sequence.

Figure 2 shows the long side of a sheet of size An for \(n=0\) to \(n=6\).

Use readings from Figure 2 to find an estimate for the common ratio of the sequence. Give your answer to 2 decimal places.

Figure 2: the long side, in mm, of a sheet of size An, for n = 0 to 6

GRAPHNotes

Show solution

e.g. \(\frac{840}{1190}\) or \(\frac{595}{840}\) or \(\frac{420}{595}\)	M1	One term divided by the term before it, using readings within 20 mm of the true values (1189, 841, 595, 420, 297, 210, 149).
0.70 or 0.71	A1	Accept answers in the range 0.69 to 0.72.

Answer: 0.71 (about)

(2)

(g)

The exact common ratio is \(\dfrac{1}{\sqrt2}\). The long side of A0 is 1189 mm.

Use \(\left(\dfrac{1}{\sqrt2}\right)^{8}\) to find the long side of a sheet of size A8. Give your answer correct to the nearest millimetre.

MULTINotes

Show solution

\(\left(\frac{1}{\sqrt2}\right)^{8}=\left(2^{-\frac12}\right)^{8}=2^{-4}=\frac{1}{16}\) or \(1189\times\left(\frac{1}{\sqrt2}\right)^{8}\)	M1	A8 is 8 steps after A0 (the 9th term).
74 (mm)	A1	Accept 74.3 before rounding. Using a power of 9 (52.6) or 7 (105.1) scores M0 A0.

Answer: 74 mm

(2)

(h)

A photocopier has a setting that enlarges a copy from size A4 to size A3. The setting is marked 141%.

Explain where the number 141% comes from.

INTERPRETNotes

Show solution

Each length of A3 is \(\sqrt2\) times the length of A4 (A4 is A3 cut in half), and \(\sqrt2=1.414\ldots\), which is 141%

Must link the scale factor \(\sqrt2\) (or the ratio \(1:\sqrt2\)) to 1.41 or 141%.

(1)

(i)

A company makes notepads with pages that have sides in the ratio \(1:2\).

A page is cut in half, parallel to its short side.

Show that the two smaller pieces are not the same shape as the original page.

CONCEPTNotes

Show solution

page \(s\) by \(2s\) cut to give pieces \(s\) by \(s\)	M1	Accept a numerical example, e.g. 10 by 20 gives 10 by 10.
ratio \(1:1\), which is not \(1:2\), so not the same shape	A1	cso. Conclusion needed.

Answer: The pieces are squares (\(1:1\)), not \(1:2\)

(2)

Show full solution

Working	Mark	Notes
Part (a)
\(x\times x\sqrt2=1\), so \(x^{2}=\frac{1}{\sqrt2}\)	M1	Area equation formed and rearranged for \(x^{2}\).
\(\frac{1}{\sqrt2}\times\frac{\sqrt2}{\sqrt2}=\frac{\sqrt2}{2}\)	A1	cso. The rationalising step must be seen.
Part (b)
\(\frac{\sqrt2}{2}=2^{\frac12}\times2^{-1}=2^{-\frac12}\) or \(\frac{1}{\sqrt2}=2^{-\frac12}\)	M1	\(x^{2}\) written as a single power of 2.
\(x=2^{-\frac14}\)	A1	Accept \(k=-\frac14\) or \(-0.25\).
841 (mm)	B1	Accept 841 from any correct method, e.g. \(\sqrt{0.7071\ldots}=0.8408\ldots\) m.
Part (c)
new sides are \(s\) and \(\frac{s\sqrt2}{2}\)	M1	Halving the long side; may be seen on a sketch.
\(\frac{s}{\frac{s\sqrt2}{2}}=\frac{2}{\sqrt2}\) or \(\frac{s\sqrt2}{2}:s=1:\frac{2}{\sqrt2}\)	M1	Long side divided by short side, or ratio scaled to the form \(1:n\).
\(\frac{2}{\sqrt2}=\frac{2\sqrt2}{2}=\sqrt2\), so the ratio is \(1:\sqrt2\)	A1	cso. Accept \(\frac{2}{\sqrt2}=\sqrt2\) shown by \(\sqrt2\times\sqrt2=2\). Decimal checks only score M1 M0 A0.
Part (d)
\(\frac{12}{\sqrt2}\)	M1
\(6\sqrt2\) cm	A1	Decimal answer 8.49 scores M1 A0.
Part (e)
\((6\sqrt2)^{2}+12^{2}\) \((=72+144)\)	M1	Pythagoras with their short side from (d).
\(\sqrt{216}\)	M1	Square root of their total.
\(6\sqrt6\) cm	A1	cao. 14.7 scores M1 M1 A0.
Part (f)
e.g. \(\frac{840}{1190}\) or \(\frac{595}{840}\) or \(\frac{420}{595}\)	M1	One term divided by the term before it, using readings within 20 mm of the true values (1189, 841, 595, 420, 297, 210, 149).
0.70 or 0.71	A1	Accept answers in the range 0.69 to 0.72.
Part (g)
\(\left(\frac{1}{\sqrt2}\right)^{8}=\left(2^{-\frac12}\right)^{8}=2^{-4}=\frac{1}{16}\) or \(1189\times\left(\frac{1}{\sqrt2}\right)^{8}\)	M1	A8 is 8 steps after A0 (the 9th term).
74 (mm)	A1	Accept 74.3 before rounding. Using a power of 9 (52.6) or 7 (105.1) scores M0 A0.
Part (h)
Each length of A3 is \(\sqrt2\) times the length of A4 (A4 is A3 cut in half), and \(\sqrt2=1.414\ldots\), which is 141%	B1	Must link the scale factor \(\sqrt2\) (or the ratio \(1:\sqrt2\)) to 1.41 or 141%.
Part (i)
page \(s\) by \(2s\) cut to give pieces \(s\) by \(s\)	M1	Accept a numerical example, e.g. 10 by 20 gives 10 by 10.
ratio \(1:1\), which is not \(1:2\), so not the same shape	A1	cso. Conclusion needed.

Answers: (a) \(x^{2}=\dfrac{1}{\sqrt2}=\dfrac{\sqrt2}{2}\) · (b) \(x=2^{-\frac14}\); 841 mm · (c) \(\dfrac{2}{\sqrt2}=\sqrt2\), so \(1:\sqrt2\) · (d) \(6\sqrt2\) cm · (e) \(6\sqrt6\) cm · (f) 0.71 (about) · (g) 74 mm · (i) The pieces are squares (\(1:1\)), not \(1:2\)

Common error (c) Halving the short side instead of the long side gives a sheet \(\frac{s}{2}\) by \(s\sqrt2\), which is long and thin. Cutting parallel to the short side halves the long side.

Exam tip (e) \((6\sqrt2)^{2}=36\times2=72\): square the 6 and the \(\sqrt2\) separately. Then look for the largest square factor of 216, which is 36.

More 1.5 questions in the question bank

1.4 Set language, notation and Venn diagrams (both tiers)

+ Add to test✓ In your testCC-EIM-14-119 marks~20 min

Streaming habits: a three-set Venn diagram survey

1.4 Set language, notation and Venn diagrams (both tiers)6.2 Probability of single and combined events1.1 Integers, fractions, decimals and percentages6.1 Averages, range and statistical diagramsFoundation

120 students at a college were asked which of three services they use.

\(\xi=\{\)the 120 students\(\}\)
\(V=\{\)students who use a video streaming service\(\}\)
\(M=\{\)students who use a music streaming service\(\}\)
\(G=\{\)students who have a gaming subscription\(\}\)

The results showed that

64 students use video streaming, 54 use music streaming and 31 have a gaming subscription
30 use video streaming and music streaming
18 use video streaming and have a gaming subscription
14 use music streaming and have a gaming subscription
8 use all three services.

Figure 1: Venn diagram for the survey of 120 students (only the number who use all three services is shown)

(a)

Complete the Venn diagram in Figure 1 to show the number of students in each region.

ROUTINENotes

Show solution

22, 10 and 6 in the regions for exactly two services (\(V\) and \(M\) only, \(V\) and \(G\) only, \(M\) and \(G\) only)	B1	From \(30-8\), \(18-8\) and \(14-8\).
e.g. \(64-22-10-8\) or \(54-22-6-8\) or \(31-10-6-8\)	M1	Correct method for at least one of the “one service only” regions. Allow ECF from their two-set regions.
24 (\(V\) only), 18 (\(M\) only) and 7 (\(G\) only)	A1	All three correct and in the correct regions.

Answer: \(V\) only 24, \(M\) only 18, \(G\) only 7; \(V\cap M\) only 22, \(V\cap G\) only 10, \(M\cap G\) only 6; centre 8

(3)

(b)

Show that 25 students use none of the three services.

PROOFNotes

Show solution

\(120-(24+18+7+22+10+6+8)\) or \(120-95\)	M1	Allow ECF from their diagram for this mark.
25	A1	cso. Their seven regions must add to 95.

Answer: 25

(2)

(c)(i)

Find \(n(M\cap G')\).

CONCEPTNotes

Show solution

\(18+22\). Allow ECF from their diagram.

Answer: 40

(1)

(c)(ii)

Describe in words the students in the set \(V\cap M\cap G'\).

CONCEPTNotes

Show solution

students who use video streaming and music streaming but do not have a gaming subscription

Must include the “not gaming” condition.

(1)

(d)

One of the 120 students is chosen at random.

Find the probability that this student uses exactly two of the three services.

ROUTINENotes

Show solution

\(\frac{22+10+6}{120}\)	M1	Allow ECF from their diagram.
\(\frac{38}{120}\) or \(\frac{19}{60}\)	A1	Accept 0.316… or 0.317 or 31.7%.

Answer: \(\dfrac{19}{60}\)

(2)

(e)

Work out the percentage of the students with a gaming subscription who also use video streaming.

Give your answer correct to 1 decimal place.

MULTINotes

Show solution

\(\frac{10+8}{31}\times100\)	M1	The denominator must be \(n(G)=31\). \(\frac{18}{120}\times100\) scores M0.
58.1(%)	A1	Accept 58.06…

Answer: 58.1%

(2)

(f)(i)

Figure 2 is an incomplete bar chart showing the number of services used by the 120 students.

Complete the bar chart.

Figure 2: an incomplete bar chart of the number of services used by the 120 students

GRAPHNotes

Show solution

bar of height 38 for 2 services	B1	From \(22+10+6\). Allow ECF from (d).
bar of height 8 for 3 services	B1	Bars must have equal width with gaps, as drawn.

Answer: Bars of height 38 (2 services) and 8 (3 services)

(2)

(f)(ii)

Work out the mean number of services used per student.

ROUTINENotes

Show solution

\(\frac{0\times25+1\times49+2\times38+3\times8}{120}\) \(\left(=\frac{149}{120}\right)\)	M1	Sum of products divided by 120. Allow ECF from (f)(i).
1.24	A1	Accept 1.241…

Answer: 1.24

(2)

(g)

Ravi says,

“The sets \(V\) and \(G\) must be disjoint, because \(n(V)+n(G)\) is less than 120.”

Explain why Ravi is wrong.

INTERPRETNotes

Show solution

Disjoint means \(V\cap G=\emptyset\), but \(n(V\cap G)=18\), so they are not disjoint

Accept any statement that 18 students (or some students) use both video streaming and gaming. The size of \(n(V)+n(G)\) does not decide whether the sets overlap.

(1)

(h)

A further 30 students are then surveyed and added to the results. Of these 30 students

9 use none of the services
12 use music streaming only
9 use all three services.

One of the 150 students is now chosen at random.

Find the probability that this student uses music streaming. Give your answer as a fraction in its simplest form.

MULTINotes

Show solution

\(54+12+9\) \((=75)\)	M1	New number in \(M\). Allow ECF from their \(n(M)\).
\(\frac{75}{150}\)	M1	Their new \(n(M)\) over 150.
\(\frac12\)	A1	cao

Answer: \(\dfrac{1}{2}\)

(3)

Show full solution

Working	Mark	Notes
Part (a)
22, 10 and 6 in the regions for exactly two services (\(V\) and \(M\) only, \(V\) and \(G\) only, \(M\) and \(G\) only)	B1	From \(30-8\), \(18-8\) and \(14-8\).
e.g. \(64-22-10-8\) or \(54-22-6-8\) or \(31-10-6-8\)	M1	Correct method for at least one of the “one service only” regions. Allow ECF from their two-set regions.
24 (\(V\) only), 18 (\(M\) only) and 7 (\(G\) only)	A1	All three correct and in the correct regions.
Part (b)
\(120-(24+18+7+22+10+6+8)\) or \(120-95\)	M1	Allow ECF from their diagram for this mark.
25	A1	cso. Their seven regions must add to 95.
Part (c)(i)
40	B1	\(18+22\). Allow ECF from their diagram.
Part (c)(ii)
students who use video streaming and music streaming but do not have a gaming subscription	B1	Must include the “not gaming” condition.
Part (d)
\(\frac{22+10+6}{120}\)	M1	Allow ECF from their diagram.
\(\frac{38}{120}\) or \(\frac{19}{60}\)	A1	Accept 0.316… or 0.317 or 31.7%.
Part (e)
\(\frac{10+8}{31}\times100\)	M1	The denominator must be \(n(G)=31\). \(\frac{18}{120}\times100\) scores M0.
58.1(%)	A1	Accept 58.06…
Part (f)(i)
bar of height 38 for 2 services	B1	From \(22+10+6\). Allow ECF from (d).
bar of height 8 for 3 services	B1	Bars must have equal width with gaps, as drawn.
Part (f)(ii)
\(\frac{0\times25+1\times49+2\times38+3\times8}{120}\) \(\left(=\frac{149}{120}\right)\)	M1	Sum of products divided by 120. Allow ECF from (f)(i).
1.24	A1	Accept 1.241…
Part (g)
Disjoint means \(V\cap G=\emptyset\), but \(n(V\cap G)=18\), so they are not disjoint	B1	Accept any statement that 18 students (or some students) use both video streaming and gaming. The size of \(n(V)+n(G)\) does not decide whether the sets overlap.
Part (h)
\(54+12+9\) \((=75)\)	M1	New number in \(M\). Allow ECF from their \(n(M)\).
\(\frac{75}{150}\)	M1	Their new \(n(M)\) over 150.
\(\frac12\)	A1	cao

Answers: (a) \(V\) only 24, \(M\) only 18, \(G\) only 7; \(V\cap M\) only 22, \(V\cap G\) only 10, \(M\cap G\) only 6; centre 8 · (b) 25 · (c)(i) 40 · (d) \(\dfrac{19}{60}\) · (e) 58.1% · (f)(i) Bars of height 38 (2 services) and 8 (3 services) · (f)(ii) 1.24 · (h) \(\dfrac{1}{2}\)

Exam tip (a) “30 use video and music” includes the 8 who use all three, so the region for video and music only is \(30-8=22\). Always fill in the centre first and work outwards.

More 1.4 questions in the question bank

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A health clinic screening: two-set Venn diagram with algebra

1.4 Set language, notation and Venn diagrams (both tiers)2.2 Linear and simultaneous equations6.4 Conditional probability and tree diagrams6.1 Averages, range and statistical diagramsFoundation

A health clinic screens 200 adults for high blood pressure and for high cholesterol.

\(\xi=\{\)the 200 adults\(\}\)
\(H=\{\)adults with high blood pressure\(\}\)
\(C=\{\)adults with high cholesterol\(\}\)

The Venn diagram in Figure 1 shows the number of adults in three of the regions.

Figure 1: the numbers of adults in three regions of the Venn diagram, in terms of x

(a)

Write down an expression, in terms of \(x\), for \(n(H)\). Simplify your answer.

ALGEBRANotes

Show solution

\(6x-8\)

From \((5x-8)+x\).

Answer: \(6x-8\)

(1)

(b)

The number of adults with high blood pressure is 38 more than the number of adults with high cholesterol.

Write down an equation in \(x\). Solve your equation to find the value of \(x\).

ALGEBRANotes

Show solution

\(6x-8=(3x-1)+38\)	M1	Accept \((5x-8)+x=x+(2x-1)+38\). Allow ECF from (a).
\(3x=45\) or \(6x-3x=37+8\)	M1	Correctly collecting the \(x\) terms and the numbers on opposite sides.
\(x=15\)	A1	cao

Answer: \(x=15\)

(3)

(c)

Show that 89 adults have neither high blood pressure nor high cholesterol.

PROOFNotes

Show solution

\(200-(67+15+29)\) or \(200-(9x-9)\) with \(x=15\)	M1	Using their \(x\) to find the three regions inside the circles, then subtracting from 200.
89	A1	cso. Regions 67, 15 and 29 (total 111) must be seen or implied.

Answer: 89

(2)

(d)

Use set notation to describe the set of adults who have high blood pressure but do not have high cholesterol.

CONCEPTNotes

Show solution

\(H\cap C'\)

Accept \(C'\cap H\). Do not accept \(H\cup C'\) or \(H'\cap C\).

Answer: \(H\cap C'\)

(1)

(e)

One of the 200 adults is chosen at random.

Find the probability that this adult has exactly one of the two conditions. Give your answer as a fraction in its simplest form.

ROUTINENotes

Show solution

\(\frac{67+29}{200}\) \(\left(=\frac{96}{200}\right)\)	M1	Allow ECF from their regions.
\(\frac{12}{25}\)	A1	cao

Answer: \(\dfrac{12}{25}\)

(2)

(f)

An adult who has high blood pressure is chosen at random.

Find the probability that this adult also has high cholesterol.

Higher onlyROUTINENotes

Show solution

denominator \(n(H)=82\)	M1	“Given high blood pressure” restricts the choice to the 82 adults in \(H\).
\(\frac{15}{82}\)	A1	Accept 0.183. \(\frac{15}{200}\) scores M0 A0.

Answer: \(\dfrac{15}{82}\)

(2)

(g)(i)

Figure 2 shows, for each age group, the number of adults screened and the number with high blood pressure.

Use Figure 2 to work out the percentage of adults aged 40–59 who have high blood pressure.

Figure 2: the number of adults screened and the number with high blood pressure, by age group

GRAPHNotes

Show solution

\(\frac{30}{60}\times100\)	M1	Both values read correctly from the chart.
50(%)	A1	cao

Answer: 50%

(2)

(g)(ii)

Ana says,

“High blood pressure is just as common in the 18–39 age group as in the 60 and over age group, because the two bars for high blood pressure are the same height.”

Is Ana correct? Explain your answer.

INTERPRETNotes

Show solution

\(\frac{26}{100}\) \((=26\%)\) and \(\frac{26}{40}\) \((=65\%)\)	M1	Both proportions (fractions, decimals or percentages) found from the chart.
No, as a much greater proportion (65% compared with 26%) of the 60 and over group has high blood pressure	A1	Conclusion with a correct comparison of proportions. A statement that the groups are different sizes, with no figures, scores M0 A0.

Answer: No: 26% of 18–39 compared with 65% of 60 and over

(2)

(h)(i)

The clinic later tests the same 200 adults for diabetes.

\(D=\{\)adults with diabetes\(\}\)

9 adults have diabetes. All of them have high blood pressure, and 4 of them also have high cholesterol.

Use set notation to write down the relationship between \(D\) and \(H\).

CONCEPTNotes

Show solution

\(D\subset H\)

Accept \(D\cap H=D\) or \(D\cap H'=\emptyset\).

Answer: \(D\subset H\)

(1)

(h)(ii)

Find \(n(H\cap C'\cap D')\).

MULTINotes

Show solution

\(67-(9-4)\)	M1	The 5 adults with diabetes and high blood pressure only are removed from the \(H\cap C'\) region. Allow ECF from their 67.
62	A1	cao

Answer: 62

(2)

Show full solution

Working	Mark	Notes
Part (a)
\(6x-8\)	B1	From \((5x-8)+x\).
Part (b)
\(6x-8=(3x-1)+38\)	M1	Accept \((5x-8)+x=x+(2x-1)+38\). Allow ECF from (a).
\(3x=45\) or \(6x-3x=37+8\)	M1	Correctly collecting the \(x\) terms and the numbers on opposite sides.
\(x=15\)	A1	cao
Part (c)
\(200-(67+15+29)\) or \(200-(9x-9)\) with \(x=15\)	M1	Using their \(x\) to find the three regions inside the circles, then subtracting from 200.
89	A1	cso. Regions 67, 15 and 29 (total 111) must be seen or implied.
Part (d)
\(H\cap C'\)	B1	Accept \(C'\cap H\). Do not accept \(H\cup C'\) or \(H'\cap C\).
Part (e)
\(\frac{67+29}{200}\) \(\left(=\frac{96}{200}\right)\)	M1	Allow ECF from their regions.
\(\frac{12}{25}\)	A1	cao
Part (f)
denominator \(n(H)=82\)	M1	“Given high blood pressure” restricts the choice to the 82 adults in \(H\).
\(\frac{15}{82}\)	A1	Accept 0.183. \(\frac{15}{200}\) scores M0 A0.
Part (g)(i)
\(\frac{30}{60}\times100\)	M1	Both values read correctly from the chart.
50(%)	A1	cao
Part (g)(ii)
\(\frac{26}{100}\) \((=26\%)\) and \(\frac{26}{40}\) \((=65\%)\)	M1	Both proportions (fractions, decimals or percentages) found from the chart.
No, as a much greater proportion (65% compared with 26%) of the 60 and over group has high blood pressure	A1	Conclusion with a correct comparison of proportions. A statement that the groups are different sizes, with no figures, scores M0 A0.
Part (h)(i)
\(D\subset H\)	B1	Accept \(D\cap H=D\) or \(D\cap H'=\emptyset\).
Part (h)(ii)
\(67-(9-4)\)	M1	The 5 adults with diabetes and high blood pressure only are removed from the \(H\cap C'\) region. Allow ECF from their 67.
62	A1	cao

Answers: (a) \(6x-8\) · (b) \(x=15\) · (c) 89 · (d) \(H\cap C'\) · (e) \(\dfrac{12}{25}\) · (f) \(\dfrac{15}{82}\) · (g)(i) 50% · (g)(ii) No: 26% of 18–39 compared with 65% of 60 and over · (h)(i) \(D\subset H\) · (h)(ii) 62

Common error (f) The adult is chosen from those with high blood pressure, so the denominator is \(n(H)=82\), not 200.

More 1.4 questions in the question bank

Topic 2 · Equations, formulae and identities

● home subchapter · ○ linked subchapter · last column: links to other topics

Setup	2.1	2.2	2.3	2.4	Other
Paving round a square pond: expanding, factorising and area	●		○		4.3 3.2
Stopping a car: braking distance and the square of the speed			○	●	3.2 1.1
Hiring an e-scooter: simultaneous equations and a break-even point		●	○		3.2 1.1
A basketball shot: the quadratic formula, completing the square and inequalities		○	●		3.2 1.3

2.1 Algebraic manipulation and factorising · 2.4 Direct and inverse proportion and rearranging formulaemerged group

+ Add to test✓ In your testCC-EIM-21-24-119 marks~20 min

Paving round a square pond: expanding, factorising and area

2.1 Algebraic manipulation and factorising4.3 Area, surface area and volume2.3 Quadratic equations and inequalities3.2 Graphs, gradients and real-life graphsFoundation

Leila is designing a garden. A square pond of side \(x\) metres is surrounded by paving. The paving is 1.5 m wide all the way round the pond, so the outer edge of the paving is also a square.

Figure 1 shows a plan view.

Figure 1: plan view of the pond and the paving (not to scale)

(a)

Find an expression, in terms of \(x\), for the perimeter of the outer edge of the paving.

Give your answer in its simplest form.

ROUTINENotes

Show solution

outer side \(x+3\), and \(4(x+3)\) or \(x+3+x+3+x+3+x+3\)	M1	Allow \(4(x+1.5+1.5)\). \(4x+6\) (one width of paving only) is M0.
\(4x+12\)	A1	cao. Must be simplified. Ignore “m” after the expression.

Answer: \(4x+12\) m

(2)

(b)

The area of the paving is \(A\) square metres.

Show that \(A=6x+9\).

PROOFNotes

Show solution

\((x+3)^2-x^2\)	M1	Outer square subtract pond. Alternative: four rectangles \(4\times1.5x\) and four corner squares \(4\times1.5^2\).
\((x+3)(x+3)=x^2+3x+3x+9\) or \(x^2+6x+9\)	M1	All four terms correct, or \(6x\) and \(9\) from the alternative method.
\(x^2+6x+9-x^2=6x+9\)	A1	cso. The line showing the \(x^2\) terms cancelling must be seen.

Answer: \(A=6x+9\)

(3)

(c)

Leila first considers a pond of side 8.5 m.

Without using a calculator, use the difference of two squares to work out the area of the paving. Show your working clearly.

No calculatorCONCEPTNotes

Show solution

\(11.5^2-8.5^2=(11.5-8.5)(11.5+8.5)\) or \(3\times20\)	M1	Must use the difference of two squares; squaring 11.5 and 8.5 directly is M0.
\(60\) m²	A1	cao. A check using \(6\times8.5+9=60\) does not by itself score the M1.

Answer: 60 m²

(2)

(d)(i)

Complete the table of values for \(A=6x+9\).

\(x\)	0	2	4	6	8
\(A\)	9		33		57

ROUTINENotes

Show solution

21 and 45

Both values needed.

Answer: 21 and 45

(1)

(d)(ii)

On the grid in Figure 2, draw the graph of \(A=6x+9\) for values of \(x\) from 0 to 8.

Figure 2: axes for the graph of A = 6x + 9

GRAPHNotes

Show solution

at least 3 points from the table plotted correctly	B1	Allow ECF from their table in (d)(i).
a single straight line from \((0,9)\) to \((8,57)\)	B1	Ruled line through all five correct points.

Answer: straight line from \((0,9)\) to \((8,57)\)

(2)

(d)(iii)

Leila has enough paving slabs to cover 42 m².

Use your graph to find the side of the largest pond that she can surround with paving 1.5 m wide.

GRAPHNotes

Show solution

\(5.5\) m

Accept 5.4 to 5.6 from a correct line; ECF from their straight line.

Answer: 5.5 m

(1)

(e)

Leila notices from the graph that each time the side of the pond increases by 1 m, the area of the paving increases by the same amount.

Explain, by referring to the shape of the paving, why the area increases by 6 m² each time.

INTERPRETNotes

Show solution

each of the 4 sides of paving gets 1 m longer and is 1.5 m wide, so the area increases by \(4\times1.5\times1=6\) m² (the corner squares do not change)

Accept: the \(x^2\) terms cancel so \(A\) is linear in \(x\) with gradient 6, provided the gradient 6 is linked to the 6 m² increase.

(1)

(f)

Leila's neighbour builds a square pond of side 4 m. The paving round it has the same width, \(w\) metres, all the way round.

The neighbour spends exactly £1200 on paving. The paving costs £25 per square metre.

Work out the value of \(w\). Show clear algebraic working.

MULTINotes

Show solution

\(1200\div25=48\) m²	B1
\((4+2w)^2-16\) or \(4w^2+16w\)	M1	Outer side \(4+2w\). Accept \(4\times4w+4w^2\).
\(4w^2+16w-48=0\) and \(w^2+4w-12=0\) factorised as \((w+6)(w-2)=0\)	M1	A correct three-term quadratic equal to zero, with a correct method to solve it (factorising, or the formula).
\(w=2\) (and \(w=-6\) rejected as a width cannot be negative)	A1	Award A0 if \(w=-6\) is given as a second answer and not rejected.

Answer: \(w=2\)

(4)

(g)(i)

For a different design, the pond is a rectangle \(2x\) metres long and \(x\) metres wide. Paving surrounds the pond so that the outer edge is also a rectangle. The total area of the pond and the paving is \((2x^2+11x+12)\) m².

Factorise fully \(2x^2+11x+12\).

Higher onlyALGEBRANotes

Show solution

\(2x^2+8x+3x+12\) or \(2x(x+4)+3(x+4)\)	M1	Two numbers with product 24 and sum 11 (8 and 3), used to split the middle term; or brackets \((2x\pm a)(x\pm b)\) with \(ab=\pm12\).
\((2x+3)(x+4)\)	A1	Either order.

Answer: \((2x+3)(x+4)\)

(2)

(g)(ii)

The paving at each end of the pond is 1.5 m wide. The paving along each long side of the pond has a different width.

Use your answer to (g)(i) to find the width of the paving along each long side of the pond.

Higher onlyINTERPRETNotes

Show solution

2 m

Outer length \(2x+3\) gives 1.5 m at each end, so the outer width is \(x+4\), and \((4)\div2=2\). ECF from a factorisation of the form \((2x+3)(x+c)\).

Answer: 2 m

(1)

Show full solution

Working	Mark	Notes
Part (a)
outer side \(x+3\), and \(4(x+3)\) or \(x+3+x+3+x+3+x+3\)	M1	Allow \(4(x+1.5+1.5)\). \(4x+6\) (one width of paving only) is M0.
\(4x+12\)	A1	cao. Must be simplified. Ignore “m” after the expression.
Part (b)
\((x+3)^2-x^2\)	M1	Outer square subtract pond. Alternative: four rectangles \(4\times1.5x\) and four corner squares \(4\times1.5^2\).
\((x+3)(x+3)=x^2+3x+3x+9\) or \(x^2+6x+9\)	M1	All four terms correct, or \(6x\) and \(9\) from the alternative method.
\(x^2+6x+9-x^2=6x+9\)	A1	cso. The line showing the \(x^2\) terms cancelling must be seen.
Part (c)
\(11.5^2-8.5^2=(11.5-8.5)(11.5+8.5)\) or \(3\times20\)	M1	Must use the difference of two squares; squaring 11.5 and 8.5 directly is M0.
\(60\) m²	A1	cao. A check using \(6\times8.5+9=60\) does not by itself score the M1.
Part (d)(i)
21 and 45	B1	Both values needed.
Part (d)(ii)
at least 3 points from the table plotted correctly	B1	Allow ECF from their table in (d)(i).
a single straight line from \((0,9)\) to \((8,57)\)	B1	Ruled line through all five correct points.
Part (d)(iii)
\(5.5\) m	B1	Accept 5.4 to 5.6 from a correct line; ECF from their straight line.
Part (e)
each of the 4 sides of paving gets 1 m longer and is 1.5 m wide, so the area increases by \(4\times1.5\times1=6\) m² (the corner squares do not change)	B1	Accept: the \(x^2\) terms cancel so \(A\) is linear in \(x\) with gradient 6, provided the gradient 6 is linked to the 6 m² increase.
Part (f)
\(1200\div25=48\) m²	B1
\((4+2w)^2-16\) or \(4w^2+16w\)	M1	Outer side \(4+2w\). Accept \(4\times4w+4w^2\).
\(4w^2+16w-48=0\) and \(w^2+4w-12=0\) factorised as \((w+6)(w-2)=0\)	M1	A correct three-term quadratic equal to zero, with a correct method to solve it (factorising, or the formula).
\(w=2\) (and \(w=-6\) rejected as a width cannot be negative)	A1	Award A0 if \(w=-6\) is given as a second answer and not rejected.
Part (g)(i)
\(2x^2+8x+3x+12\) or \(2x(x+4)+3(x+4)\)	M1	Two numbers with product 24 and sum 11 (8 and 3), used to split the middle term; or brackets \((2x\pm a)(x\pm b)\) with \(ab=\pm12\).
\((2x+3)(x+4)\)	A1	Either order.
Part (g)(ii)
2 m	B1	Outer length \(2x+3\) gives 1.5 m at each end, so the outer width is \(x+4\), and \((4)\div2=2\). ECF from a factorisation of the form \((2x+3)(x+c)\).

Answers: (a) \(4x+12\) m · (b) \(A=6x+9\) · (c) 60 m² · (d)(i) 21 and 45 · (d)(ii) straight line from \((0,9)\) to \((8,57)\) · (d)(iii) 5.5 m · (f) \(w=2\) · (g)(i) \((2x+3)(x+4)\) · (g)(ii) 2 m

Common error (a) The paving is on both sides of the pond, so the outer side is \(x+1.5+1.5=x+3\), not \(x+1.5\).

Exam tip (f) When a quadratic comes from a length, both roots must be checked against the situation: say clearly which one you reject and why.

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Stopping a car: braking distance and the square of the speed

2.4 Direct and inverse proportion and rearranging formulae3.2 Graphs, gradients and real-life graphs2.3 Quadratic equations and inequalities1.1 Integers, fractions, decimals and percentagesHigher

When a driver sees a hazard, the car travels a thinking distance before the brakes are applied and then a braking distance while it slows down to rest. The stopping distance is the thinking distance plus the braking distance.

For one car on a dry road:

the braking distance, \(d\) metres, is directly proportional to the square of the speed, \(v\) km/h;
the thinking distance, \(T\) metres, is directly proportional to \(v\).

In a road test at a speed of 50 km/h, the braking distance was 15 m and the thinking distance was 10 m.

Figure 1: the stopping distance S of the car on a dry road against its speed v

(a)

Find a formula for \(d\) in terms of \(v\).

Higher onlyROUTINENotes

Show solution

\(d=kv^2\)	M1	Must be squared; \(d=kv\) scores M0 A0 A0.
\(15=k\times50^2\)	M1	Substitution of the test values.
\(d=0.006v^2\)	A1	Accept \(d=\frac{3}{500}v^2\). The formula must be stated, not just \(k=0.006\).

Answer: \(d=0.006v^2\)

(3)

(b)

Show that if the speed of the car is doubled, the braking distance is multiplied by 4.

Higher onlyPROOFNotes

Show solution

\(k(2v)^2\) or \(0.006(2v)^2\)	M1	Or a numerical demonstration with general reasoning, e.g. \(2^2=4\). Two numerical cases alone (e.g. 15 m and 60 m) score M1 only.
\(k\times4v^2=4kv^2=4d\)	A1	cso. A conclusion linking the result to \(4d\) is needed.

Answer: \(k(2v)^2=4kv^2=4d\)

(2)

(c)

Show that the stopping distance, \(S\) metres, is given by \(S=0.006v^2+0.2v\).

MULTINotes

Show solution

\(T=cv\) and \(10=50c\), so \(c=0.2\)	M1	Thinking distance formula \(T=0.2v\) found. This mark does not depend on (a).
\(S=d+T=0.006v^2+0.2v\)	A1	cso. Both \(d=0.006v^2\) (from (a)) and \(T=0.2v\) must be seen before the given result.

Answer: \(S=0.006v^2+0.2v\)

(2)

(d)(i)

The graph of \(S=0.006v^2+0.2v\) is shown in Figure 1.

Use the graph to find an estimate for the speed of the car when its stopping distance is 60 m.

GRAPHNotes

Show solution

horizontal line drawn at \(S=60\) to meet the curve, or a reading taken at \(S=60\)	M1
85 km/h	A1	Accept any value from 83 to 87 km/h.

Answer: 85 km/h (accept 83 to 87)

(2)

(d)(ii)

Ravi says,

“If I drive at 100 km/h instead of 50 km/h, my stopping distance will double.”

Explain why Ravi is wrong. You must use values in your answer.

INTERPRETNotes

Show solution

\(S=25\) at 50 km/h and \(S=80\) at 100 km/h	M1	From the formula or read from the graph (accept 78 to 82 for the reading).
80 is more than \(2\times25=50\) (it is 3.2 times as far), so it more than doubles	A1	Accept a correct statement that \(S\) is not directly proportional to \(v\) because the graph is a curve, not a straight line through the origin, together with the two values.

Answer: 25 m becomes 80 m, which is 3.2 times as far, not twice

(2)

(e)(i)

Make \(v\) the subject of the formula \(d=0.006v^2\).

Higher onlyALGEBRANotes

Show solution

\(v^2=\frac{d}{0.006}\)	M1	Correct first step; ECF from their \(k\) in (a).
\(v=\sqrt{\frac{d}{0.006}}\)	A1	Accept \(v=\sqrt{\frac{500d}{3}}\). Ignore \(\pm\).

Answer: \(v=\sqrt{\dfrac{d}{0.006}}\)

(2)

(e)(ii)

After an accident on a dry road in a 60 km/h zone, a braking distance of 24 m is measured for this car.

Was the car going faster than 60 km/h when the brakes were applied? Give a reason for your answer.

INTERPRETNotes

Show solution

Yes, since \(v=\sqrt{\frac{24}{0.006}}=\sqrt{4000}=63.2\) km/h, which is more than 60

Value 63.2 (or 63) and a conclusion needed. Alternative: at 60 km/h, \(d=21.6\) m, which is less than 24 m.

Answer: Yes: 63.2 km/h

(1)

(f)

On a wet road the thinking distance is unchanged, but the braking distance at every speed is 40% greater than on a dry road.

Work out the speed on a wet road that gives the same stopping distance as the car has at 80 km/h on a dry road.

Give your answer correct to 1 decimal place.

Higher onlyMULTINotes

Show solution

dry: \(S=0.006\times80^2+0.2\times80=54.4\) m	B1
wet: \(0.006\times1.4=0.0084\), so \(0.0084v^2+0.2v=54.4\)	M1	Multiplier 1.4 used on the braking term only. Applying 1.4 to the whole of \(S\) is M0.
\(v=\frac{-0.2\pm\sqrt{0.2^2-4\times0.0084\times(-54.4)}}{2\times0.0084}\)	M1	Correct substitution into the formula for their three-term quadratic. Equivalent: \(21v^2+500v-136\,000=0\).
\(69.4\) km/h	A1	cao. Negative root ignored or rejected.

Answer: 69.4 km/h

(4)

Show full solution

Working	Mark	Notes
Part (a)
\(d=kv^2\)	M1	Must be squared; \(d=kv\) scores M0 A0 A0.
\(15=k\times50^2\)	M1	Substitution of the test values.
\(d=0.006v^2\)	A1	Accept \(d=\frac{3}{500}v^2\). The formula must be stated, not just \(k=0.006\).
Part (b)
\(k(2v)^2\) or \(0.006(2v)^2\)	M1	Or a numerical demonstration with general reasoning, e.g. \(2^2=4\). Two numerical cases alone (e.g. 15 m and 60 m) score M1 only.
\(k\times4v^2=4kv^2=4d\)	A1	cso. A conclusion linking the result to \(4d\) is needed.
Part (c)
\(T=cv\) and \(10=50c\), so \(c=0.2\)	M1	Thinking distance formula \(T=0.2v\) found. This mark does not depend on (a).
\(S=d+T=0.006v^2+0.2v\)	A1	cso. Both \(d=0.006v^2\) (from (a)) and \(T=0.2v\) must be seen before the given result.
Part (d)(i)
horizontal line drawn at \(S=60\) to meet the curve, or a reading taken at \(S=60\)	M1
85 km/h	A1	Accept any value from 83 to 87 km/h.
Part (d)(ii)
\(S=25\) at 50 km/h and \(S=80\) at 100 km/h	M1	From the formula or read from the graph (accept 78 to 82 for the reading).
80 is more than \(2\times25=50\) (it is 3.2 times as far), so it more than doubles	A1	Accept a correct statement that \(S\) is not directly proportional to \(v\) because the graph is a curve, not a straight line through the origin, together with the two values.
Part (e)(i)
\(v^2=\frac{d}{0.006}\)	M1	Correct first step; ECF from their \(k\) in (a).
\(v=\sqrt{\frac{d}{0.006}}\)	A1	Accept \(v=\sqrt{\frac{500d}{3}}\). Ignore \(\pm\).
Part (e)(ii)
Yes, since \(v=\sqrt{\frac{24}{0.006}}=\sqrt{4000}=63.2\) km/h, which is more than 60	B1	Value 63.2 (or 63) and a conclusion needed. Alternative: at 60 km/h, \(d=21.6\) m, which is less than 24 m.
Part (f)
dry: \(S=0.006\times80^2+0.2\times80=54.4\) m	B1
wet: \(0.006\times1.4=0.0084\), so \(0.0084v^2+0.2v=54.4\)	M1	Multiplier 1.4 used on the braking term only. Applying 1.4 to the whole of \(S\) is M0.
\(v=\frac{-0.2\pm\sqrt{0.2^2-4\times0.0084\times(-54.4)}}{2\times0.0084}\)	M1	Correct substitution into the formula for their three-term quadratic. Equivalent: \(21v^2+500v-136\,000=0\).
\(69.4\) km/h	A1	cao. Negative root ignored or rejected.

Answers: (a) \(d=0.006v^2\) · (b) \(k(2v)^2=4kv^2=4d\) · (c) \(S=0.006v^2+0.2v\) · (d)(i) 85 km/h (accept 83 to 87) · (d)(ii) 25 m becomes 80 m, which is 3.2 times as far, not twice · (e)(i) \(v=\sqrt{\dfrac{d}{0.006}}\) · (e)(ii) Yes: 63.2 km/h · (f) 69.4 km/h

Common error (a) “Proportional to the square of the speed” means \(d=kv^2\). Using \(d=kv\) gives a straight-line model, which is wrong here.

More 2.4 questions in the question bank

2.2 Linear and simultaneous equations · 2.3 Quadratic equations and inequalitiesmerged group

+ Add to test✓ In your testCC-EIM-22-23-118 marks~20 min

Hiring an e-scooter: simultaneous equations and a break-even point

2.2 Linear and simultaneous equations3.2 Graphs, gradients and real-life graphs2.3 Quadratic equations and inequalities1.1 Integers, fractions, decimals and percentagesFoundation

Two companies hire out e-scooters in a city. Each company charges a fixed unlock fee for every ride plus a cost for every minute of the ride.

Company A charges an unlock fee of £\(u\) per ride and £\(p\) per minute.

Nadia uses Company A.

On Monday she made 2 rides with a total time of 26 minutes. She paid £5.90 altogether.
On Tuesday she made 3 rides with a total time of 40 minutes. She paid £9.00 altogether.

(a)

Use algebra to work out the value of \(u\) and the value of \(p\).

ALGEBRANotes

Show solution

\(2u+26p=5.90\) and \(3u+40p=9.00\)	B1	Both equations. Accept equations in pence: \(2u+26p=590\), \(3u+40p=900\).
\(6u+78p=17.70\) and \(6u+80p=18.00\), subtracted to give \(2p=0.30\)	M1	Coefficients of one unknown matched and the equations correctly subtracted (or a correct substitution method).
\(p=0.15\)	A1
\(2u+26\times0.15=5.90\) or \(2u+3.90=5.90\)	M1	Substituting their \(p\) into one of the equations.
\(u=1\)	A1	Accept £1, £1.00 or 100p. Answers with no algebraic working score 0.

Answer: \(u=1\), \(p=0.15\) (£1 unlock fee, 15p per minute)

(5)

(b)

Write down a formula for the cost, £\(C\), of one ride of \(t\) minutes with Company A.

ROUTINENotes

Show solution

\(C=1+0.15t\)

ECF from (a). Accept \(C=0.15t+1\).

Answer: \(C=1+0.15t\)

(1)

(c)

Nadia has £7.50 to spend on one ride with Company A.

Form and solve an inequality to find the greatest whole number of minutes that her ride can last.

MODELNotes

Show solution

\(1+0.15t\le7.50\), giving \(0.15t\le6.50\)	M1	Allow \(\lt\) for this mark. ECF from (b).
43 minutes	A1	From \(t\le43.3\ldots\); the answer must be a whole number. 44 scores A0.

Answer: 43 minutes

(2)

(d)

Company B charges an unlock fee of £2.50 per ride and £0.10 per minute. The line in Figure 1 shows the cost of a ride with Company B.

On the same grid, draw the line that shows the cost of a ride with Company A, for rides of up to 50 minutes.

Figure 1: the cost of a ride with Company B

GRAPHNotes

Show solution

at least two correct points plotted, e.g. \((0,1)\), \((20,4)\), \((40,7)\)	B1	ECF from their formula in (b).
a correct straight line from \((0,1)\) to \((50,8.5)\)	B1

Answer: line from \((0,1)\) to \((50,8.5)\)

(2)

(e)

Show that the two companies charge the same amount for a ride of 30 minutes.

PROOFNotes

Show solution

\(1+0.15t=2.5+0.1t\), so \(0.05t=1.5\)	M1	Or both costs at 30 minutes: \(1+0.15\times30\) and \(2.5+0.1\times30\).
\(t=30\) (both cost £5.50)	A1	cso. For the substitution method both costs must be shown as £5.50 with a conclusion.

Answer: both cost £5.50 at 30 minutes

(2)

(f)

Nadia's daily ride to college takes 20 minutes.

Which company should she use? Use the graph to explain your answer.

INTERPRETNotes

Show solution

at 20 minutes A costs £4 and B costs £4.50, or the line for A is below the line for B to the left of where they cross	M1	Readings from the graph or from the formulae.
Company A, as it is 50p cheaper (it is cheaper for every ride shorter than 30 minutes)	A1	Conclusion with a correct reason.

Answer: Company A (£4 compared with £4.50)

(2)

(g)

Company B now gives monthly members a 40% discount on the cost per minute. The unlock fee does not change. Nadia becomes a member of Company B.

Find the length of ride, in minutes and seconds, for which the two companies now charge the same amount. Comment on whether your advice in part (f) changes.

MULTINotes

Show solution

\(0.10\times0.6=0.06\)	B1	New cost per minute £0.06 (6p).
\(1+0.15t=2.5+0.06t\)	M1	Equating the two costs with their new rate.
\(t=\frac{1.5}{0.09}=16.6\ldots\), so 16 minutes 40 seconds	A1	An answer of \(16.6\ldots\) or \(16\tfrac{2}{3}\) minutes scores A1 only when converted to 16 minutes 40 seconds.
Yes: a 20-minute ride now costs £3.70 with B but £4 with A, so she should use Company B	B1	Accept a reason based on 20 being greater than the new break-even time.

Answer: 16 minutes 40 seconds; she should now use Company B (£3.70 compared with £4)

(4)

Show full solution

Working	Mark	Notes
Part (a)
\(2u+26p=5.90\) and \(3u+40p=9.00\)	B1	Both equations. Accept equations in pence: \(2u+26p=590\), \(3u+40p=900\).
\(6u+78p=17.70\) and \(6u+80p=18.00\), subtracted to give \(2p=0.30\)	M1	Coefficients of one unknown matched and the equations correctly subtracted (or a correct substitution method).
\(p=0.15\)	A1
\(2u+26\times0.15=5.90\) or \(2u+3.90=5.90\)	M1	Substituting their \(p\) into one of the equations.
\(u=1\)	A1	Accept £1, £1.00 or 100p. Answers with no algebraic working score 0.
Part (b)
\(C=1+0.15t\)	B1	ECF from (a). Accept \(C=0.15t+1\).
Part (c)
\(1+0.15t\le7.50\), giving \(0.15t\le6.50\)	M1	Allow \(\lt\) for this mark. ECF from (b).
43 minutes	A1	From \(t\le43.3\ldots\); the answer must be a whole number. 44 scores A0.
Part (d)
at least two correct points plotted, e.g. \((0,1)\), \((20,4)\), \((40,7)\)	B1	ECF from their formula in (b).
a correct straight line from \((0,1)\) to \((50,8.5)\)	B1
Part (e)
\(1+0.15t=2.5+0.1t\), so \(0.05t=1.5\)	M1	Or both costs at 30 minutes: \(1+0.15\times30\) and \(2.5+0.1\times30\).
\(t=30\) (both cost £5.50)	A1	cso. For the substitution method both costs must be shown as £5.50 with a conclusion.
Part (f)
at 20 minutes A costs £4 and B costs £4.50, or the line for A is below the line for B to the left of where they cross	M1	Readings from the graph or from the formulae.
Company A, as it is 50p cheaper (it is cheaper for every ride shorter than 30 minutes)	A1	Conclusion with a correct reason.
Part (g)
\(0.10\times0.6=0.06\)	B1	New cost per minute £0.06 (6p).
\(1+0.15t=2.5+0.06t\)	M1	Equating the two costs with their new rate.
\(t=\frac{1.5}{0.09}=16.6\ldots\), so 16 minutes 40 seconds	A1	An answer of \(16.6\ldots\) or \(16\tfrac{2}{3}\) minutes scores A1 only when converted to 16 minutes 40 seconds.
Yes: a 20-minute ride now costs £3.70 with B but £4 with A, so she should use Company B	B1	Accept a reason based on 20 being greater than the new break-even time.

Answers: (a) \(u=1\), \(p=0.15\) (£1 unlock fee, 15p per minute) · (b) \(C=1+0.15t\) · (c) 43 minutes · (d) line from \((0,1)\) to \((50,8.5)\) · (e) both cost £5.50 at 30 minutes · (f) Company A (£4 compared with £4.50) · (g) 16 minutes 40 seconds; she should now use Company B (£3.70 compared with £4)

Exam tip (a) Check your pair in the equation you did not use for the substitution: \(3(1)+40(0.15)=3+6=9\). It works.

Common error (g) A 40% discount leaves 60% of the rate, so the new rate is \(0.10\times0.6\), not \(0.10-0.40\).

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A basketball shot: the quadratic formula, completing the square and inequalities

2.3 Quadratic equations and inequalities2.2 Linear and simultaneous equations3.2 Graphs, gradients and real-life graphs1.3 Ratio, proportion and degree of accuracyHigher

Zoe takes a shot at a basketball hoop. The ball leaves her hands at the point R and passes through the centre of the hoop, H. The centre of the hoop is 3.05 m above the floor, at a horizontal distance of 5 m from Zoe's hands.

The height of the ball, \(h\) metres above the floor, at a horizontal distance of \(x\) metres from Zoe's hands is modelled by

\[h=2.05+1.2x-0.2x^2.\]

Figure 1 shows the points R and H.

Figure 1: the release point R of the ball and the centre H of the hoop (the path of the ball is not shown)

(a)

Suppose the ball had missed the hoop and the backboard completely.

Use the quadratic formula to find the horizontal distance from Zoe's hands to the point where the model says the ball would hit the floor. Give your answer to a sensible degree of accuracy.

Higher onlyROUTINENotes

Show solution

\(x=\frac{-1.2\pm\sqrt{1.2^2-4\times(-0.2)\times2.05}}{2\times(-0.2)}\) or, from \(x^2-6x-10.25=0\), \(x=\frac{6\pm\sqrt{36+41}}{2}\)	M1	Correct substitution for \(h=0\). Discriminant \(3.08\) (or 77 for the second form).
\(x=7.387\ldots\) (\(x=-1.387\ldots\) rejected)	A1	The negative root must not be given as an answer.
7.4 m	B1	Accept 7.39 m. Award for a correct rounding of their positive root to 1 or 2 d.p. (or 2 or 3 s.f.); 7.387487… scores B0.

Answer: 7.4 m

(3)

(b)

A banner hangs across the sports hall 4 m above the floor.

Use the discriminant to show that, according to the model, the ball does not reach the height of the banner.

Higher onlyPROOFNotes

Show solution

\(2.05+1.2x-0.2x^2=4\), rearranged to \(0.2x^2-1.2x+1.95=0\) or \(x^2-6x+9.75=0\)	M1	A correct three-term quadratic equal to zero.
\(b^2-4ac=(-6)^2-4\times1\times9.75\) or \(1.44-4\times0.2\times1.95\)	M1	Correct substitution into \(b^2-4ac\) for their quadratic.
\(-3\) (or \(-0.12\)) \(\lt0\), so there are no real solutions and the ball never reaches 4 m	A1	cso. A conclusion referring to the ball must be given.

Answer: \(b^2-4ac=-3\lt0\): no real solutions

(3)

(c)(i)

The model can be written as \(h=-0.2\left(x^2-6x-10.25\right)\).

Write \(x^2-6x-10.25\) in the form \((x+p)^2+q\), where \(p\) and \(q\) are numbers.

Higher onlyALGEBRANotes

Show solution

\((x-3)^2\)	M1	Or \((x-3)^2-9-10.25\).
\((x-3)^2-19.25\)	A1	Accept \(p=-3\), \(q=-19.25\).

Answer: \((x-3)^2-19.25\)

(2)

(c)(ii)

Hence show that the greatest height reached by the ball is 3.85 m, and write down the horizontal distance at which this happens.

Higher onlyPROOFNotes

Show solution

\(h=-0.2\left[(x-3)^2-19.25\right]=3.85-0.2(x-3)^2\)	M1	ECF from (c)(i).
\((x-3)^2\ge0\), so \(h\) is greatest when \(x=3\), giving \(h=3.85\) m	A1	cso. Both \(x=3\) m and the reasoning for the maximum needed. Consistent with (b), as 3.85 < 4.

Answer: greatest height 3.85 m at \(x=3\) m

(2)

(d)

Find the set of values of \(x\) for which the ball is higher than the centre of the hoop.

Show clear algebraic working.

Higher onlyALGEBRANotes

Show solution

\(2.05+1.2x-0.2x^2\gt3.05\), giving \(x^2-6x+5\lt0\)	M1	Inequality sign reversed correctly when dividing by \(-0.2\); allow use of \(=\) for this mark.
\((x-1)(x-5)\), critical values 1 and 5	M1
\(1\lt x\lt5\)	A1	Accept \(1\lt x\lt5\) only; \(x\lt1\) or \(x\gt5\) is A0.

Answer: \(1\lt x\lt5\)

(3)

(e)

On the axes in Figure 1, sketch the path of the ball for \(0\le x\le7.4\).

Label the coordinates of the highest point and of the two points where the ball is at the same height as the centre of the hoop.

GRAPHNotes

Show solution

a \(\cap\)-shaped curve starting at R \((0,2.05)\) and reaching the \(x\)-axis at about \(x=7.4\)	B1	Smooth curve; ECF from (a).
highest point \((3,3.85)\) and points \((1,3.05)\) and \((5,3.05)\) labelled	B1	ECF from (c)(ii) and (d); all three points needed.

Answer: \(\cap\)-shaped curve through \((0,2.05)\), \((1,3.05)\), \((3,3.85)\), \((5,3.05)\), meeting the floor near \(x=7.4\)

(2)

(f)

Give one reason why the model may not describe the motion of the ball well for values of \(x\) greater than 5.

INTERPRETNotes

Show solution

e.g. the ball hits the hoop, net or backboard, so it does not continue along the curve

Accept: the model treats the ball as a point although it has a diameter; or air resistance or spin is ignored. Do not accept “the ball cannot go below 0” alone.

(1)

(g)

In the next attack, Zoe takes the same shot, and the model for the path is unchanged. A defender jumps to block it. At the moment the ball passes, the defender's raised arm lies along the straight line \(h=5.05-2x\), for \(0.95\le x\le1.4\).

By solving simultaneous equations, find the coordinates of the point where the path of the ball meets the line of the arm, and hence state whether the shot is blocked. Give a reason for your answer.

Higher onlyMULTINotes

Show solution

\(2.05+1.2x-0.2x^2=5.05-2x\)	M1	Eliminating \(h\).
\(0.2x^2-3.2x+3=0\) or \(x^2-16x+15=0\)	M1	A correct three-term quadratic equal to zero.
\((x-1)(x-15)=0\), so \(x=1\) or \(x=15\)	M1	Correct method to solve their quadratic (formula allowed).
\((1,3.05)\); \(x=1\) is in the range \(0.95\le x\le1.4\) (\(x=15\) is not), so the ball hits the arm and the shot is blocked	A1	cso. Coordinates and conclusion with reason needed.

Answer: \((1,\ 3.05)\); blocked, since \(x=1\) lies on the arm

(4)

Show full solution

Working	Mark	Notes
Part (a)
\(x=\frac{-1.2\pm\sqrt{1.2^2-4\times(-0.2)\times2.05}}{2\times(-0.2)}\) or, from \(x^2-6x-10.25=0\), \(x=\frac{6\pm\sqrt{36+41}}{2}\)	M1	Correct substitution for \(h=0\). Discriminant \(3.08\) (or 77 for the second form).
\(x=7.387\ldots\) (\(x=-1.387\ldots\) rejected)	A1	The negative root must not be given as an answer.
7.4 m	B1	Accept 7.39 m. Award for a correct rounding of their positive root to 1 or 2 d.p. (or 2 or 3 s.f.); 7.387487… scores B0.
Part (b)
\(2.05+1.2x-0.2x^2=4\), rearranged to \(0.2x^2-1.2x+1.95=0\) or \(x^2-6x+9.75=0\)	M1	A correct three-term quadratic equal to zero.
\(b^2-4ac=(-6)^2-4\times1\times9.75\) or \(1.44-4\times0.2\times1.95\)	M1	Correct substitution into \(b^2-4ac\) for their quadratic.
\(-3\) (or \(-0.12\)) \(\lt0\), so there are no real solutions and the ball never reaches 4 m	A1	cso. A conclusion referring to the ball must be given.
Part (c)(i)
\((x-3)^2\)	M1	Or \((x-3)^2-9-10.25\).
\((x-3)^2-19.25\)	A1	Accept \(p=-3\), \(q=-19.25\).
Part (c)(ii)
\(h=-0.2\left[(x-3)^2-19.25\right]=3.85-0.2(x-3)^2\)	M1	ECF from (c)(i).
\((x-3)^2\ge0\), so \(h\) is greatest when \(x=3\), giving \(h=3.85\) m	A1	cso. Both \(x=3\) m and the reasoning for the maximum needed. Consistent with (b), as 3.85 < 4.
Part (d)
\(2.05+1.2x-0.2x^2\gt3.05\), giving \(x^2-6x+5\lt0\)	M1	Inequality sign reversed correctly when dividing by \(-0.2\); allow use of \(=\) for this mark.
\((x-1)(x-5)\), critical values 1 and 5	M1
\(1\lt x\lt5\)	A1	Accept \(1\lt x\lt5\) only; \(x\lt1\) or \(x\gt5\) is A0.
Part (e)
a \(\cap\)-shaped curve starting at R \((0,2.05)\) and reaching the \(x\)-axis at about \(x=7.4\)	B1	Smooth curve; ECF from (a).
highest point \((3,3.85)\) and points \((1,3.05)\) and \((5,3.05)\) labelled	B1	ECF from (c)(ii) and (d); all three points needed.
Part (f)
e.g. the ball hits the hoop, net or backboard, so it does not continue along the curve	B1	Accept: the model treats the ball as a point although it has a diameter; or air resistance or spin is ignored. Do not accept “the ball cannot go below 0” alone.
Part (g)
\(2.05+1.2x-0.2x^2=5.05-2x\)	M1	Eliminating \(h\).
\(0.2x^2-3.2x+3=0\) or \(x^2-16x+15=0\)	M1	A correct three-term quadratic equal to zero.
\((x-1)(x-15)=0\), so \(x=1\) or \(x=15\)	M1	Correct method to solve their quadratic (formula allowed).
\((1,3.05)\); \(x=1\) is in the range \(0.95\le x\le1.4\) (\(x=15\) is not), so the ball hits the arm and the shot is blocked	A1	cso. Coordinates and conclusion with reason needed.

Answers: (a) 7.4 m · (b) \(b^2-4ac=-3\lt0\): no real solutions · (c)(i) \((x-3)^2-19.25\) · (c)(ii) greatest height 3.85 m at \(x=3\) m · (d) \(1\lt x\lt5\) · (e) \(\cap\)-shaped curve through \((0,2.05)\), \((1,3.05)\), \((3,3.85)\), \((5,3.05)\), meeting the floor near \(x=7.4\) · (g) \((1,\ 3.05)\); blocked, since \(x=1\) lies on the arm

Exam tip (d) Dividing by \(-0.2\) reverses the inequality. A quick sketch of the \(\cup\)-shaped curve \(y=x^2-6x+5\) shows it is below the axis between its roots.

More 2.3 questions in the question bank

Topic 3 · Sequences, functions and graphs

● home subchapter · ○ linked subchapter · last column: links to other topics

Setup	3.1	3.2	3.3	3.4	Other
Weekly savings plans: linear sequences, graphs and equations	●	○			2.2 2.3
A metro train between stations: speed-time and distance-time graphs		●			1.3 4.3 2.2
A weather station: temperature conversions as functions		○	●		2.4 2.2
A drone's vertical test flight: differentiation and motion		○		●	2.3 2.1

3.1 Sequences and nth terms · 3.2 Graphs, gradients and real-life graphsmerged group

+ Add to test✓ In your testCC-EIM-31-32-117 marks~20 min

Weekly savings plans: linear sequences, graphs and equations

3.1 Sequences and nth terms3.2 Graphs, gradients and real-life graphs2.2 Linear and simultaneous equations2.3 Quadratic equations and inequalitiesFoundation

Ana starts a savings plan. She saves £7 in week 1, £10 in week 2, £13 in week 3, and so on. Each week she saves £3 more than she saved the week before.

(a)

Ben says, “The amount, in £, that Ana saves in week \(n\) is \(3n + 7\).”

Explain why Ben is wrong, and write down a correct expression, in terms of \(n\), for the amount, in £, that Ana saves in week \(n\).

CONCEPTNotes

Show solution

\(3(1) + 7 = 10\), but Ana saves £7 in week 1 (or \(3n + 7\) gives the amount for week \(n + 1\))	B1	Any correct reason that tests a value of \(n\) or refers to the first term. Do not accept “the 7 is wrong” without a reason.
\(3n + 4\)	B1	Accept \(4 + 3n\).

Answer: \(3n + 4\); \(3n + 7\) gives £10, not £7, in week 1

(2)

(b)

Ana can save at most £70 in any one week.

By solving an inequality, find the last week in which Ana can follow her plan.

ALGEBRANotes

Show solution

\(3n + 4 \le 70\) leading to \(n \le 22\)	M1	Allow \(3n + 4 = 70\) or \(3n + 4 \lt 70\). Allow ECF from their expression in (a).
Week 22	A1	In week 22 she saves exactly £70, which is allowed. Answer 21 scores M1 A0.

Answer: week 22

(2)

(c)

Ben also saves some money each week. The points in Figure 1 lie on a straight line.

Use Figure 1 to find an expression, in terms of \(n\), for the amount, in £, that Ben saves in week \(n\).

Figure 1: the amount Ben saves in each of weeks 1 to 12

GRAPHNotes

Show solution

\(\frac{42 - 20}{12 - 1}\) or any correct rise ÷ run from two points, e.g. \(\frac{30 - 20}{6 - 1}\)	M1	Values read correctly from the graph. Or differences between consecutive points: 20, 22, 24, …
Gradient (common difference) 2, or \(2n\) seen	A1
\(2n + 18\)	A1	Accept \(18 + 2n\).

Answer: \(2n + 18\)

(3)

(d)

By forming and solving an equation, show that Ana and Ben save the same amount as each other in week 14.

PROOFNotes

Show solution

\(3n + 4 = 2n + 18\)	M1	Allow ECF from their expressions in (a) and (c).
\(n = 18 - 4 = 14\)	A1	cso. Substituting \(n = 14\) into both expressions (46 and 46) without forming and solving an equation scores M0 A0.

Answer: \(3n + 4 = 2n + 18\), so \(n = 14\) (both save £46)

(2)

(e)

Explain why Ana saves more than Ben in every week after week 14.

INTERPRETNotes

Show solution

Ana’s amount goes up by £3 each week and Ben’s by only £2, so after they are equal in week 14 Ana’s amount gets further ahead each week

Accept: Ana’s points would lie on a steeper line than Ben’s, so once the lines meet Ana’s line stays above.

(1)

(f)

Ben changes his plan. In his new plan he saves £3 more each week than the week before, the same increase as Ana. In week 10 of the new plan he saves £45.

Find an expression for the amount, in £, that Ben saves in week \(n\) of his new plan. Explain why Ana and Ben would then never save the same amount in any week.

MODELNotes

Show solution

\(3n + c\) with \(3(10) + c = 45\), or \(45 - 9 \times 3\) (= 18 for week 1)	M1
\(3n + 15\)	A1
Both amounts go up by £3 each week, so Ben always saves £11 more than Ana (their points lie on parallel lines, which never meet)	B1	Allow ECF from their expression with gradient 3. A reason is needed: “the lines do not cross” alone is not enough.

Answer: \(3n + 15\); Ben always saves £11 more (parallel lines)

(3)

(g)(i)

Their friend Dev tries a different plan. He saves 50p in week 1, and each week he saves twice as much as he did the week before.

Find an expression, in terms of \(n\), for the amount, in £, that Dev saves in week \(n\).

Higher onlyROUTINENotes

Show solution

A geometric sequence with first term 0.5 and common ratio 2 identified, e.g. \(a \times 2^{n-1}\) seen	M1	Or terms 0.5, 1, 2, 4, … with a ratio of 2 stated.
\(0.5 \times 2^{n-1}\)	A1	Accept \(2^{n-2}\) or \(\frac{1}{2} \times 2^{n-1}\). An answer in pence must be stated as such, e.g. \(50 \times 2^{n-1}\) pence.

Answer: \(0.5 \times 2^{n-1}\)

(2)

(g)(ii)

Find the first week in which Dev saves more than Ana in that week.

Higher onlyMULTINotes

Show solution

Compares the two amounts for the same week, e.g. week 6: Dev £16, Ana £22, or week 7: Dev £32, Ana £25	M1	At least one correct comparison for week 6 or week 7. Allow ECF from (a) and (g)(i).
Week 7	A1	Both week 6 and week 7 must be checked, or a full list of both sequences up to week 7.

Answer: week 7

(2)

Show full solution

Working	Mark	Notes
Part (a)
\(3(1) + 7 = 10\), but Ana saves £7 in week 1 (or \(3n + 7\) gives the amount for week \(n + 1\))	B1	Any correct reason that tests a value of \(n\) or refers to the first term. Do not accept “the 7 is wrong” without a reason.
\(3n + 4\)	B1	Accept \(4 + 3n\).
Part (b)
\(3n + 4 \le 70\) leading to \(n \le 22\)	M1	Allow \(3n + 4 = 70\) or \(3n + 4 \lt 70\). Allow ECF from their expression in (a).
Week 22	A1	In week 22 she saves exactly £70, which is allowed. Answer 21 scores M1 A0.
Part (c)
\(\frac{42 - 20}{12 - 1}\) or any correct rise ÷ run from two points, e.g. \(\frac{30 - 20}{6 - 1}\)	M1	Values read correctly from the graph. Or differences between consecutive points: 20, 22, 24, …
Gradient (common difference) 2, or \(2n\) seen	A1
\(2n + 18\)	A1	Accept \(18 + 2n\).
Part (d)
\(3n + 4 = 2n + 18\)	M1	Allow ECF from their expressions in (a) and (c).
\(n = 18 - 4 = 14\)	A1	cso. Substituting \(n = 14\) into both expressions (46 and 46) without forming and solving an equation scores M0 A0.
Part (e)
Ana’s amount goes up by £3 each week and Ben’s by only £2, so after they are equal in week 14 Ana’s amount gets further ahead each week	B1	Accept: Ana’s points would lie on a steeper line than Ben’s, so once the lines meet Ana’s line stays above.
Part (f)
\(3n + c\) with \(3(10) + c = 45\), or \(45 - 9 \times 3\) (= 18 for week 1)	M1
\(3n + 15\)	A1
Both amounts go up by £3 each week, so Ben always saves £11 more than Ana (their points lie on parallel lines, which never meet)	B1	Allow ECF from their expression with gradient 3. A reason is needed: “the lines do not cross” alone is not enough.
Part (g)(i)
A geometric sequence with first term 0.5 and common ratio 2 identified, e.g. \(a \times 2^{n-1}\) seen	M1	Or terms 0.5, 1, 2, 4, … with a ratio of 2 stated.
\(0.5 \times 2^{n-1}\)	A1	Accept \(2^{n-2}\) or \(\frac{1}{2} \times 2^{n-1}\). An answer in pence must be stated as such, e.g. \(50 \times 2^{n-1}\) pence.
Part (g)(ii)
Compares the two amounts for the same week, e.g. week 6: Dev £16, Ana £22, or week 7: Dev £32, Ana £25	M1	At least one correct comparison for week 6 or week 7. Allow ECF from (a) and (g)(i).
Week 7	A1	Both week 6 and week 7 must be checked, or a full list of both sequences up to week 7.

Answers: (a) \(3n + 4\); \(3n + 7\) gives £10, not £7, in week 1 · (b) week 22 · (c) \(2n + 18\) · (d) \(3n + 4 = 2n + 18\), so \(n = 14\) (both save £46) · (f) \(3n + 15\); Ben always saves £11 more (parallel lines) · (g)(i) \(0.5 \times 2^{n-1}\) · (g)(ii) week 7

Common error (a) Using the first term as the number added on. The constant in \(3n + c\) is the first term minus the common difference: \(7 - 3 = 4\). Check by putting \(n = 1\) into your expression.

Exam tip (g)(ii) When a geometric sequence and a linear sequence are compared, list both, week by week, near where they cross. Check the week before and the week after the change, so you are sure which week is the first.

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A metro train between stations: speed-time and distance-time graphs

3.2 Graphs, gradients and real-life graphs1.3 Ratio, proportion and degree of accuracy4.3 Area, surface area and volume2.2 Linear and simultaneous equationsFoundation

A metro train travels along a straight track from station A to station B.

The train starts from rest at A and speeds up at a constant rate for 40 seconds, until its speed is 20 m/s.

It then travels at 20 m/s for 50 seconds.

It then slows down at a constant rate and stops at B, 20 seconds later.

Figure 1: axes for the speed-time graph of the journey from A to B

(a)

On the axes in Figure 1, draw the speed-time graph for the journey from A to B.

GRAPHNotes

Show solution

Straight line from (0, 0) to (40, 20)	B1
Horizontal line from (40, 20) to (90, 20)	B1
Straight line from (90, 20) to (110, 0)	B1	Lines must be ruled and must end at the correct points.

Answer: Lines joining (0, 0), (40, 20), (90, 20) and (110, 0)

(3)

(b)

Work out the acceleration of the train during the first 40 seconds.

ROUTINENotes

Show solution

\(\frac{20}{40}\)	M1	Gradient of the first section.
0.5 m/s²	A1	Accept \(\frac{1}{2}\).

Answer: 0.5 m/s²

(2)

(c)

Show that the distance between station A and station B is 1.6 km.

PROOFNotes

Show solution

One correct area, e.g. \(\frac{1}{2} \times 40 \times 20\) (\(=400\)) or \(50 \times 20\) (\(=1000\)) or \(\frac{1}{2} \times 20 \times 20\) (\(=200\))	M1
\(\frac{1}{2}(110 + 50) \times 20\) or \(400 + 1000 + 200\)	M1	Complete method for the total area.
1600 m = 1.6 km	A1	cso. The conversion from metres to kilometres must be seen or stated.

Answer: Area = 1600 m = 1.6 km

(3)

(d)

Work out the average speed of the train for the journey from A to B.

Give your answer in km/h, correct to 3 significant figures.

MULTINotes

Show solution

\(\frac{1600}{110}\) (\(=14.5…\) m/s) or \(\frac{1.6}{110} \times 3600\)	M1	Total distance ÷ total time.
\(14.54… \times 3.6\) or \(14.54… \times \frac{3600}{1000}\)	M1	Correct conversion to km/h (may come first).
52.4 km/h	A1	Accept 52.36… rounded to 52.4 only.

Answer: 52.4 km/h

(3)

(e)

Figure 2 shows the distance-time graph for the same journey.

Explain what the shape of the graph shows about the motion of the train between 0 and 40 seconds and between 40 and 90 seconds.

Figure 2: the distance-time graph of the journey from A to B

INTERPRETNotes

Show solution

0 to 40 s: the curve gets steeper, so the speed is increasing	B1	Must link the steepness (gradient) to the speed.
40 to 90 s: a straight line, so the train moves at a constant speed	B1	Accept: constant gradient, so constant speed (20 m/s).

(2)

(f)

By drawing a tangent on Figure 2, estimate the speed of the train 20 seconds after it leaves A.

Higher onlyGRAPHNotes

Show solution

Tangent drawn at \(t = 20\) and a gradient method, e.g. \(\frac{200 - 0}{30 - 10}\)	M1	The tangent must touch the curve at \(t = 20\) without crossing it.
Answer in the range 9 to 11 m/s	A1	The exact value is 10 m/s, which can be checked from the speed-time graph.

Answer: about 10 m/s

(2)

(g)

On another day, a speed limit means that the train must not travel faster than 16 m/s.

The train speeds up from rest at A to 16 m/s, travels at 16 m/s, and then slows down and stops at B. It speeds up and slows down at the same constant rates as before.

Work out how many seconds longer the journey from A to B takes on this day.

MULTINotes

Show solution

32 s to speed up and 16 s to slow down	B1	From \(16 \div 0.5\) and \(16 \div 1\); the slowing rate is \(20 \div 20 = 1\) m/s². Allow ECF from (b).
\(\frac{1}{2} \times 32 \times 16\) (\(=256\)) and \(\frac{1}{2} \times 16 \times 16\) (\(=128\))	M1	Both distances, using their times.
\(256 + 16t + 128 = 1600\) or \(\frac{1600 - 384}{16}\)	M1	Forms and solves an equation for the time \(t\) at 16 m/s (\(t = 76\)).
\(32 + 76 + 16\) (\(=124\)) and \(124 - 110\)	M1	Total time on this day compared with 110 s.
14 seconds	A1	cao

Answer: 14 seconds

(5)

Show full solution

Working	Mark	Notes
Part (a)
Straight line from (0, 0) to (40, 20)	B1
Horizontal line from (40, 20) to (90, 20)	B1
Straight line from (90, 20) to (110, 0)	B1	Lines must be ruled and must end at the correct points.
Part (b)
\(\frac{20}{40}\)	M1	Gradient of the first section.
0.5 m/s²	A1	Accept \(\frac{1}{2}\).
Part (c)
One correct area, e.g. \(\frac{1}{2} \times 40 \times 20\) (\(=400\)) or \(50 \times 20\) (\(=1000\)) or \(\frac{1}{2} \times 20 \times 20\) (\(=200\))	M1
\(\frac{1}{2}(110 + 50) \times 20\) or \(400 + 1000 + 200\)	M1	Complete method for the total area.
1600 m = 1.6 km	A1	cso. The conversion from metres to kilometres must be seen or stated.
Part (d)
\(\frac{1600}{110}\) (\(=14.5…\) m/s) or \(\frac{1.6}{110} \times 3600\)	M1	Total distance ÷ total time.
\(14.54… \times 3.6\) or \(14.54… \times \frac{3600}{1000}\)	M1	Correct conversion to km/h (may come first).
52.4 km/h	A1	Accept 52.36… rounded to 52.4 only.
Part (e)
0 to 40 s: the curve gets steeper, so the speed is increasing	B1	Must link the steepness (gradient) to the speed.
40 to 90 s: a straight line, so the train moves at a constant speed	B1	Accept: constant gradient, so constant speed (20 m/s).
Part (f)
Tangent drawn at \(t = 20\) and a gradient method, e.g. \(\frac{200 - 0}{30 - 10}\)	M1	The tangent must touch the curve at \(t = 20\) without crossing it.
Answer in the range 9 to 11 m/s	A1	The exact value is 10 m/s, which can be checked from the speed-time graph.
Part (g)
32 s to speed up and 16 s to slow down	B1	From \(16 \div 0.5\) and \(16 \div 1\); the slowing rate is \(20 \div 20 = 1\) m/s². Allow ECF from (b).
\(\frac{1}{2} \times 32 \times 16\) (\(=256\)) and \(\frac{1}{2} \times 16 \times 16\) (\(=128\))	M1	Both distances, using their times.
\(256 + 16t + 128 = 1600\) or \(\frac{1600 - 384}{16}\)	M1	Forms and solves an equation for the time \(t\) at 16 m/s (\(t = 76\)).
\(32 + 76 + 16\) (\(=124\)) and \(124 - 110\)	M1	Total time on this day compared with 110 s.
14 seconds	A1	cao

Answers: (a) Lines joining (0, 0), (40, 20), (90, 20) and (110, 0) · (b) 0.5 m/s² · (c) Area = 1600 m = 1.6 km · (d) 52.4 km/h · (f) about 10 m/s · (g) 14 seconds

Common error (d) Averaging the speeds of the three sections does not give the average speed, because the sections last for different times. Always use total distance ÷ total time.

More 3.2 questions in the question bank

3.3 Function notation, composite and inverse functions · 3.4 Differentiation and stationary pointsmerged group

+ Add to test✓ In your testCC-EIM-33-34-120 marks~20 min

A weather station: temperature conversions as functions

3.3 Function notation, composite and inverse functions3.2 Graphs, gradients and real-life graphs2.4 Direct and inverse proportion and rearranging formulae2.2 Linear and simultaneous equationsHigher

A weather station measures the air temperature, \(x\) °C.

Its display converts the temperature to degrees Fahrenheit (°F) using the function

\(f(x) = 1.8x + 32\)

and to kelvin (K) using the function

\(g(x) = x + 273\)

Figure 1 shows the graph of \(y = f(x)\) for \(-60 \le x \le 40\), and the line \(y = x\).

Figure 1: the graph of y = f(x) for −60 ≤ x ≤ 40, and the line y = x

(a)

Find \(f(-15)\).

ROUTINENotes

Show solution

Answer: 5

(1)

(b)

One afternoon the display shows 104 °F.

Solve \(f(x) = 104\) to find the temperature in °C.

ROUTINENotes

Show solution

\(1.8x + 32 = 104\) or \(1.8x = 72\)	M1
\(x = 40\)	A1	Accept 40 °C.

Answer: 40 °C

(2)

(c)

Find \(f^{-1}(x)\).

ALGEBRANotes

Show solution

\(y = 1.8x + 32\) rearranged to \(x = \frac{y - 32}{1.8}\), or \(y - 32 = 1.8x\)	M1	Or the steps “subtract 32, then divide by 1.8” applied in that order.
\(f^{-1}(x) = \frac{x - 32}{1.8}\)	A1	Accept \(\frac{5(x - 32)}{9}\) or \(\frac{5x - 160}{9}\). Must be in terms of \(x\). \(\frac{x}{1.8} - 32\) scores M0 A0.

Answer: \(f^{-1}(x) = \dfrac{x - 32}{1.8}\)

(2)

(d)

On Figure 1, draw the graph of \(y = f^{-1}(x)\).

GRAPHNotes

Show solution

A straight line through \((32, 0)\) and \((0, -17.8)\) approximately	B1	Any two correct points of \(y = \frac{x - 32}{1.8}\), e.g. \((104, 40)\), \((50, 10)\).
Fully correct line: the reflection of \(y = f(x)\) in \(y = x\), from \((-76, -60)\) to \((104, 40)\)	B1	Accept a line drawn across the whole grid, provided it passes through the correct points.

Answer: The line through \((-40, -40)\), \((32, 0)\) and \((104, 40)\)

(2)

(e)

Show that the graphs of \(y = f(x)\) and \(y = f^{-1}(x)\) meet at the point \((-40, -40)\).

PROOFNotes

Show solution

\(1.8x + 32 = \frac{x - 32}{1.8}\)	M1	Or \(1.8x + 32 = x\), with a statement that the graphs meet on the line \(y = x\). Allow ECF from their \(f^{-1}\).
\(3.24x + 57.6 = x - 32\) or \(2.24x = -89.6\) or \(0.8x = -32\)	M1	Correct method to collect the \(x\) terms.
\(x = -40\) and \(y = f(-40) = -72 + 32 = -40\)	A1	cso. Both coordinates must be shown. Alternative method: M1 for \(f(-40) = -72 + 32 = -40\), M1 for \(f^{-1}(-40) = \frac{-72}{1.8} = -40\), A1 for the conclusion that \((-40, -40)\) lies on both graphs.

Answer: \(x = -40\), \(y = -40\)

(3)

(f)

Find \(fg^{-1}(x)\). Give your answer in its simplest form.

ALGEBRANotes

Show solution

\(g^{-1}(x) = x - 273\)	B1
\(1.8(x - 273) + 32\)	M1	Their \(g^{-1}(x)\) substituted into \(f\).
\(1.8x - 459.4\)	A1	cao

Answer: \(fg^{-1}(x) = 1.8x - 459.4\)

(3)

(g)

The input of \(fg^{-1}\) is a temperature in kelvin. No temperature can be lower than 0 K (absolute zero).

State, with a reason, a suitable domain for \(fg^{-1}\) in this context, and find the range of \(fg^{-1}\) for this domain.

INTERPRETNotes

Show solution

\(x \ge 0\), because a temperature in kelvin cannot be negative (0 K is the lowest possible temperature)	B1	Reason needed.
\(fg^{-1}(x) \ge -459.4\)	B1	Allow ECF from (f). Accept range \(\ge -459.4\) °F.

Answer: \(x \ge 0\); \(fg^{-1}(x) \ge -459.4\)

(2)

(h)(i)

The station fits a new sensor. The sensor reading is \(x\) °C. A correction function \(h\) changes the sensor reading into the true temperature in °C.

The display now shows the temperature in °F as

\(fh(x) = 1.71x + 34.7\)

Find \(h(x)\).

ALGEBRANotes

Show solution

\(1.8h(x) + 32 = 1.71x + 34.7\)	M1	Writes the composite using \(f\).
\(1.8h(x) = 1.71x + 2.7\)	M1	Correct rearrangement towards \(h(x)\).
\(h(x) = 0.95x + 1.5\)	A1	cao

Answer: \(h(x) = 0.95x + 1.5\)

(3)

(h)(ii)

Find the sensor reading for which the true temperature is the same as the sensor reading.

ALGEBRANotes

Show solution

\(0.95x + 1.5 = x\)	M1	Allow ECF from (h)(i) if linear.
\(x = 30\)	A1	Accept 30 °C.

Answer: 30 °C

(2)

Show full solution

Working	Mark	Notes
Part (a)
5	B1
Part (b)
\(1.8x + 32 = 104\) or \(1.8x = 72\)	M1
\(x = 40\)	A1	Accept 40 °C.
Part (c)
\(y = 1.8x + 32\) rearranged to \(x = \frac{y - 32}{1.8}\), or \(y - 32 = 1.8x\)	M1	Or the steps “subtract 32, then divide by 1.8” applied in that order.
\(f^{-1}(x) = \frac{x - 32}{1.8}\)	A1	Accept \(\frac{5(x - 32)}{9}\) or \(\frac{5x - 160}{9}\). Must be in terms of \(x\). \(\frac{x}{1.8} - 32\) scores M0 A0.
Part (d)
A straight line through \((32, 0)\) and \((0, -17.8)\) approximately	B1	Any two correct points of \(y = \frac{x - 32}{1.8}\), e.g. \((104, 40)\), \((50, 10)\).
Fully correct line: the reflection of \(y = f(x)\) in \(y = x\), from \((-76, -60)\) to \((104, 40)\)	B1	Accept a line drawn across the whole grid, provided it passes through the correct points.
Part (e)
\(1.8x + 32 = \frac{x - 32}{1.8}\)	M1	Or \(1.8x + 32 = x\), with a statement that the graphs meet on the line \(y = x\). Allow ECF from their \(f^{-1}\).
\(3.24x + 57.6 = x - 32\) or \(2.24x = -89.6\) or \(0.8x = -32\)	M1	Correct method to collect the \(x\) terms.
\(x = -40\) and \(y = f(-40) = -72 + 32 = -40\)	A1	cso. Both coordinates must be shown. Alternative method: M1 for \(f(-40) = -72 + 32 = -40\), M1 for \(f^{-1}(-40) = \frac{-72}{1.8} = -40\), A1 for the conclusion that \((-40, -40)\) lies on both graphs.
Part (f)
\(g^{-1}(x) = x - 273\)	B1
\(1.8(x - 273) + 32\)	M1	Their \(g^{-1}(x)\) substituted into \(f\).
\(1.8x - 459.4\)	A1	cao
Part (g)
\(x \ge 0\), because a temperature in kelvin cannot be negative (0 K is the lowest possible temperature)	B1	Reason needed.
\(fg^{-1}(x) \ge -459.4\)	B1	Allow ECF from (f). Accept range \(\ge -459.4\) °F.
Part (h)(i)
\(1.8h(x) + 32 = 1.71x + 34.7\)	M1	Writes the composite using \(f\).
\(1.8h(x) = 1.71x + 2.7\)	M1	Correct rearrangement towards \(h(x)\).
\(h(x) = 0.95x + 1.5\)	A1	cao
Part (h)(ii)
\(0.95x + 1.5 = x\)	M1	Allow ECF from (h)(i) if linear.
\(x = 30\)	A1	Accept 30 °C.

Answers: (a) 5 · (b) 40 °C · (c) \(f^{-1}(x) = \dfrac{x - 32}{1.8}\) · (d) The line through \((-40, -40)\), \((32, 0)\) and \((104, 40)\) · (e) \(x = -40\), \(y = -40\) · (f) \(fg^{-1}(x) = 1.8x - 459.4\) · (g) \(x \ge 0\); \(fg^{-1}(x) \ge -459.4\) · (h)(i) \(h(x) = 0.95x + 1.5\) · (h)(ii) 30 °C

Common error (c) Undoing the steps in the wrong order gives \(\dfrac{x}{1.8} - 32\). The function multiplies by 1.8 and then adds 32, so the inverse must subtract 32 first and then divide by 1.8.

Exam tip (f) \(fg^{-1}(x)\) means \(f(g^{-1}(x))\): find \(g^{-1}(x)\) first, then put the whole of it into \(f\). Here it changes a temperature in kelvin into degrees Fahrenheit.

More 3.3 questions in the question bank

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A drone's vertical test flight: differentiation and motion

3.4 Differentiation and stationary points2.3 Quadratic equations and inequalities3.2 Graphs, gradients and real-life graphs2.1 Algebraic manipulation and factorisingHigher

During a test flight, a drone takes off from the flat roof of a building and moves vertically upwards. It then starts to come back down.

The height of the drone above the roof, \(t\) seconds after take-off, is \(h\) metres, where

\(h = 1.5t^2 - 0.1t^3\) for \(0 \le t \le 12\)

Figure 1 shows the graph of \(h\) against \(t\). The tangent to the curve at the point P, where \(t = 5\), has been drawn.

Figure 1: the graph of h against t for 0 ≤ t ≤ 12, with the tangent to the curve at P, where t = 5

(a)

Find \(\dfrac{dh}{dt}\).

ROUTINENotes

Show solution

One term correct: \(3t\) or \(-0.3t^2\)	M1
\(\frac{dh}{dt} = 3t - 0.3t^2\)	A1

Answer: \(\dfrac{dh}{dt} = 3t - 0.3t^2\)

(2)

(b)

Find an expression for the acceleration of the drone, in m/s², at time \(t\) seconds.

ROUTINENotes

Show solution

Differentiates their \(\frac{dh}{dt}\), at least one term correct	M1	Allow ECF from (a).
\(3 - 0.6t\)	A1

Answer: \(3 - 0.6t\)

(2)

(c)(i)

The drone is momentarily at rest when it stops rising.

Find the value of \(t\) when this happens.

ALGEBRANotes

Show solution

\(3t - 0.3t^2 = 0\)	M1	Their \(\frac{dh}{dt}\) set equal to 0.
\(0.3t(10 - t) = 0\) or \(3t(1 - 0.1t) = 0\)	M1	Correct factorisation or other valid method of solution.
\(t = 10\)	A1	\(t = 0\) (take-off) must be rejected or not given as the answer.

Answer: \(t = 10\)

(3)

(c)(ii)

Show that the greatest height of the drone above the roof is 50 m.

PROOFNotes

Show solution

At \(t = 10\), \(\frac{d^2h}{dt^2} = 3 - 6 = -3 \lt 0\), so a maximum	M1	Or the gradient test, e.g. \(\frac{dh}{dt} \gt 0\) at \(t = 9\) and \(\lt 0\) at \(t = 11\), with a conclusion.
\(1.5 \times 10^2 - 0.1 \times 10^3\)	M1	Substitutes their \(t\) from (c)(i) into \(h\).
\(150 - 100 = 50\)	A1	cso. Both the maximum test and the substitution must be seen.

Answer: At \(t = 10\): \(\dfrac{d^2h}{dt^2} = -3 \lt 0\) and \(h = 50\) m

(3)

(d)

Use the tangent at P in Figure 1 to estimate the velocity of the drone when \(t = 5\).

Then use your answer to part (a) to find its exact velocity when \(t = 5\).

GRAPHNotes

Show solution

\(\frac{40 - 10}{7 - 3}\) or any correct rise ÷ run from two points on the tangent	M1	Points read correctly from the drawn tangent.
Estimate in the range 7 to 8 m/s	A1
\(3(5) - 0.3(5)^2 = 7.5\) m/s	B1	Allow ECF from (a).

Answer: Estimate about 7.5 m/s; exact 7.5 m/s

(3)

(e)

Use your answer to part (b) to explain when the drone is speeding up and when it is slowing down while it is rising, for \(0 \lt t \lt 10\).

INTERPRETNotes

Show solution

For \(0 \lt t \lt 5\), \(3 - 0.6t \gt 0\), so the drone is speeding up	B1	Must refer to the sign of the acceleration.
For \(5 \lt t \lt 10\), \(3 - 0.6t \lt 0\) while the drone is still rising, so it is slowing down	B1	Accept: it rises fastest at \(t = 5\), where the acceleration is 0.

Answer: Speeding up for \(0 \lt t \lt 5\); slowing down for \(5 \lt t \lt 10\)

(2)

(f)

For a second test flight the drone is programmed differently. Its height above the roof is now

\(h = 1.5t^2 - kt^3\)

where \(k\) is a positive constant.

Safety rules say that the drone must not rise more than 72 m above the roof.

Find the least possible value of \(k\).

MULTINotes

Show solution

\(3t - 3kt^2 = 0\)	M1	Differentiates and sets equal to 0.
\(t = \frac{1}{k}\)	A1	From \(3t(1 - kt) = 0\), rejecting \(t = 0\).
\(1.5\left(\frac{1}{k}\right)^2 - k\left(\frac{1}{k}\right)^3 = 72\) or \(\frac{1}{2k^2} = 72\)	M1	Substitutes their \(t\) into \(h\) and sets equal to 72 (or uses \(\le 72\)).
\(k = \frac{1}{12}\)	A1	Accept 0.0833 (3 s.f.). \(-\frac{1}{12}\) must be rejected. The greatest height \(\frac{1}{2k^2}\) gets smaller as \(k\) increases, so \(k \ge \frac{1}{12}\).

Answer: \(k = \dfrac{1}{12}\)

(4)

Show full solution

Working	Mark	Notes
Part (a)
One term correct: \(3t\) or \(-0.3t^2\)	M1
\(\frac{dh}{dt} = 3t - 0.3t^2\)	A1
Part (b)
Differentiates their \(\frac{dh}{dt}\), at least one term correct	M1	Allow ECF from (a).
\(3 - 0.6t\)	A1
Part (c)(i)
\(3t - 0.3t^2 = 0\)	M1	Their \(\frac{dh}{dt}\) set equal to 0.
\(0.3t(10 - t) = 0\) or \(3t(1 - 0.1t) = 0\)	M1	Correct factorisation or other valid method of solution.
\(t = 10\)	A1	\(t = 0\) (take-off) must be rejected or not given as the answer.
Part (c)(ii)
At \(t = 10\), \(\frac{d^2h}{dt^2} = 3 - 6 = -3 \lt 0\), so a maximum	M1	Or the gradient test, e.g. \(\frac{dh}{dt} \gt 0\) at \(t = 9\) and \(\lt 0\) at \(t = 11\), with a conclusion.
\(1.5 \times 10^2 - 0.1 \times 10^3\)	M1	Substitutes their \(t\) from (c)(i) into \(h\).
\(150 - 100 = 50\)	A1	cso. Both the maximum test and the substitution must be seen.
Part (d)
\(\frac{40 - 10}{7 - 3}\) or any correct rise ÷ run from two points on the tangent	M1	Points read correctly from the drawn tangent.
Estimate in the range 7 to 8 m/s	A1
\(3(5) - 0.3(5)^2 = 7.5\) m/s	B1	Allow ECF from (a).
Part (e)
For \(0 \lt t \lt 5\), \(3 - 0.6t \gt 0\), so the drone is speeding up	B1	Must refer to the sign of the acceleration.
For \(5 \lt t \lt 10\), \(3 - 0.6t \lt 0\) while the drone is still rising, so it is slowing down	B1	Accept: it rises fastest at \(t = 5\), where the acceleration is 0.
Part (f)
\(3t - 3kt^2 = 0\)	M1	Differentiates and sets equal to 0.
\(t = \frac{1}{k}\)	A1	From \(3t(1 - kt) = 0\), rejecting \(t = 0\).
\(1.5\left(\frac{1}{k}\right)^2 - k\left(\frac{1}{k}\right)^3 = 72\) or \(\frac{1}{2k^2} = 72\)	M1	Substitutes their \(t\) into \(h\) and sets equal to 72 (or uses \(\le 72\)).
\(k = \frac{1}{12}\)	A1	Accept 0.0833 (3 s.f.). \(-\frac{1}{12}\) must be rejected. The greatest height \(\frac{1}{2k^2}\) gets smaller as \(k\) increases, so \(k \ge \frac{1}{12}\).

Answers: (a) \(\dfrac{dh}{dt} = 3t - 0.3t^2\) · (b) \(3 - 0.6t\) · (c)(i) \(t = 10\) · (c)(ii) At \(t = 10\): \(\dfrac{d^2h}{dt^2} = -3 \lt 0\) and \(h = 50\) m · (d) Estimate about 7.5 m/s; exact 7.5 m/s · (e) Speeding up for \(0 \lt t \lt 5\); slowing down for \(5 \lt t \lt 10\) · (f) \(k = \dfrac{1}{12}\)

Exam tip (e) A negative acceleration does not mean that the drone is moving down. For \(5 \lt t \lt 10\) the velocity is still positive, so the drone is still rising, but more and more slowly.

More 3.4 questions in the question bank

Topic 4 · Geometry and trigonometry

● home subchapter · ○ linked subchapter · last column: links to other topics

Setup	4.1	4.2	4.3	4.4	Other
A football's panels: polygons, angles and tiling	●	○	○		3.2
A roundabout layout: circle theorems and a scale model	○	○	○	●	3.2
Siting a wind turbine: depression, bearings and swept area		●	○		1.3 3.2
A grain silo: volume, surface area and filling		○	●		3.2 1.3

4.1 Angles, polygons and circle properties · 4.4 Similar shapes and circle theoremsmerged group

+ Add to test✓ In your testCC-EIM-41-44-119 marks~20 min

A football's panels: polygons, angles and tiling

4.1 Angles, polygons and circle properties4.2 Pythagoras' theorem and trigonometry4.3 Area, surface area and volume3.2 Graphs, gradients and real-life graphsFoundation

A traditional football is stitched together from 32 leather panels: 12 regular pentagons and 20 regular hexagons. Every panel has sides of length 4.5 cm.

At every corner of the ball, one pentagon and two hexagons meet. Figure 1 shows the three panels at one corner, \(P\), laid out flat on a table.

Figure 1: one pentagon and two hexagons from the football laid out flat around the corner point P

(a)

Work out the size of each interior angle of a regular pentagon.

ROUTINENotes

Show solution

\((5-2)\times180\) \((=540)\) and \(\div5\), or \(180-\frac{360}{5}\)	M1	Exterior angle \(72^\circ\) seen implies the method if subtracted from 180.
\(108^\circ\)	A1	cao

Answer: \(108^\circ\)

(2)

(b)

Show that each interior angle of a regular hexagon is \(120^\circ\).

PROOFNotes

Show solution

\((6-2)\times180=720\), or exterior angle \(\frac{360}{6}=60\)	M1
\(720\div6=120\), or \(180-60=120\)	A1	cso. Both steps must be shown; starting from 120 and checking is not enough unless the angle sum is also used.

Answer: \(720^\circ\div6=120^\circ\)

(2)

(c)(i)

Work out the size of the gap, marked ? in Figure 1, between the panels when they are laid flat around \(P\).

ROUTINENotes

Show solution

\(360-(108+120+120)\) or \(360-348\)	M1	Allow ECF from (a).
\(12^\circ\)	A1

Answer: \(12^\circ\)

(2)

(c)(ii)

Use your answer to (c)(i) to explain why the panels at a corner cannot lie flat when they are stitched together, and what this means for the shape of the ball.

INTERPRETNotes

Show solution

The three angles add up to less than \(360^\circ\) (angles around a point on a flat surface add up to \(360^\circ\)), so closing the \(12^\circ\) gap forces the panels to bend out of the flat: the surface curves, which is how the panels make a ball

Must refer to the total being less than 360 (or the gap) and to the surface bending or curving.

(1)

(d)(i)

Figure 2 shows the interior angle of a regular polygon plotted against its number of sides, \(n\).

Rosa says,

“If you double the number of sides of a regular polygon, you double the size of its interior angle.”

Use Figure 2 to show that Rosa is wrong.

Figure 2: the size of each interior angle of a regular polygon, plotted against its number of sides

GRAPHNotes

Show solution

Two correct readings for \(n\) and \(2n\) compared, e.g. \(n=4\): \(90^\circ\) and \(n=8\): \(135^\circ\), with \(2\times90=180\)	M1	Or \(n=5\): 108 and \(n=10\): 144, or \(n=6\): 120 and \(n=12\): 150. \(n=3\) to \(n=6\) (60 to 120) does double, so it cannot be used as the counter-example.
\(135\ne180\), so doubling the number of sides does not double the interior angle: Rosa is wrong	A1	cso. One pair that fails, with a conclusion, is enough.

Answer: e.g. 4 sides: \(90^\circ\), 8 sides: \(135^\circ\), not \(180^\circ\)

(2)

(d)(ii)

Regular polygons of one type can cover a flat floor with no gaps and no overlaps only if a whole number of their interior angles make exactly \(360^\circ\).

Use Figure 2 to find all the regular polygons that can cover a flat floor in this way. Give a reason why no regular polygon with more than 6 sides can do so.

INTERPRETNotes

Show solution

\(360\div\) interior angle tried for the polygons, e.g. \(360\div60=6\), \(360\div90=4\), \(360\div108=3.33\ldots\), \(360\div120=3\)	M1	At least three trials, including one that fails.
Triangle, square and hexagon (\(n=3,4,6\)) only	A1	Accept the three angles 60, 90, 120 identified.
For \(n>6\) each angle is more than \(120^\circ\) but less than \(180^\circ\), so \(360\div\) angle is between 2 and 3 and cannot be a whole number	B1	Accept: two angles make less than 360 and three make more than 360.

Answer: Equilateral triangles, squares and regular hexagons

(3)

(e)

A designer wants a flat pattern in which one regular pentagon and two identical regular polygons meet at each point with no gap.

Show that this is not possible.

MULTINotes

Show solution

\((360-108)\div2\) \((=126)\)	M1	Allow ECF from (a).
\(180-126=54\) and \(360\div54\)	M1	Or \((n-2)\times180=126n\) set up.
\(6.66\ldots\), not a whole number, so no regular polygon has an interior angle of \(126^\circ\)	A1	cso. Accept reading Figure 2: 126 lies between 120 (\(n=6\)) and 128.6 (\(n=7\)), so no polygon fits, after the 126 is found.

Answer: Each polygon would need an angle of \(126^\circ\), giving \(n=6.66\ldots\), not a whole number

(3)

(f)

Figure 3 shows one hexagonal panel. It is made of six equilateral triangles such as \(OAB\).

Work out the area of one hexagonal panel. Give your answer correct to 3 significant figures.

Figure 3: one hexagonal panel ABCDEF with centre O; triangle OAB is equilateral

MULTINotes

Show solution

Height of triangle \(OAB\): \(4.5\sin60^\circ\) or \(\sqrt{4.5^2-2.25^2}\) \((=3.897\ldots)\)	M1	Or \(2.25\tan60^\circ\).
\(\frac{1}{2}\times4.5\times3.897\ldots\) \((=8.768\ldots)\)	M1	Area of one triangle with their height (not 4.5).
\(6\times8.768\ldots\)	M1
\(52.6\) cm\(^2\)	A1	Accept 52.6 or 52.61... Using \(\frac12\times4.5\times4.5\) for the triangle scores M0 M0 M1 A0.

Answer: \(52.6\) cm\(^2\)

(4)

Show full solution

Working	Mark	Notes
Part (a)
\((5-2)\times180\) \((=540)\) and \(\div5\), or \(180-\frac{360}{5}\)	M1	Exterior angle \(72^\circ\) seen implies the method if subtracted from 180.
\(108^\circ\)	A1	cao
Part (b)
\((6-2)\times180=720\), or exterior angle \(\frac{360}{6}=60\)	M1
\(720\div6=120\), or \(180-60=120\)	A1	cso. Both steps must be shown; starting from 120 and checking is not enough unless the angle sum is also used.
Part (c)(i)
\(360-(108+120+120)\) or \(360-348\)	M1	Allow ECF from (a).
\(12^\circ\)	A1
Part (c)(ii)
The three angles add up to less than \(360^\circ\) (angles around a point on a flat surface add up to \(360^\circ\)), so closing the \(12^\circ\) gap forces the panels to bend out of the flat: the surface curves, which is how the panels make a ball	B1	Must refer to the total being less than 360 (or the gap) and to the surface bending or curving.
Part (d)(i)
Two correct readings for \(n\) and \(2n\) compared, e.g. \(n=4\): \(90^\circ\) and \(n=8\): \(135^\circ\), with \(2\times90=180\)	M1	Or \(n=5\): 108 and \(n=10\): 144, or \(n=6\): 120 and \(n=12\): 150. \(n=3\) to \(n=6\) (60 to 120) does double, so it cannot be used as the counter-example.
\(135\ne180\), so doubling the number of sides does not double the interior angle: Rosa is wrong	A1	cso. One pair that fails, with a conclusion, is enough.
Part (d)(ii)
\(360\div\) interior angle tried for the polygons, e.g. \(360\div60=6\), \(360\div90=4\), \(360\div108=3.33\ldots\), \(360\div120=3\)	M1	At least three trials, including one that fails.
Triangle, square and hexagon (\(n=3,4,6\)) only	A1	Accept the three angles 60, 90, 120 identified.
For \(n>6\) each angle is more than \(120^\circ\) but less than \(180^\circ\), so \(360\div\) angle is between 2 and 3 and cannot be a whole number	B1	Accept: two angles make less than 360 and three make more than 360.
Part (e)
\((360-108)\div2\) \((=126)\)	M1	Allow ECF from (a).
\(180-126=54\) and \(360\div54\)	M1	Or \((n-2)\times180=126n\) set up.
\(6.66\ldots\), not a whole number, so no regular polygon has an interior angle of \(126^\circ\)	A1	cso. Accept reading Figure 2: 126 lies between 120 (\(n=6\)) and 128.6 (\(n=7\)), so no polygon fits, after the 126 is found.
Part (f)
Height of triangle \(OAB\): \(4.5\sin60^\circ\) or \(\sqrt{4.5^2-2.25^2}\) \((=3.897\ldots)\)	M1	Or \(2.25\tan60^\circ\).
\(\frac{1}{2}\times4.5\times3.897\ldots\) \((=8.768\ldots)\)	M1	Area of one triangle with their height (not 4.5).
\(6\times8.768\ldots\)	M1
\(52.6\) cm\(^2\)	A1	Accept 52.6 or 52.61... Using \(\frac12\times4.5\times4.5\) for the triangle scores M0 M0 M1 A0.

Answers: (a) \(108^\circ\) · (b) \(720^\circ\div6=120^\circ\) · (c)(i) \(12^\circ\) · (d)(i) e.g. 4 sides: \(90^\circ\), 8 sides: \(135^\circ\), not \(180^\circ\) · (d)(ii) Equilateral triangles, squares and regular hexagons · (e) Each polygon would need an angle of \(126^\circ\), giving \(n=6.66\ldots\), not a whole number · (f) \(52.6\) cm\(^2\)

Common error (e) Dividing 360 by the interior angle does not give the number of sides. Find the exterior angle first: \(180-126=54\), then \(n=\frac{360}{54}\).

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A roundabout layout: circle theorems and a scale model

4.4 Similar shapes and circle theorems4.1 Angles, polygons and circle properties4.2 Pythagoras' theorem and trigonometry4.3 Area, surface area and volume3.2 Graphs, gradients and real-life graphsHigher

Figure 1 shows the plan of a circular traffic island with centre \(O\) and radius 10 m.

Two straight kerbs, \(PA\) and \(PB\), touch the island at \(A\) and \(B\), so \(PA\) and \(PB\) are tangents to the circle. Angle \(APB=44^\circ\).

Lamp posts stand on the edge of the island at \(A\), \(Q\), \(B\) and \(R\). \(Q\) is on the major arc \(AB\) and \(R\) is on the minor arc \(AB\).

Figure 1: the circular island, centre O, with the kerbs PA and PB touching it at A and B

(a)

Show that the two kerbs have the same length, \(PA=PB\). Give reasons for your working.

PROOFNotes

Show solution

Angle \(OAP=\) angle \(OBP=90^\circ\) because the angle between a tangent and a radius is \(90^\circ\)	B1	Reason must be given.
\(PA^2=OP^2-OA^2\) and \(PB^2=OP^2-OB^2\)	M1	Pythagoras in both right-angled triangles. Or triangles \(OAP\) and \(OBP\) compared side by side for RHS.
\(OA=OB\) (radii) and \(OP\) is common, so \(PA^2=PB^2\) and \(PA=PB\)	A1	cso. Accept congruent triangles by RHS (right angle, common hypotenuse \(OP\), equal radii), hence \(PA=PB\).

Answer: \(PA=PB\)

(3)

(b)

Work out the size of angle \(AQB\). Give a reason for each stage of your working.

MULTINotes

Show solution

Angle \(PAB=\frac{180-44}{2}\) \((=68^\circ)\)	M1	Or angle \(AOB=360-90-90-44\) \((=136^\circ)\) from quadrilateral \(OAPB\).
Angle \(AQB=68^\circ\)	A1	From \(136\div2\) on the alternative route.
Reasons: base angles of an isosceles triangle are equal (\(PA=PB\)) and the alternate segment theorem	B1	Alternative route: angles in a quadrilateral add up to \(360^\circ\) (with tangent and radius at \(90^\circ\)) and the angle at the centre is twice the angle at the circumference. Both reasons needed.

Answer: \(68^\circ\)

(3)

(c)

Work out the size of angle \(ARB\). Give a reason for your answer.

CONCEPTNotes

Show solution

\(112^\circ\)	B1	Allow \(180-\) their (b).
Opposite angles of a cyclic quadrilateral (\(AQBR\)) add up to \(180^\circ\)	B1	The words cyclic quadrilateral and opposite angles are both needed.

Answer: \(112^\circ\)

(2)

(d)(i)

Figure 2 shows the island on a coordinate grid, in metres. A straight footpath crosses the island from \(C\) to \(D\). \(C\) and \(D\) are at grid points.

Find the equation of the perpendicular bisector of \(CD\).

Figure 2: the island on a coordinate grid (units in metres), with the footpath CD

GRAPHNotes

Show solution

Gradient of \(CD=\frac{23-15}{26-10}=\frac{1}{2}\)	M1	From \(C(10,15)\) and \(D(26,23)\) read from the grid.
Gradient \(-2\) used with the midpoint \((18,19)\)	M1	e.g. \(19=-2(18)+c\) or \(y-19=-2(x-18)\).
\(y=-2x+55\)	A1	Accept any equivalent, e.g. \(2x+y=55\).

Answer: \(y=-2x+55\)

(3)

(d)(ii)

Explain why the centre \(O\) of the island must lie on this line.

INTERPRETNotes

Show solution

The perpendicular from the centre to a chord bisects the chord, so the perpendicular bisector of the chord \(CD\) passes through the centre

Accept the check \(-2(20)+55=15\) only together with the circle fact.

(1)

(e)(i)

A fountain stands at the centre of the island. A scale model of the fountain is mathematically similar to the real fountain. The model is 24 cm tall and the real fountain is 1.2 m tall.

The basin of the model holds 6.4 litres of water.

Work out how many litres of water the basin of the real fountain holds.

MULTINotes

Show solution

\(6.4\times\left(\frac{120}{24}\right)^3\) or \(6.4\times125\)	M1	Linear scale factor 5 must be cubed; \(6.4\times5\) or \(6.4\times25\) scores M0.
800 litres	A1

Answer: 800 litres

(2)

(e)(ii)

Mosaic tiles cover 900 cm\(^2\) of the surface of the model.

Work out the area of the surface of the real fountain covered by mosaic tiles. Give your answer in m\(^2\).

ROUTINENotes

Show solution

\(900\times5^2\) \((=22\,500\text{ cm}^2)\)	M1	Or \(0.09\times25\) after converting 900 cm\(^2\) to 0.09 m\(^2\).
\(2.25\) m\(^2\)	A1	22 500 cm\(^2\) not converted scores M1 A0.

Answer: \(2.25\) m\(^2\)

(2)

(f)

The island, apart from a circular fountain base of radius 2 m at the centre, is planted with flowers.

The council increases the radius of the island from 10 m to 12.5 m. The fountain base does not change.

Work out the percentage increase in the planted area.

MULTINotes

Show solution

\(\pi\times10^2-\pi\times2^2\) \((=96\pi=301.5\ldots)\)	M1
\(\pi\times12.5^2-\pi\times2^2\) \((=152.25\pi=478.3\ldots)\)	M1
\(\frac{478.3-301.5}{301.5}\times100\) or \(\frac{152.25}{96}\)	M1	Dependent on at least one previous M1.
\(58.6\)%	A1	Accept 58.5 to 58.6 (58.59...). Ignoring the fountain gives 56.25% and scores M0 M0 M1 A0.

Answer: \(58.6\)%

(4)

Show full solution

Working	Mark	Notes
Part (a)
Angle \(OAP=\) angle \(OBP=90^\circ\) because the angle between a tangent and a radius is \(90^\circ\)	B1	Reason must be given.
\(PA^2=OP^2-OA^2\) and \(PB^2=OP^2-OB^2\)	M1	Pythagoras in both right-angled triangles. Or triangles \(OAP\) and \(OBP\) compared side by side for RHS.
\(OA=OB\) (radii) and \(OP\) is common, so \(PA^2=PB^2\) and \(PA=PB\)	A1	cso. Accept congruent triangles by RHS (right angle, common hypotenuse \(OP\), equal radii), hence \(PA=PB\).
Part (b)
Angle \(PAB=\frac{180-44}{2}\) \((=68^\circ)\)	M1	Or angle \(AOB=360-90-90-44\) \((=136^\circ)\) from quadrilateral \(OAPB\).
Angle \(AQB=68^\circ\)	A1	From \(136\div2\) on the alternative route.
Reasons: base angles of an isosceles triangle are equal (\(PA=PB\)) and the alternate segment theorem	B1	Alternative route: angles in a quadrilateral add up to \(360^\circ\) (with tangent and radius at \(90^\circ\)) and the angle at the centre is twice the angle at the circumference. Both reasons needed.
Part (c)
\(112^\circ\)	B1	Allow \(180-\) their (b).
Opposite angles of a cyclic quadrilateral (\(AQBR\)) add up to \(180^\circ\)	B1	The words cyclic quadrilateral and opposite angles are both needed.
Part (d)(i)
Gradient of \(CD=\frac{23-15}{26-10}=\frac{1}{2}\)	M1	From \(C(10,15)\) and \(D(26,23)\) read from the grid.
Gradient \(-2\) used with the midpoint \((18,19)\)	M1	e.g. \(19=-2(18)+c\) or \(y-19=-2(x-18)\).
\(y=-2x+55\)	A1	Accept any equivalent, e.g. \(2x+y=55\).
Part (d)(ii)
The perpendicular from the centre to a chord bisects the chord, so the perpendicular bisector of the chord \(CD\) passes through the centre	B1	Accept the check \(-2(20)+55=15\) only together with the circle fact.
Part (e)(i)
\(6.4\times\left(\frac{120}{24}\right)^3\) or \(6.4\times125\)	M1	Linear scale factor 5 must be cubed; \(6.4\times5\) or \(6.4\times25\) scores M0.
800 litres	A1
Part (e)(ii)
\(900\times5^2\) \((=22\,500\text{ cm}^2)\)	M1	Or \(0.09\times25\) after converting 900 cm\(^2\) to 0.09 m\(^2\).
\(2.25\) m\(^2\)	A1	22 500 cm\(^2\) not converted scores M1 A0.
Part (f)
\(\pi\times10^2-\pi\times2^2\) \((=96\pi=301.5\ldots)\)	M1
\(\pi\times12.5^2-\pi\times2^2\) \((=152.25\pi=478.3\ldots)\)	M1
\(\frac{478.3-301.5}{301.5}\times100\) or \(\frac{152.25}{96}\)	M1	Dependent on at least one previous M1.
\(58.6\)%	A1	Accept 58.5 to 58.6 (58.59...). Ignoring the fountain gives 56.25% and scores M0 M0 M1 A0.

Answers: (a) \(PA=PB\) · (b) \(68^\circ\) · (c) \(112^\circ\) · (d)(i) \(y=-2x+55\) · (e)(i) 800 litres · (e)(ii) \(2.25\) m\(^2\) · (f) \(58.6\)%

Exam tip (b) For the alternate segment theorem, find the angle between the tangent and the chord first, then look for the angle on the other side of that chord, here at Q.

More 4.4 questions in the question bank

4.2 Pythagoras' theorem and trigonometry · 4.3 Area, surface area and volumemerged group

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Siting a wind turbine: depression, bearings and swept area

4.2 Pythagoras' theorem and trigonometry4.3 Area, surface area and volume1.3 Ratio, proportion and degree of accuracy3.2 Graphs, gradients and real-life graphsFoundation

A wind turbine stands on level ground. Its tower is vertical, with the hub \(H\) at the top, 100 m above the foot \(F\) of the tower. Each of its three blades is 48 m long.

A technician at the hub looks down at a substation \(S\) on the ground. The angle of depression of \(S\) from \(H\) is \(38^\circ\), as shown in Figure 1.

Figure 1: the turbine and the substation S (not to scale). One blade points vertically down to its tip D; the dashed circle is the path of the blade tips and the other two blades are not shown.

(a)

Show that the distance \(FS\) is 128 m, correct to the nearest metre.

PROOFNotes

Show solution

\(\tan38^\circ=\frac{100}{FS}\) or \(FS=\frac{100}{\tan38^\circ}\)	M1	Angle \(HSF=38^\circ\) (alternate angles) may be used, or \(FS=100\tan52^\circ\).
\(127.99\ldots\) (m), which is 128 m to the nearest metre	A1	cso. A value such as 127.99 must be seen before rounding.

Answer: \(\frac{100}{\tan38^\circ}=127.99\ldots\approx128\) m

(2)

(b)(i)

One blade points vertically downwards, so its tip \(D\) is directly below \(H\).

Work out the angle of elevation of \(D\) from \(S\). Give your answer correct to 1 decimal place.

ROUTINENotes

Show solution

\(\tan^{-1}\left(\frac{52}{128}\right)\)	M1	\(100-48=52\); allow 127.99… for 128.
\(22.1^\circ\)	A1	22.10… or 22.11…

Answer: \(22.1^\circ\)

(2)

(b)(ii)

A survey drone flies in a straight line from \(S\) to \(D\).

Work out the distance the drone flies. Give your answer correct to 3 significant figures.

ROUTINENotes

Show solution

\(\sqrt{128^2+52^2}\)	M1	Or \(\frac{52}{\sin22.11\ldots^\circ}\) using (b)(i).
\(138\) m	A1	138.15…

Answer: \(138\) m

(2)

(c)

As the blades turn, they sweep out a circle of radius 48 m. The area of this circle is called the swept area.

Work out the swept area. Give your answer correct to 3 significant figures.

ROUTINENotes

Show solution

\(\pi\times48^2\)	M1	\(\pi\times96^2\) or \(2\pi\times48\) scores M0.
\(7240\) m\(^2\)	A1	7238.2…

Answer: \(7240\) m\(^2\)

(2)

(d)

The blade length is 48 m, correct to the nearest metre.

Write down the error interval for the blade length, \(L\) metres. Explain why a swept area of 7238.2 m\(^2\) claims more accuracy than the data allows.

INTERPRETNotes

Show solution

\(47.5\le L\lt48.5\)	B1	Both inequality signs correct.
\(L\) could be anything in this interval, so the swept area could be anywhere from about 7090 m\(^2\) to about 7390 m\(^2\); giving it to 1 decimal place is not justified	B1	Must link the uncertainty in \(L\) to the area. Numerical bounds are not required.

Answer: \(47.5\le L\lt48.5\)

(2)

(e)(i)

Figure 2 shows how the swept area, \(A\) m\(^2\), depends on the blade length, \(L\) m.

Use the graph to estimate the blade length that gives a swept area of 10 000 m\(^2\).

Figure 2: swept area A against blade length L

GRAPHNotes

Show solution

\(56\) to \(57\) m

56.4…

Answer: about \(56.5\) m

(1)

(e)(ii)

Omar says,

“The swept area is directly proportional to the blade length.”

Use Figure 2 to explain why Omar is wrong.

INTERPRETNotes

Show solution

Correct readings for a length and double that length, e.g. \(L=20\): about 1260 and \(L=40\): about 5030	B1	Accept readings within half a small square.
Doubling \(L\) multiplies \(A\) by about 4, not 2; or the graph is a curve, not a straight line through the origin	B1	A conclusion is needed.

(2)

(f)

The turbine is refitted with longer blades, each 60 m long.

Work out the percentage increase in the swept area.

MULTINotes

Show solution

\(\pi\times60^2\) \((=11\,309.7\ldots)\)	M1
\(\frac{11\,309.7\ldots-7238.2\ldots}{7238.2\ldots}\times100\) or \(\frac{60^2}{48^2}=1.5625\)	M1	Allow ECF from (c).
\(56.25\)%	A1	Accept 56.2 or 56.3. The answer 25% (from \(\frac{60-48}{48}\)) scores M0 M0 A0.

Answer: \(56.25\)%

(3)

(g)(i)

The wind farm has two more turbines, \(B\) and \(C\). Figure 3 shows their positions, seen from above, relative to turbine \(A\).

The bearing of \(A\) from \(B\) is \(238^\circ\). Write down the bearing of \(B\) from \(A\).

Figure 3: turbines A, B and C seen from above (not to scale)

ROUTINENotes

Show solution

\(058^\circ\)

Accept \(58^\circ\).

Answer: \(058^\circ\)

(1)

(g)(ii)

\(B\) is 640 m from \(A\). \(C\) is 480 m from \(A\) on a bearing of \(163^\circ\).

Work out the distance \(BC\). Give your answer correct to 3 significant figures.

Higher onlyMULTINotes

Show solution

Angle \(BAC=163-58=105^\circ\)	B1	Allow ECF from (g)(i).
\(BC^2=640^2+480^2-2\times640\times480\times\cos105^\circ\)	M1	Cosine rule with their angle.
\(894\) m	A1	893.87…

Answer: \(894\) m

(3)

Show full solution

Working	Mark	Notes
Part (a)
\(\tan38^\circ=\frac{100}{FS}\) or \(FS=\frac{100}{\tan38^\circ}\)	M1	Angle \(HSF=38^\circ\) (alternate angles) may be used, or \(FS=100\tan52^\circ\).
\(127.99\ldots\) (m), which is 128 m to the nearest metre	A1	cso. A value such as 127.99 must be seen before rounding.
Part (b)(i)
\(\tan^{-1}\left(\frac{52}{128}\right)\)	M1	\(100-48=52\); allow 127.99… for 128.
\(22.1^\circ\)	A1	22.10… or 22.11…
Part (b)(ii)
\(\sqrt{128^2+52^2}\)	M1	Or \(\frac{52}{\sin22.11\ldots^\circ}\) using (b)(i).
\(138\) m	A1	138.15…
Part (c)
\(\pi\times48^2\)	M1	\(\pi\times96^2\) or \(2\pi\times48\) scores M0.
\(7240\) m\(^2\)	A1	7238.2…
Part (d)
\(47.5\le L\lt48.5\)	B1	Both inequality signs correct.
\(L\) could be anything in this interval, so the swept area could be anywhere from about 7090 m\(^2\) to about 7390 m\(^2\); giving it to 1 decimal place is not justified	B1	Must link the uncertainty in \(L\) to the area. Numerical bounds are not required.
Part (e)(i)
\(56\) to \(57\) m	B1	56.4…
Part (e)(ii)
Correct readings for a length and double that length, e.g. \(L=20\): about 1260 and \(L=40\): about 5030	B1	Accept readings within half a small square.
Doubling \(L\) multiplies \(A\) by about 4, not 2; or the graph is a curve, not a straight line through the origin	B1	A conclusion is needed.
Part (f)
\(\pi\times60^2\) \((=11\,309.7\ldots)\)	M1
\(\frac{11\,309.7\ldots-7238.2\ldots}{7238.2\ldots}\times100\) or \(\frac{60^2}{48^2}=1.5625\)	M1	Allow ECF from (c).
\(56.25\)%	A1	Accept 56.2 or 56.3. The answer 25% (from \(\frac{60-48}{48}\)) scores M0 M0 A0.
Part (g)(i)
\(058^\circ\)	B1	Accept \(58^\circ\).
Part (g)(ii)
Angle \(BAC=163-58=105^\circ\)	B1	Allow ECF from (g)(i).
\(BC^2=640^2+480^2-2\times640\times480\times\cos105^\circ\)	M1	Cosine rule with their angle.
\(894\) m	A1	893.87…

Answers: (a) \(\frac{100}{\tan38^\circ}=127.99\ldots\approx128\) m · (b)(i) \(22.1^\circ\) · (b)(ii) \(138\) m · (c) \(7240\) m\(^2\) · (d) \(47.5\le L\lt48.5\) · (e)(i) about \(56.5\) m · (f) \(56.25\)% · (g)(i) \(058^\circ\) · (g)(ii) \(894\) m

Common error (b)(ii) Keep the full value of an angle from your calculator if you reuse it. Rounding 22.11...° to 22° before dividing gives 138.8 m, which is not correct to 3 significant figures.

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A grain silo: volume, surface area and filling

4.3 Area, surface area and volume4.2 Pythagoras' theorem and trigonometry3.2 Graphs, gradients and real-life graphs1.3 Ratio, proportion and degree of accuracyFoundation

A farm stores wheat in a metal silo. The silo is a cylinder of radius 2.5 m and height 9 m, standing on top of a hopper in the shape of a cone. The cone has radius 2.5 m and vertical height 3 m, with its point at the bottom, as shown in Figure 1.

Figure 1: the silo, a cylinder on top of a cone-shaped hopper (not to scale)

(a)

Show that the volume of the cylinder is 177 m\(^3\), correct to 3 significant figures.

PROOFNotes

Show solution

\(\pi\times2.5^2\times9\)	M1
\(176.7\ldots\) (m\(^3\)), which is 177 to 3 s.f.	A1	cso. Accept \(56.25\pi\) evaluated; a value such as 176.7 must be seen.

Answer: \(56.25\pi=176.7\ldots\approx177\) m\(^3\)

(2)

(b)

Work out the total capacity of the silo in litres. Give your answer correct to 3 significant figures.

MULTINotes

Show solution

\(\frac{1}{3}\times\pi\times2.5^2\times3\) \((=19.63\ldots)\)	M1	Volume of the cone.
\((176.7\ldots+19.63\ldots)\times1000\)	M1	Their total volume in m\(^3\) multiplied by 1000.
\(196\,000\) litres	A1	196 349.5…; accept 196 300 to 196 400 before rounding. Using the rounded 177 for the cylinder gives 196 635…, which rounds to 197 000 and scores A0.

Answer: \(196\,000\) litres

(3)

(c)

Work out the slant height of the cone. Give your answer correct to 3 significant figures.

ROUTINENotes

Show solution

\(\sqrt{2.5^2+3^2}\)	M1
\(3.91\) m	A1	3.905...

Answer: \(3.91\) m

(2)

(d)

The curved surface of the cylinder and the curved surface of the cone are made from sheet metal.

Work out the total area of sheet metal needed. Give your answer correct to 3 significant figures.

MULTINotes

Show solution

\(2\times\pi\times2.5\times9\) \((=141.3\ldots)\)	M1
\(\pi\times2.5\times3.905\ldots\) \((=30.67\ldots)\)	M1	Allow ECF from (c). Using the vertical height 3 scores M0.
\(172\) m\(^2\)	A1	172.04...

Answer: \(172\) m\(^2\)

(3)

(e)

The sheet metal is to be painted. One tin of paint covers 50 m\(^2\) and costs £38.

Lena works out \(172\div50=3.44\) and says,

“3.44 rounds to 3, so the paint will cost \(3\times\)£38 = £114.”

Comment on Lena's method and work out the correct cost of the paint.

INTERPRETNotes

Show solution

Three tins cover only 150 m\(^2\), less than 172 m\(^2\), so the number of tins must be rounded up, not to the nearest whole number	B1
4 tins, costing £152	B1	Allow ECF from their (d).

Answer: £152 (4 tins)

(2)

(f)(i)

Wheat is poured into the empty silo at a constant rate of 1.5 m\(^3\) per minute. Figure 2 shows how the depth of wheat, \(d\) metres, measured from the point of the cone, changes with time, \(t\) minutes.

Use the graph to find the depth of wheat after 1 hour, and the time taken for the depth to reach 9 m.

Figure 2: depth of grain in the silo against time while it is filled from empty

GRAPHNotes

Show solution

\(6.4\) to \(6.8\) m	B1	6.58...
\(90\) to \(93\) minutes	B1	91.6...

Answer: about \(6.6\) m; about \(92\) minutes

(2)

(f)(ii)

The graph is curved for the first 13 minutes and is a straight line after that.

Explain why, with reference to the shape of the silo.

INTERPRETNotes

Show solution

In the cone the circle at the surface of the wheat gets wider as the depth increases, so each cubic metre raises the depth less and less: the gradient decreases	B1
In the cylinder the cross-section is the same at every depth, so the depth increases at a constant rate: a straight line	B1	Accept constant gradient.

(2)

(g)

The hopper is replaced by a frustum of a cone, shown in Figure 3. The frustum has top radius 2.5 m, bottom radius 0.5 m and vertical height 2.4 m.

Work out the volume of the frustum. Give your answer correct to 3 significant figures.

Figure 3: the new hopper, a frustum of a cone (not to scale)

Higher onlyMULTINotes

Show solution

Height of the full cone \(=3\) m (or small cone \(=0.6\) m), e.g. from \(\frac{h}{h+2.4}=\frac{0.5}{2.5}\)	M1	Similar triangles or scale factor 5.
\(\frac{1}{3}\pi\times2.5^2\times3-\frac{1}{3}\pi\times0.5^2\times0.6\)	M1	With their heights.
\(19.5\) m\(^3\)	A1	19.47...

Answer: \(19.5\) m\(^3\)

(3)

Show full solution

Working	Mark	Notes
Part (a)
\(\pi\times2.5^2\times9\)	M1
\(176.7\ldots\) (m\(^3\)), which is 177 to 3 s.f.	A1	cso. Accept \(56.25\pi\) evaluated; a value such as 176.7 must be seen.
Part (b)
\(\frac{1}{3}\times\pi\times2.5^2\times3\) \((=19.63\ldots)\)	M1	Volume of the cone.
\((176.7\ldots+19.63\ldots)\times1000\)	M1	Their total volume in m\(^3\) multiplied by 1000.
\(196\,000\) litres	A1	196 349.5…; accept 196 300 to 196 400 before rounding. Using the rounded 177 for the cylinder gives 196 635…, which rounds to 197 000 and scores A0.
Part (c)
\(\sqrt{2.5^2+3^2}\)	M1
\(3.91\) m	A1	3.905...
Part (d)
\(2\times\pi\times2.5\times9\) \((=141.3\ldots)\)	M1
\(\pi\times2.5\times3.905\ldots\) \((=30.67\ldots)\)	M1	Allow ECF from (c). Using the vertical height 3 scores M0.
\(172\) m\(^2\)	A1	172.04...
Part (e)
Three tins cover only 150 m\(^2\), less than 172 m\(^2\), so the number of tins must be rounded up, not to the nearest whole number	B1
4 tins, costing £152	B1	Allow ECF from their (d).
Part (f)(i)
\(6.4\) to \(6.8\) m	B1	6.58...
\(90\) to \(93\) minutes	B1	91.6...
Part (f)(ii)
In the cone the circle at the surface of the wheat gets wider as the depth increases, so each cubic metre raises the depth less and less: the gradient decreases	B1
In the cylinder the cross-section is the same at every depth, so the depth increases at a constant rate: a straight line	B1	Accept constant gradient.
Part (g)
Height of the full cone \(=3\) m (or small cone \(=0.6\) m), e.g. from \(\frac{h}{h+2.4}=\frac{0.5}{2.5}\)	M1	Similar triangles or scale factor 5.
\(\frac{1}{3}\pi\times2.5^2\times3-\frac{1}{3}\pi\times0.5^2\times0.6\)	M1	With their heights.
\(19.5\) m\(^3\)	A1	19.47...

Answers: (a) \(56.25\pi=176.7\ldots\approx177\) m\(^3\) · (b) \(196\,000\) litres · (c) \(3.91\) m · (d) \(172\) m\(^2\) · (e) £152 (4 tins) · (f)(i) about \(6.6\) m; about \(92\) minutes · (g) \(19.5\) m\(^3\)

Common error (d) The curved surface of a cone is \(\pi rl\), where \(l\) is the slant height, not the vertical height.

More 4.3 questions in the question bank

Topic 5 · Vectors and transformation geometry

● home subchapter · ○ linked subchapter · last column: links to other topics

Setup	5.1	5.2	Other
A sailing regatta: column vectors on a race course	●	○	4.2 3.2
Designing a patchwork quilt block: transformations and fabric areas	○	●	4.3 1.3

5.1 Vectors and vector geometry · 5.2 Transformations described in wordsmerged group

+ Add to test✓ In your testCC-EIM-51-52-120 marks~20 min

A sailing regatta: column vectors on a race course

5.1 Vectors and vector geometry4.2 Pythagoras' theorem and trigonometry3.2 Graphs, gradients and real-life graphs5.2 Transformations described in wordsFoundation

A sailing club sets out a triangular race course. Figure 1 shows the course on a grid of squares. Each square represents 100 m by 100 m. The positive \(x\)-direction is due east and the positive \(y\)-direction is due north.

The boats start at the committee boat \(S\). They sail from \(S\) to buoy \(A\), then from \(A\) to buoy \(B\), and then from \(B\) back to \(S\).

\(\overrightarrow{SA}=\mathbf{p}\) and \(\overrightarrow{AB}=\mathbf{q}\).

Figure 1: the race course. Each square represents 100 m by 100 m.

(a)

Use Figure 1 to write down \(\mathbf{p}\) and \(\mathbf{q}\) as column vectors.

GRAPHNotes

Show solution

\(\mathbf{p}=\begin{pmatrix}6\\8\end{pmatrix}\)	B1	Coordinates such as \((6,8)\) score B0.
\(\mathbf{q}=\begin{pmatrix}6\\-4\end{pmatrix}\)	B1

Answer: \(\mathbf{p}=\begin{pmatrix}6\\8\end{pmatrix}\), \(\mathbf{q}=\begin{pmatrix}6\\-4\end{pmatrix}\)

(2)

(b)

Work out \(\overrightarrow{BS}\) as a column vector.

ROUTINENotes

Show solution

\(\overrightarrow{BS}=-(\mathbf{p}+\mathbf{q})\) or \(\mathbf{p}+\mathbf{q}=\begin{pmatrix}12\\4\end{pmatrix}\) or \(\begin{pmatrix}2-14\\1-5\end{pmatrix}\)	M1	Allow ECF from (a).
\(\begin{pmatrix}-12\\-4\end{pmatrix}\)	A1	\(\begin{pmatrix}12\\4\end{pmatrix}\) as the final answer scores M1 A0.

Answer: \(\begin{pmatrix}-12\\-4\end{pmatrix}\)

(2)

(c)

A camera boat is anchored at the midpoint \(M\) of \(AB\).

Find the coordinates of \(M\).

ROUTINENotes

Show solution

\(\frac{8+14}{2}\) or \(\frac{9+5}{2}\)	M1	A correct method for either coordinate, using \(A(8,9)\) and \(B(14,5)\).
\((11,7)\)	A1

Answer: \((11,7)\)

(2)

(d)

Show that the first leg of the course, from \(S\) to \(A\), is 1 km long.

PROOFNotes

Show solution

\(6^2+8^2\) \((=100)\)	M1	Pythagoras with the two components of \(\mathbf{p}\); allow ECF from (a).
\(\sqrt{100}=10\) squares, and \(10\times100\) \((=1000\text{ m})\)	M1	Square root and conversion to metres.
\(1000\text{ m}=1\text{ km}\)	A1	cso. The conversion to kilometres must be stated.

Answer: \(\sqrt{6^2+8^2}=10\) squares \(=1000\) m \(=1\) km

(3)

(e)

Work out the bearing of buoy \(A\) from \(S\).

Give your answer to the nearest degree.

MULTINotes

Show solution

\(\tan^{-1}\left(\frac{6}{8}\right)\) \((=36.86\ldots)\)	M1	Or \(\sin^{-1}\left(\frac{6}{10}\right)\) or \(\cos^{-1}\left(\frac{8}{10}\right)\). \(\tan^{-1}\left(\frac{8}{6}\right)=53.1\) is the angle from east and scores M0 unless \(90-53.1\) follows.
\(037^\circ\)	A1	Accept 37°.

Answer: \(037^\circ\)

(2)

(f)

Before the second race the wind changes direction. The race officer moves the committee boat and both buoys by the translation \(\begin{pmatrix}-2\\3\end{pmatrix}\).

Megan says, “The vectors \(\mathbf{p}\) and \(\mathbf{q}\) for the legs of the course will be different now.”

Explain why Megan is wrong.

INTERPRETNotes

Show solution

a translation moves every point (\(S\), \(A\) and \(B\)) by the same vector \(\begin{pmatrix}-2\\3\end{pmatrix}\)	B1	Or the new positions found, e.g. \(S'(0,4)\), \(A'(6,12)\), \(B'(12,8)\).
so each leg keeps the same length and direction, e.g. \(\overrightarrow{S'A'}=\begin{pmatrix}6-0\\12-4\end{pmatrix}=\begin{pmatrix}6\\8\end{pmatrix}=\mathbf{p}\)	B1	A conclusion that \(\mathbf{p}\) and \(\mathbf{q}\) are unchanged is needed, either in words or from a correct check of a new vector.

Answer: A translation moves every point by the same vector, so each leg keeps its length and direction and \(\mathbf{p}\) and \(\mathbf{q}\) do not change.

(2)

(g)(i)

For the third race the course is back in the position shown in Figure 1, but buoy \(B\) is replaced by a new buoy \(C\), where \(\overrightarrow{SC}=\frac12\mathbf{p}+\mathbf{q}\).

Find the coordinates of \(C\).

MULTINotes

Show solution

\(\frac12\mathbf{p}=\begin{pmatrix}3\\4\end{pmatrix}\) or \(\begin{pmatrix}3\\4\end{pmatrix}+\begin{pmatrix}6\\-4\end{pmatrix}\)	M1	Allow ECF from (a).
\(\overrightarrow{SC}=\begin{pmatrix}9\\0\end{pmatrix}\)	A1
\((11,1)\)	B1	ft: \((2,1)\) plus their \(\overrightarrow{SC}\). \((9,0)\) as the answer scores B0.

Answer: \((11,1)\)

(3)

(g)(ii)

Write down the bearing of \(S\) from \(C\).

ROUTINENotes

Show solution

\(270^\circ\)

ft their \(C\) from (g)(i): \(270^\circ\) only if their \(C\) has the same \(y\)-coordinate as \(S\).

Answer: \(270^\circ\)

(1)

(h)

A small turning buoy \(D\) is placed so that \(\overrightarrow{SD}=\frac23\mathbf{p}+\frac13\mathbf{q}\). \(M\) is the midpoint of \(AB\), as in part (c).

Use vectors to show that \(S\), \(D\) and \(M\) lie on a straight line.

Higher onlyPROOFNotes

Show solution

\(\overrightarrow{SM}=\mathbf{p}+\frac12\mathbf{q}\)	M1	Route \(S\to A\to M\). Or in columns: \(\overrightarrow{SM}=\begin{pmatrix}9\\6\end{pmatrix}\) and \(\overrightarrow{SD}=\begin{pmatrix}6\\4\end{pmatrix}\).
\(\overrightarrow{DM}=\overrightarrow{SM}-\overrightarrow{SD}=\frac13\mathbf{p}+\frac16\mathbf{q}\)	M1	Or \(\overrightarrow{DM}=\begin{pmatrix}3\\2\end{pmatrix}\); or \(\overrightarrow{SM}=\frac32\overrightarrow{SD}\) seen.
\(\overrightarrow{DM}=\frac12\overrightarrow{SD}\), so \(SD\) and \(DM\) are parallel and share the point \(D\): \(S\), \(D\) and \(M\) are on a straight line	A1	cso. Both the multiple and the shared point must be stated.

Answer: \(\overrightarrow{DM}=\frac13\mathbf{p}+\frac16\mathbf{q}=\frac12\overrightarrow{SD}\), parallel with the common point \(D\), so collinear

(3)

Show full solution

Working	Mark	Notes
Part (a)
\(\mathbf{p}=\begin{pmatrix}6\\8\end{pmatrix}\)	B1	Coordinates such as \((6,8)\) score B0.
\(\mathbf{q}=\begin{pmatrix}6\\-4\end{pmatrix}\)	B1
Part (b)
\(\overrightarrow{BS}=-(\mathbf{p}+\mathbf{q})\) or \(\mathbf{p}+\mathbf{q}=\begin{pmatrix}12\\4\end{pmatrix}\) or \(\begin{pmatrix}2-14\\1-5\end{pmatrix}\)	M1	Allow ECF from (a).
\(\begin{pmatrix}-12\\-4\end{pmatrix}\)	A1	\(\begin{pmatrix}12\\4\end{pmatrix}\) as the final answer scores M1 A0.
Part (c)
\(\frac{8+14}{2}\) or \(\frac{9+5}{2}\)	M1	A correct method for either coordinate, using \(A(8,9)\) and \(B(14,5)\).
\((11,7)\)	A1
Part (d)
\(6^2+8^2\) \((=100)\)	M1	Pythagoras with the two components of \(\mathbf{p}\); allow ECF from (a).
\(\sqrt{100}=10\) squares, and \(10\times100\) \((=1000\text{ m})\)	M1	Square root and conversion to metres.
\(1000\text{ m}=1\text{ km}\)	A1	cso. The conversion to kilometres must be stated.
Part (e)
\(\tan^{-1}\left(\frac{6}{8}\right)\) \((=36.86\ldots)\)	M1	Or \(\sin^{-1}\left(\frac{6}{10}\right)\) or \(\cos^{-1}\left(\frac{8}{10}\right)\). \(\tan^{-1}\left(\frac{8}{6}\right)=53.1\) is the angle from east and scores M0 unless \(90-53.1\) follows.
\(037^\circ\)	A1	Accept 37°.
Part (f)
a translation moves every point (\(S\), \(A\) and \(B\)) by the same vector \(\begin{pmatrix}-2\\3\end{pmatrix}\)	B1	Or the new positions found, e.g. \(S'(0,4)\), \(A'(6,12)\), \(B'(12,8)\).
so each leg keeps the same length and direction, e.g. \(\overrightarrow{S'A'}=\begin{pmatrix}6-0\\12-4\end{pmatrix}=\begin{pmatrix}6\\8\end{pmatrix}=\mathbf{p}\)	B1	A conclusion that \(\mathbf{p}\) and \(\mathbf{q}\) are unchanged is needed, either in words or from a correct check of a new vector.
Part (g)(i)
\(\frac12\mathbf{p}=\begin{pmatrix}3\\4\end{pmatrix}\) or \(\begin{pmatrix}3\\4\end{pmatrix}+\begin{pmatrix}6\\-4\end{pmatrix}\)	M1	Allow ECF from (a).
\(\overrightarrow{SC}=\begin{pmatrix}9\\0\end{pmatrix}\)	A1
\((11,1)\)	B1	ft: \((2,1)\) plus their \(\overrightarrow{SC}\). \((9,0)\) as the answer scores B0.
Part (g)(ii)
\(270^\circ\)	B1	ft their \(C\) from (g)(i): \(270^\circ\) only if their \(C\) has the same \(y\)-coordinate as \(S\).
Part (h)
\(\overrightarrow{SM}=\mathbf{p}+\frac12\mathbf{q}\)	M1	Route \(S\to A\to M\). Or in columns: \(\overrightarrow{SM}=\begin{pmatrix}9\\6\end{pmatrix}\) and \(\overrightarrow{SD}=\begin{pmatrix}6\\4\end{pmatrix}\).
\(\overrightarrow{DM}=\overrightarrow{SM}-\overrightarrow{SD}=\frac13\mathbf{p}+\frac16\mathbf{q}\)	M1	Or \(\overrightarrow{DM}=\begin{pmatrix}3\\2\end{pmatrix}\); or \(\overrightarrow{SM}=\frac32\overrightarrow{SD}\) seen.
\(\overrightarrow{DM}=\frac12\overrightarrow{SD}\), so \(SD\) and \(DM\) are parallel and share the point \(D\): \(S\), \(D\) and \(M\) are on a straight line	A1	cso. Both the multiple and the shared point must be stated.

Answers: (a) \(\mathbf{p}=\begin{pmatrix}6\\8\end{pmatrix}\), \(\mathbf{q}=\begin{pmatrix}6\\-4\end{pmatrix}\) · (b) \(\begin{pmatrix}-12\\-4\end{pmatrix}\) · (c) \((11,7)\) · (d) \(\sqrt{6^2+8^2}=10\) squares \(=1000\) m \(=1\) km · (e) \(037^\circ\) · (f) A translation moves every point by the same vector, so each leg keeps its length and direction and \(\mathbf{p}\) and \(\mathbf{q}\) do not change. · (g)(i) \((11,1)\) · (g)(ii) \(270^\circ\) · (h) \(\overrightarrow{DM}=\frac13\mathbf{p}+\frac16\mathbf{q}=\frac12\overrightarrow{SD}\), parallel with the common point \(D\), so collinear

Common error (b) \(\overrightarrow{BS}\) is the journey from \(B\) to \(S\), so it is end minus start, \(S-B\). The vector \(\mathbf{p}+\mathbf{q}\) goes the other way, from \(S\) to \(B\).

Exam tip (g)(i) A vector says how far and which way, not where you are. To turn \(\overrightarrow{SC}\) into a position, add it to the coordinates of the start point \(S\).

More 5.1 questions in the question bank

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Designing a patchwork quilt block: transformations and fabric areas

5.2 Transformations described in words5.1 Vectors and vector geometry4.3 Area, surface area and volume1.3 Ratio, proportion and degree of accuracyFoundation

A quilt maker designs a patchwork block on a coordinate grid. Figure 2 shows three patches of fabric.

Patch \(A\) is a trapezium with vertices \(K(2,-3)\), \(L(5,-3)\), \(M(5,-2)\) and \(N(2,-1)\). Patches \(B\) and \(C\) are images of patch \(A\) under two different transformations.

Figure 2: patch A and the patches B and C on the design grid

(a)

Describe fully the single transformation that maps patch \(A\) onto patch \(B\).

CONCEPTNotes

Show solution

reflection	B1	Any mention of a second transformation scores B0 B0.
in the line \(y=x\)	B1	Accept “in \(y=x\)”. Do not accept “in the diagonal” without the equation.

Answer: Reflection in the line \(y=x\)

(2)

(b)(i)

Patch \(A\) is rotated \(90^\circ\) clockwise about the origin \(O\) to give patch \(D\).

On Figure 2, draw patch \(D\).

ROUTINENotes

Show solution

2 or 3 vertices correct, or a correct \(90^\circ\) clockwise rotation about a different centre, or a \(90^\circ\) anticlockwise rotation about \(O\)	B1
fully correct: vertices \((-3,-2)\), \((-3,-5)\), \((-2,-5)\), \((-1,-2)\)	B1	Images of \(K\), \(L\), \(M\), \(N\) in that order. The shape must be closed.

Answer: Vertices \((-3,-2)\), \((-3,-5)\), \((-2,-5)\), \((-1,-2)\)

(2)

(b)(ii)

Nadia says, “Patch \(B\) is also a rotation of patch \(A\). It just has a different centre from patch \(D\).”

Explain why Nadia is wrong.

INTERPRETNotes

Show solution

\(B\) is a mirror image of \(A\): going round \(A\) from \(K\) to \(L\) to \(M\) to \(N\) is anticlockwise, but going round the matching vertices of \(B\) is clockwise	B1	Or: \(B\) is \(A\) flipped over, e.g. the slanted side is on the opposite side of the shape. Or the matching vertices \((-3,2)\), \((-3,5)\), \((-2,5)\), \((-1,2)\) identified and compared.
a rotation only turns a shape and never flips it, so no rotation (about any centre) can map \(A\) onto \(B\)	B1	A conclusion about every possible centre is needed. “\(B\) is a reflection” alone, with no reason, scores B0 B0.

(2)

(c)

Describe fully the single transformation that maps patch \(A\) onto patch \(C\).

GRAPHNotes

Show solution

enlargement	B1	Any mention of a second transformation scores B0 B0 B0.
scale factor 2	B1	Accept “×2”. For example \(KL=3\) and the matching side of \(C\) is 6.
centre \((0,-7)\)	B1	From lines through matching vertices, e.g. \(K(2,-3)\) and \((4,1)\), \(L(5,-3)\) and \((10,1)\).

Answer: Enlargement, scale factor 2, centre \((0,-7)\)

(3)

(d)

On the design grid, each unit represents 4 cm.

Show that the area of fabric needed for patch \(A\) is 72 cm\(^2\).

PROOFNotes

Show solution

\(\frac12(2+1)\times3\) \((=4.5)\) square units, or \(3\times1+\frac12\times3\times1\)	M1	Trapezium with parallel sides \(KN=2\) and \(LM=1\) a distance 3 apart. Or with lengths in cm: \(\frac12(8+4)\times12\).
\(4.5\times4^2\) or \(4.5\times16\)	M1	Using 1 square unit \(=16\) cm\(^2\). \(4.5\times4\) scores M0. Implied by the cm method in the first row.
\(=72\) (cm\(^2\))	A1	cso. Full working must be seen; 72 alone scores no marks.

Answer: \(\frac12(2+1)\times3=4.5\) square units; \(4.5\times16=72\) cm\(^2\)

(3)

(e)

Work out the area of fabric needed for patch \(C\).

MULTINotes

Show solution

\(72\times2^2\)	M1	Allow ECF from their scale factor in (c). Or the area of \(C\) worked out directly: \(\frac12(4+2)\times6=18\) square units \(\times16\). \(72\times2\) scores M0.
288 (cm\(^2\))	A1

Answer: 288 cm\(^2\)

(2)

(f)

For a cushion cover, the quilt maker uses the same design at a smaller size. Patch \(A\) now needs 40.5 cm\(^2\) of fabric.

Work out the length that each grid unit now represents.

MULTINotes

Show solution

\(\frac{40.5}{4.5}\) \((=9)\) and \(\sqrt{9}\)	M1	Or \(\frac{40.5}{72}=\frac{9}{16}\), so the length scale factor is \(\frac34\) and \(\frac34\times4\). Allow ECF from 4.5 in (d).
3 cm	A1	9 cm (no square root) scores M0 A0.

Answer: 3 cm

(2)

(g)(i)

In the finished quilt, the quilt maker repeats patch \(A\) by translating it by the vector \(\mathbf{a}=\begin{pmatrix}3\\1\end{pmatrix}\) and then translating the image by the vector \(\mathbf{b}=\begin{pmatrix}4\\3\end{pmatrix}\).

Write down, as a column vector, the single translation that has the same effect.

ROUTINENotes

Show solution

\(\begin{pmatrix}7\\4\end{pmatrix}\)

Accept \(\mathbf{a}+\mathbf{b}=\begin{pmatrix}7\\4\end{pmatrix}\). Coordinates \((7,4)\) score B0.

Answer: \(\begin{pmatrix}7\\4\end{pmatrix}\)

(1)

(g)(ii)

Each grid unit represents 4 cm, as in part (d).

Work out the distance, in cm, that each point of patch \(A\) moves under the single translation in part (g)(i).

Give your answer correct to 3 significant figures.

Higher onlyMULTINotes

Show solution

\(\sqrt{7^2+4^2}\) \((=\sqrt{65}=8.06\ldots)\)	M1	Allow ECF from (g)(i). Or with lengths in cm: \(\sqrt{28^2+16^2}\).
\(\sqrt{65}\times4\) \((=32.24\ldots)\)	M1	Implied by the cm method in the first row.
32.2 (cm)	A1	32.25 or 32.24 alone scores A0; the answer must be given to 3 s.f.

Answer: 32.2 cm

(3)

Show full solution

Working	Mark	Notes
Part (a)
reflection	B1	Any mention of a second transformation scores B0 B0.
in the line \(y=x\)	B1	Accept “in \(y=x\)”. Do not accept “in the diagonal” without the equation.
Part (b)(i)
2 or 3 vertices correct, or a correct \(90^\circ\) clockwise rotation about a different centre, or a \(90^\circ\) anticlockwise rotation about \(O\)	B1
fully correct: vertices \((-3,-2)\), \((-3,-5)\), \((-2,-5)\), \((-1,-2)\)	B1	Images of \(K\), \(L\), \(M\), \(N\) in that order. The shape must be closed.
Part (b)(ii)
\(B\) is a mirror image of \(A\): going round \(A\) from \(K\) to \(L\) to \(M\) to \(N\) is anticlockwise, but going round the matching vertices of \(B\) is clockwise	B1	Or: \(B\) is \(A\) flipped over, e.g. the slanted side is on the opposite side of the shape. Or the matching vertices \((-3,2)\), \((-3,5)\), \((-2,5)\), \((-1,2)\) identified and compared.
a rotation only turns a shape and never flips it, so no rotation (about any centre) can map \(A\) onto \(B\)	B1	A conclusion about every possible centre is needed. “\(B\) is a reflection” alone, with no reason, scores B0 B0.
Part (c)
enlargement	B1	Any mention of a second transformation scores B0 B0 B0.
scale factor 2	B1	Accept “×2”. For example \(KL=3\) and the matching side of \(C\) is 6.
centre \((0,-7)\)	B1	From lines through matching vertices, e.g. \(K(2,-3)\) and \((4,1)\), \(L(5,-3)\) and \((10,1)\).
Part (d)
\(\frac12(2+1)\times3\) \((=4.5)\) square units, or \(3\times1+\frac12\times3\times1\)	M1	Trapezium with parallel sides \(KN=2\) and \(LM=1\) a distance 3 apart. Or with lengths in cm: \(\frac12(8+4)\times12\).
\(4.5\times4^2\) or \(4.5\times16\)	M1	Using 1 square unit \(=16\) cm\(^2\). \(4.5\times4\) scores M0. Implied by the cm method in the first row.
\(=72\) (cm\(^2\))	A1	cso. Full working must be seen; 72 alone scores no marks.
Part (e)
\(72\times2^2\)	M1	Allow ECF from their scale factor in (c). Or the area of \(C\) worked out directly: \(\frac12(4+2)\times6=18\) square units \(\times16\). \(72\times2\) scores M0.
288 (cm\(^2\))	A1
Part (f)
\(\frac{40.5}{4.5}\) \((=9)\) and \(\sqrt{9}\)	M1	Or \(\frac{40.5}{72}=\frac{9}{16}\), so the length scale factor is \(\frac34\) and \(\frac34\times4\). Allow ECF from 4.5 in (d).
3 cm	A1	9 cm (no square root) scores M0 A0.
Part (g)(i)
\(\begin{pmatrix}7\\4\end{pmatrix}\)	B1	Accept \(\mathbf{a}+\mathbf{b}=\begin{pmatrix}7\\4\end{pmatrix}\). Coordinates \((7,4)\) score B0.
Part (g)(ii)
\(\sqrt{7^2+4^2}\) \((=\sqrt{65}=8.06\ldots)\)	M1	Allow ECF from (g)(i). Or with lengths in cm: \(\sqrt{28^2+16^2}\).
\(\sqrt{65}\times4\) \((=32.24\ldots)\)	M1	Implied by the cm method in the first row.
32.2 (cm)	A1	32.25 or 32.24 alone scores A0; the answer must be given to 3 s.f.

Answers: (a) Reflection in the line \(y=x\) · (b)(i) Vertices \((-3,-2)\), \((-3,-5)\), \((-2,-5)\), \((-1,-2)\) · (c) Enlargement, scale factor 2, centre \((0,-7)\) · (d) \(\frac12(2+1)\times3=4.5\) square units; \(4.5\times16=72\) cm\(^2\) · (e) 288 cm\(^2\) · (f) 3 cm · (g)(i) \(\begin{pmatrix}7\\4\end{pmatrix}\) · (g)(ii) 32.2 cm

Common error (b)(i) For a turn of \(90^\circ\) clockwise about \(O\), the point \((x,y)\) goes to \((y,-x)\). Check one vertex: \(L(5,-3)\) is 5 right and 3 down from \(O\), so after a quarter turn clockwise it is 5 down and 3 left, at \((-3,-5)\).

Exam tip (c) To find the centre of an enlargement, join at least two pairs of matching vertices with straight lines and extend them backwards until they cross. Check with a third pair.

More 5.2 questions in the question bank

Topic 6 · Statistics and probability

● home subchapter · ○ linked subchapter · last column: links to other topics

Setup	6.1	6.2	6.3	6.4	Other
A sprint club: 100 m times as grouped data	●	○			1.1 1.3
Testing smartphone batteries: cumulative frequency and histograms	○	○	●		1.1
A penalty shoot-out: relative frequency and tree diagrams	○	●			1.1 1.4
A music playlist on shuffle: dependent events and conditional probability		○		●	2.3 1.4

6.1 Averages, range and statistical diagrams · 6.3 Cumulative frequency and histogramsmerged group

+ Add to test✓ In your testCC-EIM-61-63-118 marks~20 min

A sprint club: 100 m times as grouped data

6.1 Averages, range and statistical diagrams1.1 Integers, fractions, decimals and percentages1.3 Ratio, proportion and degree of accuracy6.2 Probability of single and combined eventsFoundation

A sprint club records the best 100 m time, \(t\) seconds, of each of its 40 athletes this season. The table shows the results.

Time \(t\) (seconds)	Frequency
\(11.0\lt t\le 11.5\)	3
\(11.5\lt t\le 12.0\)	7
\(12.0\lt t\le 12.5\)	12
\(12.5\lt t\le 13.0\)	10
\(13.0\lt t\le 13.5\)	6
\(13.5\lt t\le 14.0\)	2

(a)

Write down the modal class.

ROUTINENotes

Show solution

\(12.0\lt t\le 12.5\)

Do not accept 12 (the frequency).

Answer: \(12.0\lt t\le 12.5\)

(1)

(b)

Show that the total of frequency \(\times\) midpoint for the 40 athletes is 497.5

PROOFNotes

Show solution

midpoints 11.25, 11.75, 12.25, 12.75, 13.25, 13.75	M1	At least 4 correct midpoints used.
\(3\times11.25+7\times11.75+12\times12.25+10\times12.75+6\times13.25+2\times13.75\)	M1	Products \(33.75+82.25+147+127.5+79.5+27.5\) seen or implied.
\(=497.5\)	A1	cso. All six products, or the full sum, must be seen.

Answer: \(33.75+82.25+147+127.5+79.5+27.5=497.5\)

(3)

(c)

Work out an estimate for the mean time.

Give your answer correct to 3 significant figures.

ROUTINENotes

Show solution

\(\frac{497.5}{40}\) \((=12.4375)\)	M1	Dividing by 6 (the number of classes) scores M0.
12.4 (seconds)	A1	12.4375 or 12.44 alone scores A0.

Answer: 12.4 seconds

(2)

(d)

To enter the regional championship, an athlete needs a time of 12.5 seconds or less.

Work out the percentage of the 40 athletes who can enter.

ROUTINENotes

Show solution

\(\frac{3+7+12}{40}\times100\) or \(\frac{22}{40}\)	M1
55(%)	A1

Answer: 55%

(2)

(e)

On the axes in Figure 3, draw a frequency polygon for the information in the table.

Figure 3: axes for the frequency polygon

GRAPHNotes

Show solution

points plotted at the class midpoints: \((11.25,3)\), \((11.75,7)\), \((12.25,12)\), \((12.75,10)\), \((13.25,6)\), \((13.75,2)\)	B1	At least 5 points correct; points at the class ends score B0.
all 6 points correct and joined in order with straight line segments	B1	Joining the first point to the last, or drawing bars, scores B0 for this mark.

Answer: Points \((11.25,3)\), \((11.75,7)\), \((12.25,12)\), \((12.75,10)\), \((13.25,6)\), \((13.75,2)\) joined with straight lines

(2)

(f)(i)

The coach measures the reaction time at the start, in seconds, of 11 of the athletes. Figure 4 shows each athlete's reaction time and 100 m time, with a line of best fit.

Describe the correlation shown in Figure 4.

Figure 4: reaction time and 100 m time for 11 athletes, with a line of best fit

CONCEPTNotes

Show solution

positive (correlation)

Accept strong positive. Do not accept a description of the relationship without the word positive.

Answer: Positive correlation

(1)

(f)(ii)

Ravi wants to join the club. His reaction time is 0.30 seconds. The coach uses the line of best fit to predict Ravi's 100 m time.

Give two reasons why this prediction may not be reliable.

INTERPRETNotes

Show solution

0.30 s is outside the range of the reaction times (0.13 s to 0.24 s), so this is extrapolation and the pattern may not continue	B1
a second valid reason, e.g. the line is based on only 11 athletes; or the points do not lie exactly on the line, so even inside the range the prediction is only an estimate; or other factors such as fitness also affect the time	B1	Two different reasons are needed for B1 B1.

Answer: It is extrapolation (0.30 s is outside 0.13–0.24 s), and the line is based on only 11 athletes that do not all lie on it

(2)

(g)(i)

Eight new athletes join the club. Each of them has a best time in the class \(12.5\lt t\le 13.0\).

Work out an estimate for the mean time of all 48 athletes.

MULTINotes

Show solution

\(\frac{497.5+8\times12.75}{48}\) \((=\frac{599.5}{48})\)	M1	Or \(\frac{497.5+102}{40+8}\).
12.5 (seconds)	A1	Accept 12.49 or 12.489… An answer from dividing by 40 scores M0 A0.

Answer: 12.5 seconds (12.489…)

(2)

(g)(ii)

One of the 48 athletes is chosen at random.

Find the probability that this athlete's best time is more than 13.0 seconds.

ROUTINENotes

Show solution

\(\frac{8}{48}\)

Accept \(\frac16\) or 0.167 (3 s.f.). \(\frac{8}{40}\) scores B0.

Answer: \(\frac{8}{48}=\frac16\)

(1)

(h)

The reaction times, in seconds, of the 11 athletes in Figure 4, in order, are

0.13 0.14 0.15 0.15 0.16 0.17 0.18 0.19 0.21 0.22 0.24

Find the interquartile range of these reaction times.

Higher onlyROUTINENotes

Show solution

\(Q_1=0.15\) (3rd value) or \(Q_3=0.21\) (9th value)	M1	Positions \(\frac{11+1}{4}=3\) and \(\frac{3(11+1)}{4}=9\) may be seen.
0.06 (seconds)	A1

Answer: 0.06 seconds

(2)

Show full solution

Working	Mark	Notes
Part (a)
\(12.0\lt t\le 12.5\)	B1	Do not accept 12 (the frequency).
Part (b)
midpoints 11.25, 11.75, 12.25, 12.75, 13.25, 13.75	M1	At least 4 correct midpoints used.
\(3\times11.25+7\times11.75+12\times12.25+10\times12.75+6\times13.25+2\times13.75\)	M1	Products \(33.75+82.25+147+127.5+79.5+27.5\) seen or implied.
\(=497.5\)	A1	cso. All six products, or the full sum, must be seen.
Part (c)
\(\frac{497.5}{40}\) \((=12.4375)\)	M1	Dividing by 6 (the number of classes) scores M0.
12.4 (seconds)	A1	12.4375 or 12.44 alone scores A0.
Part (d)
\(\frac{3+7+12}{40}\times100\) or \(\frac{22}{40}\)	M1
55(%)	A1
Part (e)
points plotted at the class midpoints: \((11.25,3)\), \((11.75,7)\), \((12.25,12)\), \((12.75,10)\), \((13.25,6)\), \((13.75,2)\)	B1	At least 5 points correct; points at the class ends score B0.
all 6 points correct and joined in order with straight line segments	B1	Joining the first point to the last, or drawing bars, scores B0 for this mark.
Part (f)(i)
positive (correlation)	B1	Accept strong positive. Do not accept a description of the relationship without the word positive.
Part (f)(ii)
0.30 s is outside the range of the reaction times (0.13 s to 0.24 s), so this is extrapolation and the pattern may not continue	B1
a second valid reason, e.g. the line is based on only 11 athletes; or the points do not lie exactly on the line, so even inside the range the prediction is only an estimate; or other factors such as fitness also affect the time	B1	Two different reasons are needed for B1 B1.
Part (g)(i)
\(\frac{497.5+8\times12.75}{48}\) \((=\frac{599.5}{48})\)	M1	Or \(\frac{497.5+102}{40+8}\).
12.5 (seconds)	A1	Accept 12.49 or 12.489… An answer from dividing by 40 scores M0 A0.
Part (g)(ii)
\(\frac{8}{48}\)	B1	Accept \(\frac16\) or 0.167 (3 s.f.). \(\frac{8}{40}\) scores B0.
Part (h)
\(Q_1=0.15\) (3rd value) or \(Q_3=0.21\) (9th value)	M1	Positions \(\frac{11+1}{4}=3\) and \(\frac{3(11+1)}{4}=9\) may be seen.
0.06 (seconds)	A1

Answers: (a) \(12.0\lt t\le 12.5\) · (b) \(33.75+82.25+147+127.5+79.5+27.5=497.5\) · (c) 12.4 seconds · (d) 55% · (e) Points \((11.25,3)\), \((11.75,7)\), \((12.25,12)\), \((12.75,10)\), \((13.25,6)\), \((13.75,2)\) joined with straight lines · (f)(i) Positive correlation · (f)(ii) It is extrapolation (0.30 s is outside 0.13–0.24 s), and the line is based on only 11 athletes that do not all lie on it · (g)(i) 12.5 seconds (12.489…) · (g)(ii) \(\frac{8}{48}=\frac16\) · (h) 0.06 seconds

Common error (c) Divide by the total frequency, 40, not by the 6 classes. A mean that is not between 11.0 and 14.0 seconds cannot be right.

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Testing smartphone batteries: cumulative frequency and histograms

6.3 Cumulative frequency and histograms6.1 Averages, range and statistical diagrams1.1 Integers, fractions, decimals and percentages6.2 Probability of single and combined eventsHigher

A testing laboratory measures the battery life, \(h\) hours, of 80 phones of Model X. Each phone plays video continuously from a full charge until the battery is empty.

The table shows the results.

Battery life \(h\) (hours)	Frequency	Cumulative frequency
\(10\lt h\le 14\)	6
\(14\lt h\le 16\)	10
\(16\lt h\le 18\)	22
\(18\lt h\le 20\)	24
\(20\lt h\le 24\)	14
\(24\lt h\le 28\)	4

(a)

Complete the cumulative frequency column in the table.

ROUTINENotes

Show solution

at least 4 correct cumulative frequencies	B1
6, 16, 38, 62, 76, 80	B1	All six correct.

Answer: 6, 16, 38, 62, 76, 80

(2)

(b)

On the grid in Figure 5, draw a cumulative frequency graph for your table.

Figure 5: axes for the cumulative frequency graph

GRAPHNotes

Show solution

points plotted at the upper class boundaries: \((14,6)\), \((16,16)\), \((18,38)\), \((20,62)\), \((24,76)\), \((28,80)\)	B1	At least 5 points correct, ft their table. Points at midpoints score B0.
all points joined with a smooth curve or straight line segments, starting from \((10,0)\)	B1

Answer: Curve through \((10,0)\), \((14,6)\), \((16,16)\), \((18,38)\), \((20,62)\), \((24,76)\), \((28,80)\)

(2)

(c)

Use your graph to find an estimate for the median battery life and an estimate for the interquartile range.

MULTINotes

Show solution

median 18.2 (hours)	B1	Readings come from the student's own curve: accept 17.9 to 18.5 from a reading at 40, or ft a correct reading of their curve.
readings at 20 and 60: \(Q_1\approx16.4\) and \(Q_3\approx19.8\)	M1	Readings from the student's own curve at 20 and 60: accept \(Q_1\) 16.1 to 16.7 and \(Q_3\) 19.6 to 20.1 for a curve through the correct points, or ft correct readings of their curve.
IQR 3.5 (hours)	A1	Accept 3.1 to 3.9, or ft their \(Q_3-Q_1\) from correct readings of their curve.

Answer: Median about 18.2 hours; IQR about 3.5 hours

(3)

(d)

Use your graph to find an estimate for the percentage of the phones that had a battery life of more than 19 hours.

MULTINotes

Show solution

\(80-50\) \((=30)\)	M1	Reading from the student's own curve at \(h=19\) (accept 48 to 52 for a curve through the correct points, or ft a correct reading of their curve), subtracted from 80.
37.5(%)	A1	Accept 35% to 40%, or ft their reading.

Answer: About 37.5%

(2)

(e)

For Model X, the shortest battery life was 11.5 hours and the longest was 27.0 hours.

On Figure 6, draw a box plot for the battery lives of Model X.

Figure 6: box plot of battery life for 80 phones of Model Y, with space for Model X

ROUTINENotes

Show solution

box from \(Q_1\) to \(Q_3\) with a line at the median, ft their (c)	B1
whiskers from 11.5 to the box and from the box to 27.0	B1

Answer: Whiskers 11.5 and 27.0; box from about 16.4 to 19.8 with median about 18.2

(2)

(f)

Figure 6 also shows a box plot for the battery lives of 80 phones of a rival model, Model Y.

Make two comparisons between the battery lives of Model X and Model Y.

INTERPRETNotes

Show solution

Model X has the higher median (about 18.2 h compared with 17 h), so its batteries typically last longer	B1	ft their median. Must be in context.
Model Y has the smaller IQR (3 h compared with about 3.5 h), so its battery lives are more consistent	B1	ft their IQR. Accept a comparison of ranges (10.5 h and 15.5 h) as the measure of spread.

Answer: Model X batteries typically last longer (higher median); Model Y battery lives are more consistent (smaller IQR)

(2)

(g)

The results for Model X are also to be shown in a histogram.

Show that an estimate of the number of Model X phones with a battery life between 15 and 17 hours is 16.

PROOFNotes

Show solution

frequency densities \(\frac{10}{2}=5\) and \(\frac{22}{2}=11\)	M1	For the classes \(14\lt h\le16\) and \(16\lt h\le18\).
\(1\times5+1\times11\) or \(\frac12\times10+\frac12\times22\)	M1	Half of each class, assuming the values are spread evenly.
\(5+11=16\)	A1	cso.

Answer: \(1\times5+1\times11=16\)

(3)

(h)

After a software update, the 80 phones of Model X are tested again. Every phone's battery life is now 1.5 hours longer than before.

Write down the new median and the new interquartile range.

CONCEPTNotes

Show solution

median 19.7 (hours)	B1	ft their median \(+1.5\).
IQR 3.5 (hours), unchanged	B1	ft their IQR, unchanged.

Answer: Median about 19.7 hours; IQR unchanged, about 3.5 hours

(2)

(i)

Use the original results for Model X to find an estimate for the probability that a phone of this model has a battery life of more than 20 hours.

Priya buys two of these phones. Work out an estimate for the probability that both phones have a battery life of more than 20 hours.

MULTINotes

Show solution

\(\frac{18}{80}\times\frac{18}{80}\)	M1	\(\frac{14+4}{80}\) \((=\frac{9}{40})\) seen or implied.
\(\frac{81}{1600}\)	A1	Accept 0.0506 or better (0.050625).

Answer: \(\frac{9}{40}\); both: \(\frac{81}{1600}\approx0.0506\)

(2)

Show full solution

Working	Mark	Notes
Part (a)
at least 4 correct cumulative frequencies	B1
6, 16, 38, 62, 76, 80	B1	All six correct.
Part (b)
points plotted at the upper class boundaries: \((14,6)\), \((16,16)\), \((18,38)\), \((20,62)\), \((24,76)\), \((28,80)\)	B1	At least 5 points correct, ft their table. Points at midpoints score B0.
all points joined with a smooth curve or straight line segments, starting from \((10,0)\)	B1
Part (c)
median 18.2 (hours)	B1	Readings come from the student's own curve: accept 17.9 to 18.5 from a reading at 40, or ft a correct reading of their curve.
readings at 20 and 60: \(Q_1\approx16.4\) and \(Q_3\approx19.8\)	M1	Readings from the student's own curve at 20 and 60: accept \(Q_1\) 16.1 to 16.7 and \(Q_3\) 19.6 to 20.1 for a curve through the correct points, or ft correct readings of their curve.
IQR 3.5 (hours)	A1	Accept 3.1 to 3.9, or ft their \(Q_3-Q_1\) from correct readings of their curve.
Part (d)
\(80-50\) \((=30)\)	M1	Reading from the student's own curve at \(h=19\) (accept 48 to 52 for a curve through the correct points, or ft a correct reading of their curve), subtracted from 80.
37.5(%)	A1	Accept 35% to 40%, or ft their reading.
Part (e)
box from \(Q_1\) to \(Q_3\) with a line at the median, ft their (c)	B1
whiskers from 11.5 to the box and from the box to 27.0	B1
Part (f)
Model X has the higher median (about 18.2 h compared with 17 h), so its batteries typically last longer	B1	ft their median. Must be in context.
Model Y has the smaller IQR (3 h compared with about 3.5 h), so its battery lives are more consistent	B1	ft their IQR. Accept a comparison of ranges (10.5 h and 15.5 h) as the measure of spread.
Part (g)
frequency densities \(\frac{10}{2}=5\) and \(\frac{22}{2}=11\)	M1	For the classes \(14\lt h\le16\) and \(16\lt h\le18\).
\(1\times5+1\times11\) or \(\frac12\times10+\frac12\times22\)	M1	Half of each class, assuming the values are spread evenly.
\(5+11=16\)	A1	cso.
Part (h)
median 19.7 (hours)	B1	ft their median \(+1.5\).
IQR 3.5 (hours), unchanged	B1	ft their IQR, unchanged.
Part (i)
\(\frac{18}{80}\times\frac{18}{80}\)	M1	\(\frac{14+4}{80}\) \((=\frac{9}{40})\) seen or implied.
\(\frac{81}{1600}\)	A1	Accept 0.0506 or better (0.050625).

Answers: (a) 6, 16, 38, 62, 76, 80 · (b) Curve through \((10,0)\), \((14,6)\), \((16,16)\), \((18,38)\), \((20,62)\), \((24,76)\), \((28,80)\) · (c) Median about 18.2 hours; IQR about 3.5 hours · (d) About 37.5% · (e) Whiskers 11.5 and 27.0; box from about 16.4 to 19.8 with median about 18.2 · (f) Model X batteries typically last longer (higher median); Model Y battery lives are more consistent (smaller IQR) · (g) \(1\times5+1\times11=16\) · (h) Median about 19.7 hours; IQR unchanged, about 3.5 hours · (i) \(\frac{9}{40}\); both: \(\frac{81}{1600}\approx0.0506\)

Exam tip (c) With \(n=80\), read the median at 40 and the quartiles at 20 and 60 on the cumulative frequency axis. Draw the reading lines on your graph.

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6.2 Probability of single and combined events · 6.4 Conditional probability and tree diagramsmerged group

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A penalty shoot-out: relative frequency and tree diagrams

6.2 Probability of single and combined events6.1 Averages, range and statistical diagrams1.1 Integers, fractions, decimals and percentages1.4 Set language, notation and Venn diagrams (both tiers)Foundation

Ana and Bea take the penalties for a women's football team.

Ana went to 20 training sessions. In each session she took exactly 5 penalties against the team's goalkeeper. The bar chart in Figure 1 shows the number of goals Ana scored in each session.

Figure 1: the number of goals Ana scored in each of her 20 training sessions

(a)(i)

Write down the number of sessions in which Ana scored fewer than 4 goals.

GRAPHNotes

Show solution

From \(1 + 4\) read from the bar chart.

Answer: 5

(1)

(a)(ii)

Work out the mean number of goals Ana scored per session.

ROUTINENotes

Show solution

at least three correct products from \(2\times1,\ 3\times4,\ 4\times9,\ 5\times6\) (\(2, 12, 36, 30\))	M1	Accept a correct table of goals \(\times\) frequency.
\(80 \div 20\)	M1	Their \(\sum fx\) divided by 20. Dividing by 4 or by 6 (the number of bars or categories) scores M0.
4	A1

Answer: 4 goals

(3)

(b)

Show that the relative frequency with which Ana scored a penalty in training is 0.8

ROUTINENotes

Show solution

\(\frac{80}{100}\) or \(\frac{80}{20\times5}\) or \(\frac{4}{5}\)	M1	Total goals over total penalties; allow their 80 from (a)(ii) seen with \(20\times5 = 100\).
\(0.8\)	A1	cso: the total of 100 penalties must be seen or clearly implied.

Answer: \(\frac{80}{100} = 0.8\)

(2)

(c)

In training, Bea took 40 penalties and scored 34 of them.

Work out the percentage of her penalties that Bea scored.

ROUTINENotes

Show solution

\(\frac{34}{40}\times100\)	M1	Or \(34 \div 40 = 0.85\).
85%	A1

Answer: 85%

(2)

(d)(i)

In a match, the team has a penalty shoot-out. Ana takes one penalty and then Bea takes one penalty.

For each player, use her relative frequency from training as the probability that she scores. Assume the two penalties are independent.

Complete the probability tree diagram in Figure 2.

Figure 2: probability tree diagram for the two penalties

ROUTINENotes

Show solution

Ana misses: \(0.2\)	B1
Bea: \(0.85\) and \(0.15\) on both pairs of branches	B1	All four correct. Accept equivalent fractions or percentages consistently used; allow ECF from (c).

Answer: Ana: 0.8, 0.2; Bea: 0.85, 0.15 on each pair

(2)

(d)(ii)

Show that the probability that exactly one of the two players scores is 0.29

PROOFNotes

Show solution

\(0.8\times0.15\) \((=0.12)\) or \(0.2\times0.85\) \((=0.17)\)	M1	One correct path for exactly one goal.
\(0.8\times0.15 + 0.2\times0.85\)	M1	Both paths multiplied and added.
\(0.12 + 0.17 = 0.29\)	A1	cso. Both path values must be seen. \(1 - 0.68 - 0.03 = 0.29\) with both 0.68 and 0.03 shown from correct products is also fine.

Answer: \(0.12 + 0.17 = 0.29\)

(3)

(d)(iii)

The tree diagram assumes that the two penalties are independent.

Give one reason why this may not be true in a real shoot-out.

INTERPRETNotes

Show solution

a reason linking Bea's chance of scoring to Ana's result, e.g. if Ana misses, Bea is under more pressure (or more confident if Ana scores), so the probability that Bea scores changes

Do not accept a reason that is not about one penalty affecting the other, such as “the goalkeeper might be good”.

(1)

(e)

There are 22 players in the team's squad.

8 of these players practise penalties, 10 practise free kicks and 7 practise neither.

One of the 22 players is chosen at random.

Find the probability that this player practises free kicks but does not practise penalties.

MULTINotes

Show solution

\(22 - 7 = 15\) and \(8 + 10 - 15 = 3\) (practise both)	M1	May be seen as 3 in the intersection of a Venn diagram.
\(10 - 3 = 7\)	M1	Free kicks only; ft their 3.
\(\frac{7}{22}\)	A1	Accept 0.318 (3 s.f.). \(\frac{10}{22}\) scores M0 M0 A0.

Answer: \(\frac{7}{22}\)

(3)

(f)

In the final, the team plays against a goalkeeper who saves more penalties.

Against this goalkeeper, the probability that Ana scores is 0.7 and the probability that Bea scores is 0.75. The two penalties are still independent.

Work out the probability that at least one of the two players scores.

MULTINotes

Show solution

\(0.3\times0.25\) \((=0.075)\)	M1	Or one correct product from \(0.7\times0.75,\ 0.7\times0.25,\ 0.3\times0.75\).
\(1 - 0.3\times0.25\)	M1	Or \(0.525 + 0.175 + 0.225\).
\(0.925\)	A1	Accept \(\frac{37}{40}\).

Answer: \(0.925\)

(3)

Show full solution

Working	Mark	Notes
Part (a)(i)
5	B1	From \(1 + 4\) read from the bar chart.
Part (a)(ii)
at least three correct products from \(2\times1,\ 3\times4,\ 4\times9,\ 5\times6\) (\(2, 12, 36, 30\))	M1	Accept a correct table of goals \(\times\) frequency.
\(80 \div 20\)	M1	Their \(\sum fx\) divided by 20. Dividing by 4 or by 6 (the number of bars or categories) scores M0.
4	A1
Part (b)
\(\frac{80}{100}\) or \(\frac{80}{20\times5}\) or \(\frac{4}{5}\)	M1	Total goals over total penalties; allow their 80 from (a)(ii) seen with \(20\times5 = 100\).
\(0.8\)	A1	cso: the total of 100 penalties must be seen or clearly implied.
Part (c)
\(\frac{34}{40}\times100\)	M1	Or \(34 \div 40 = 0.85\).
85%	A1
Part (d)(i)
Ana misses: \(0.2\)	B1
Bea: \(0.85\) and \(0.15\) on both pairs of branches	B1	All four correct. Accept equivalent fractions or percentages consistently used; allow ECF from (c).
Part (d)(ii)
\(0.8\times0.15\) \((=0.12)\) or \(0.2\times0.85\) \((=0.17)\)	M1	One correct path for exactly one goal.
\(0.8\times0.15 + 0.2\times0.85\)	M1	Both paths multiplied and added.
\(0.12 + 0.17 = 0.29\)	A1	cso. Both path values must be seen. \(1 - 0.68 - 0.03 = 0.29\) with both 0.68 and 0.03 shown from correct products is also fine.
Part (d)(iii)
a reason linking Bea's chance of scoring to Ana's result, e.g. if Ana misses, Bea is under more pressure (or more confident if Ana scores), so the probability that Bea scores changes	B1	Do not accept a reason that is not about one penalty affecting the other, such as “the goalkeeper might be good”.
Part (e)
\(22 - 7 = 15\) and \(8 + 10 - 15 = 3\) (practise both)	M1	May be seen as 3 in the intersection of a Venn diagram.
\(10 - 3 = 7\)	M1	Free kicks only; ft their 3.
\(\frac{7}{22}\)	A1	Accept 0.318 (3 s.f.). \(\frac{10}{22}\) scores M0 M0 A0.
Part (f)
\(0.3\times0.25\) \((=0.075)\)	M1	Or one correct product from \(0.7\times0.75,\ 0.7\times0.25,\ 0.3\times0.75\).
\(1 - 0.3\times0.25\)	M1	Or \(0.525 + 0.175 + 0.225\).
\(0.925\)	A1	Accept \(\frac{37}{40}\).

Answers: (a)(i) 5 · (a)(ii) 4 goals · (b) \(\frac{80}{100} = 0.8\) · (c) 85% · (d)(i) Ana: 0.8, 0.2; Bea: 0.85, 0.15 on each pair · (d)(ii) \(0.12 + 0.17 = 0.29\) · (e) \(\frac{7}{22}\) · (f) \(0.925\)

Common error (a)(ii) Dividing the total number of goals by the number of bars gives the wrong mean. Divide \(\sum fx\) by the total frequency, which is the number of sessions, 20.

Exam tip (e) “10 practise free kicks” is the whole free-kick circle, including the players who practise both. Find the overlap first, then subtract it.

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A music playlist on shuffle: dependent events and conditional probability

6.4 Conditional probability and tree diagrams2.3 Quadratic equations and inequalities6.2 Probability of single and combined events1.4 Set language, notation and Venn diagrams (both tiers)Higher

Rui has a playlist of 21 songs.

Artist	Kira	Milo	Tess	Joe
Number of songs	6	8	4	3

He plays the playlist on shuffle. The first song is chosen at random from the 21 songs. Each song after that is chosen at random from the songs that have not yet been played, so no song is repeated.

(a)

Complete the probability tree diagram in Figure 1 for the first two songs.

Figure 1: probability tree diagram for the first two songs

ROUTINENotes

Show solution

First song, not Kira: \(\frac{15}{21}\)	B1
Second song: \(\frac{15}{20}\) on the upper “not Kira” branch, and \(\frac{6}{20}\) and \(\frac{14}{20}\) on the lower pair	B1	All three correct; accept equivalent fractions.

Answer: \(\frac{6}{21}, \frac{15}{21}\); \(\frac{5}{20}, \frac{15}{20}\); \(\frac{6}{20}, \frac{14}{20}\)

(2)

(b)

Work out the probability that exactly one of the first two songs is by Kira.

ROUTINENotes

Show solution

\(\frac{6}{21}\times\frac{15}{20}\) or \(\frac{15}{21}\times\frac{6}{20}\) \(\left(=\frac{90}{420}\right)\)	M1	ft their tree from (a).
\(\frac{6}{21}\times\frac{15}{20} + \frac{15}{21}\times\frac{6}{20}\)	M1
\(\frac{3}{7}\)	A1	Accept \(\frac{180}{420}\) or any equivalent fraction, or 0.429 (3 s.f.).

Answer: \(\frac{3}{7}\)

(3)

(c)

Given that at least one of the first two songs is by Kira, work out the probability that both of them are by Kira.

MULTINotes

Show solution

\(P(\text{both}) = \frac{6}{21}\times\frac{5}{20} = \frac{1}{14}\) and \(P(\text{at least one}) = 1 - \frac{15}{21}\times\frac{14}{20} = \frac{1}{2}\)	M1	Both probabilities needed; \(P(\text{at least one})\) may be found as \(\frac{1}{14} + \frac{3}{7}\), ft from (b).
\(\frac{1}{14} \div \frac{1}{2}\)	M1	Their \(P(\text{both})\) divided by their \(P(\text{at least one})\).
\(\frac{1}{7}\)	A1	Accept 0.143 (3 s.f.). \(\frac{1}{14}\) alone scores M0 M0 A0.

Answer: \(\frac{1}{7}\)

(3)

(d)(i)

Rui started the playlist on shuffle 84 times. The bar chart in Figure 2 shows the artist of the first song each time.

Use the bar chart to work out the relative frequency with which the first song was by Tess or by Joe.

Figure 2: the artist of the first song in each of 84 shuffle sessions

GRAPHNotes

Show solution

\(\frac{29}{84}\)

From \(\frac{17 + 12}{84}\). Accept 0.345 (3 s.f.).

Answer: \(\frac{29}{84}\)

(1)

(d)(ii)

Rui says,

“The shuffle chooses songs by Tess and Joe more often than it should.”

Does the bar chart support what Rui says? Give a reason for your answer.

INTERPRETNotes

Show solution

No, with a reason, e.g. the probability is \(\frac{7}{21} = \frac{1}{3}\) (expected 28 sessions), which is very close to \(\frac{29}{84}\) (29 sessions)

A comparison with \(\frac{7}{21}\) or with an expected number of 28 is needed. Allow ft from their (d)(i).

Answer: No: \(\frac{29}{84} \approx 0.345\) is close to \(\frac{1}{3}\)

(1)

(e)

For the 21 songs in the playlist, \(P\) is the set of pop songs and \(B\) is the set of songs released before 2010.

12 songs are in \(P\), 8 songs are in \(B\) and 5 songs are in neither set.

One of the 21 songs is chosen at random. Given that it is not a pop song, find the probability that it was released before 2010.

MULTINotes

Show solution

\(n(P\cup B) = 21 - 5 = 16\) and \(n(P\cap B) = 12 + 8 - 16 = 4\)	M1	May be seen in a Venn diagram.
\(n(P') = 21 - 12 = 9\) and \(n(B\cap P') = 8 - 4 = 4\)	M1	Or \(\frac{4/21}{9/21}\); ft their 4.
\(\frac{4}{9}\)	A1	\(\frac{4}{21}\) (wrong total) scores M1 M1 A0.

Answer: \(\frac{4}{9}\)

(3)

(f)

Rui changes the setting so that every song is chosen at random from all 21 songs, so a song can now be repeated.

Work out the probability that neither of the first two songs is by Kira. Give your answer as a fraction in its simplest form.

ROUTINENotes

Show solution

\(\frac{15}{21}\times\frac{15}{21}\)	M1	The two songs are now independent.
\(\frac{25}{49}\)	A1	\(\frac{15}{21}\times\frac{14}{20} = \frac{1}{2}\) is the no-repeat answer and scores M0 A0 here.

Answer: \(\frac{25}{49}\)

(2)

(g)(i)

Rui's friend Zoe has a different playlist of \(n\) songs, which she plays on shuffle with no song repeated. 4 of her songs are by Kira.

The probability that neither of the first two songs is by Kira is \(\dfrac{1}{3}\)

Show that \(n^2 - 13n + 30 = 0\)

PROOFNotes

Show solution

\(\frac{n-4}{n}\times\frac{n-5}{n-1}\)	M1	Correct product for neither song by Kira.
\(3(n-4)(n-5) = n(n-1)\) or \(3n^2 - 27n + 60 = n^2 - n\)	M1	Equation set equal to \(\frac{1}{3}\) and fractions cleared correctly.
\(2n^2 - 26n + 60 = 0\) leading to \(n^2 - 13n + 30 = 0\)	A1	cso: every line of the expansion shown.

Answer: \(n^2 - 13n + 30 = 0\)

(3)

(g)(ii)

Find the number of songs in Zoe's playlist.

You must give a reason for any value of \(n\) that you reject.

ALGEBRANotes

Show solution

\((n-10)(n-3) = 0\)	M1	Or correct substitution into the quadratic formula.
\(n = 10\), with \(n = 3\) rejected because the playlist has at least 4 songs (the songs by Kira)	A1	A reason for rejecting 3 is needed.

Answer: 10 songs

(2)

Show full solution

Working	Mark	Notes
Part (a)
First song, not Kira: \(\frac{15}{21}\)	B1
Second song: \(\frac{15}{20}\) on the upper “not Kira” branch, and \(\frac{6}{20}\) and \(\frac{14}{20}\) on the lower pair	B1	All three correct; accept equivalent fractions.
Part (b)
\(\frac{6}{21}\times\frac{15}{20}\) or \(\frac{15}{21}\times\frac{6}{20}\) \(\left(=\frac{90}{420}\right)\)	M1	ft their tree from (a).
\(\frac{6}{21}\times\frac{15}{20} + \frac{15}{21}\times\frac{6}{20}\)	M1
\(\frac{3}{7}\)	A1	Accept \(\frac{180}{420}\) or any equivalent fraction, or 0.429 (3 s.f.).
Part (c)
\(P(\text{both}) = \frac{6}{21}\times\frac{5}{20} = \frac{1}{14}\) and \(P(\text{at least one}) = 1 - \frac{15}{21}\times\frac{14}{20} = \frac{1}{2}\)	M1	Both probabilities needed; \(P(\text{at least one})\) may be found as \(\frac{1}{14} + \frac{3}{7}\), ft from (b).
\(\frac{1}{14} \div \frac{1}{2}\)	M1	Their \(P(\text{both})\) divided by their \(P(\text{at least one})\).
\(\frac{1}{7}\)	A1	Accept 0.143 (3 s.f.). \(\frac{1}{14}\) alone scores M0 M0 A0.
Part (d)(i)
\(\frac{29}{84}\)	B1	From \(\frac{17 + 12}{84}\). Accept 0.345 (3 s.f.).
Part (d)(ii)
No, with a reason, e.g. the probability is \(\frac{7}{21} = \frac{1}{3}\) (expected 28 sessions), which is very close to \(\frac{29}{84}\) (29 sessions)	B1	A comparison with \(\frac{7}{21}\) or with an expected number of 28 is needed. Allow ft from their (d)(i).
Part (e)
\(n(P\cup B) = 21 - 5 = 16\) and \(n(P\cap B) = 12 + 8 - 16 = 4\)	M1	May be seen in a Venn diagram.
\(n(P') = 21 - 12 = 9\) and \(n(B\cap P') = 8 - 4 = 4\)	M1	Or \(\frac{4/21}{9/21}\); ft their 4.
\(\frac{4}{9}\)	A1	\(\frac{4}{21}\) (wrong total) scores M1 M1 A0.
Part (f)
\(\frac{15}{21}\times\frac{15}{21}\)	M1	The two songs are now independent.
\(\frac{25}{49}\)	A1	\(\frac{15}{21}\times\frac{14}{20} = \frac{1}{2}\) is the no-repeat answer and scores M0 A0 here.
Part (g)(i)
\(\frac{n-4}{n}\times\frac{n-5}{n-1}\)	M1	Correct product for neither song by Kira.
\(3(n-4)(n-5) = n(n-1)\) or \(3n^2 - 27n + 60 = n^2 - n\)	M1	Equation set equal to \(\frac{1}{3}\) and fractions cleared correctly.
\(2n^2 - 26n + 60 = 0\) leading to \(n^2 - 13n + 30 = 0\)	A1	cso: every line of the expansion shown.
Part (g)(ii)
\((n-10)(n-3) = 0\)	M1	Or correct substitution into the quadratic formula.
\(n = 10\), with \(n = 3\) rejected because the playlist has at least 4 songs (the songs by Kira)	A1	A reason for rejecting 3 is needed.

Answers: (a) \(\frac{6}{21}, \frac{15}{21}\); \(\frac{5}{20}, \frac{15}{20}\); \(\frac{6}{20}, \frac{14}{20}\) · (b) \(\frac{3}{7}\) · (c) \(\frac{1}{7}\) · (d)(i) \(\frac{29}{84}\) · (d)(ii) No: \(\frac{29}{84} \approx 0.345\) is close to \(\frac{1}{3}\) · (e) \(\frac{4}{9}\) · (f) \(\frac{25}{49}\) · (g)(i) \(n^2 - 13n + 30 = 0\) · (g)(ii) 10 songs

Common error (b) Using \(\frac{6}{21}\) for the second song as well treats the songs as if they could repeat. After one song has been played there are only 20 songs left to choose from.

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