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Cambridge IGCSE Maths 0580 · Concept clusters

Cambridge 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. 26 questions · 495 marks · two per subchapter or merged group · every part with a worked solution · All courses

Topic

Subchapter

Level

Skill

Topic 1 · Number

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

Setup	1.1	1.2	1.3	1.4	Other
Festival stage lights: flashing patterns, primes and fractions of a show	●	○		○	—
An online sports-kit shop: profit, sales, reverse percentages and packing orders	○	●		○	—
Talking to a Mars rover: standard form, indices and signal delays		○	●	○	—
Saving for a gap year: ratio, interest and exchange rates		○	○	●	—

1.1 Types of number, factors, HCF and LCM · 1.2 Fractions, decimals and percentagesmerged group

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Festival stage lights: flashing patterns, primes and fractions of a show

1.1 Types of number, factors, HCF and LCM1.2 Fractions, decimals and percentages1.4 Estimation, rounding, bounds, ratio, proportion and moneyCore

At a music festival, three stage lights A, B and C flash during a band’s set.

Light A flashes every 12 seconds.
Light B flashes every 18 seconds.
Light C flashes every 45 seconds.

All three lights flash together at the very start of the set. Figure 1 shows the flashes in the first 60 seconds.

Figure 1: the flashes of lights A, B and C during the first 60 seconds of the set (each dot is one flash)

(a)

Write 45 as a product of its prime factors.

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a correct first split or division, e.g. $45 = 5\times9$ or $45\div3=15$	M1	A correct factor tree or repeated division with at most one slip.
$3^2\times5$	A1	Accept $3\times3\times5$. Do not accept a list 3, 3, 5 or an answer with + signs.

Answer: $3^2\times5$

(2)

(b)(i)

Find how many seconds after the start of the set all three lights next flash together.

No calculatorMULTINotes

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$12 = 2^2\times3$ and $18 = 2\times3^2$	B1	Both correct. May be seen in a Venn diagram.
LCM $=2^2\times3^2\times5$ or listing multiples of 45 and testing them against 12 and 18	M1	Uses the highest power of each prime from 12, 18 and 45. Allow ECF from (a).
180	A1	Accept 3 minutes. 3 (the HCF) scores M0 A0.

Answer: 180 seconds (3 minutes)

(3)

(b)(ii)

The set lasts exactly 90 minutes.

Find the number of times that all three lights flash together during the set. Do not count the flash at the very start, but do count a flash at the very end.

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$90\times60\div180$ or $5400\div180$	M1	Allow ECF from their (b)(i): their $5400\div$ their LCM, or $90\div3$.
30	A1	31 (counting the start) scores M1 A0.

Answer: 30

(2)

(c)

Sam says:

“The HCF of 12, 18 and 45 is 3, so all three lights flash together every 3 seconds.”

Explain why Sam is wrong.

No calculatorINTERPRETNotes

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lights flash together only at times that are multiples of 12, 18 and 45, i.e. common multiples; 3 is a common factor, not a common multiple	B1	Any clear statement that a common multiple (LCM) is needed.
e.g. light A does not flash at 3 seconds: its first flash after the start is at 12 seconds (Figure 1 shows no flash at 3 s)	B1	Any correct counter-example from the data or the figure.

(2)

(d)

During the set, a laser display is used for $\frac{1}{4}$ of the time and a smoke effect is used for $\frac{2}{9}$ of the time. The two effects are never used at the same time.

Show that, for $\dfrac{19}{36}$ of the set, neither effect is used.

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$\frac{1}{4}=\frac{9}{36}$ and $\frac{2}{9}=\frac{8}{36}$	M1	Both fractions over a common denominator (36 is the LCM of 4 and 9). Any correct common denominator, e.g. 72, is fine.
$1-\frac{17}{36}$	M1	Subtracting the total of the two effects from 1, e.g. $\frac{36}{36}-\frac{9}{36}-\frac{8}{36}$.
$\frac{19}{36}$	A1	cso. Decimal working (e.g. 0.25 + 0.222) scores at most M0 M1 A0.

Answer: $\frac{19}{36}$ (shown)

(3)

(e)(i)

Figure 2 shows the energy used by each of the four lighting rigs during the set.

Find the fraction of the total energy that was used by the back rig. Give your answer in its simplest form.

Figure 2: the energy used by each of the four lighting rigs during the set

No calculatorGRAPHNotes

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18, 24, 12 and 6 read from the graph, or a total of 60	B1
$\frac{24}{60}$	M1	Their back-rig value over their total.
$\frac{2}{5}$	A1	0.4 or 40% alone scores A0. ECF for the correct simplest form of their fraction.

Answer: $\frac{2}{5}$

(3)

(e)(ii)

Write the ratio

energy used by the back rig : energy used by the front rig : energy used by the strobe rig

in its simplest form.

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$24 : 18 : 6$ or a correct partly simplified ratio, e.g. $12 : 9 : 3$	M1	Values from their readings in (e)(i).
$4 : 3 : 1$	A1	Must be in this order.

Answer: $4 : 3 : 1$

(2)

(f)

For the next set, light C is re-programmed to flash every 40 seconds. Lights A and B do not change. All three lights flash together at the start of the set.

Find how many seconds after the start all three lights next flash together.

No calculatorMULTINotes

Show solution

$40 = 2^3\times5$	B1
LCM $=2^3\times3^2\times5$	M1	Highest powers of 2, 3 and 5 from $2^2\times3$, $2\times3^2$ and $2^3\times5$; or listing multiples of 40 tested against 12 and 18.
360	A1	Accept 6 minutes.

Answer: 360 seconds (6 minutes)

(3)

Show full solution

Working	Mark	Notes
Part (a)
a correct first split or division, e.g. $45 = 5\times9$ or $45\div3=15$	M1	A correct factor tree or repeated division with at most one slip.
$3^2\times5$	A1	Accept $3\times3\times5$. Do not accept a list 3, 3, 5 or an answer with + signs.
Part (b)(i)
$12 = 2^2\times3$ and $18 = 2\times3^2$	B1	Both correct. May be seen in a Venn diagram.
LCM $=2^2\times3^2\times5$ or listing multiples of 45 and testing them against 12 and 18	M1	Uses the highest power of each prime from 12, 18 and 45. Allow ECF from (a).
180	A1	Accept 3 minutes. 3 (the HCF) scores M0 A0.
Part (b)(ii)
$90\times60\div180$ or $5400\div180$	M1	Allow ECF from their (b)(i): their $5400\div$ their LCM, or $90\div3$.
30	A1	31 (counting the start) scores M1 A0.
Part (c)
lights flash together only at times that are multiples of 12, 18 and 45, i.e. common multiples; 3 is a common factor, not a common multiple	B1	Any clear statement that a common multiple (LCM) is needed.
e.g. light A does not flash at 3 seconds: its first flash after the start is at 12 seconds (Figure 1 shows no flash at 3 s)	B1	Any correct counter-example from the data or the figure.
Part (d)
$\frac{1}{4}=\frac{9}{36}$ and $\frac{2}{9}=\frac{8}{36}$	M1	Both fractions over a common denominator (36 is the LCM of 4 and 9). Any correct common denominator, e.g. 72, is fine.
$1-\frac{17}{36}$	M1	Subtracting the total of the two effects from 1, e.g. $\frac{36}{36}-\frac{9}{36}-\frac{8}{36}$.
$\frac{19}{36}$	A1	cso. Decimal working (e.g. 0.25 + 0.222) scores at most M0 M1 A0.
Part (e)(i)
18, 24, 12 and 6 read from the graph, or a total of 60	B1
$\frac{24}{60}$	M1	Their back-rig value over their total.
$\frac{2}{5}$	A1	0.4 or 40% alone scores A0. ECF for the correct simplest form of their fraction.
Part (e)(ii)
$24 : 18 : 6$ or a correct partly simplified ratio, e.g. $12 : 9 : 3$	M1	Values from their readings in (e)(i).
$4 : 3 : 1$	A1	Must be in this order.
Part (f)
$40 = 2^3\times5$	B1
LCM $=2^3\times3^2\times5$	M1	Highest powers of 2, 3 and 5 from $2^2\times3$, $2\times3^2$ and $2^3\times5$; or listing multiples of 40 tested against 12 and 18.
360	A1	Accept 6 minutes.

Answers: (a) $3^2\times5$ · (b)(i) 180 seconds (3 minutes) · (b)(ii) 30 · (d) $\frac{19}{36}$ (shown) · (e)(i) $\frac{2}{5}$ · (e)(ii) $4 : 3 : 1$ · (f) 360 seconds (6 minutes)

Exam tip (c) Ask yourself whether you are splitting something into equal groups (HCF) or waiting for things that repeat to line up again (LCM). Repeating flashes always need the LCM.

Common error (f) Multiplying the numbers together ($12\times18\times40$) gives a common multiple but not the lowest one, because the numbers share factors.

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An online sports-kit shop: profit, sales, reverse percentages and packing orders

1.2 Fractions, decimals and percentages1.1 Types of number, factors, HCF and LCM1.4 Estimation, rounding, bounds, ratio, proportion and moneyCore

A small online shop sells football shirts. The shop buys each shirt for $16 and normally sells it for $26.

(a)

Calculate the percentage profit the shop makes on one shirt sold at the normal price.

ROUTINENotes

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$\frac{26-16}{16}\times100$ or $\frac{10}{16}\times100$	M1	Or $\frac{26}{16}\times100-100$. Dividing by 26 (giving 38.5%) scores M0.
62.5	A1	Accept 62.5% or 62½%.

Answer: 62.5%

(2)

(b)(i)

Figure 1 shows the number of shirts the shop sold in each month from January to June.

Find the percentage increase in the number of shirts sold from February to April.

Figure 1: the number of shirts sold by the shop in each month from January to June

GRAPHNotes

Show solution

75 and 120 read from the graph	B1	Both values.
$\frac{120-75}{75}\times100$	M1	Their April minus their February, divided by their February.
60	A1	Accept 60%.

Answer: 60%

(3)

(b)(ii)

Find the fraction of all the shirts sold from January to June that were sold in April. Give your answer in its simplest form.

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$\frac{120}{630}$	M1	Total $60+75+90+120+150+135=630$. Allow ECF from their readings.
$\frac{4}{21}$	A1	Simplest form required (divide by the HCF, 30). 0.190… scores A0.

Answer: $\frac{4}{21}$

(2)

(c)

In an end-of-season sale, all prices are reduced by 35%. A training jacket costs $28.60 in the sale.

Show that the price of the jacket before the sale was $44.

MULTINotes

Show solution

$28.60\div0.65$ or $65\% = 28.60$ leading to $1\% = 0.44$	M1	$28.60\times1.35$ or $28.60\times0.35$ added on scores M0. Checking $44\times0.65=28.60$ alone is not a method: M0.
44	A1	cso

Answer: $44 (shown)

(2)

(d)

The shop packs 120 shirts and 168 pairs of socks into identical gift boxes. Every box contains the same number of shirts and the same number of pairs of socks, and nothing is left over.

Find the greatest possible number of boxes and the contents of each box.

No calculatorMULTINotes

Show solution

$120=2^3\times3\times5$ and $168=2^3\times3\times7$, or listing factors of both	M1	Shows a method for the HCF.
24 boxes	A1
5 shirts and 7 pairs of socks	B1	Follow through: $120\div$ their HCF and $168\div$ their HCF, if both are whole numbers.

Answer: 24 boxes, each with 5 shirts and 7 pairs of socks

(3)

(e)

After the sale ends, the shirts go back to the normal price of $26. The 35% reduction is replaced by the offer “buy 3 shirts, pay for 2”.

Kai wants to buy 3 shirts. He says the new offer is a better deal than the 35% reduction was.

Is Kai correct? Show all your working.

MULTINotes

Show solution

$2\times26 = 52$	M1	Cost with the new offer.
$3\times26\times0.65 = 50.70$ or $78\times0.65$	M1	Cost of 3 shirts with 35% off.
No, with $52 and $50.70 compared (the 35% reduction was $1.30 cheaper)	A1	Both values must be correct. Alternative: “pay for 2 of 3” is a reduction of $\frac{1}{3}$, about 33.3%, which is less than 35%, with values seen.

Answer: No: $52 with the offer, $50.70 with 35% off

(3)

(f)

At the start of the new season, the normal price of a shirt, $26, is increased by 20%. A month later, the new price is decreased by 20%.

Lucia says the price is now back to $26.

Show that Lucia is wrong, and explain why the price does not return to $26.

Extended onlyINTERPRETNotes

Show solution

$26\times1.2$ $(=31.20)$	M1
$31.20\times0.8 = 24.96$	A1	Accept $26\times0.96$. Must reach $24.96.
the 20% decrease is 20% of the larger amount, $31.20, so it is a bigger decrease ($6.24) than the increase ($5.20)	B1	A comparison of the bases of the two percentages is needed; “percentages don’t cancel” alone is not enough.

Answer: $24.96, not $26

(3)

(g)

The proportion of shirts that customers return is $0.0\dot{4}$.

Write $0.0\dot{4}$ as a fraction in its simplest form.

Extended onlyNo calculatorALGEBRANotes

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e.g. $x=0.0444\ldots$, $100x=4.444\ldots$, $10x=0.444\ldots$ and $90x=4$	M1	Two multiples of $x$ whose recurring parts line up, subtracted. Or $\frac{4}{90}$ from a correct method.
$\frac{2}{45}$	A1	$\frac{4}{90}$ not simplified scores M1 A0.

Answer: $\frac{2}{45}$

(2)

Show full solution

Working	Mark	Notes
Part (a)
$\frac{26-16}{16}\times100$ or $\frac{10}{16}\times100$	M1	Or $\frac{26}{16}\times100-100$. Dividing by 26 (giving 38.5%) scores M0.
62.5	A1	Accept 62.5% or 62½%.
Part (b)(i)
75 and 120 read from the graph	B1	Both values.
$\frac{120-75}{75}\times100$	M1	Their April minus their February, divided by their February.
60	A1	Accept 60%.
Part (b)(ii)
$\frac{120}{630}$	M1	Total $60+75+90+120+150+135=630$. Allow ECF from their readings.
$\frac{4}{21}$	A1	Simplest form required (divide by the HCF, 30). 0.190… scores A0.
Part (c)
$28.60\div0.65$ or $65\% = 28.60$ leading to $1\% = 0.44$	M1	$28.60\times1.35$ or $28.60\times0.35$ added on scores M0. Checking $44\times0.65=28.60$ alone is not a method: M0.
44	A1	cso
Part (d)
$120=2^3\times3\times5$ and $168=2^3\times3\times7$, or listing factors of both	M1	Shows a method for the HCF.
24 boxes	A1
5 shirts and 7 pairs of socks	B1	Follow through: $120\div$ their HCF and $168\div$ their HCF, if both are whole numbers.
Part (e)
$2\times26 = 52$	M1	Cost with the new offer.
$3\times26\times0.65 = 50.70$ or $78\times0.65$	M1	Cost of 3 shirts with 35% off.
No, with $52 and $50.70 compared (the 35% reduction was $1.30 cheaper)	A1	Both values must be correct. Alternative: “pay for 2 of 3” is a reduction of $\frac{1}{3}$, about 33.3%, which is less than 35%, with values seen.
Part (f)
$26\times1.2$ $(=31.20)$	M1
$31.20\times0.8 = 24.96$	A1	Accept $26\times0.96$. Must reach $24.96.
the 20% decrease is 20% of the larger amount, $31.20, so it is a bigger decrease ($6.24) than the increase ($5.20)	B1	A comparison of the bases of the two percentages is needed; “percentages don’t cancel” alone is not enough.
Part (g)
e.g. $x=0.0444\ldots$, $100x=4.444\ldots$, $10x=0.444\ldots$ and $90x=4$	M1	Two multiples of $x$ whose recurring parts line up, subtracted. Or $\frac{4}{90}$ from a correct method.
$\frac{2}{45}$	A1	$\frac{4}{90}$ not simplified scores M1 A0.

Answers: (a) 62.5% · (b)(i) 60% · (b)(ii) $\frac{4}{21}$ · (c) $44 (shown) · (d) 24 boxes, each with 5 shirts and 7 pairs of socks · (e) No: $52 with the offer, $50.70 with 35% off · (f) $24.96, not $26 · (g) $\frac{2}{45}$

Common error (c) Adding 35% of the sale price back on gives $38.61, not the original price: the 35% was taken off the larger, original price, so divide by the multiplier 0.65.

More 1.2 questions in the question bank

1.3 Indices, standard form and surds · 1.4 Estimation, rounding, bounds, ratio, proportion and moneymerged group

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Talking to a Mars rover: standard form, indices and signal delays

1.3 Indices, standard form and surds1.4 Estimation, rounding, bounds, ratio, proportion and money1.2 Fractions, decimals and percentagesCore

Mission control on Earth sends radio signals to a rover on Mars. The signals travel at $3.0\times10^{8}$ m/s.

At closest approach, the distance between Earth and Mars is 56 000 000 km.
At greatest separation, the distance is $4.0\times10^{8}$ km.

1 km = 1000 m. The one-way signal delay is the time a signal takes to travel from Earth to Mars.

(a)

Write the closest distance in metres, in standard form.

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$5.6\times10^{10}$

$56\times10^{9}$ is not standard form: B0.

Answer: $5.6\times10^{10}$ m

(1)

(b)

By rounding each number to 1 significant figure, work out an estimate of the one-way signal delay, in seconds, at closest approach.

Show the rounded numbers you use. State whether your estimate is greater than or less than the actual delay, and give a reason.

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$\frac{6\times10^{10}}{3\times10^{8}}$	M1	Rounded values must be seen. Allow ECF from (a).
200	A1	Accept $2\times10^2$.
greater, because the distance was rounded up (5.6 to 6) and the speed was not changed	B1	Needs the reason; “greater” alone scores B0.

Answer: 200 s, an overestimate

(3)

(c)

Calculate the actual one-way signal delay at closest approach. Give your answer in seconds, correct to 3 significant figures.

ROUTINENotes

Show solution

$\frac{5.6\times10^{10}}{3.0\times10^{8}}$	M1	Allow ECF from (a).
187	A1	186.66… seen and then rounded correctly. 186 scores A0.

Answer: 187 s

(2)

(d)

Show that, at greatest separation, the one-way signal delay is 22 minutes, correct to the nearest minute.

MULTINotes

Show solution

$\frac{4.0\times10^{11}}{3.0\times10^{8}}$ $(=1333.3\ldots\text{ s})$	M1	Must use $4.0\times10^{11}$ m (or $4.0\times10^{8}\times10^{3}$).
$1333.3\ldots\div60 = 22.2\ldots$, so 22 minutes	A1	cso. Both 1333.3… s (or 22.2…) and the division by 60 must be seen.

Answer: 22 minutes (shown)

(2)

(e)

Write the delay at closest approach as a percentage of the delay at greatest separation.

MULTINotes

Show solution

$\frac{186.66\ldots}{1333.33\ldots}\times100$ or $\frac{5.6\times10^{7}}{4.0\times10^{8}}\times100$	M1	Delays are in proportion to distances, so either ratio is fine. Allow ECF from (c) and (d).
14	A1	Accept 14% or 14.0%. Answers from rounded values in the range 13.9 to 14.1 score A1.

Answer: 14%

(2)

(f)(i)

One photograph from the rover contains $2^{25}$ bits of data. The rover sends data to a satellite at a rate of $2^{15}$ bits per second.

Find the time taken to send one photograph. Give your answer as a power of 2 and as an ordinary number of seconds.

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Show solution

$2^{25}\div2^{15} = 2^{10}$	M1	Subtracting the indices.
$2^{10}$ and 1024	A1	Both forms needed.

Answer: $2^{10}$ seconds = 1024 seconds

(2)

(f)(ii)

Find the number of bits the rover sends in $\frac{1}{8}$ of a second. Give your answer as a power of 2.

No calculatorALGEBRANotes

Show solution

$\frac{1}{8}=2^{-3}$ and $2^{15}\times2^{-3}$, or $2^{15}\div2^{3}$	M1
$2^{12}$	A1	4096 alone scores M1 A0.

Answer: $2^{12}$

(2)

(g)

Figure 1 shows how the distance between Earth and Mars changes during the first 24 months of the mission.

Use Figure 1 to work out an estimate of the one-way signal delay 18 months after the start of the mission. Give your answer in minutes.

Figure 1: the distance between Earth and Mars during the first 24 months of the mission

GRAPHNotes

Show solution

a distance between 160 and 175 million km read from the graph	B1
$\frac{\text{their reading}\times10^{9}}{3.0\times10^{8}}\div60$, e.g. $\frac{1.67\times10^{11}}{3.0\times10^{8}}\div60$	M1	1 million km $=10^{6}$ km $=10^{9}$ m. Division by 60 (seconds to minutes) must be seen.
answer in the range 8.9 to 9.7	A1	From a reading in range; e.g. 167 million km gives 9.28 minutes.

Answer: about 9.3 minutes

(3)

(h)

The closest distance, $5.6\times10^{7}$ km, is correct to 2 significant figures. The speed, $3.0\times10^{8}$ m/s, is also correct to 2 significant figures.

Calculate the lower bound and the upper bound of the one-way signal delay at closest approach. Give your answers in seconds, correct to 3 significant figures.

Extended onlyMULTINotes

Show solution

$5.55\times10^{7}$ or $5.65\times10^{7}$ km, and $2.95\times10^{8}$ or $3.05\times10^{8}$ m/s	B1	At least one correct bound for each quantity.
$\frac{5.65\times10^{10}}{2.95\times10^{8}}$ or $\frac{5.55\times10^{10}}{3.05\times10^{8}}$	M1	Upper bound of distance over lower bound of speed, or lower over upper.
182 and 192	A1	Both values (181.967… and 191.525…).

Answer: lower bound 182 s, upper bound 192 s

(3)

Show full solution

Working	Mark	Notes
Part (a)
$5.6\times10^{10}$	B1	$56\times10^{9}$ is not standard form: B0.
Part (b)
$\frac{6\times10^{10}}{3\times10^{8}}$	M1	Rounded values must be seen. Allow ECF from (a).
200	A1	Accept $2\times10^2$.
greater, because the distance was rounded up (5.6 to 6) and the speed was not changed	B1	Needs the reason; “greater” alone scores B0.
Part (c)
$\frac{5.6\times10^{10}}{3.0\times10^{8}}$	M1	Allow ECF from (a).
187	A1	186.66… seen and then rounded correctly. 186 scores A0.
Part (d)
$\frac{4.0\times10^{11}}{3.0\times10^{8}}$ $(=1333.3\ldots\text{ s})$	M1	Must use $4.0\times10^{11}$ m (or $4.0\times10^{8}\times10^{3}$).
$1333.3\ldots\div60 = 22.2\ldots$, so 22 minutes	A1	cso. Both 1333.3… s (or 22.2…) and the division by 60 must be seen.
Part (e)
$\frac{186.66\ldots}{1333.33\ldots}\times100$ or $\frac{5.6\times10^{7}}{4.0\times10^{8}}\times100$	M1	Delays are in proportion to distances, so either ratio is fine. Allow ECF from (c) and (d).
14	A1	Accept 14% or 14.0%. Answers from rounded values in the range 13.9 to 14.1 score A1.
Part (f)(i)
$2^{25}\div2^{15} = 2^{10}$	M1	Subtracting the indices.
$2^{10}$ and 1024	A1	Both forms needed.
Part (f)(ii)
$\frac{1}{8}=2^{-3}$ and $2^{15}\times2^{-3}$, or $2^{15}\div2^{3}$	M1
$2^{12}$	A1	4096 alone scores M1 A0.
Part (g)
a distance between 160 and 175 million km read from the graph	B1
$\frac{\text{their reading}\times10^{9}}{3.0\times10^{8}}\div60$, e.g. $\frac{1.67\times10^{11}}{3.0\times10^{8}}\div60$	M1	1 million km $=10^{6}$ km $=10^{9}$ m. Division by 60 (seconds to minutes) must be seen.
answer in the range 8.9 to 9.7	A1	From a reading in range; e.g. 167 million km gives 9.28 minutes.
Part (h)
$5.55\times10^{7}$ or $5.65\times10^{7}$ km, and $2.95\times10^{8}$ or $3.05\times10^{8}$ m/s	B1	At least one correct bound for each quantity.
$\frac{5.65\times10^{10}}{2.95\times10^{8}}$ or $\frac{5.55\times10^{10}}{3.05\times10^{8}}$	M1	Upper bound of distance over lower bound of speed, or lower over upper.
182 and 192	A1	Both values (181.967… and 191.525…).

Answers: (a) $5.6\times10^{10}$ m · (b) 200 s, an overestimate · (c) 187 s · (d) 22 minutes (shown) · (e) 14% · (f)(i) $2^{10}$ seconds = 1024 seconds · (f)(ii) $2^{12}$ · (g) about 9.3 minutes · (h) lower bound 182 s, upper bound 192 s

Exam tip (b) In an estimate, write down every rounded number before you calculate. Then look at which way each number moved to decide whether the estimate is too big or too small.

Common error (h) For the upper bound of a quotient, divide the upper bound of the distance by the lower bound of the speed. Using both upper bounds does not give the greatest value.

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Saving for a gap year: ratio, interest and exchange rates

1.4 Estimation, rounding, bounds, ratio, proportion and money1.3 Indices, standard form and surds1.2 Fractions, decimals and percentagesCore

Lena and Tomás work in a café during the summer. Lena works for 84 hours and Tomás works for 60 hours. They are paid $1728 altogether, and they share it in the ratio of the hours they worked. They both save their shares for a gap-year trip.

(a)

Show that Lena receives $1008.

No calculatorROUTINENotes

Show solution

$1728\div(84+60)$ $(=12)$ or $84:60=7:5$ and $1728\div12$ $(=144)$	M1	Finding the value of one part.
$12\times84=1008$ or $7\times144=1008$	A1	cso

Answer: $1008 (shown)

(2)

(b)

Next summer, Tomás works 75 hours at the same rate of pay.

Work out how much he earns.

No calculatorROUTINENotes

Show solution

$1728\div144$ or $720\div60$ $(=12)$	M1	Pay per hour; or $720\times\frac{75}{60}$.
900	A1	Accept $900.00.

Answer: $900

(2)

(c)

Lena invests $1000 of her share in an account that pays simple interest at 3.5% per year.

Calculate the value of her account after 4 years.

ROUTINENotes

Show solution

$\frac{1000\times3.5\times4}{100}$ $(=140)$	M1
1140	A1	140 alone (the interest, not the value) scores M1 A0.

Answer: $1140

(2)

(d)

Tomás invests $700 of his share in an account that pays compound interest at 4% per year.

Show that the value of his account after 3 years is $787.40, correct to the nearest cent.

Extended onlyMULTINotes

Show solution

$700\times1.04^{3}$	M1	Or year-by-year: 728, 757.12, …
$787.4048$ leading to $787.40$	A1	cso. The unrounded value must be seen.

Answer: $787.40 (shown)

(2)

(e)(i)

Figure 1 shows the value of each account at the end of each year. Interest is added only at the end of each year.

Lena needs more than $1200 for her trip. Use Figure 1 to find the least number of complete years after which her account is worth more than $1200.

Figure 1: the value of each account at the end of each year (points joined with straight lines)

GRAPHNotes

Show solution

values about $1175 after 5 years and about $1210 after 6 years read from the graph, or the line crossing $1200 between 5 and 6 years	M1	Or $1000+35n\gt1200$ worked out.
6 years	A1	5.7 or 5.71 scores M1 A0: interest is added only at the end of a year.

Answer: 6 years

(2)

(e)(ii)

Explain why the points for Lena’s account lie on a straight line.

No calculatorINTERPRETNotes

Show solution

simple interest is worked out on the original $1000 only, so the same amount ($35) is added every year

Must mention the same amount each year.

(1)

(f)

The value of Tomás’s account after $n$ years is $700\times1.04^{n}$ dollars.

Use the laws of indices to explain why his account increases by the same percentage from the end of year 5 to the end of year 8 as it does in the first 3 years. Find this percentage.

Extended onlyINTERPRETNotes

Show solution

$\frac{700\times1.04^{8}}{700\times1.04^{5}}=1.04^{8-5}=1.04^{3}$	M1
the multiplier over the first 3 years is also $\frac{700\times1.04^{3}}{700}=1.04^{3}$, so the percentage increase is the same	A1	A conclusion comparing both multipliers is needed.
12.5	B1	Accept 12.4864% or 12.49%. From $1.04^{3}=1.124864$.

Answer: 12.5% (3 s.f.)

(3)

(g)

After 4 years, Lena takes all $1140 out of her account and changes it into euros. The exchange rate is $1 = €0.92.

A rail pass cost €480 last year. This year the price has increased by 8%.

Lena buys the rail pass. Work out how many euros she has left.

MULTINotes

Show solution

$1140\times0.92$ $(=1048.80)$	M1	Allow ECF from (c).
$480\times1.08$ $(=518.40)$	M1
530.40	A1	Money to 2 d.p.: €530.40. Allow ECF from (c): their $1140\times0.92-518.40$.

Answer: €530.40

(3)

(h)

Suppose instead that, after the first 2 years, the bank reduces the simple interest rate on Lena’s $1000 to 2.5% per year.

Calculate the value of her account after 5 years in total.

MULTINotes

Show solution

$\frac{1000\times3.5\times2}{100}$ $(=70)$	M1	Interest for the first 2 years.
$\frac{1000\times2.5\times3}{100}$ $(=75)$	M1	Interest for the last 3 years, on the original $1000.
1145	A1	Accept $1145.00. Using 3.5% for 5 years ($1175) scores M0 M0 A0; 145 alone (the interest) scores M1 M1 A0.

Answer: $1145

(3)

Show full solution

Working	Mark	Notes
Part (a)
$1728\div(84+60)$ $(=12)$ or $84:60=7:5$ and $1728\div12$ $(=144)$	M1	Finding the value of one part.
$12\times84=1008$ or $7\times144=1008$	A1	cso
Part (b)
$1728\div144$ or $720\div60$ $(=12)$	M1	Pay per hour; or $720\times\frac{75}{60}$.
900	A1	Accept $900.00.
Part (c)
$\frac{1000\times3.5\times4}{100}$ $(=140)$	M1
1140	A1	140 alone (the interest, not the value) scores M1 A0.
Part (d)
$700\times1.04^{3}$	M1	Or year-by-year: 728, 757.12, …
$787.4048$ leading to $787.40$	A1	cso. The unrounded value must be seen.
Part (e)(i)
values about $1175 after 5 years and about $1210 after 6 years read from the graph, or the line crossing $1200 between 5 and 6 years	M1	Or $1000+35n\gt1200$ worked out.
6 years	A1	5.7 or 5.71 scores M1 A0: interest is added only at the end of a year.
Part (e)(ii)
simple interest is worked out on the original $1000 only, so the same amount ($35) is added every year	B1	Must mention the same amount each year.
Part (f)
$\frac{700\times1.04^{8}}{700\times1.04^{5}}=1.04^{8-5}=1.04^{3}$	M1
the multiplier over the first 3 years is also $\frac{700\times1.04^{3}}{700}=1.04^{3}$, so the percentage increase is the same	A1	A conclusion comparing both multipliers is needed.
12.5	B1	Accept 12.4864% or 12.49%. From $1.04^{3}=1.124864$.
Part (g)
$1140\times0.92$ $(=1048.80)$	M1	Allow ECF from (c).
$480\times1.08$ $(=518.40)$	M1
530.40	A1	Money to 2 d.p.: €530.40. Allow ECF from (c): their $1140\times0.92-518.40$.
Part (h)
$\frac{1000\times3.5\times2}{100}$ $(=70)$	M1	Interest for the first 2 years.
$\frac{1000\times2.5\times3}{100}$ $(=75)$	M1	Interest for the last 3 years, on the original $1000.
1145	A1	Accept $1145.00. Using 3.5% for 5 years ($1175) scores M0 M0 A0; 145 alone (the interest) scores M1 M1 A0.

Answers: (a) $1008 (shown) · (b) $900 · (c) $1140 · (d) $787.40 (shown) · (e)(i) 6 years · (f) 12.5% (3 s.f.) · (g) €530.40 · (h) $1145

Common error (h) With simple interest, the new rate is still worked out on the original $1000, not on the value after 2 years. Only compound interest uses the value already reached.

More 1.4 questions in the question bank

Topic 2 · Algebra and graphs

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

Setup	2.1	2.2	2.3	2.4	2.5	Other
Resurfacing a five-a-side pitch: expanding, factorising and area	●	○				5.1
Comparing taxi and ride-share tariffs: forming and solving equations	○	●				3.2
A basketball shot: quadratic models, graphs and inequalities			●	○		5.1
Seats in a stadium stand: sequences, nth terms and graphs		○	○	●		—
Designing a rainwater tank: functions and differentiation to save material				○	●	5.3 1.3
A delivery drone's flight: function notation, gradients and stationary points		○		○	●	—

2.1 Expanding, factorising and algebraic fractions · 2.2 Linear and simultaneous equations and rearranging formulaemerged group

+ Add to test✓ In your testCC-CIM-21-22-119 marks~20 min

Resurfacing a five-a-side pitch: expanding, factorising and area

2.1 Expanding, factorising and algebraic fractions2.2 Linear and simultaneous equations and rearranging formulae5.1 Perimeter and area of plane shapesCore

A sports club is building a rectangular five-a-side football pitch. The pitch is $(2x+6)$ metres long and $(x+4)$ metres wide.

The pitch is surrounded on all four sides by a border of artificial turf that is 2 metres wide, as shown in Figure 1.

All lengths are in metres.

Figure 1: the pitch and its 2 m border (not to scale)

(a)

Find an expression for the area of the pitch. Give your answer in the form $ax^2+bx+c$.

No calculatorROUTINENotes

Show solution

$(2x+6)(x+4)$ expanded with at least 3 of the 4 terms $2x^2$, $8x$, $6x$, $24$ correct	M1
$2x^2+14x+24$	A1

Answer: $2x^2+14x+24$ m²

(2)

(b)

Show that the area of the border is $(12x+56)$ square metres.

No calculatorPROOFNotes

Show solution

Outer rectangle $(2x+10)$ by $(x+8)$	B1	Seen or used. Lengths $2x+8$ and $x+6$ (adding the width only once) score B0.
$(2x+10)(x+8)-(2x^2+14x+24)$ with $(2x+10)(x+8)=2x^2+26x+80$	M1	Subtracting their (a) from the expanded outer area. Alternative: strips $2\times2(2x+6)+2\times2(x+8)$ or equivalent split of the border into rectangles.
$2x^2+26x+80-2x^2-14x-24=12x+56$	A1	cso. Every step must be shown, with no sign errors.

Answer: $12x+56$ (shown)

(3)

(c)

Factorise $12x+56$ completely.

No calculatorROUTINENotes

Show solution

$4(3x+14)$

$2(6x+28)$ is not complete and scores B0.

Answer: $4(3x+14)$

(1)

(d)

The area of the border is 200 m².

Form an equation in $x$ and solve it to find the length and the width of the pitch.

No calculatorMODELNotes

Show solution

$12x+56=200$	M1	Or $4(3x+14)=200$. Allow a correct first step such as $12x=144$.
$x=12$	A1
length 30 m and width 16 m	A1	Both needed, with units. Allow ECF from their $x$ only if $x$ is positive.

Answer: $x=12$; 30 m by 16 m

(3)

(e)

The artificial turf for the border costs $24 per square metre. The club spends $6240 on turf for the border.

Use Figure 2 to find the value of $x$.

Figure 2: the area of the border, in m², for values of $x$ from 0 to 20

GRAPHNotes

Show solution

$6240\div24=260$	M1	Area of the border found.
$x=17$	A1	Read from the graph at a border area of 260 m². Accept answers from $16.8$ to $17.2$.

Answer: $x=17$

(2)

(f)

The pitch and its border must fit on a rectangular plot of land that is 46 m long and 30 m wide. The length of the pitch runs along the length of the plot.

Explain why $x$ cannot be greater than 18.

INTERPRETNotes

Show solution

The total length is $2x+10$, and $2x+10=46$ when $x=18$, so a larger $x$ makes the total length more than 46 m	B1	Accept $2x+10\le46$ leading to $x\le18$.
When $x=18$ the total width is $18+8=26$ m, which is less than 30 m, so it is the length that limits $x$	B1	Must refer to the width being within 30 m.

(2)

(g)

The club now decides to make the border $w$ metres wide on all four sides, instead of 2 metres.

Find an expression, in terms of $x$ and $w$, for the area of the border. Give your answer factorised completely.

No calculatorALGEBRANotes

Show solution

$(2x+6+2w)(x+4+2w)-(2x^2+14x+24)$	M1	Or strips: $2w(2x+6)+2w(x+4+2w)$, or equivalent.
$6wx+4w^2+20w$	A1	Any order of terms.
$2w(3x+2w+10)$	A1	Check: $w=2$ gives $4(3x+14)$. $w(6x+4w+20)$ or $2(3wx+2w^2+10w)$ is not complete and scores A0.

Answer: $2w(3x+2w+10)$

(3)

(h)

The fraction of the total area (pitch and border, with the 2 m border) that is border is $\dfrac{12x+56}{2x^2+26x+80}$.

Simplify this fraction fully.

Extended onlyNo calculatorALGEBRANotes

Show solution

Numerator $4(3x+14)$	B1
Denominator $2(x^2+13x+40)=2(x+5)(x+8)$	M1	Common factor 2 taken out and the quadratic factorised.
$\frac{2(3x+14)}{(x+5)(x+8)}$	A1	Accept $\frac{6x+28}{(x+5)(x+8)}$ or $\frac{6x+28}{x^2+13x+40}$. Check: $x=12$ gives $\frac{5}{17}$.

Answer: $\dfrac{2(3x+14)}{(x+5)(x+8)}$

(3)

Show full solution

Working	Mark	Notes
Part (a)
$(2x+6)(x+4)$ expanded with at least 3 of the 4 terms $2x^2$, $8x$, $6x$, $24$ correct	M1
$2x^2+14x+24$	A1
Part (b)
Outer rectangle $(2x+10)$ by $(x+8)$	B1	Seen or used. Lengths $2x+8$ and $x+6$ (adding the width only once) score B0.
$(2x+10)(x+8)-(2x^2+14x+24)$ with $(2x+10)(x+8)=2x^2+26x+80$	M1	Subtracting their (a) from the expanded outer area. Alternative: strips $2\times2(2x+6)+2\times2(x+8)$ or equivalent split of the border into rectangles.
$2x^2+26x+80-2x^2-14x-24=12x+56$	A1	cso. Every step must be shown, with no sign errors.
Part (c)
$4(3x+14)$	B1	$2(6x+28)$ is not complete and scores B0.
Part (d)
$12x+56=200$	M1	Or $4(3x+14)=200$. Allow a correct first step such as $12x=144$.
$x=12$	A1
length 30 m and width 16 m	A1	Both needed, with units. Allow ECF from their $x$ only if $x$ is positive.
Part (e)
$6240\div24=260$	M1	Area of the border found.
$x=17$	A1	Read from the graph at a border area of 260 m². Accept answers from $16.8$ to $17.2$.
Part (f)
The total length is $2x+10$, and $2x+10=46$ when $x=18$, so a larger $x$ makes the total length more than 46 m	B1	Accept $2x+10\le46$ leading to $x\le18$.
When $x=18$ the total width is $18+8=26$ m, which is less than 30 m, so it is the length that limits $x$	B1	Must refer to the width being within 30 m.
Part (g)
$(2x+6+2w)(x+4+2w)-(2x^2+14x+24)$	M1	Or strips: $2w(2x+6)+2w(x+4+2w)$, or equivalent.
$6wx+4w^2+20w$	A1	Any order of terms.
$2w(3x+2w+10)$	A1	Check: $w=2$ gives $4(3x+14)$. $w(6x+4w+20)$ or $2(3wx+2w^2+10w)$ is not complete and scores A0.
Part (h)
Numerator $4(3x+14)$	B1
Denominator $2(x^2+13x+40)=2(x+5)(x+8)$	M1	Common factor 2 taken out and the quadratic factorised.
$\frac{2(3x+14)}{(x+5)(x+8)}$	A1	Accept $\frac{6x+28}{(x+5)(x+8)}$ or $\frac{6x+28}{x^2+13x+40}$. Check: $x=12$ gives $\frac{5}{17}$.

Answers: (a) $2x^2+14x+24$ m² · (b) $12x+56$ (shown) · (c) $4(3x+14)$ · (d) $x=12$; 30 m by 16 m · (e) $x=17$ · (g) $2w(3x+2w+10)$ · (h) $\dfrac{2(3x+14)}{(x+5)(x+8)}$

Common error (b) The outer rectangle is 4 m longer and 4 m wider than the pitch, because the 2 m border is on both ends and on both sides. Adding only 2 m to each dimension gives the wrong outer area.

Exam tip (h) Take out the common factor first: $2x^2+26x+80=2(x^2+13x+40)$. Only factors that multiply the whole numerator and the whole denominator can be cancelled.

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Comparing taxi and ride-share tariffs: forming and solving equations

2.2 Linear and simultaneous equations and rearranging formulae2.1 Expanding, factorising and algebraic fractions3.2 Equation of a straight line, parallel and perpendicular linesCore

Two companies offer journeys across a city.

RideGo charges a fixed fee of $$f$ for each journey plus $$r$ for each kilometre travelled.

One week Lena made 3 journeys with RideGo. The total distance was 20 km and she paid $38.50 altogether.
In the same week Omar made 5 journeys with RideGo. The total distance was 26 km and he paid $53.90 altogether.

MetroTaxi also charges a fixed fee for each journey plus an amount for each kilometre. Figure 1 shows the cost of a MetroTaxi journey.

Figure 1: the cost, $C$ dollars, of a MetroTaxi journey of $d$ kilometres

(a)

Write down two equations in $f$ and $r$. Solve them to find the fixed fee and the charge per kilometre for RideGo.

MODELNotes

Show solution

$3f+20r=38.5$	B1	Each journey has its own fixed fee, so $3f$.
$5f+26r=53.9$	B1
Correct method to eliminate one letter, e.g. $15f+100r=192.5$ and $15f+78r=161.7$, then subtract to get $22r=30.8$	M1	Allow one arithmetic slip. Or eliminate $r$, or use substitution.
$f=3.5$ and $r=1.4$	A1	Fixed fee $3.50, $1.40 per km. Both needed.

Answer: fixed fee $3.50; $1.40 per km

(4)

(b)(i)

The total cost, $$T$, of $n$ RideGo journeys with a total distance of $D$ km is $T=3.5n+1.4D$.

Rearrange this formula to make $n$ the subject.

No calculatorALGEBRANotes

Show solution

$T-1.4D=3.5n$	M1	First step correct.
$n=\frac{T-1.4D}{3.5}$	A1	Accept $n=\frac{10T-14D}{35}$, $n=\frac{2T-2.8D}{7}$ or equivalent. $n=\frac{T}{3.5}-1.4D$ scores A0.

Answer: $n=\dfrac{T-1.4D}{3.5}$

(2)

(b)(ii)

In another week Priya travelled a total distance of 15 km with RideGo and paid $35.00 altogether.

Use your formula to find the number of journeys she made.

ROUTINENotes

Show solution

$n=\frac{35-1.4\times15}{3.5}=4$ journeys

Allow ECF from a formula in (b)(i) of the form $n=\frac{T-aD}{b}$ if it gives a whole number.

Answer: 4 journeys

(1)

(c)

Use Figure 1 to find the fixed fee and the charge per kilometre for MetroTaxi.

Write down a formula for the cost, $$C$, of a MetroTaxi journey of $d$ km.

GRAPHNotes

Show solution

Fixed fee $8 (the intercept on the cost axis)	B1
Gradient from two points on the line, e.g. $\frac{30-8}{20-0}$	M1	Any two points read correctly from the line.
$C=1.1d+8$	A1	Charge $1.10 per km. Accept $C=8+1.1d$.

Answer: $C=1.1d+8$

(3)

(d)

The cost, $$C$, of one RideGo journey of $d$ km is $C=1.4d+3.5$.

For a journey of $d$ km, the amount saved by choosing RideGo instead of MetroTaxi is $$S$.

Show that $S=0.3(15-d)$.

No calculatorPROOFNotes

Show solution

$S=(1.1d+8)-(1.4d+3.5)$	M1	MetroTaxi cost minus RideGo cost, with the brackets.
$1.1d+8-1.4d-3.5=4.5-0.3d$	M1	Correct removal of the brackets and collection of terms.
$4.5-0.3d=0.3(15-d)$	A1	cso. The common factor 0.3 must be taken out (or $0.3(15-d)$ expanded and matched) with no sign errors.

Answer: $S=0.3(15-d)$ (shown)

(3)

(e)

Use the formula in (d) to explain which company a passenger should choose. Your answer should refer to short and long journeys.

INTERPRETNotes

Show solution

$S=0$ when $d=15$: both companies cost the same ($24.50) for a 15 km journey	B1
For $d\lt 15$, $S$ is positive so RideGo is cheaper; for $d\gt 15$, $S$ is negative so MetroTaxi is cheaper	B1	Both cases needed, with a reason based on the sign of $S$ (or on comparing the costs).

(2)

(f)

At night, RideGo multiplies its charge per kilometre by 1.5. The fixed fee does not change. MetroTaxi charges stay the same.

Find the length of journey for which the two companies charge the same amount at night.

MODELNotes

Show solution

Night rate $1.4\times1.5=2.1$, so $C=2.1d+3.5$	B1
$2.1d+3.5=1.1d+8$	M1	Allow ECF from their night formula and their (c).
4.5 km	A1

Answer: 4.5 km

(3)

Show full solution

Working	Mark	Notes
Part (a)
$3f+20r=38.5$	B1	Each journey has its own fixed fee, so $3f$.
$5f+26r=53.9$	B1
Correct method to eliminate one letter, e.g. $15f+100r=192.5$ and $15f+78r=161.7$, then subtract to get $22r=30.8$	M1	Allow one arithmetic slip. Or eliminate $r$, or use substitution.
$f=3.5$ and $r=1.4$	A1	Fixed fee $3.50, $1.40 per km. Both needed.
Part (b)(i)
$T-1.4D=3.5n$	M1	First step correct.
$n=\frac{T-1.4D}{3.5}$	A1	Accept $n=\frac{10T-14D}{35}$, $n=\frac{2T-2.8D}{7}$ or equivalent. $n=\frac{T}{3.5}-1.4D$ scores A0.
Part (b)(ii)
$n=\frac{35-1.4\times15}{3.5}=4$ journeys	B1	Allow ECF from a formula in (b)(i) of the form $n=\frac{T-aD}{b}$ if it gives a whole number.
Part (c)
Fixed fee $8 (the intercept on the cost axis)	B1
Gradient from two points on the line, e.g. $\frac{30-8}{20-0}$	M1	Any two points read correctly from the line.
$C=1.1d+8$	A1	Charge $1.10 per km. Accept $C=8+1.1d$.
Part (d)
$S=(1.1d+8)-(1.4d+3.5)$	M1	MetroTaxi cost minus RideGo cost, with the brackets.
$1.1d+8-1.4d-3.5=4.5-0.3d$	M1	Correct removal of the brackets and collection of terms.
$4.5-0.3d=0.3(15-d)$	A1	cso. The common factor 0.3 must be taken out (or $0.3(15-d)$ expanded and matched) with no sign errors.
Part (e)
$S=0$ when $d=15$: both companies cost the same ($24.50) for a 15 km journey	B1
For $d\lt 15$, $S$ is positive so RideGo is cheaper; for $d\gt 15$, $S$ is negative so MetroTaxi is cheaper	B1	Both cases needed, with a reason based on the sign of $S$ (or on comparing the costs).
Part (f)
Night rate $1.4\times1.5=2.1$, so $C=2.1d+3.5$	B1
$2.1d+3.5=1.1d+8$	M1	Allow ECF from their night formula and their (c).
4.5 km	A1

Answers: (a) fixed fee $3.50; $1.40 per km · (b)(i) $n=\dfrac{T-1.4D}{3.5}$ · (b)(ii) 4 journeys · (c) $C=1.1d+8$ · (d) $S=0.3(15-d)$ (shown) · (f) 4.5 km

Common error (a) Lena made 3 journeys, so she paid the fixed fee 3 times: the equation is $3f+20r=38.5$, not $f+20r=38.5$.

Common error (d) The minus sign in front of the second bracket applies to both terms: $-(1.4d+3.5)=-1.4d-3.5$.

More 2.2 questions in the question bank

2.3 Quadratic equations and inequalities · 2.4 Sequences and graphs of functionsmerged group

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A basketball shot: quadratic models, graphs and inequalities

2.3 Quadratic equations and inequalities2.4 Sequences and graphs of functions5.1 Perimeter and area of plane shapesExtended

Dina shoots a basketball towards a hoop. The height, $h$ metres, of the ball above the floor $t$ seconds after it leaves her hands is modelled by

$h=2+7t-5t^2$.

The ring of the hoop is 3.05 m above the floor.

(a)

Complete the table of values for $h=2+7t-5t^2$.

$t$ (s)	0	0.2	0.4	0.6	0.8	1.0	1.2	1.4	1.6
$h$ (m)	2	3.2		4.4	4.4	4		2

No calculatorROUTINENotes

Show solution

$h=4$ when $t=0.4$	B1	Accept 4.0.
$h=3.2$ when $t=1.2$ and $h=0.4$ when $t=1.6$	B1	Both needed. The table is symmetrical about $t=0.7$, which is a useful check.

Answer: 4, 3.2, 0.4

(2)

(b)(i)

The graph of $h=2+7t-5t^2$ is drawn in Figure 1.

Use the graph to find the greatest height of the ball.

Figure 1: the graph of $h=2+7t-5t^2$ for $0\le t\le1.6$

GRAPHNotes

Show solution

4.45 m

Accept any value from 4.4 to 4.5.

Answer: about 4.45 m

(1)

(b)(ii)

Use the graph to find the two times when the ball is at the same height as the ring of the hoop.

GRAPHNotes

Show solution

$t=0.17$	B1	Accept 0.15 to 0.2. Line $h=3.05$ drawn or implied.
$t=1.23$	B1	Accept 1.2 to 1.25.

Answer: about 0.17 s and 1.23 s

(2)

(c)

Use the quadratic formula to find the two times when the ball is at the height of the ring. Give your answers correct to 3 significant figures.

Extended onlyROUTINENotes

Show solution

$5t^2-7t+1.05=0$	M1	Or $-5t^2+7t-1.05=0$, or any multiple, e.g. $100t^2-140t+21=0$.
$t=\frac{7\pm\sqrt{(-7)^2-4(5)(1.05)}}{2(5)}$	M1	Correct substitution into the formula for their equation; $b^2-4ac=28$.
$t=0.171$ and $t=1.23$	A1	Both needed. Answers from the graph alone, or without the formula, score 0.

Answer: 0.171 s and 1.23 s

(3)

(d)

The ball goes through the hoop.

Explain which of your two answers to (c) is the time when this happens.

INTERPRETNotes

Show solution

$t=1.23$ s, because at $t=0.171$ the ball is still rising (it reaches its greatest height at about $t=0.7$), but it must be coming down to drop through the hoop

Answer with a reason based on the ball rising/falling or the graph. Allow ECF from (c) for the larger root.

Answer: 1.23 s

(1)

(e)

By solving a quadratic inequality, show that the ball is more than 4 m above the floor for 0.6 seconds.

Extended onlyNo calculatorPROOFNotes

Show solution

$2+7t-5t^2\gt 4$, so $5t^2-7t+2\lt 0$	M1	Correct rearrangement, with the inequality sign reversed correctly if both sides are multiplied by $-1$.
$(5t-2)(t-1)\lt 0$, critical values $t=0.4$ and $t=1$	M1	Factorising or the formula.
$0.4\lt t\lt 1$, so the time is $1-0.4=0.6$ s	A1	cso. The interval must be between the roots (the curve $y=5t^2-7t+2$ is U shaped). Reading 0.4 and 1 from the table alone scores 0.

Answer: $0.4\lt t\lt 1$, so 0.6 s (shown)

(3)

(f)(i)

Behind the ring is a rectangular backboard. Dina aims her shot so that the ball drops into the ring without hitting the backboard.

The backboard has a perimeter of 5.7 m and an area of 1.89 m². Its width, $w$ metres, is greater than its height.

Form an equation in $w$ and solve it to find the width and the height of the backboard.

Extended onlyMODELNotes

Show solution

Height $=2.85-w$, so $w(2.85-w)=1.89$	M1	Or equivalent, e.g. $w^2-2.85w+1.89=0$.
$w=\frac{2.85\pm\sqrt{2.85^2-4(1)(1.89)}}{2}$, giving $w=1.8$ or $w=1.05$	M1	Correct method to solve their quadratic (formula, or factorising $100w^2-285w+189=(20w-21)(5w-9)$).
width 1.8 m, height 1.05 m	A1	Both values, correctly assigned (the width is the larger).

Answer: width 1.8 m, height 1.05 m

(3)

(f)(ii)

The lower edge of the backboard is 2.90 m above the floor.

Use your answers to (b)(i) and (f)(i) to explain whether the ball goes higher than the top edge of the backboard.

Extended onlyINTERPRETNotes

Show solution

Top edge $2.90+1.05=3.95$ m, which is less than the greatest height (about 4.45 m), so yes, the ball goes higher than the top of the backboard

Both heights and a conclusion needed. Allow ECF from their height in (f)(i) and their (b)(i).

Answer: Yes (3.95 m < 4.45 m)

(1)

(g)

For her next shot Dina throws the ball more gently, so the model becomes $h=2+6t-5t^2$.

Use the discriminant to determine whether the ball now reaches a height of 4 m.

Extended onlyNo calculatorINTERPRETNotes

Show solution

$2+6t-5t^2=4$, so $5t^2-6t+2=0$	M1	Or $-5t^2+6t-2=0$.
$b^2-4ac=(-6)^2-4(5)(2)=36-40$	M1	For their equation.
$-4\lt 0$, so there are no real solutions: the ball does not reach 4 m	A1	Conclusion in context needed.

Answer: No, the discriminant is $-4$

(3)

Show full solution

Working	Mark	Notes
Part (a)
$h=4$ when $t=0.4$	B1	Accept 4.0.
$h=3.2$ when $t=1.2$ and $h=0.4$ when $t=1.6$	B1	Both needed. The table is symmetrical about $t=0.7$, which is a useful check.
Part (b)(i)
4.45 m	B1	Accept any value from 4.4 to 4.5.
Part (b)(ii)
$t=0.17$	B1	Accept 0.15 to 0.2. Line $h=3.05$ drawn or implied.
$t=1.23$	B1	Accept 1.2 to 1.25.
Part (c)
$5t^2-7t+1.05=0$	M1	Or $-5t^2+7t-1.05=0$, or any multiple, e.g. $100t^2-140t+21=0$.
$t=\frac{7\pm\sqrt{(-7)^2-4(5)(1.05)}}{2(5)}$	M1	Correct substitution into the formula for their equation; $b^2-4ac=28$.
$t=0.171$ and $t=1.23$	A1	Both needed. Answers from the graph alone, or without the formula, score 0.
Part (d)
$t=1.23$ s, because at $t=0.171$ the ball is still rising (it reaches its greatest height at about $t=0.7$), but it must be coming down to drop through the hoop	B1	Answer with a reason based on the ball rising/falling or the graph. Allow ECF from (c) for the larger root.
Part (e)
$2+7t-5t^2\gt 4$, so $5t^2-7t+2\lt 0$	M1	Correct rearrangement, with the inequality sign reversed correctly if both sides are multiplied by $-1$.
$(5t-2)(t-1)\lt 0$, critical values $t=0.4$ and $t=1$	M1	Factorising or the formula.
$0.4\lt t\lt 1$, so the time is $1-0.4=0.6$ s	A1	cso. The interval must be between the roots (the curve $y=5t^2-7t+2$ is U shaped). Reading 0.4 and 1 from the table alone scores 0.
Part (f)(i)
Height $=2.85-w$, so $w(2.85-w)=1.89$	M1	Or equivalent, e.g. $w^2-2.85w+1.89=0$.
$w=\frac{2.85\pm\sqrt{2.85^2-4(1)(1.89)}}{2}$, giving $w=1.8$ or $w=1.05$	M1	Correct method to solve their quadratic (formula, or factorising $100w^2-285w+189=(20w-21)(5w-9)$).
width 1.8 m, height 1.05 m	A1	Both values, correctly assigned (the width is the larger).
Part (f)(ii)
Top edge $2.90+1.05=3.95$ m, which is less than the greatest height (about 4.45 m), so yes, the ball goes higher than the top of the backboard	B1	Both heights and a conclusion needed. Allow ECF from their height in (f)(i) and their (b)(i).
Part (g)
$2+6t-5t^2=4$, so $5t^2-6t+2=0$	M1	Or $-5t^2+6t-2=0$.
$b^2-4ac=(-6)^2-4(5)(2)=36-40$	M1	For their equation.
$-4\lt 0$, so there are no real solutions: the ball does not reach 4 m	A1	Conclusion in context needed.

Answers: (a) 4, 3.2, 0.4 · (b)(i) about 4.45 m · (b)(ii) about 0.17 s and 1.23 s · (c) 0.171 s and 1.23 s · (d) 1.23 s · (e) $0.4\lt t\lt 1$, so 0.6 s (shown) · (f)(i) width 1.8 m, height 1.05 m · (f)(ii) Yes (3.95 m < 4.45 m) · (g) No, the discriminant is $-4$

Common error (c) Rearrange to $\ldots=0$ before using the formula. With $2+7t-5t^2=3.05$ the constant becomes $1.05$ (or $-1.05$), not 2 or 3.05.

Exam tip (g) A negative discriminant means the line $h=4$ never meets the curve: the greatest height is now below 4 m (it is 3.8 m).

More 2.3 questions in the question bank

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Seats in a stadium stand: sequences, nth terms and graphs

2.4 Sequences and graphs of functions2.3 Quadratic equations and inequalities2.2 Linear and simultaneous equations and rearranging formulaeCore

The rows of seats in a new stand at a stadium are numbered 1, 2, 3, … starting from the front.

The numbers of seats in the rows form a sequence. Row 1 has 24 seats, row 2 has 28 seats, row 3 has 32 seats, and so on:

24, 28, 32, 36, …

(a)

Find an expression, in terms of $n$, for the number of seats in row $n$.

No calculatorROUTINENotes

Show solution

$4n+k$ with $k\ne0$, or $4n$ seen	M1	The common difference 4 used as the coefficient of $n$.
$4n+20$	A1	$n+4$ scores M0 A0.

Answer: $4n+20$

(2)

(b)

Determine whether a row of this stand can hold exactly 150 seats. Show your working.

No calculatorROUTINENotes

Show solution

$4n+20=150$, so $4n=130$	M1	Allow ECF from their (a).
$n=32.5$, which is not a whole number, so no row holds exactly 150 seats	A1	Conclusion with reason needed. Listing terms 148, 152 with a conclusion also scores M1 A1.

Answer: No ($n=32.5$)

(2)

(c)

Figure 1 shows the number of seats in each of the first 10 rows, plotted against the row number.

The points lie on a straight line. Write down the gradient of this line and state what it represents. Write down the value where the line, extended, would meet the vertical axis.

Figure 1: the number of seats in each of the first 10 rows

No calculatorGRAPHNotes

Show solution

Gradient 4: each row has 4 more seats than the row in front of it	B1	Value and meaning both needed.
20 (the constant in $4n+20$)	B1

Answer: gradient 4; meets the vertical axis at 20

(2)

(d)

Show that the first 10 rows hold a total of 420 seats.

No calculatorPROOFNotes

Show solution

$24+28+32+36+40+44+48+52+56+60$	M1	All ten terms, with the 10th term 60 (from $4(10)+20$). Or pairing: $5\times(24+60)$.
$=420$	A1	cso.

Answer: 420 (shown)

(2)

(e)

The total numbers of seats in the first 1, 2, 3, 4, … rows form the sequence

24, 52, 84, 120, …

Find an expression, in terms of $n$, for the total number of seats in the first $n$ rows.

Extended onlyNo calculatorALGEBRANotes

Show solution

First differences 28, 32, 36 and second differences 4, so the expression starts $2n^2$	M1	Half of the second difference.
Subtracting $2n^2$: $24-2=22$, $52-8=44$, $84-18=66$, $120-32=88$, giving $22n$	M1	Or forming and solving equations in $a$, $b$, $c$ from $an^2+bn+c$.
$2n^2+22n$	A1	Check: $n=10$ gives $200+220=420$.

Answer: $2n^2+22n$

(3)

(f)

The total number of seats in the first $n$ rows is $2n^2+22n$. The whole stand holds 1240 seats.

Form an equation and solve it by factorising to find the number of rows in the stand.

No calculatorMODELNotes

Show solution

$2n^2+22n=1240$, so $n^2+11n-620=0$	M1	Or $2n^2+22n-1240=0$.
$(n+31)(n-20)=0$	M1	Or $(2n+62)(n-20)=0$, or $2(n+31)(n-20)=0$.
$n=20$ rows, since $n=-31$ is impossible (the number of rows cannot be negative)	A1	The negative root must be rejected.

Answer: 20 rows

(3)

(g)(i)

Figure 2 shows the total number of seats in the first $n$ rows, for $n=1$ to 19.

For a cup match the club opens rows starting from the front. It needs at least 1000 seats.

Use Figure 2 to find the smallest number of rows that must be opened.

Figure 2: the total number of seats in the first $n$ rows

GRAPHNotes

Show solution

Line at 1000 used: the total for 17 rows is below 1000 and the total for 18 rows is above 1000	M1	Accept 17 rows giving about 950 and 18 rows giving about 1050 read from the graph.
18 rows	A1	17.5 or 17 scores A0.

Answer: 18 rows

(2)

(g)(ii)

Explain why the points in Figure 2 do not lie on a straight line.

INTERPRETNotes

Show solution

Each new row adds more seats than the row before (4 more each time), so the total increases by a larger amount each time

Accept: the first differences are not constant, or the total is a quadratic in $n$. Not just 'it is a curve'.

(1)

(h)

The stadium designs a second stand. Row 1 again has 24 seats, but now each row has 6 more seats than the row in front of it. The last row has 132 seats.

Find the number of rows in the second stand.

No calculatorMODELNotes

Show solution

$n$th term $6n+18$	B1
$6n+18=132$	M1	Allow ECF from their $n$th term of the form $6n+k$.
19 rows	A1

Answer: 19 rows

(3)

Show full solution

Working	Mark	Notes
Part (a)
$4n+k$ with $k\ne0$, or $4n$ seen	M1	The common difference 4 used as the coefficient of $n$.
$4n+20$	A1	$n+4$ scores M0 A0.
Part (b)
$4n+20=150$, so $4n=130$	M1	Allow ECF from their (a).
$n=32.5$, which is not a whole number, so no row holds exactly 150 seats	A1	Conclusion with reason needed. Listing terms 148, 152 with a conclusion also scores M1 A1.
Part (c)
Gradient 4: each row has 4 more seats than the row in front of it	B1	Value and meaning both needed.
20 (the constant in $4n+20$)	B1
Part (d)
$24+28+32+36+40+44+48+52+56+60$	M1	All ten terms, with the 10th term 60 (from $4(10)+20$). Or pairing: $5\times(24+60)$.
$=420$	A1	cso.
Part (e)
First differences 28, 32, 36 and second differences 4, so the expression starts $2n^2$	M1	Half of the second difference.
Subtracting $2n^2$: $24-2=22$, $52-8=44$, $84-18=66$, $120-32=88$, giving $22n$	M1	Or forming and solving equations in $a$, $b$, $c$ from $an^2+bn+c$.
$2n^2+22n$	A1	Check: $n=10$ gives $200+220=420$.
Part (f)
$2n^2+22n=1240$, so $n^2+11n-620=0$	M1	Or $2n^2+22n-1240=0$.
$(n+31)(n-20)=0$	M1	Or $(2n+62)(n-20)=0$, or $2(n+31)(n-20)=0$.
$n=20$ rows, since $n=-31$ is impossible (the number of rows cannot be negative)	A1	The negative root must be rejected.
Part (g)(i)
Line at 1000 used: the total for 17 rows is below 1000 and the total for 18 rows is above 1000	M1	Accept 17 rows giving about 950 and 18 rows giving about 1050 read from the graph.
18 rows	A1	17.5 or 17 scores A0.
Part (g)(ii)
Each new row adds more seats than the row before (4 more each time), so the total increases by a larger amount each time	B1	Accept: the first differences are not constant, or the total is a quadratic in $n$. Not just 'it is a curve'.
Part (h)
$n$th term $6n+18$	B1
$6n+18=132$	M1	Allow ECF from their $n$th term of the form $6n+k$.
19 rows	A1

Answers: (a) $4n+20$ · (b) No ($n=32.5$) · (c) gradient 4; meets the vertical axis at 20 · (d) 420 (shown) · (e) $2n^2+22n$ · (f) 20 rows · (g)(i) 18 rows · (h) 19 rows

Common error (e) The coefficient of $n^2$ is half the second difference: $4\div2=2$, not 4.

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2.5 Functions and differentiation

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Designing a rainwater tank: functions and differentiation to save material

2.5 Functions and differentiation5.3 Surface area and volume of solids1.3 Indices, standard form and surds2.4 Sequences and graphs of functionsExtended

A company makes closed cylindrical rainwater tanks from sheet metal. Each tank must hold $1.5$ m³ of water.

A tank has radius $r$ metres and height $h$ metres. The total area of sheet metal used for the curved surface, the base and the lid is $A$ m².

(a)

Show that $h = \dfrac{1.5}{\pi r^2}$ and hence that $A = 2\pi r^2 + \dfrac{3}{r}$.

No calculatorPROOFNotes

Show solution

$\pi r^2 h = 1.5$, so $h = \frac{1.5}{\pi r^2}$	M1	Volume of a cylinder set equal to 1.5.
$A = 2\pi r^2 + 2\pi r h$	M1	Two circular ends plus the curved surface; may be seen with $h$ already replaced.
$A = 2\pi r^2 + 2\pi r\times\frac{1.5}{\pi r^2} = 2\pi r^2 + \frac{3}{r}$	A1	cso. The substitution and the cancelling to $\frac{3}{r}$ must be seen.

Answer: $A = 2\pi r^2 + \dfrac{3}{r}$

(3)

(b)(i)

The function $A(r) = 2\pi r^2 + \dfrac{3}{r}$ gives the area of sheet metal for a tank of radius $r$ metres.

Calculate $A(0.8)$, giving your answer correct to 3 significant figures.

Extended onlyROUTINENotes

Show solution

$2\pi(0.8)^2 + \frac{3}{0.8}$	M1	Correct substitution.
$7.77$	A1	Accept 7.771… .

Answer: $7.77$ m²

(2)

(b)(ii)

The supplier charges $C(x) = 25x + 60$ dollars for an order of $x$ m² of sheet metal.

Find $C(A(0.8))$ and state what it represents.

Extended onlyROUTINENotes

Show solution

$\$254.28$ (or $\$254$), the cost of the sheet metal for a tank of radius 0.8 m

Allow ECF: $25\times$ their (b)(i) $+\,60$. Both the value and the meaning needed. Accept $\$254.25$ (from 7.77).

Answer: $\$254.28$

(1)

(c)

Write down $\dfrac{3}{r}$ in the form $3r^{n}$, and hence find $\dfrac{dA}{dr}$.

Extended onlyNo calculatorROUTINENotes

Show solution

$3r^{-1}$	B1
$\frac{dA}{dr} = 4\pi r - 3r^{-2}$	B1	Accept $4\pi r - \frac{3}{r^2}$. Allow ECF for the second term from an incorrect power in the first mark.

Answer: $\dfrac{dA}{dr} = 4\pi r - \dfrac{3}{r^2}$

(2)

(d)

Figure 1 shows the graph of $A$ against $r$. A tangent to the curve has been drawn at the point where $r = 1.2$.

Use the tangent to find an estimate of the gradient of the curve at $r = 1.2$.

Figure 1: the graph of $A$ against $r$ for $0.2 \le r \le 1.4$, with the tangent drawn at $r = 1.2$

Extended onlyGRAPHNotes

Show solution

$\frac{\text{rise}}{\text{run}}$ using two points on the tangent, e.g. $\frac{14.1 - 6.3}{1.4 - 0.8}$	M1	Points should be read from the tangent, not the curve, and be well apart.
$13$	A1	Accept answers from 12 to 14. (The exact value from (c) is 12.996….)

Answer: about $13$

(2)

(e)

Find the value of $r$ for which $A$ has a stationary value. Show that this value of $r$ gives a minimum.

Extended onlyMULTINotes

Show solution

$4\pi r - \frac{3}{r^2} = 0$ leading to $r^3 = \frac{3}{4\pi}$	M1	Their derivative from (c) set equal to 0 and rearranged.
$r = 0.620$	A1	Accept 0.6203… .
$\frac{d^2A}{dr^2} = 4\pi + \frac{6}{r^3}$, evaluated or its sign considered at their $r$	M1	Alternative: gradient values either side, e.g. at $r = 0.6$ and $r = 0.65$, with signs.
$37.7 \gt 0$ (or positive because $r \gt 0$), so a minimum	A1	Conclusion needed. For the alternative: negative then positive, so a minimum.

Answer: $r = 0.620$ m, a minimum

(4)

(f)

Find the least possible area of sheet metal for one tank.

Extended onlyROUTINENotes

Show solution

$2\pi(0.6203\ldots)^2 + \frac{3}{0.6203\ldots}$	M1	Their $r$ from (e) substituted into $A$.
$7.25$ m²	A1	Accept 7.253… to 7.254.

Answer: $7.25$ m²

(2)

(g)

Look at the shape of the graph in Figure 1 near the minimum point.

Suggest one reason why the company might choose a radius of $0.5$ m instead of the value found in (e).

Extended onlyINTERPRETNotes

Show solution

A valid reason, e.g. the curve is very flat near the minimum, so $r = 0.5$ uses only a little more metal ($A(0.5) = 7.57$), and a narrower, taller tank takes up less ground space

Accept other practical reasons linked to the context, e.g. fitting a narrow space beside a wall, or standard sheet widths. Not just “it is cheaper”.

(1)

(h)

The company also makes an open tank, with no lid, that holds $1.5$ m³.

Find the radius of the open tank that uses the least sheet metal.

Extended onlyMODELNotes

Show solution

$A = \pi r^2 + \frac{3}{r}$	B1	One circular end only.
$2\pi r - \frac{3}{r^2} = 0$	M1	Their new $A$ differentiated and set equal to 0.
$r = 0.782$	A1	Accept 0.7816… . From $r^3 = \frac{3}{2\pi}$.

Answer: $r = 0.782$ m

(3)

Show full solution

Working	Mark	Notes
Part (a)
$\pi r^2 h = 1.5$, so $h = \frac{1.5}{\pi r^2}$	M1	Volume of a cylinder set equal to 1.5.
$A = 2\pi r^2 + 2\pi r h$	M1	Two circular ends plus the curved surface; may be seen with $h$ already replaced.
$A = 2\pi r^2 + 2\pi r\times\frac{1.5}{\pi r^2} = 2\pi r^2 + \frac{3}{r}$	A1	cso. The substitution and the cancelling to $\frac{3}{r}$ must be seen.
Part (b)(i)
$2\pi(0.8)^2 + \frac{3}{0.8}$	M1	Correct substitution.
$7.77$	A1	Accept 7.771… .
Part (b)(ii)
$\$254.28$ (or $\$254$), the cost of the sheet metal for a tank of radius 0.8 m	B1	Allow ECF: $25\times$ their (b)(i) $+\,60$. Both the value and the meaning needed. Accept $\$254.25$ (from 7.77).
Part (c)
$3r^{-1}$	B1
$\frac{dA}{dr} = 4\pi r - 3r^{-2}$	B1	Accept $4\pi r - \frac{3}{r^2}$. Allow ECF for the second term from an incorrect power in the first mark.
Part (d)
$\frac{\text{rise}}{\text{run}}$ using two points on the tangent, e.g. $\frac{14.1 - 6.3}{1.4 - 0.8}$	M1	Points should be read from the tangent, not the curve, and be well apart.
$13$	A1	Accept answers from 12 to 14. (The exact value from (c) is 12.996….)
Part (e)
$4\pi r - \frac{3}{r^2} = 0$ leading to $r^3 = \frac{3}{4\pi}$	M1	Their derivative from (c) set equal to 0 and rearranged.
$r = 0.620$	A1	Accept 0.6203… .
$\frac{d^2A}{dr^2} = 4\pi + \frac{6}{r^3}$, evaluated or its sign considered at their $r$	M1	Alternative: gradient values either side, e.g. at $r = 0.6$ and $r = 0.65$, with signs.
$37.7 \gt 0$ (or positive because $r \gt 0$), so a minimum	A1	Conclusion needed. For the alternative: negative then positive, so a minimum.
Part (f)
$2\pi(0.6203\ldots)^2 + \frac{3}{0.6203\ldots}$	M1	Their $r$ from (e) substituted into $A$.
$7.25$ m²	A1	Accept 7.253… to 7.254.
Part (g)
A valid reason, e.g. the curve is very flat near the minimum, so $r = 0.5$ uses only a little more metal ($A(0.5) = 7.57$), and a narrower, taller tank takes up less ground space	B1	Accept other practical reasons linked to the context, e.g. fitting a narrow space beside a wall, or standard sheet widths. Not just “it is cheaper”.
Part (h)
$A = \pi r^2 + \frac{3}{r}$	B1	One circular end only.
$2\pi r - \frac{3}{r^2} = 0$	M1	Their new $A$ differentiated and set equal to 0.
$r = 0.782$	A1	Accept 0.7816… . From $r^3 = \frac{3}{2\pi}$.

Answers: (a) $A = 2\pi r^2 + \dfrac{3}{r}$ · (b)(i) $7.77$ m² · (b)(ii) $\$254.28$ · (c) $\dfrac{dA}{dr} = 4\pi r - \dfrac{3}{r^2}$ · (d) about $13$ · (e) $r = 0.620$ m, a minimum · (f) $7.25$ m² · (h) $r = 0.782$ m

Common error (a) A closed tank has two circular ends. Using $\pi r^2 + 2\pi r h$ is the formula for an open tank, which is only right in part (h).

Exam tip (e) Keep the full calculator value of $r$ (0.62035…) in the memory and use it in (f); a rounded radius can change the third significant figure of the area.

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A delivery drone's flight: function notation, gradients and stationary points

2.5 Functions and differentiation2.4 Sequences and graphs of functions2.2 Linear and simultaneous equations and rearranging formulaeExtended

During a test flight, the height of a delivery drone above the ground, $h$ metres, $t$ minutes after take-off is modelled by

$h(t) = t^3 - 12t^2 + 36t$, for $0 \le t \le 6$.

The drone's display shows heights in feet. It converts a height of $x$ metres to feet using the function $f(x) = 3.28x$.

(a)(i)

Find $h(2)$ and $f(h(2))$.

Extended onlyROUTINENotes

Show solution

$h(2) = 8 - 48 + 72 = 32$	B1
$f(h(2)) = 3.28\times32 = 104.96$	B1	Accept 105. Allow ECF: $3.28\times$ their $h(2)$.

Answer: $32$; $104.96$

(2)

(a)(ii)

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

Extended onlyNo calculatorALGEBRANotes

Show solution

$f^{-1}(x) = \frac{x}{3.28}$

Accept equivalent forms. $\frac{1}{3.28x}$ scores B0.

Answer: $f^{-1}(x) = \dfrac{x}{3.28}$

(1)

(b)

Find $\dfrac{dh}{dt}$.

Extended onlyNo calculatorROUTINENotes

Show solution

Two of the three terms $3t^2$, $-24t$, $+36$ correct	M1
$\frac{dh}{dt} = 3t^2 - 24t + 36$	A1

Answer: $3t^2 - 24t + 36$

(2)

(c)

Show that, at $t = 6$, the drone is back on the ground and its vertical speed is zero.

Extended onlyNo calculatorPROOFNotes

Show solution

$h(6) = 216 - 432 + 216 = 0$	M1	Substitution into $h$ with the values seen.
$\frac{dh}{dt} = 3(36) - 24(6) + 36 = 108 - 144 + 36 = 0$	M1	Substitution into their $\frac{dh}{dt}$ with the values seen.
Both equal 0, so the drone is at ground level with zero vertical speed	A1	cso. Both results and a conclusion.

Answer: $h(6) = 0$ and $\dfrac{dh}{dt} = 0$

(3)

(d)

Find the coordinates of the other stationary point of $h(t)$ and determine whether it is a maximum or a minimum.

Extended onlyNo calculatorMULTINotes

Show solution

$3t^2 - 24t + 36 = 0$ leading to $(t-2)(t-6) = 0$, $t = 2$	M1	Their derivative set equal to 0 and solved.
$\frac{d^2h}{dt^2} = 6t - 24$, which is $-12$ at $t = 2$	M1	Or the sign of $\frac{dh}{dt}$ either side of $t = 2$.
Maximum at $(2, 32)$	A1	Coordinates and the type needed.

Answer: $(2, 32)$, a maximum

(3)

(e)

The flight controller uses the tangent to the curve at $t = 1$ to make a short-term prediction of the height.

Find the equation of this tangent in the form $h = mt + c$.

Extended onlyNo calculatorALGEBRANotes

Show solution

Gradient $= 3 - 24 + 36 = 15$	M1	Their $\frac{dh}{dt}$ at $t = 1$.
$h - 25 = 15(t - 1)$	M1	Using the point $(1, 25)$ with their gradient. Or $25 = 15(1) + c$.
$h = 15t + 10$	A1

Answer: $h = 15t + 10$

(3)

(f)

Figure 1 shows the graph of $h(t)$.

Use the graph to find the two values of $t$ for which $h(t) = 20$.

Figure 1: the graph of $h(t) = t^3 - 12t^2 + 36t$ for $0 \le t \le 6$

GRAPHNotes

Show solution

$t = 0.7$	B1	Accept 0.6 to 0.8.
$t = 3.7$	B1	Accept 3.6 to 3.8.

Answer: $t = 0.7$ and $t = 3.7$

(2)

(g)

Find the value of $\dfrac{dh}{dt}$ when $t = 4$ and explain what it tells you about the drone.

Extended onlyNo calculatorINTERPRETNotes

Show solution

$\frac{dh}{dt} = 48 - 96 + 36 = -12$	B1	Allow ECF from their (b).
The drone is coming down: its height is decreasing at 12 metres per minute	B1	Must mention descending (or height decreasing) and the rate with units.

Answer: $-12$

(2)

(h)

On a second test, the drone records data only between $t = 3$ and $t = 5$.

Find the greatest height recorded on this test. Give a reason for your answer.

Extended onlyCONCEPTNotes

Show solution

$\frac{dh}{dt} \lt 0$ for $2 \lt t \lt 6$ (or the graph falls), so the greatest value is at $t = 3$	M1	Reason based on the gradient, the maximum at $t = 2$ or the graph.
$h(3) = 27 - 108 + 108 = 27$ m	A1	32 m scores M0 A0: $t = 2$ is outside the new interval.

Answer: $27$ m

(2)

Show full solution

Working	Mark	Notes
Part (a)(i)
$h(2) = 8 - 48 + 72 = 32$	B1
$f(h(2)) = 3.28\times32 = 104.96$	B1	Accept 105. Allow ECF: $3.28\times$ their $h(2)$.
Part (a)(ii)
$f^{-1}(x) = \frac{x}{3.28}$	B1	Accept equivalent forms. $\frac{1}{3.28x}$ scores B0.
Part (b)
Two of the three terms $3t^2$, $-24t$, $+36$ correct	M1
$\frac{dh}{dt} = 3t^2 - 24t + 36$	A1
Part (c)
$h(6) = 216 - 432 + 216 = 0$	M1	Substitution into $h$ with the values seen.
$\frac{dh}{dt} = 3(36) - 24(6) + 36 = 108 - 144 + 36 = 0$	M1	Substitution into their $\frac{dh}{dt}$ with the values seen.
Both equal 0, so the drone is at ground level with zero vertical speed	A1	cso. Both results and a conclusion.
Part (d)
$3t^2 - 24t + 36 = 0$ leading to $(t-2)(t-6) = 0$, $t = 2$	M1	Their derivative set equal to 0 and solved.
$\frac{d^2h}{dt^2} = 6t - 24$, which is $-12$ at $t = 2$	M1	Or the sign of $\frac{dh}{dt}$ either side of $t = 2$.
Maximum at $(2, 32)$	A1	Coordinates and the type needed.
Part (e)
Gradient $= 3 - 24 + 36 = 15$	M1	Their $\frac{dh}{dt}$ at $t = 1$.
$h - 25 = 15(t - 1)$	M1	Using the point $(1, 25)$ with their gradient. Or $25 = 15(1) + c$.
$h = 15t + 10$	A1
Part (f)
$t = 0.7$	B1	Accept 0.6 to 0.8.
$t = 3.7$	B1	Accept 3.6 to 3.8.
Part (g)
$\frac{dh}{dt} = 48 - 96 + 36 = -12$	B1	Allow ECF from their (b).
The drone is coming down: its height is decreasing at 12 metres per minute	B1	Must mention descending (or height decreasing) and the rate with units.
Part (h)
$\frac{dh}{dt} \lt 0$ for $2 \lt t \lt 6$ (or the graph falls), so the greatest value is at $t = 3$	M1	Reason based on the gradient, the maximum at $t = 2$ or the graph.
$h(3) = 27 - 108 + 108 = 27$ m	A1	32 m scores M0 A0: $t = 2$ is outside the new interval.

Answers: (a)(i) $32$; $104.96$ · (a)(ii) $f^{-1}(x) = \dfrac{x}{3.28}$ · (b) $3t^2 - 24t + 36$ · (c) $h(6) = 0$ and $\dfrac{dh}{dt} = 0$ · (d) $(2, 32)$, a maximum · (e) $h = 15t + 10$ · (f) $t = 0.7$ and $t = 3.7$ · (g) $-12$ · (h) $27$ m

Common error (g) A negative gradient does not mean the drone is below the ground. It means the height is decreasing; the drone is still 16 m up at $t = 4$.

More 2.5 questions in the question bank

Topic 3 · Coordinate geometry

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

Setup	3.1	3.2	Other
Building a zip line between two towers: gradient, midpoint and length	●	○	6.1
Planning city cycle lanes on a coordinate map	○	●	2.2

3.1 Gradient, midpoint and length of a line segment · 3.2 Equation of a straight line, parallel and perpendicular linesmerged group

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

Building a zip line between two towers: gradient, midpoint and length

3.1 Gradient, midpoint and length of a line segment3.2 Equation of a straight line, parallel and perpendicular lines6.1 Pythagoras' theorem and right-angled trigonometryCore

An adventure park is building a zip line. On the plan, $x$ is the horizontal distance in metres from a fence post $O$, and $y$ is the height in metres above the ground.

The cable starts at the top of a launch tower, $A(5, 20)$, and ends at a landing platform, $B(85, 4)$. The ground is horizontal and the cable is modelled as a straight line segment $AB$.

(a)

Find the gradient of $AB$.

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$\frac{4 - 20}{85 - 5}$	M1	Same order on top and bottom.
$-0.2$	A1	Accept $-\frac{1}{5}$ or $-\frac{16}{80}$. $0.2$ scores M1 A0.

Answer: $-0.2$

(2)

(b)

A support pole is placed vertically from the ground to the midpoint, $M$, of the cable.

Find the coordinates of $M$ and write down the height of the pole.

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$\left(\frac{5 + 85}{2}, \frac{20 + 4}{2}\right) = (45, 12)$	B1	Both coordinates correct.
Height $12$ m	B1	Allow ECF: their $y$-coordinate of $M$.

Answer: $M(45, 12)$; $12$ m

(2)

(c)

Find the equation of the line through $A$ and $B$ in the form $y = mx + c$.

No calculatorALGEBRANotes

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$20 = -0.2(5) + c$ or $4 = -0.2(85) + c$	M1	Their gradient from (a) with a point on the line.
$y = -0.2x + 21$	A1	Accept $y = -\frac{1}{5}x + 21$.

Answer: $y = -0.2x + 21$

(2)

(d)

On the grid in Figure 1, plot the points $A$ and $B$ and join them with a ruled straight line to show the cable $AB$.

Use your line to find the value of $x$ where the cable is 10 m above the ground.

Figure 1: grid for plotting the zip line

GRAPHNotes

Show solution

$A$ and $B$ plotted correctly and joined with a ruled line	B1	Within half a small square.
$x = 55$	B1	Accept 53 to 57. Check: $-0.2(55) + 21 = 10$.

Answer: $x = 55$

(2)

(e)

Show that the length of the cable is 81.6 m, correct to 3 significant figures.

PROOFNotes

Show solution

Horizontal distance $80$ and vertical drop $16$	M1	Both differences seen, or in the formula.
$\sqrt{80^2 + 16^2} = \sqrt{6656}$	M1	Pythagoras or the length formula.
$81.58\ldots$, so $81.6$ m	A1	cso. A value with more than 3 s.f. must be seen before rounding.

Answer: $81.6$ m

(3)

(f)

Calculate the angle that the cable makes with the horizontal.

ROUTINENotes

Show solution

$\tan\theta = \frac{16}{80}$	M1	Or $\sin\theta = \frac{16}{81.58\ldots}$ or $\cos\theta = \frac{80}{81.58\ldots}$.
$11.3^\circ$	A1	Accept 11.30… .

Answer: $11.3^\circ$

(2)

(g)

For a zip line used by young children, the safety rule is that the cable must drop at least 6 m and at most 12 m for every 100 m travelled horizontally.

Is this cable suitable for young children? Give a reason for your answer.

No calculatorINTERPRETNotes

Show solution

The cable drops $20$ m per $100$ m (or compares $0.2$ with $0.06$ to $0.12$)	B1	Correct use of their gradient from (a).
Not suitable, because it is too steep: $20 \gt 12$	B1	Conclusion with a reason. Dependent on a correct comparison.

Answer: No: it drops 20 m per 100 m, which is too steep

(2)

(h)

The launch tower is not changed. The landing platform stays at $x = 85$, but its height is changed to $k$ metres, so that the cable drops 10 m for every 100 m travelled horizontally.

Find $k$.

No calculatorALGEBRANotes

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$\frac{k - 20}{85 - 5} = -0.1$	M1	Gradient formula with the unknown, set equal to $-0.1$.
$k = 12$	A1

Answer: $k = 12$

(2)

Show full solution

Working	Mark	Notes
Part (a)
$\frac{4 - 20}{85 - 5}$	M1	Same order on top and bottom.
$-0.2$	A1	Accept $-\frac{1}{5}$ or $-\frac{16}{80}$. $0.2$ scores M1 A0.
Part (b)
$\left(\frac{5 + 85}{2}, \frac{20 + 4}{2}\right) = (45, 12)$	B1	Both coordinates correct.
Height $12$ m	B1	Allow ECF: their $y$-coordinate of $M$.
Part (c)
$20 = -0.2(5) + c$ or $4 = -0.2(85) + c$	M1	Their gradient from (a) with a point on the line.
$y = -0.2x + 21$	A1	Accept $y = -\frac{1}{5}x + 21$.
Part (d)
$A$ and $B$ plotted correctly and joined with a ruled line	B1	Within half a small square.
$x = 55$	B1	Accept 53 to 57. Check: $-0.2(55) + 21 = 10$.
Part (e)
Horizontal distance $80$ and vertical drop $16$	M1	Both differences seen, or in the formula.
$\sqrt{80^2 + 16^2} = \sqrt{6656}$	M1	Pythagoras or the length formula.
$81.58\ldots$, so $81.6$ m	A1	cso. A value with more than 3 s.f. must be seen before rounding.
Part (f)
$\tan\theta = \frac{16}{80}$	M1	Or $\sin\theta = \frac{16}{81.58\ldots}$ or $\cos\theta = \frac{80}{81.58\ldots}$.
$11.3^\circ$	A1	Accept 11.30… .
Part (g)
The cable drops $20$ m per $100$ m (or compares $0.2$ with $0.06$ to $0.12$)	B1	Correct use of their gradient from (a).
Not suitable, because it is too steep: $20 \gt 12$	B1	Conclusion with a reason. Dependent on a correct comparison.
Part (h)
$\frac{k - 20}{85 - 5} = -0.1$	M1	Gradient formula with the unknown, set equal to $-0.1$.
$k = 12$	A1

Answers: (a) $-0.2$ · (b) $M(45, 12)$; $12$ m · (c) $y = -0.2x + 21$ · (d) $x = 55$ · (e) $81.6$ m · (f) $11.3^\circ$ · (g) No: it drops 20 m per 100 m, which is too steep · (h) $k = 12$

Exam tip (f) Use the two given distances, 16 and 80, with tan rather than the rounded length 81.6, so that rounding cannot affect your angle.

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Planning city cycle lanes on a coordinate map

3.2 Equation of a straight line, parallel and perpendicular lines3.1 Gradient, midpoint and length of a line segment2.2 Linear and simultaneous equations and rearranging formulaeCore

A city map uses coordinates in kilometres, with the town hall at the origin $O$. The $x$-axis points east and the $y$-axis points north.

A straight main road passes through the points $(0, 3)$ and $(4, 11)$. A straight river path has equation $3x + y = 18$. A school is at $S(6, 5)$.

Figure 1 shows the road, the river path and the school.

Figure 1: the road and the river path on the city map (1 unit = 1 km)

(a)

Find the equation of the road in the form $y = mx + c$.

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$m = \frac{11 - 3}{4 - 0} = 2$	M1
$y = 2x + 3$	A1

Answer: $y = 2x + 3$

(2)

(b)

Find the equation of the river path in the form $y = mx + c$ and write down its gradient.

No calculatorALGEBRANotes

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$y = -3x + 18$	B1
Gradient $-3$	B1	Allow ECF from their rearrangement. 3 scores B0.

Answer: $y = -3x + 18$; gradient $-3$

(2)

(c)

The road meets the river path at the junction $J$.

By solving a pair of simultaneous equations, find the coordinates of $J$.

No calculatorALGEBRANotes

Show solution

$2x + 3 = -3x + 18$ or $y - 2x = 3$ and $y + 3x = 18$ subtracted	M1	Correct method to eliminate one letter.
$x = 3$	A1
$y = 9$	A1	Check: $3(3) + 9 = 18$.

Answer: $J(3, 9)$

(3)

(d)

A new cycle lane is to be built through the school $S(6, 5)$, parallel to the road.

Find the equation of the cycle lane.

No calculatorROUTINENotes

Show solution

$5 = 2(6) + c$	M1	Gradient 2 (ECF from (a)) with the point $S$.
$y = 2x - 7$	A1

Answer: $y = 2x - 7$

(2)

(e)

On Figure 1, plot the cycle lane as a ruled straight line for $3.5 \le x \le 8$.

The cycle lane meets the river path at the junction $K$. Use your graph to write down the coordinates of $K$.

GRAPHNotes

Show solution

Ruled line through $(3.5, 0)$ and $(8, 9)$, passing through $S$	B1	Parallel to the road, within half a small square.
$K(5, 3)$	B1	Allow ECF from their line. Check: $3(5) + 3 = 18$ and $2(5) - 7 = 3$.

Answer: $K(5, 3)$

(2)

(f)

Show that the distance from $J$ to $K$ along the river path is 6.32 km, correct to 3 significant figures.

PROOFNotes

Show solution

Differences $5 - 3 = 2$ and $3 - 9 = -6$	M1	Both differences, or in the formula.
$\sqrt{2^2 + (-6)^2} = \sqrt{40}$	M1
$6.324\ldots$, so $6.32$ km	A1	cso. A value with more than 3 s.f. must be seen before rounding.

Answer: $6.32$ km

(3)

(g)

Explain how you can tell from the equations alone, without drawing, that the cycle lane and the road never meet.

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Both have gradient $2$	B1	Accept “same $m$”.
The intercepts $3$ and $-7$ are different, so the lines are parallel and never meet	B1	Must refer to the different intercepts (or that they are not the same line).

(2)

(h)

Instead of the cycle lane, the council considers a straight footbridge path from the school $S$ that meets the river path at right angles.

Find the equation of the line of this footbridge path.

Extended onlyNo calculatorROUTINENotes

Show solution

Gradient $\frac{1}{3}$	M1	Negative reciprocal of their gradient from (b).
$y = \frac{1}{3}x + 3$	A1	From $5 = \frac{1}{3}(6) + c$. Accept $3y = x + 9$.

Answer: $y = \dfrac{1}{3}x + 3$

(2)

Show full solution

Working	Mark	Notes
Part (a)
$m = \frac{11 - 3}{4 - 0} = 2$	M1
$y = 2x + 3$	A1
Part (b)
$y = -3x + 18$	B1
Gradient $-3$	B1	Allow ECF from their rearrangement. 3 scores B0.
Part (c)
$2x + 3 = -3x + 18$ or $y - 2x = 3$ and $y + 3x = 18$ subtracted	M1	Correct method to eliminate one letter.
$x = 3$	A1
$y = 9$	A1	Check: $3(3) + 9 = 18$.
Part (d)
$5 = 2(6) + c$	M1	Gradient 2 (ECF from (a)) with the point $S$.
$y = 2x - 7$	A1
Part (e)
Ruled line through $(3.5, 0)$ and $(8, 9)$, passing through $S$	B1	Parallel to the road, within half a small square.
$K(5, 3)$	B1	Allow ECF from their line. Check: $3(5) + 3 = 18$ and $2(5) - 7 = 3$.
Part (f)
Differences $5 - 3 = 2$ and $3 - 9 = -6$	M1	Both differences, or in the formula.
$\sqrt{2^2 + (-6)^2} = \sqrt{40}$	M1
$6.324\ldots$, so $6.32$ km	A1	cso. A value with more than 3 s.f. must be seen before rounding.
Part (g)
Both have gradient $2$	B1	Accept “same $m$”.
The intercepts $3$ and $-7$ are different, so the lines are parallel and never meet	B1	Must refer to the different intercepts (or that they are not the same line).
Part (h)
Gradient $\frac{1}{3}$	M1	Negative reciprocal of their gradient from (b).
$y = \frac{1}{3}x + 3$	A1	From $5 = \frac{1}{3}(6) + c$. Accept $3y = x + 9$.

Answers: (a) $y = 2x + 3$ · (b) $y = -3x + 18$; gradient $-3$ · (c) $J(3, 9)$ · (d) $y = 2x - 7$ · (e) $K(5, 3)$ · (f) $6.32$ km · (h) $y = \dfrac{1}{3}x + 3$

Common error (f) Squaring $-6$ gives $36$, not $-36$. Adding the differences, $2 + 6 = 8$, walks along two sides of the triangle instead of across it.

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Topic 4 · Geometry

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

Setup	4.1	4.2	4.3	Other
A hexagonal garden pavilion: polygon angles, symmetry and similar tiles	●	○		5.1
Geometry of an observation wheel: circle theorems and similar triangles	○	○	●	—

4.1 Angle properties of lines, triangles and polygons · 4.2 Similarity, congruence and symmetry · 4.3 Circle theoremsmerged group

+ Add to test✓ In your testCC-CIM-41-42-43-119 marks~20 min

A hexagonal garden pavilion: polygon angles, symmetry and similar tiles

4.1 Angle properties of lines, triangles and polygons4.2 Similarity, congruence and symmetry5.1 Perimeter and area of plane shapesCore

A garden pavilion has a floor in the shape of a regular hexagon $ABCDEF$ with sides of 4 m. O is the centre of the hexagon.

The floor is divided into six triangles by lines from O to each vertex. Three of the triangles are covered with blue tiles and three with white tiles, as shown in Figure 1.

Figure 1: plan of the floor ABCDEF, with the triangles OAB, OCD and OEF in blue tiles and the other three in white tiles

(a)

Work out the size of one interior angle of the regular hexagon.

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$360 \div 6$ $(= 60)$ or $\frac{(6-2) \times 180}{6}$	M1	Exterior angle method or angle sum method.
$120^\circ$	A1

Answer: $120^\circ$

(2)

(b)(i)

Write down the number of lines of symmetry of the floor design in Figure 1, including the colours of the tiles.

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6 (the answer for a plain regular hexagon) scores B0: lines through opposite vertices swap blue and white triangles.

Answer: 3

(1)

(b)(ii)

Write down the order of rotational symmetry of the floor design, including the colours of the tiles.

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Turns of $120^\circ$ map blue onto blue.

Answer: 3

(1)

(c)(i)

The six triangles OAB, OBC, OCD, ODE, OEF and OFA are congruent, and $OA = OB$.

Show that triangle $OAB$ is equilateral. Give a reason for each step.

No calculatorPROOFNotes

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angle $AOB = 360 \div 6 = 60^\circ$, angles around a point add up to $360^\circ$	B1	The six angles at O are equal because the triangles are congruent.
$(180 - 60) \div 2$, base angles of an isosceles triangle are equal and angles in a triangle add up to $180^\circ$	M1	Alternative: angle $OAB = 120 \div 2$ using the interior angle from (a), since OA bisects angle FAB by symmetry.
all three angles are $60^\circ$, so triangle $OAB$ is equilateral	A1	cso. Both reasons needed for full marks.

(3)

(c)(ii)

The perpendicular height of triangle $OAB$, from O to AB, is 3.46 m.

Calculate the area of the floor.

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$6 \times \frac{1}{2} \times 4 \times 3.46$	M1	One triangle $= 6.92\ \text{m}^2$ seen with intention to multiply by 6 is enough.
$41.52\ \text{m}^2$	A1	Accept 41.5.

Answer: $41.52\ \text{m}^2$

(2)

(d)

The path around the pavilion is paved with tiles in the shape of regular octagons and squares, all with the same side length. Figure 2 shows two octagons and one square meeting at the point V.

Show that two regular octagons and one square fit together exactly around the point V, with no gaps.

Figure 2: two regular octagons and a square meeting at the point V

No calculatorPROOFNotes

Show solution

exterior angle of octagon $= 360 \div 8 = 45$ or $\frac{(8-2) \times 180}{8}$	M1
interior angle $= 135^\circ$	A1
$135 + 135 + 90 = 360$, which is the angle sum around a point, so there is no gap	A1	cso. The total $360$ and a conclusion are both needed.

(3)

(e)

The designer suggests using only regular pentagons for the path instead.

Explain why regular pentagons alone cannot fit together around a point without gaps or overlaps.

No calculatorINTERPRETNotes

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interior angle of a regular pentagon $= 180 - \frac{360}{5} = 108^\circ$	B1
$360 \div 108$ is not a whole number, or $3 \times 108 = 324 \lt 360$ and $4 \times 108 = 432 \gt 360$	B1	Must link to 360 around a point. Dep on 108 or a correct alternative argument.

(2)

(f)

Figure 3 shows the triangular frame at the front of the roof. XY and XZ are the two rafters and MN is a horizontal beam. M lies on XY and N lies on XZ. MN is parallel to YZ and $XY = XZ$.

Angle $XMN = 32^\circ$.

Find angle $YXZ$. Give a reason for each step of your working.

Figure 3: the front frame of the roof. MN is parallel to YZ and XY = XZ

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Show solution

angle $XYZ = 32^\circ$, corresponding angles (MN parallel to YZ) are equal	B1
angle $XZY = 32^\circ$ and $180 - 32 - 32$, base angles of an isosceles triangle are equal and angles in a triangle add up to $180^\circ$	M1	Alternative: angle $XNM = 32^\circ$ (isosceles triangle XMN) then $180 - 2 \times 32$.
$116^\circ$	A1	With reasons stated.

Answer: $116^\circ$

(3)

(g)

A planter in the garden has a top in the shape of a regular hexagon with sides of 0.8 m. It is mathematically similar to the floor of the pavilion.

Use your answer to part (c)(ii) to calculate the area of the top of the planter.

Extended onlyNo calculatorMULTINotes

Show solution

$41.52 \times \left(\frac{0.8}{4}\right)^2$ or $41.52 \div 25$	M1	Allow ECF from (c)(ii). Using $\times 0.2$ (length factor) scores M0.
$1.66\ \text{m}^2$	A1	Accept 1.6608. Answers from 41.5 or 41.6 rounding to 1.66 also score.

Answer: $1.66\ \text{m}^2$

(2)

Show full solution

Working	Mark	Notes
Part (a)
$360 \div 6$ $(= 60)$ or $\frac{(6-2) \times 180}{6}$	M1	Exterior angle method or angle sum method.
$120^\circ$	A1
Part (b)(i)
3	B1	6 (the answer for a plain regular hexagon) scores B0: lines through opposite vertices swap blue and white triangles.
Part (b)(ii)
3	B1	Turns of $120^\circ$ map blue onto blue.
Part (c)(i)
angle $AOB = 360 \div 6 = 60^\circ$, angles around a point add up to $360^\circ$	B1	The six angles at O are equal because the triangles are congruent.
$(180 - 60) \div 2$, base angles of an isosceles triangle are equal and angles in a triangle add up to $180^\circ$	M1	Alternative: angle $OAB = 120 \div 2$ using the interior angle from (a), since OA bisects angle FAB by symmetry.
all three angles are $60^\circ$, so triangle $OAB$ is equilateral	A1	cso. Both reasons needed for full marks.
Part (c)(ii)
$6 \times \frac{1}{2} \times 4 \times 3.46$	M1	One triangle $= 6.92\ \text{m}^2$ seen with intention to multiply by 6 is enough.
$41.52\ \text{m}^2$	A1	Accept 41.5.
Part (d)
exterior angle of octagon $= 360 \div 8 = 45$ or $\frac{(8-2) \times 180}{8}$	M1
interior angle $= 135^\circ$	A1
$135 + 135 + 90 = 360$, which is the angle sum around a point, so there is no gap	A1	cso. The total $360$ and a conclusion are both needed.
Part (e)
interior angle of a regular pentagon $= 180 - \frac{360}{5} = 108^\circ$	B1
$360 \div 108$ is not a whole number, or $3 \times 108 = 324 \lt 360$ and $4 \times 108 = 432 \gt 360$	B1	Must link to 360 around a point. Dep on 108 or a correct alternative argument.
Part (f)
angle $XYZ = 32^\circ$, corresponding angles (MN parallel to YZ) are equal	B1
angle $XZY = 32^\circ$ and $180 - 32 - 32$, base angles of an isosceles triangle are equal and angles in a triangle add up to $180^\circ$	M1	Alternative: angle $XNM = 32^\circ$ (isosceles triangle XMN) then $180 - 2 \times 32$.
$116^\circ$	A1	With reasons stated.
Part (g)
$41.52 \times \left(\frac{0.8}{4}\right)^2$ or $41.52 \div 25$	M1	Allow ECF from (c)(ii). Using $\times 0.2$ (length factor) scores M0.
$1.66\ \text{m}^2$	A1	Accept 1.6608. Answers from 41.5 or 41.6 rounding to 1.66 also score.

Answers: (a) $120^\circ$ · (b)(i) 3 · (b)(ii) 3 · (c)(ii) $41.52\ \text{m}^2$ · (f) $116^\circ$ · (g) $1.66\ \text{m}^2$

Common error (b)(ii) A plain regular hexagon has 6 lines of symmetry and order 6, but the colours count too. A turn of $60^\circ$ moves every blue triangle onto a white one, so it is not a symmetry of this design.

Exam tip (g) Lengths scale by $k = 0.2$, so areas scale by $k^2 = 0.04$. Check: the planter area should be much smaller than $0.2$ of the floor.

More 4.1 questions in the question bank

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Geometry of an observation wheel: circle theorems and similar triangles

4.3 Circle theorems4.1 Angle properties of lines, triangles and polygons4.2 Similarity, congruence and symmetryExtended

An observation wheel is modelled as a circle with centre O. P is the lowest point of the wheel. The boarding platform SPT is a straight line that is a tangent to the circle at P.

Cabins A, C and D are points on the circle. A straight strut joins P to A, and angle $TPA = 60^\circ$.

Figure 1 shows the wheel.

Figure 1: the wheel, centre O. SPT is the platform, a tangent at P. PQ and AC are straight struts

(a)

Write down the size of angle $OPT$. Give a reason for your answer.

No calculatorCONCEPTNotes

Show solution

$90^\circ$, the tangent is perpendicular to the radius

Both value and reason needed.

Answer: $90^\circ$

(1)

(b)

Show that angle $POA = 120^\circ$. Give a reason for each step.

No calculatorPROOFNotes

Show solution

angle $OPA = 90 - 60 = 30^\circ$	M1	Using the right angle from (a).
angle $OAP = 30^\circ$ and $180 - 30 - 30$, triangle $OPA$ is isosceles because $OP = OA$ (radii), angles in a triangle add up to $180^\circ$	M1
angle $POA = 120^\circ$ with the reasons above	A1	cso. Alternative (Extended route): angle in the alternate segment $= 60^\circ$, angle at the centre $= 2 \times 60$, with both theorems named.

(3)

(c)

Q is the highest point of the wheel, so PQ is a diameter.

Find angle $AQP$. Give reasons for your answer.

No calculatorROUTINENotes

Show solution

angle $PAQ = 90^\circ$, the angle in a semicircle is $90^\circ$	B1
angle $AQP = 180 - 90 - 30 = 60^\circ$	B1	Allow ECF from their angle $OPA$ in (b). Also accept $60^\circ$ from the alternate segment theorem, with that reason.

Answer: $60^\circ$

(2)

(d)

Cabin D is on the minor arc PA.

Aisha says: “Angle $PDA = 60^\circ$, because the angle between a tangent and a chord equals the angle in the alternate segment.”

Explain why Aisha is wrong, and find the correct size of angle $PDA$.

Extended onlyNo calculatorINTERPRETNotes

Show solution

D is on the same side of the chord PA as angle $TPA$, so D is not in the alternate segment (Q and C are)	B1
angle $PDA = 180 - 60 = 120^\circ$, opposite angles of the cyclic quadrilateral $PDAQ$ add up to $180^\circ$	B1	Allow ECF from (c).

Answer: $120^\circ$

(2)

(e)(i)

The struts AC and PQ cross at X.

Explain why triangle $XAP$ is similar to triangle $XQC$.

Extended onlyNo calculatorINTERPRETNotes

Show solution

angle $AXP = $ angle $QXC$, vertically opposite angles are equal	B1
angle $XAP = $ angle $XQC$ (both are angles $CAP$, $CQP$ standing on chord $PC$), angles in the same segment are equal	B1	Alternative: angle $XPA = $ angle $XCQ$, both on chord $AQ$. Two pairs of equal angles is enough.

(2)

(e)(ii)

$XP = 77$ m, $XQ = 35$ m and $XA = 49$ m, as shown in Figure 2.

Find the length of the strut section $XC$.

Figure 2: triangles XAP and XQC formed by the struts

Extended onlyNo calculatorGRAPHNotes

Show solution

$\frac{XC}{XP} = \frac{XQ}{XA}$ or $XC = \frac{77 \times 35}{49}$	M1	Correct matching of sides: XA with XQ and XP with XC.
55 m	A1

Answer: 55 m

(2)

(f)

The wheel turns so that a different cabin is at the position of A. The new strut makes angle $TPA = 70^\circ$. D is still on the minor arc PA.

Find the new size of angle $POA$ and of angle $PDA$.

Extended onlyNo calculatorROUTINENotes

Show solution

angle $POA = 140^\circ$	B1	From $2 \times 70$ or $180 - 2 \times 20$.
angle $PDA = 110^\circ$	B1	From $180 - 70$.

Answer: $POA = 140^\circ$, $PDA = 110^\circ$

(2)

Show full solution

Working	Mark	Notes
Part (a)
$90^\circ$, the tangent is perpendicular to the radius	B1	Both value and reason needed.
Part (b)
angle $OPA = 90 - 60 = 30^\circ$	M1	Using the right angle from (a).
angle $OAP = 30^\circ$ and $180 - 30 - 30$, triangle $OPA$ is isosceles because $OP = OA$ (radii), angles in a triangle add up to $180^\circ$	M1
angle $POA = 120^\circ$ with the reasons above	A1	cso. Alternative (Extended route): angle in the alternate segment $= 60^\circ$, angle at the centre $= 2 \times 60$, with both theorems named.
Part (c)
angle $PAQ = 90^\circ$, the angle in a semicircle is $90^\circ$	B1
angle $AQP = 180 - 90 - 30 = 60^\circ$	B1	Allow ECF from their angle $OPA$ in (b). Also accept $60^\circ$ from the alternate segment theorem, with that reason.
Part (d)
D is on the same side of the chord PA as angle $TPA$, so D is not in the alternate segment (Q and C are)	B1
angle $PDA = 180 - 60 = 120^\circ$, opposite angles of the cyclic quadrilateral $PDAQ$ add up to $180^\circ$	B1	Allow ECF from (c).
Part (e)(i)
angle $AXP = $ angle $QXC$, vertically opposite angles are equal	B1
angle $XAP = $ angle $XQC$ (both are angles $CAP$, $CQP$ standing on chord $PC$), angles in the same segment are equal	B1	Alternative: angle $XPA = $ angle $XCQ$, both on chord $AQ$. Two pairs of equal angles is enough.
Part (e)(ii)
$\frac{XC}{XP} = \frac{XQ}{XA}$ or $XC = \frac{77 \times 35}{49}$	M1	Correct matching of sides: XA with XQ and XP with XC.
55 m	A1
Part (f)
angle $POA = 140^\circ$	B1	From $2 \times 70$ or $180 - 2 \times 20$.
angle $PDA = 110^\circ$	B1	From $180 - 70$.

Answers: (a) $90^\circ$ · (c) $60^\circ$ · (d) $120^\circ$ · (e)(ii) 55 m · (f) $POA = 140^\circ$, $PDA = 110^\circ$

Common error (d) The alternate segment is on the other side of the chord from the tangent angle. A point on the minor arc between P and A sees the chord PA at $180^\circ$ minus the alternate segment angle.

More 4.3 questions in the question bank

Topic 5 · Mensuration

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

Setup	5.1	5.2	5.3	Other
An ice-cream cone with a scoop: cones, hemispheres and sector nets	○	○	●	—
An athletics track: semicircles, staggered starts and resurfacing	○	●	○	—

5.1 Perimeter and area of plane shapes · 5.2 Circles, arcs and sectors · 5.3 Surface area and volume of solidsmerged group

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An ice-cream cone with a scoop: cones, hemispheres and sector nets

5.3 Surface area and volume of solids5.2 Circles, arcs and sectors5.1 Perimeter and area of plane shapesExtended

An ice-cream cone is a hollow cone with radius 2.5 cm and slant height 6.5 cm. The cone is completely filled with ice cream, and a hemisphere of ice cream with radius 2.5 cm sits on top, as shown in Figure 1.

Ignore the thickness of the wafer.

Figure 1: the cone filled with ice cream and topped with a hemisphere (not to scale)

(a)

Show that the perpendicular height of the cone is 6 cm.

Extended onlyPROOFNotes

Show solution

$h^2 = 6.5^2 - 2.5^2$ or $h = \sqrt{42.25 - 6.25}$	M1	Using $l^2 = r^2 + h^2$.
$h = \sqrt{36} = 6$	A1	cso. $36$ must be seen.

Answer: 6 cm

(2)

(b)

Calculate the total volume of ice cream.

Extended onlyMULTINotes

Show solution

$\frac{1}{3} \times \pi \times 2.5^2 \times 6$ $(= 12.5\pi = 39.27\ldots)$	M1	Cone volume with the perpendicular height. Using 6.5 scores M0.
$\frac{2}{3} \times \pi \times 2.5^3$ $(= 32.72\ldots)$	M1	Hemisphere volume.
$72.0\ \text{cm}^3$	A1	Accept 71.99 or 72 from $\frac{275\pi}{12}$.

Answer: $72.0\ \text{cm}^3$

(3)

(c)

Calculate the curved surface area of the cone.

Extended onlyROUTINENotes

Show solution

$\pi \times 2.5 \times 6.5$	M1
$51.1\ \text{cm}^2$	A1	Accept $16.25\pi$ or 51.05.

Answer: $51.1\ \text{cm}^2$

(2)

(d)

The cone is made by rolling up a sector of a circle of radius 6.5 cm. The arc of the sector becomes the circumference of the top of the cone.

Calculate the angle of the sector.

Extended onlyMULTINotes

Show solution

circumference of the top $= 2 \times \pi \times 2.5$ $(= 5\pi = 15.70\ldots)$	B1
$\frac{\theta}{360} \times 2 \times \pi \times 6.5 = 5\pi$	M1	Or $\theta = \frac{5\pi}{13\pi} \times 360$. Alternative: $\frac{\theta}{360} \times \pi \times 6.5^2 = $ their (c).
$138.5^\circ$	A1	Accept 138.46 or better. $\frac{1800}{13}$ is exact.

Answer: $138.5^\circ$

(3)

(e)

The sectors are cut from rectangular wafer sheets measuring 30 cm by 20 cm. Six sectors are cut from each sheet.

Show that the area of wafer left over from one sheet is $294\ \text{cm}^2$, correct to 3 significant figures.

Extended onlyPROOFNotes

Show solution

$30 \times 20 = 600$	B1
$600 - 6 \times 51.05\ldots$ or $600 - 6 \times \frac{138.46\ldots}{360} \times \pi \times 6.5^2$	M1	The area of one sector equals the curved surface area from (c). Allow ECF from (c) or (d) for this mark only.
$293.69\ldots = 294$	A1	cso. A more accurate value than 294 (for example 293.7) must be seen.

(3)

(f)(i)

The shop also sells single hemispherical scoops of different sizes. Figure 2 shows the volume of a scoop plotted against its radius.

Use the graph to find

the volume of a scoop of radius 3.5 cm,
the radius of a scoop of volume $50\ \text{cm}^3$.

Figure 2: the volume, V cm³, of a hemispherical scoop of radius r cm

Extended onlyGRAPHNotes

Show solution

$88$ to $92\ \text{cm}^3$	B1	Exact value $89.8$.
$2.8$ to $3.0$ cm	B1	Exact value $2.88$.

Answer: about $90\ \text{cm}^3$; about 2.9 cm

(2)

(f)(ii)

Mo says: “A scoop with twice the radius has twice the volume.”

Use values from the graph to show that Mo is wrong, and explain why.

Extended onlyINTERPRETNotes

Show solution

a correct pair of readings, for example $r = 2$, $V \approx 17$ and $r = 4$, $V \approx 134$, showing the volume is about 8 times as large	B1	Any pair r and 2r from the graph with a ratio of about 8 (accept 7 to 9).
volume depends on $r^3$ ($V = \frac{2}{3}\pi r^3$), so doubling $r$ multiplies $V$ by $2^3 = 8$	B1

(2)

(g)

The shop changes to cylindrical tubs of radius 2.5 cm. Each tub holds the same volume of ice cream as your answer to part (b).

Calculate the depth of ice cream in a tub.

ROUTINENotes

Show solution

$\pi \times 2.5^2 \times h = 72.0$	M1	Or $h = \frac{72.0}{\pi \times 2.5^2}$. Allow ECF from (b).
3.67 cm	A1	Accept 3.66 to 3.67. Exactly $\frac{11}{3}$ from $\frac{275\pi}{12} \div 6.25\pi$.

Answer: 3.67 cm

(2)

Show full solution

Working	Mark	Notes
Part (a)
$h^2 = 6.5^2 - 2.5^2$ or $h = \sqrt{42.25 - 6.25}$	M1	Using $l^2 = r^2 + h^2$.
$h = \sqrt{36} = 6$	A1	cso. $36$ must be seen.
Part (b)
$\frac{1}{3} \times \pi \times 2.5^2 \times 6$ $(= 12.5\pi = 39.27\ldots)$	M1	Cone volume with the perpendicular height. Using 6.5 scores M0.
$\frac{2}{3} \times \pi \times 2.5^3$ $(= 32.72\ldots)$	M1	Hemisphere volume.
$72.0\ \text{cm}^3$	A1	Accept 71.99 or 72 from $\frac{275\pi}{12}$.
Part (c)
$\pi \times 2.5 \times 6.5$	M1
$51.1\ \text{cm}^2$	A1	Accept $16.25\pi$ or 51.05.
Part (d)
circumference of the top $= 2 \times \pi \times 2.5$ $(= 5\pi = 15.70\ldots)$	B1
$\frac{\theta}{360} \times 2 \times \pi \times 6.5 = 5\pi$	M1	Or $\theta = \frac{5\pi}{13\pi} \times 360$. Alternative: $\frac{\theta}{360} \times \pi \times 6.5^2 = $ their (c).
$138.5^\circ$	A1	Accept 138.46 or better. $\frac{1800}{13}$ is exact.
Part (e)
$30 \times 20 = 600$	B1
$600 - 6 \times 51.05\ldots$ or $600 - 6 \times \frac{138.46\ldots}{360} \times \pi \times 6.5^2$	M1	The area of one sector equals the curved surface area from (c). Allow ECF from (c) or (d) for this mark only.
$293.69\ldots = 294$	A1	cso. A more accurate value than 294 (for example 293.7) must be seen.
Part (f)(i)
$88$ to $92\ \text{cm}^3$	B1	Exact value $89.8$.
$2.8$ to $3.0$ cm	B1	Exact value $2.88$.
Part (f)(ii)
a correct pair of readings, for example $r = 2$, $V \approx 17$ and $r = 4$, $V \approx 134$, showing the volume is about 8 times as large	B1	Any pair r and 2r from the graph with a ratio of about 8 (accept 7 to 9).
volume depends on $r^3$ ($V = \frac{2}{3}\pi r^3$), so doubling $r$ multiplies $V$ by $2^3 = 8$	B1
Part (g)
$\pi \times 2.5^2 \times h = 72.0$	M1	Or $h = \frac{72.0}{\pi \times 2.5^2}$. Allow ECF from (b).
3.67 cm	A1	Accept 3.66 to 3.67. Exactly $\frac{11}{3}$ from $\frac{275\pi}{12} \div 6.25\pi$.

Answers: (a) 6 cm · (b) $72.0\ \text{cm}^3$ · (c) $51.1\ \text{cm}^2$ · (d) $138.5^\circ$ · (f)(i) about $90\ \text{cm}^3$; about 2.9 cm · (g) 3.67 cm

Common error (b) The cone volume uses the perpendicular height (6 cm), not the slant height (6.5 cm). The slant height belongs in the curved surface area $\pi r l$.

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An athletics track: semicircles, staggered starts and resurfacing

5.2 Circles, arcs and sectors5.1 Perimeter and area of plane shapes5.3 Surface area and volume of solidsCore

An athletics track has two straights, each 84.4 m long, joined by two semicircular ends. The inner edge of the track is shown in Figure 1. The radius of each semicircle of the inner edge is 36.5 m.

The track has 8 lanes, each 1.22 m wide. Distances in a lane are measured along a running line. In lane 1 the running line is 0.3 m outside the inner edge, so its semicircles have radius 36.8 m. The straight parts of every running line are 84.4 m long.

Figure 1: plan of the track. The inner edge (teal) is two straights and two semicircles of radius 36.5 m

(a)

Show that the length of one lap along the running line of lane 1 is 400 m, correct to the nearest metre.

PROOFNotes

Show solution

$2 \times \pi \times 36.8$ or $\pi \times 73.6$ $(= 231.2\ldots)$	M1	The two semicircles together make one full circle.
$2 \times 84.4 + 2 \times \pi \times 36.8$	M1
$400.02\ldots = 400$	A1	cso. A value such as 400.02 must be seen before rounding.

(3)

(b)

Assume that the radius of the running line in each lane is 1.22 m more than in the lane inside it.

Calculate how much further a runner in lane 4 runs in one lap than a runner in lane 1.

MULTINotes

Show solution

radius in lane 4 $= 36.8 + 3 \times 1.22 = 40.46$	B1	Or the increase in radius $3 \times 1.22 = 3.66$.
$2 \times \pi \times 40.46 + 168.8 - 400.02$ or $2 \times \pi \times 3.66$	M1
23.0 m	A1	Accept 22.99 to 23.0.

Answer: 23.0 m

(3)

(c)(i)

Figure 2 shows the lap length for each of the eight lanes.

Use the graph to find the gradient of the line.

Figure 2: the length of one lap along the running line of each lane

GRAPHNotes

Show solution

$\frac{454 - 400}{8 - 1}$ or another correct pair of readings	M1	Readings must be from the line (lane 8 between 452 and 456).
$7.5$ to $7.9$	A1	Exact value $2\pi \times 1.22 = 7.67$.

Answer: about 7.7 m per lane

(2)

(c)(ii)

Explain why the points lie on a straight line, and state what the gradient represents.

INTERPRETNotes

Show solution

each lane adds the same 1.22 m to the radius and the straights do not change, so the lap length goes up by the same amount, $2 \times \pi \times 1.22$, every time	B1	Must mention a constant increase.
the gradient is the extra distance run in one lap for each lane further out	B1	In metres per lane.

(2)

(d)

The grass infield is the region inside the inner edge of the track.

Calculate the area of the infield.

MULTINotes

Show solution

$84.4 \times 73$ $(= 6161.2)$	M1	Width of the rectangle $= 2 \times 36.5$.
$\pi \times 36.5^2$ $(= 4185.4\ldots)$	M1	Two semicircles make one circle.
$10\,300\ \text{m}^2$	A1	Accept 10 346 or 10 347.

Answer: $10\,300\ \text{m}^2$

(3)

(e)

The whole track (all eight lanes) is resurfaced with a layer of rubber 13 mm thick.

Calculate the volume of rubber needed. Give your answer in $\text{m}^3$.

MULTINotes

Show solution

outer radius $= 36.5 + 8 \times 1.22 = 46.26$ or track width $= 9.76$	B1
$\pi \times (46.26^2 - 36.5^2) + 2 \times 84.4 \times 9.76$ $(= 4185\ldots)$	M1	Ring from the two ends plus two rectangles on the straights. $\pi (46.26 - 36.5)^2$ scores M0.
their area $\times 0.013$	M1	13 mm $= 0.013$ m. Dep on an attempt at an area.
$54.4\ \text{m}^3$	A1	Accept 54.40 to 54.41.

Answer: $54.4\ \text{m}^3$

(4)

(f)

A different stadium has straights of 80 m. One lap along the running line of lane 1 must still be exactly 400 m.

Calculate the radius of the running line of lane 1 on the semicircular ends.

ROUTINENotes

Show solution

$2 \times 80 + 2 \times \pi \times r = 400$	M1	Or $r = \frac{240}{2\pi}$.
38.2 m	A1	Accept 38.19 to 38.2.

Answer: 38.2 m

(2)

Show full solution

Working	Mark	Notes
Part (a)
$2 \times \pi \times 36.8$ or $\pi \times 73.6$ $(= 231.2\ldots)$	M1	The two semicircles together make one full circle.
$2 \times 84.4 + 2 \times \pi \times 36.8$	M1
$400.02\ldots = 400$	A1	cso. A value such as 400.02 must be seen before rounding.
Part (b)
radius in lane 4 $= 36.8 + 3 \times 1.22 = 40.46$	B1	Or the increase in radius $3 \times 1.22 = 3.66$.
$2 \times \pi \times 40.46 + 168.8 - 400.02$ or $2 \times \pi \times 3.66$	M1
23.0 m	A1	Accept 22.99 to 23.0.
Part (c)(i)
$\frac{454 - 400}{8 - 1}$ or another correct pair of readings	M1	Readings must be from the line (lane 8 between 452 and 456).
$7.5$ to $7.9$	A1	Exact value $2\pi \times 1.22 = 7.67$.
Part (c)(ii)
each lane adds the same 1.22 m to the radius and the straights do not change, so the lap length goes up by the same amount, $2 \times \pi \times 1.22$, every time	B1	Must mention a constant increase.
the gradient is the extra distance run in one lap for each lane further out	B1	In metres per lane.
Part (d)
$84.4 \times 73$ $(= 6161.2)$	M1	Width of the rectangle $= 2 \times 36.5$.
$\pi \times 36.5^2$ $(= 4185.4\ldots)$	M1	Two semicircles make one circle.
$10\,300\ \text{m}^2$	A1	Accept 10 346 or 10 347.
Part (e)
outer radius $= 36.5 + 8 \times 1.22 = 46.26$ or track width $= 9.76$	B1
$\pi \times (46.26^2 - 36.5^2) + 2 \times 84.4 \times 9.76$ $(= 4185\ldots)$	M1	Ring from the two ends plus two rectangles on the straights. $\pi (46.26 - 36.5)^2$ scores M0.
their area $\times 0.013$	M1	13 mm $= 0.013$ m. Dep on an attempt at an area.
$54.4\ \text{m}^3$	A1	Accept 54.40 to 54.41.
Part (f)
$2 \times 80 + 2 \times \pi \times r = 400$	M1	Or $r = \frac{240}{2\pi}$.
38.2 m	A1	Accept 38.19 to 38.2.

Answers: (b) 23.0 m · (c)(i) about 7.7 m per lane · (d) $10\,300\ \text{m}^2$ · (e) $54.4\ \text{m}^3$ · (f) 38.2 m

Exam tip (e) Convert the thickness to metres before multiplying: $13\ \text{mm} = 1.3\ \text{cm} = 0.013\ \text{m}$. Then the volume comes out directly in $\text{m}^3$.

More 5.2 questions in the question bank

Topic 6 · Trigonometry

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

Setup	6.1	6.2	6.3	Other
Inspecting a wind turbine: Pythagoras, elevation and depression	●	○		4.1
A sailing regatta course: bearings, distances and timing	○	●		1.4
A tidal harbour: sine and cosine rules and a depth graph	○		●	2.4
A glass pyramid atrium: 3D trigonometry and triangle areas	○		●	5.3

6.1 Pythagoras' theorem and right-angled trigonometry · 6.2 Bearingsmerged group

+ Add to test✓ In your testCC-CIM-61-62-119 marks~20 min

Inspecting a wind turbine: Pythagoras, elevation and depression

6.1 Pythagoras' theorem and right-angled trigonometry6.2 Bearings4.1 Angle properties of lines, triangles and polygonsCore

An engineer inspects a wind turbine on level ground. The tower is vertical. Its base is at T and the centre of the hub, H, is at the top of the tower. Each of the three blades is 40 m long, measured from H.

The engineer stands at E, 60 m from T. Her eye is 1.6 m above the ground. From her eye, the angle of elevation of H is 53°.

A straight safety cable joins H to an anchor G on the ground, 35 m from T, on the other side of the tower from E. A maintenance van is parked at V, 90 m from T, on the same side as E. The points G, T, E and V lie in a straight line.

Figure 1: side view of the turbine, with T, G, E and V on level ground (not to scale)

(a)

Calculate the height of H above the ground.

MULTINotes

Show solution

$\tan 53^\circ = \frac{x}{60}$ or $x = 60\tan 53^\circ$ $(= 79.62\ldots)$	M1	$x$ is the height of H above eye level.
$79.62\ldots + 1.6$	M1	Adding the eye height to their vertical distance.
$81.2$ m	A1	Accept 81.22… . 79.6 m alone scores M1 M0 A0.

Answer: 81.2 m

(3)

(b)

Show that the length of the safety cable GH is 88.4 m, correct to 3 significant figures.

PROOFNotes

Show solution

$GH^2 = 81.22\ldots^2 + 35^2$ $(= 7822.\ldots)$	M1	Use of their height from (a) with 35 in Pythagoras.
$GH = \sqrt{7822.\ldots} = 88.44\ldots = 88.4$ m	A1	cso. A value to at least 4 s.f. (88.44 or 88.42 from 81.2) must be seen before rounding.

Answer: 88.4 m

(2)

(c)(i)

Calculate the angle of depression of the van V from H.

ROUTINENotes

Show solution

$\tan x = \frac{81.22\ldots}{90}$	M1	Allow ECF from (a).
$42.1^\circ$	A1	42.06… or 42.05…

Answer: 42.1°

(2)

(c)(ii)

Explain why the angle of elevation of H from V is equal to your answer to part (c)(i).

CONCEPTNotes

Show solution

The horizontal line through H is parallel to the ground, so the two angles are alternate angles (and so are equal).

Must mention parallel (horizontal) lines and alternate angles.

(1)

(d)(i)

The engineer walks in a straight line towards T. Figure 2 shows how the angle of elevation, $\theta$, of H from her eye changes with her distance, $d$ metres, from T.

Use Figure 2 to find her distance from T when $\theta = 70^\circ$.

Figure 2: the angle of elevation θ of H from the engineer's eye against her distance d from T

GRAPHNotes

Show solution

$29$ m

Accept 28 to 30.

Answer: 29 m

(1)

(d)(ii)

She stops when she is 25 m closer to T than her starting point E.

Calculate the angle of elevation of H from her eye at this point.

MULTINotes

Show solution

$\tan\theta = \frac{79.62\ldots}{35}$	M1	Distance $60 - 25 = 35$ and the height above eye level. Allow ECF from (a) minus 1.6.
$66.3^\circ$	A1	66.26… or 66.27… . Using 81.2 gives 66.7° and scores M1 A0.

Answer: 66.3°

(2)

(e)

The angle of 53° was measured to the nearest degree.

Calculate the height of H above the ground if the angle had been 53.5°. Use your answer to explain whether it is sensible to give the height of H as 81.2 m.

INTERPRETNotes

Show solution

$60\tan 53.5^\circ + 1.6$	M1
$82.7$ m	A1	82.68…
A change of only 0.5° in the angle changes the height by about 1.5 m, so giving the height to the nearest 0.1 m (3 s.f.) is not reliable; about 81 m is more sensible.	B1	Comment must refer to the size of the change compared with the accuracy of 81.2 m.

Answer: 82.7 m; the height is only reliable to about the nearest metre or two

(3)

(f)(i)

A second turbine has its base at Q, which is 150 m west and 360 m north of T.

Calculate the bearing of Q from T.

MULTINotes

Show solution

$\tan^{-1}\left(\frac{150}{360}\right) = 22.6\ldots^\circ$	M1	Or $\tan^{-1}\left(\frac{360}{150}\right) = 67.4\ldots^\circ$ (angle from the west line).
$337^\circ$	A1	$360 - 22.6$ or $270 + 67.4$. Accept 337.4°.

Answer: 337°

(2)

(f)(ii)

Write down the bearing of T from Q.

No calculatorROUTINENotes

Show solution

$157^\circ$

Allow ECF: their (f)(i) − 180°.

Answer: 157°

(1)

(g)

In Figure 1, one blade points upwards and makes an angle of 30° with the vertical through H.

Without using a calculator, show that the tip of this blade is $20\sqrt3$ m above H.

Extended onlyNo calculatorPROOFNotes

Show solution

$40\cos 30^\circ$ or $40\sin 60^\circ$	M1	The vertical height is adjacent to the 30° angle.
$40\times\frac{\sqrt3}{2} = 20\sqrt3$	A1	cso. The exact value $\frac{\sqrt3}{2}$ must be seen; a decimal such as 34.64 scores M1 A0.

Answer: $20\sqrt3$ m

(2)

Show full solution

Working	Mark	Notes
Part (a)
$\tan 53^\circ = \frac{x}{60}$ or $x = 60\tan 53^\circ$ $(= 79.62\ldots)$	M1	$x$ is the height of H above eye level.
$79.62\ldots + 1.6$	M1	Adding the eye height to their vertical distance.
$81.2$ m	A1	Accept 81.22… . 79.6 m alone scores M1 M0 A0.
Part (b)
$GH^2 = 81.22\ldots^2 + 35^2$ $(= 7822.\ldots)$	M1	Use of their height from (a) with 35 in Pythagoras.
$GH = \sqrt{7822.\ldots} = 88.44\ldots = 88.4$ m	A1	cso. A value to at least 4 s.f. (88.44 or 88.42 from 81.2) must be seen before rounding.
Part (c)(i)
$\tan x = \frac{81.22\ldots}{90}$	M1	Allow ECF from (a).
$42.1^\circ$	A1	42.06… or 42.05…
Part (c)(ii)
The horizontal line through H is parallel to the ground, so the two angles are alternate angles (and so are equal).	B1	Must mention parallel (horizontal) lines and alternate angles.
Part (d)(i)
$29$ m	B1	Accept 28 to 30.
Part (d)(ii)
$\tan\theta = \frac{79.62\ldots}{35}$	M1	Distance $60 - 25 = 35$ and the height above eye level. Allow ECF from (a) minus 1.6.
$66.3^\circ$	A1	66.26… or 66.27… . Using 81.2 gives 66.7° and scores M1 A0.
Part (e)
$60\tan 53.5^\circ + 1.6$	M1
$82.7$ m	A1	82.68…
A change of only 0.5° in the angle changes the height by about 1.5 m, so giving the height to the nearest 0.1 m (3 s.f.) is not reliable; about 81 m is more sensible.	B1	Comment must refer to the size of the change compared with the accuracy of 81.2 m.
Part (f)(i)
$\tan^{-1}\left(\frac{150}{360}\right) = 22.6\ldots^\circ$	M1	Or $\tan^{-1}\left(\frac{360}{150}\right) = 67.4\ldots^\circ$ (angle from the west line).
$337^\circ$	A1	$360 - 22.6$ or $270 + 67.4$. Accept 337.4°.
Part (f)(ii)
$157^\circ$	B1	Allow ECF: their (f)(i) − 180°.
Part (g)
$40\cos 30^\circ$ or $40\sin 60^\circ$	M1	The vertical height is adjacent to the 30° angle.
$40\times\frac{\sqrt3}{2} = 20\sqrt3$	A1	cso. The exact value $\frac{\sqrt3}{2}$ must be seen; a decimal such as 34.64 scores M1 A0.

Answers: (a) 81.2 m · (b) 88.4 m · (c)(i) 42.1° · (d)(i) 29 m · (d)(ii) 66.3° · (e) 82.7 m; the height is only reliable to about the nearest metre or two · (f)(i) 337° · (f)(ii) 157° · (g) $20\sqrt3$ m

Common error (a) The 53° angle is measured at the engineer's eye, not at ground level, so the tangent gives the height of H above her eye. Add the 1.6 m at the end.

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A sailing regatta course: bearings, distances and timing

6.2 Bearings6.1 Pythagoras' theorem and right-angled trigonometry1.4 Estimation, rounding, bounds, ratio, proportion and moneyCore

A sailing club sets out a race course with three buoys, S, A and B. The start buoy is S.

Buoy A is 3.2 km from S on a bearing of 050°.
Buoy B is due east of S, and the bearing of B from A is 140°.

The yachts sail the course S → A → B → S (Figure 1).

Figure 1: the regatta course S → A → B → S

(a)

Show that angle SAB $= 90^\circ$.

No calculatorPROOFNotes

Show solution

Bearing of S from A $= 050^\circ + 180^\circ = 230^\circ$	M1	Or the angle between north at A and AS is $180^\circ - 50^\circ = 130^\circ$ (co-interior angles).
$230^\circ - 140^\circ = 90^\circ$	A1	cso. Or, from the M1 alternative: angle SAB $= 360^\circ - 130^\circ - 140^\circ = 90^\circ$ (angles around A). Any complete correct angle argument; a reason (back bearing or co-interior angles) must be seen.

(2)

(b)

Write down the size of angle ASB.

No calculatorROUTINENotes

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$40^\circ$

$90^\circ - 50^\circ$

Answer: 40°

(1)

(c)

Calculate the distance AB.

ROUTINENotes

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$\tan 40^\circ = \frac{AB}{3.2}$	M1	Or $AB = 3.2\tan 40^\circ$. Allow ECF from (b).
$2.69$ km	A1	2.685…

Answer: 2.69 km

(2)

(d)

Show that the total length of the course is 10.1 km, correct to 3 significant figures.

MULTINotes

Show solution

$SB = \frac{3.2}{\cos 40^\circ}$ $(= 4.177\ldots)$	M1	Or $SB = \sqrt{3.2^2 + 2.685\ldots^2}$.
$3.2 + 2.685\ldots + 4.177\ldots$	M1	Adding the three legs.
$10.06\ldots = 10.1$ km	A1	cso. 10.06 must be seen before rounding.

Answer: 10.1 km

(3)

(e)

A rescue boat waits at R, the point on SB directly south of A.

Calculate the distance SR.

ROUTINENotes

Show solution

$3.2\sin 50^\circ$	M1	SR is the distance east of S travelled on the first leg. Or $3.2\cos 40^\circ$.
$2.45$ km	A1	2.451…

Answer: 2.45 km

(2)

(f)

A yacht completes the course at an average speed of 6.5 knots, where 1 knot $= 1.852$ km/h.

Calculate the time it takes, giving your answer to the nearest minute.

MULTINotes

Show solution

$6.5\times 1.852$ $(= 12.038$ km/h$)$	M1
$\frac{10.06\ldots}{12.038}$ $(= 0.8358\ldots$ h$)$ $\times 60$	M1	Distance ÷ their speed, converted to minutes. Using 10.1 km is fine.
$50$ minutes	A1	50.1… or 50.3… seen, then rounded to 50.

Answer: 50 minutes

(3)

(g)(i)

Figure 2 shows the time, $t$ minutes, taken to complete the course at an average speed of $v$ knots.

One crew wants to finish in 45 minutes. Use Figure 2 to find the average speed they need.

Figure 2: time t to complete the course against average speed v

GRAPHNotes

Show solution

$7.2$ knots

Accept 7.1 to 7.4.

Answer: 7.2 knots

(1)

(g)(ii)

Use two points on the graph in Figure 2 to show that $t$ is inversely proportional to $v$. State what happens to the time when the average speed is doubled.

INTERPRETNotes

Show solution

Two readings with the same product, e.g. $5\times 65.2 \approx 10\times 32.6 \approx 326$	B1	Any two correct readings with $v\times t$ approximately constant; or a correct explanation that the distance is fixed so $vt$ is constant.
The time halves.	B1

Answer: $vt$ is constant (about 326); the time halves

(2)

(h)(i)

The wind changes direction. Buoy B is moved along the line due east of S so that the bearing of B from A is now 160°. Buoy A does not move.

Find angle SAB and angle SBA, and explain why triangle SAB is now isosceles.

No calculatorCONCEPTNotes

Show solution

Angle SAB $= 230^\circ - 160^\circ = 70^\circ$	B1
Angle SBA $= 180^\circ - 40^\circ - 70^\circ = 70^\circ$; two angles are equal, so $SB = SA$ and the triangle is isosceles	B1	Must state the equal angles. Allow ECF from (b).

Answer: SAB = 70°, SBA = 70°

(2)

(h)(ii)

Pythagoras' theorem can no longer be used in triangle SAB. Calculate the new distance AB.

MULTINotes

Show solution

$2\times 3.2\sin 20^\circ$ or $\frac{3.2\cos 50^\circ}{\cos 20^\circ}$	M1	Splitting the isosceles triangle into two right-angled triangles, or using the 2.057 km north of A with the 20° angle at A. Sine rule (Extended) is also fine.
$2.19$ km	A1	2.188…

Answer: 2.19 km

(2)

Show full solution

Working	Mark	Notes
Part (a)
Bearing of S from A $= 050^\circ + 180^\circ = 230^\circ$	M1	Or the angle between north at A and AS is $180^\circ - 50^\circ = 130^\circ$ (co-interior angles).
$230^\circ - 140^\circ = 90^\circ$	A1	cso. Or, from the M1 alternative: angle SAB $= 360^\circ - 130^\circ - 140^\circ = 90^\circ$ (angles around A). Any complete correct angle argument; a reason (back bearing or co-interior angles) must be seen.
Part (b)
$40^\circ$	B1	$90^\circ - 50^\circ$
Part (c)
$\tan 40^\circ = \frac{AB}{3.2}$	M1	Or $AB = 3.2\tan 40^\circ$. Allow ECF from (b).
$2.69$ km	A1	2.685…
Part (d)
$SB = \frac{3.2}{\cos 40^\circ}$ $(= 4.177\ldots)$	M1	Or $SB = \sqrt{3.2^2 + 2.685\ldots^2}$.
$3.2 + 2.685\ldots + 4.177\ldots$	M1	Adding the three legs.
$10.06\ldots = 10.1$ km	A1	cso. 10.06 must be seen before rounding.
Part (e)
$3.2\sin 50^\circ$	M1	SR is the distance east of S travelled on the first leg. Or $3.2\cos 40^\circ$.
$2.45$ km	A1	2.451…
Part (f)
$6.5\times 1.852$ $(= 12.038$ km/h$)$	M1
$\frac{10.06\ldots}{12.038}$ $(= 0.8358\ldots$ h$)$ $\times 60$	M1	Distance ÷ their speed, converted to minutes. Using 10.1 km is fine.
$50$ minutes	A1	50.1… or 50.3… seen, then rounded to 50.
Part (g)(i)
$7.2$ knots	B1	Accept 7.1 to 7.4.
Part (g)(ii)
Two readings with the same product, e.g. $5\times 65.2 \approx 10\times 32.6 \approx 326$	B1	Any two correct readings with $v\times t$ approximately constant; or a correct explanation that the distance is fixed so $vt$ is constant.
The time halves.	B1
Part (h)(i)
Angle SAB $= 230^\circ - 160^\circ = 70^\circ$	B1
Angle SBA $= 180^\circ - 40^\circ - 70^\circ = 70^\circ$; two angles are equal, so $SB = SA$ and the triangle is isosceles	B1	Must state the equal angles. Allow ECF from (b).
Part (h)(ii)
$2\times 3.2\sin 20^\circ$ or $\frac{3.2\cos 50^\circ}{\cos 20^\circ}$	M1	Splitting the isosceles triangle into two right-angled triangles, or using the 2.057 km north of A with the 20° angle at A. Sine rule (Extended) is also fine.
$2.19$ km	A1	2.188…

Answers: (b) 40° · (c) 2.69 km · (d) 10.1 km · (e) 2.45 km · (f) 50 minutes · (g)(i) 7.2 knots · (g)(ii) $vt$ is constant (about 326); the time halves · (h)(i) SAB = 70°, SBA = 70° · (h)(ii) 2.19 km

Exam tip (h)(ii) A triangle with no right angle can often still be solved at Core level: an isosceles triangle splits into two right-angled triangles by the line from its apex to the midpoint of its base.

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6.3 Sine rule, cosine rule, trigonometric graphs and 3D problems

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A tidal harbour: sine and cosine rules and a depth graph

6.3 Sine rule, cosine rule, trigonometric graphs and 3D problems2.4 Sequences and graphs of functions6.1 Pythagoras' theorem and right-angled trigonometryExtended

The entrance to a small harbour is marked by a lighthouse L and a buoy B. The end of the harbour wall is at H. The points H, L and B are at sea level.

$HL = 420$ m, $HB = 650$ m and angle $LHB = 38^\circ$ (Figure 1).

Figure 1: the harbour wall H, the lighthouse L and the buoy B

(a)

Calculate the distance LB.

Extended onlyROUTINENotes

Show solution

$LB^2 = 420^2 + 650^2 - 2\times 420\times 650\times\cos 38^\circ$	M1	$(= 168\,646\ldots)$
$411$ m	A1	410.6…

Answer: 411 m

(2)

(b)(i)

Calculate angle HBL.

Extended onlyROUTINENotes

Show solution

$\frac{\sin B}{420} = \frac{\sin 38^\circ}{410.6\ldots}$	M1	Allow ECF from (a). Or cosine rule with three sides.
$39.0^\circ$	A1	39.02…; 38.98… from 411 is fine.

Answer: 39.0°

(2)

(b)(ii)

The equation in part (b)(i) is also satisfied by an obtuse angle. Explain why angle HBL cannot be obtuse.

Extended onlyINTERPRETNotes

Show solution

Angle HLB is opposite the longest side HB (650 m), so it is the largest angle. Angle HBL is opposite HL (420 m), which is shorter than HB, so angle HBL is smaller than angle HLB. A triangle can have only one obtuse angle, and it must be the largest angle, so angle HBL is acute.

Or: if angle HBL were $141.0^\circ$, angle HLB would be $180^\circ - 38^\circ - 141.0^\circ = 1.0^\circ$, the smallest angle, although it is opposite the longest side, which is impossible. The angle sum alone is not a valid reason here, because $38^\circ + 141.0^\circ = 179.0^\circ$, which is less than $180^\circ$.

(1)

(c)(i)

A boat sails in a straight line from H directly towards B.

Calculate the shortest distance between the boat and the lighthouse during this journey.

MODELNotes

Show solution

$420\sin 38^\circ$	M1	Perpendicular distance from L to HB, in a right-angled triangle with hypotenuse HL.
$259$ m	A1	258.5…

Answer: 259 m

(2)

(c)(ii)

Calculate the area of triangle HLB.

Extended onlyROUTINENotes

Show solution

$\frac{1}{2}\times 420\times 650\times\sin 38^\circ$	M1	Or $\frac{1}{2}\times 650\times$ their (c)(i).
$84\,000$ m²	A1	84 037.… Accept 84 040.

Answer: 84 000 m²

(2)

(d)

The depth of water, $d$ metres, at the harbour entrance $t$ hours after midnight is modelled by

$d = 6 + 2.5\sin x$, where $x = 30t$ and $x$ is in degrees, for $0 \le t \le 12$.

Complete the table. Give your values correct to 2 decimal places where necessary.

$t$	0	1	2	3	4	5	6	7	8	9	10	11	12
$d$	6		8.17	8.5	8.17	7.25	6			3.5	3.83	4.75	6

Extended onlyROUTINENotes

Show solution

$7.25$ (at $t = 1$) and $4.75$ (at $t = 7$)	B1
$3.83$ (at $t = 8$)	B1	3.834…

Answer: 7.25, 4.75, 3.83

(2)

(e)

Show that the greatest depth is 8.5 m and that it occurs at 03:00.

Extended onlyPROOFNotes

Show solution

Greatest value of $\sin x$ is 1, at $x = 90^\circ$	M1	Reasoning from the sine graph; reading the table alone is not enough.
$d = 6 + 2.5 = 8.5$ and $t = \frac{90}{30} = 3$, so 03:00	A1	cso

(2)

(f)

Figure 2 shows the graph of $d$ against $t$.

A fishing boat needs a depth of at least 4 m to pass through the entrance. Use Figure 2 to find the times between which the boat cannot pass through.

Figure 2: the depth of water d at the harbour entrance, t hours after midnight

Extended onlyGRAPHNotes

Show solution

$t \approx 7.8$ (about 07:45)	B1	Accept 7.7 to 7.9 (07:40 to 07:55).
$t \approx 10.2$ (about 10:15)	B1	Accept 10.1 to 10.3 (10:05 to 10:20).

Answer: between about 07:45 and 10:15

(2)

(g)

By solving a trigonometric equation, find the two times between 00:00 and 12:00 when the depth is exactly 7.5 m. Give your answers to the nearest minute.

Extended onlyALGEBRANotes

Show solution

$6 + 2.5\sin x = 7.5$ so $\sin x = 0.6$	M1
$x = 36.87\ldots^\circ$ and $x = 180^\circ - 36.87\ldots^\circ = 143.13\ldots^\circ$	A1	Both values needed.
$t = 1.229\ldots$ and $t = 4.771\ldots$: 01:14 and 04:46	A1	Both times, to the nearest minute.

Answer: 01:14 and 04:46

(3)

(h)

During a storm, a surge raises the water level by 0.8 m at all times, so that $d = 6.8 + 2.5\sin x$.

Explain whether the fishing boat in part (f) can now pass through the entrance at any time between 00:00 and 12:00.

Extended onlyMODELNotes

Show solution

Least depth $= 6.8 - 2.5 = 4.3$ m (when $\sin x = -1$, at 09:00)	B1
$4.3 \gt 4$, so yes, the boat can pass through at any time	B1	Conclusion must follow from a least depth above 4 m.

Answer: Yes: the least depth is 4.3 m

(2)

Show full solution

Working	Mark	Notes
Part (a)
$LB^2 = 420^2 + 650^2 - 2\times 420\times 650\times\cos 38^\circ$	M1	$(= 168\,646\ldots)$
$411$ m	A1	410.6…
Part (b)(i)
$\frac{\sin B}{420} = \frac{\sin 38^\circ}{410.6\ldots}$	M1	Allow ECF from (a). Or cosine rule with three sides.
$39.0^\circ$	A1	39.02…; 38.98… from 411 is fine.
Part (b)(ii)
Angle HLB is opposite the longest side HB (650 m), so it is the largest angle. Angle HBL is opposite HL (420 m), which is shorter than HB, so angle HBL is smaller than angle HLB. A triangle can have only one obtuse angle, and it must be the largest angle, so angle HBL is acute.	B1	Or: if angle HBL were $141.0^\circ$, angle HLB would be $180^\circ - 38^\circ - 141.0^\circ = 1.0^\circ$, the smallest angle, although it is opposite the longest side, which is impossible. The angle sum alone is not a valid reason here, because $38^\circ + 141.0^\circ = 179.0^\circ$, which is less than $180^\circ$.
Part (c)(i)
$420\sin 38^\circ$	M1	Perpendicular distance from L to HB, in a right-angled triangle with hypotenuse HL.
$259$ m	A1	258.5…
Part (c)(ii)
$\frac{1}{2}\times 420\times 650\times\sin 38^\circ$	M1	Or $\frac{1}{2}\times 650\times$ their (c)(i).
$84\,000$ m²	A1	84 037.… Accept 84 040.
Part (d)
$7.25$ (at $t = 1$) and $4.75$ (at $t = 7$)	B1
$3.83$ (at $t = 8$)	B1	3.834…
Part (e)
Greatest value of $\sin x$ is 1, at $x = 90^\circ$	M1	Reasoning from the sine graph; reading the table alone is not enough.
$d = 6 + 2.5 = 8.5$ and $t = \frac{90}{30} = 3$, so 03:00	A1	cso
Part (f)
$t \approx 7.8$ (about 07:45)	B1	Accept 7.7 to 7.9 (07:40 to 07:55).
$t \approx 10.2$ (about 10:15)	B1	Accept 10.1 to 10.3 (10:05 to 10:20).
Part (g)
$6 + 2.5\sin x = 7.5$ so $\sin x = 0.6$	M1
$x = 36.87\ldots^\circ$ and $x = 180^\circ - 36.87\ldots^\circ = 143.13\ldots^\circ$	A1	Both values needed.
$t = 1.229\ldots$ and $t = 4.771\ldots$: 01:14 and 04:46	A1	Both times, to the nearest minute.
Part (h)
Least depth $= 6.8 - 2.5 = 4.3$ m (when $\sin x = -1$, at 09:00)	B1
$4.3 \gt 4$, so yes, the boat can pass through at any time	B1	Conclusion must follow from a least depth above 4 m.

Answers: (a) 411 m · (b)(i) 39.0° · (c)(i) 259 m · (c)(ii) 84 000 m² · (d) 7.25, 4.75, 3.83 · (f) between about 07:45 and 10:15 · (g) 01:14 and 04:46 · (h) Yes: the least depth is 4.3 m

Common error (b)(ii) Here $38^\circ + 141.0^\circ = 179.0^\circ$, which is less than $180^\circ$, so the angle sum alone does not rule out the obtuse angle. Use the fact that the largest angle is opposite the longest side.

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A glass pyramid atrium: 3D trigonometry and triangle areas

6.3 Sine rule, cosine rule, trigonometric graphs and 3D problems6.1 Pythagoras' theorem and right-angled trigonometry5.3 Surface area and volume of solidsExtended

The glass roof of a hotel atrium is a pyramid VABCD. Its base ABCD is a horizontal square of side 18 m, and the apex V is vertically above O, the centre of the base. Each sloping edge VA, VB, VC and VD is 15 m long. M is the midpoint of AB (Figure 1).

The four triangular faces are made of glass.

Figure 1: the pyramid roof VABCD; O is the centre of the base and M is the midpoint of AB (not to scale)

(a)

Show that the height VO of the pyramid is 7.94 m, correct to 3 significant figures.

Extended onlyPROOFNotes

Show solution

$AC = \sqrt{18^2 + 18^2}$ $(= 25.45\ldots)$ or $AO = 12.72\ldots$	M1
$VO^2 = 15^2 - 12.72\ldots^2$ $(= 63)$	M1	Pythagoras in triangle VOA, right-angled at O.
$VO = \sqrt{63} = 7.937\ldots = 7.94$ m	A1	cso. 7.937 or $\sqrt{63}$ must be seen.

Answer: 7.94 m

(3)

(b)

Calculate the angle between the edge VA and the base ABCD.

Extended onlyROUTINENotes

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$\cos\theta = \frac{12.72\ldots}{15}$	M1	Or $\sin\theta = \frac{7.937\ldots}{15}$ or $\tan\theta = \frac{7.937\ldots}{12.72\ldots}$; the angle is VAO.
$31.9^\circ$	A1	31.94…

Answer: 31.9°

(2)

(c)

Calculate the angle between the face VAB and the base ABCD.

Extended onlyMULTINotes

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$\tan(\text{VMO}) = \frac{7.937\ldots}{9}$	M1	Angle VMO, with $OM = 9$. Or $VM = \sqrt{15^2 - 9^2} = 12$ and $\cos(\text{VMO}) = \frac{9}{12}$.
$41.4^\circ$	A1	41.40…

Answer: 41.4°

(2)

(d)(i)

Use the cosine rule to calculate angle AVB.

Extended onlyROUTINENotes

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$\cos(\text{AVB}) = \frac{15^2 + 15^2 - 18^2}{2\times 15\times 15}$ $(= 0.28)$	M1
$73.7^\circ$	A1	73.73…

Answer: 73.7°

(2)

(d)(ii)

Calculate the total area of glass in the four faces.

Extended onlyMULTINotes

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$\frac{1}{2}\times 15\times 15\times\sin 73.7\ldots^\circ$ $(= 108)$	M1	Allow ECF from (d)(i). Or $\frac{1}{2}\times 18\times 12$ using $VM = 12$.
$432$ m²	A1	Four faces; accept 431.9.

Answer: 432 m²

(2)

(e)(i)

Figure 2 shows the graphs of $y = \sin x$ and $y = \cos x$ for $0^\circ \le x \le 180^\circ$.

Use the graph of $y = \cos x$ to find an approximate value of angle AVB from your value of $\cos(\text{AVB})$ in part (d)(i), and explain why the cosine rule gives only one possible angle.

Figure 2: the graphs of y = sin x and y = cos x for 0° ≤ x ≤ 180°

Extended onlyGRAPHNotes

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About $74^\circ$ read where $y = 0.28$ meets the cosine curve	B1	Accept 72° to 76°.
The line $y = 0.28$ meets the cosine curve only once between $0^\circ$ and $180^\circ$ (cos decreases all the way), so there is only one angle	B1

Answer: about 74°; only one intersection

(2)

(e)(ii)

A student instead uses the sine rule in triangle VAB, with angle $VAB = 53.1^\circ$, and obtains $\sin(\text{AVB}) = 0.960$.

Use Figure 2 to explain why this equation on its own gives two possible angles, and state, with a reason, which one is correct.

Extended onlyINTERPRETNotes

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The line $y = 0.96$ meets the sine curve twice: $73.7^\circ$ and $180^\circ - 73.7^\circ = 106.3^\circ$	B1	Accept graph readings of about 74° and 106°.
Triangle VAB is isosceles ($VA = VB$), so angle $VBA = 53.1^\circ$; $106.3^\circ + 53.1^\circ + 53.1^\circ \gt 180^\circ$, so angle AVB $= 73.7^\circ$	B1	Or: $\cos(\text{AVB}) = 0.28$ from (d)(i) is positive, so AVB is acute.

Answer: 73.7° (106.3° is impossible)

(2)

(f)

Calculate the volume of air inside the atrium roof.

Extended onlyROUTINENotes

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$\frac{1}{3}\times 18^2\times 7.937\ldots$	M1	$\frac13\times$ base area $\times$ perpendicular height.
$857$ m³	A1	857.2…; 858 from 7.94 is fine.

Answer: 857 m³

(2)

(g)

The architect considers a taller roof: the base stays the same but each sloping edge is lengthened to 17 m.

Calculate the new angle between the face VAB and the base, and compare it with your answer to part (c).

Extended onlyMULTINotes

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$VM = \sqrt{17^2 - 9^2}$ $(= 14.42\ldots)$ or $VO = \sqrt{17^2 - 12.72\ldots^2}$ $(= 11.26\ldots)$	M1
$\cos(\text{VMO}) = \frac{9}{14.42\ldots}$ or $\tan(\text{VMO}) = \frac{11.26\ldots}{9}$	M1
$51.4^\circ$, about $10^\circ$ steeper than before	A1	51.38…; comparison needed. Allow ECF from (c) for the comparison.

Answer: 51.4°, about 10° greater

(3)

Show full solution

Working	Mark	Notes
Part (a)
$AC = \sqrt{18^2 + 18^2}$ $(= 25.45\ldots)$ or $AO = 12.72\ldots$	M1
$VO^2 = 15^2 - 12.72\ldots^2$ $(= 63)$	M1	Pythagoras in triangle VOA, right-angled at O.
$VO = \sqrt{63} = 7.937\ldots = 7.94$ m	A1	cso. 7.937 or $\sqrt{63}$ must be seen.
Part (b)
$\cos\theta = \frac{12.72\ldots}{15}$	M1	Or $\sin\theta = \frac{7.937\ldots}{15}$ or $\tan\theta = \frac{7.937\ldots}{12.72\ldots}$; the angle is VAO.
$31.9^\circ$	A1	31.94…
Part (c)
$\tan(\text{VMO}) = \frac{7.937\ldots}{9}$	M1	Angle VMO, with $OM = 9$. Or $VM = \sqrt{15^2 - 9^2} = 12$ and $\cos(\text{VMO}) = \frac{9}{12}$.
$41.4^\circ$	A1	41.40…
Part (d)(i)
$\cos(\text{AVB}) = \frac{15^2 + 15^2 - 18^2}{2\times 15\times 15}$ $(= 0.28)$	M1
$73.7^\circ$	A1	73.73…
Part (d)(ii)
$\frac{1}{2}\times 15\times 15\times\sin 73.7\ldots^\circ$ $(= 108)$	M1	Allow ECF from (d)(i). Or $\frac{1}{2}\times 18\times 12$ using $VM = 12$.
$432$ m²	A1	Four faces; accept 431.9.
Part (e)(i)
About $74^\circ$ read where $y = 0.28$ meets the cosine curve	B1	Accept 72° to 76°.
The line $y = 0.28$ meets the cosine curve only once between $0^\circ$ and $180^\circ$ (cos decreases all the way), so there is only one angle	B1
Part (e)(ii)
The line $y = 0.96$ meets the sine curve twice: $73.7^\circ$ and $180^\circ - 73.7^\circ = 106.3^\circ$	B1	Accept graph readings of about 74° and 106°.
Triangle VAB is isosceles ($VA = VB$), so angle $VBA = 53.1^\circ$; $106.3^\circ + 53.1^\circ + 53.1^\circ \gt 180^\circ$, so angle AVB $= 73.7^\circ$	B1	Or: $\cos(\text{AVB}) = 0.28$ from (d)(i) is positive, so AVB is acute.
Part (f)
$\frac{1}{3}\times 18^2\times 7.937\ldots$	M1	$\frac13\times$ base area $\times$ perpendicular height.
$857$ m³	A1	857.2…; 858 from 7.94 is fine.
Part (g)
$VM = \sqrt{17^2 - 9^2}$ $(= 14.42\ldots)$ or $VO = \sqrt{17^2 - 12.72\ldots^2}$ $(= 11.26\ldots)$	M1
$\cos(\text{VMO}) = \frac{9}{14.42\ldots}$ or $\tan(\text{VMO}) = \frac{11.26\ldots}{9}$	M1
$51.4^\circ$, about $10^\circ$ steeper than before	A1	51.38…; comparison needed. Allow ECF from (c) for the comparison.

Answers: (a) 7.94 m · (b) 31.9° · (c) 41.4° · (d)(i) 73.7° · (d)(ii) 432 m² · (e)(i) about 74°; only one intersection · (e)(ii) 73.7° (106.3° is impossible) · (f) 857 m³ · (g) 51.4°, about 10° greater

Common error (c) The angle between a face and the base is measured along the line VM, which meets AB at right angles. Using VA (as in part (b)) gives the angle of the edge, not the face.

More 6.3 questions in the question bank

Topic 7 · Transformations and vectors

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

Setup	7.1	7.2	Other
Moving a video-game sprite: describing transformations on a grid	●		4.2 3.2
A robot vacuum cleaner's path: position vectors and vector geometry	○	●	3.1

7.1 Reflections, rotations, translations and enlargements described in words · 7.2 Vectors, magnitude and vector geometrymerged group

+ Add to test✓ In your testCC-CIM-71-72-119 marks~20 min

Moving a video-game sprite: describing transformations on a grid

7.1 Reflections, rotations, translations and enlargements described in words4.2 Similarity, congruence and symmetry3.2 Equation of a straight line, parallel and perpendicular linesCore

A game designer programs the moves of a spaceship sprite. On a coordinate grid the sprite is the triangle S with vertices $(1, 2)$, $(1, 5)$ and $(4, 3)$.

Figure 1 shows S and two other positions of the sprite, labelled T and U.

Figure 1: the sprite S and two other positions of the sprite, T and U

(a)

On the grid in Figure 1, draw the image of S after a reflection in the line $x = -1$. Label the image A.

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two vertices correct, or a correct reflection of S in a different vertical line	B1	Vertices of A: (−3, 2), (−3, 5) and (−6, 3).
triangle A drawn with vertices (−3, 2), (−3, 5) and (−6, 3)	B1	Each $x$-coordinate becomes $2(-1) - x = -2 - x$; the $y$-coordinates do not change.

Answer: A has vertices (−3, 2), (−3, 5) and (−6, 3)

(2)

(b)

Describe fully the single transformation that maps S onto T.

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reflection	B1	Only one transformation may be named; reflection plus anything else scores 0.
in the line $y = -x$	B1	Accept $x + y = 0$. Check: $(1, 2) \to (-2, -1)$, i.e. $(x, y) \to (-y, -x)$.

Answer: Reflection in the line $y = -x$

(2)

(c)

Describe fully the single transformation that maps S onto U.

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rotation	B1
$90^\circ$ clockwise	B1	Accept $270^\circ$ anticlockwise or $-90^\circ$.
centre $(0, -1)$	B1	Found by perpendicular bisectors, e.g. of (1, 2) to (3, −2) and of (1, 5) to (6, −2), or by tracing paper. Check: (4, 3) is 4 right and 4 up from the centre; its image (4, −5) is 4 right and 4 down.

Answer: Rotation, $90^\circ$ clockwise, centre $(0, -1)$

(3)

(d)

In one level, the sprite S is reflected in the line $x = -1$ to give A. Then A is reflected in the line $x = 2$.

Show that the final image is the image of S under the translation $\begin{pmatrix}6\\0\end{pmatrix}$.

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reflecting A in $x = 2$: $(x, y) \to (4 - x, y)$, giving at least two of (7, 2), (7, 5), (10, 3)	M1	Use of the vertices of A from part (a). Allow a correct general method such as $x \to -2 - x \to 4 - (-2 - x)$.
translating S by $\begin{pmatrix}6\\0\end{pmatrix}$: (1, 2) → (7, 2), (1, 5) → (7, 5), (4, 3) → (10, 3), or $4 - (-2 - x) = x + 6$	M1	Dependent on the first M1: the translated vertices must be compared with the image after both reflections. Translating S by $\begin{pmatrix}6\\0\end{pmatrix}$ on its own scores M0. Or the general form $x + 6$ with $y$ unchanged.
all three vertices agree (or $x \to x + 6$, $y \to y$), so the result is the translation $\begin{pmatrix}6\\0\end{pmatrix}$	A1	cso. A conclusion must be stated.

Answer: Final image (7, 2), (7, 5), (10, 3) = S translated by $\begin{pmatrix}6\\0\end{pmatrix}$

(3)

(e)

The order of the two reflections in part (d) is now swapped: S is reflected first in the line $x = 2$, and that image is then reflected in the line $x = -1$.

Find the single transformation that maps S onto the new final image. Does the sprite end in the same position as in part (d)?

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reflecting in $x = 2$ first: (3, 2), (3, 5), (0, 3); then in $x = -1$: (−5, 2), (−5, 5), (−2, 3)	M1	Or the general form $x \to 4 - x \to -2 - (4 - x) = x - 6$.
translation $\begin{pmatrix}-6\\0\end{pmatrix}$; not the same position (it is 12 units to the left of the result in (d))	A1	Both the vector and the conclusion needed.

Answer: Translation $\begin{pmatrix}-6\\0\end{pmatrix}$; no, the final position is different

(2)

(f)(i)

When the sprite collects a power-up, S is enlarged with scale factor 2 and centre $(2, 4)$ to give the image E.

Find the coordinates of the vertices of E.

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two vertices correct	B1	From the centre (2, 4): (1, 2) is $(-1, -2)$ away, doubled to $(-2, -4)$, giving (0, 0).
(0, 0), (0, 6) and (6, 2)	B1	Multiplying the coordinates of S by 2 gives (2, 4), (2, 10), (8, 6): B0 B0 (the centre is not the origin).

Answer: (0, 0), (0, 6) and (6, 2)

(2)

(f)(ii)

Explain why E is similar to S but is not congruent to S.

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the angles of E are the same as the angles of S / all lengths are multiplied by the same factor 2, so E is the same shape (similar)	B1	e.g. the vertical side of S is 3 units, the vertical side of E is 6 units.
E is twice as long in every direction, so it is not the same size and cannot be congruent	B1	Congruent shapes must have equal matching sides.

(2)

(g)

When the power-up wears off, E is enlarged with scale factor $-\dfrac{1}{2}$ and centre $(2, 2)$ to give the image F.

Find the coordinates of the vertices of F, and explain why F is congruent to S.

Extended onlyNo calculatorMULTINotes

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two vertices of F correct	B1	From (2, 2): (0, 0) is $(-2, -2)$ away; $-\frac{1}{2}$ of that is $(1, 1)$, giving (3, 3). Allow ECF from (f)(i).
(3, 3), (3, 0) and (0, 2)	B1
lengths are multiplied by 2 and then by $\frac{1}{2}$, so every side of F equals the matching side of S (e.g. vertical side 3)	B1	Accept a check of the side lengths. (F is S turned through $180^\circ$.)

Answer: (3, 3), (3, 0) and (0, 2); every length is multiplied by $2 \times \frac{1}{2} = 1$

(3)

Show full solution

Working	Mark	Notes
Part (a)
two vertices correct, or a correct reflection of S in a different vertical line	B1	Vertices of A: (−3, 2), (−3, 5) and (−6, 3).
triangle A drawn with vertices (−3, 2), (−3, 5) and (−6, 3)	B1	Each $x$-coordinate becomes $2(-1) - x = -2 - x$; the $y$-coordinates do not change.
Part (b)
reflection	B1	Only one transformation may be named; reflection plus anything else scores 0.
in the line $y = -x$	B1	Accept $x + y = 0$. Check: $(1, 2) \to (-2, -1)$, i.e. $(x, y) \to (-y, -x)$.
Part (c)
rotation	B1
$90^\circ$ clockwise	B1	Accept $270^\circ$ anticlockwise or $-90^\circ$.
centre $(0, -1)$	B1	Found by perpendicular bisectors, e.g. of (1, 2) to (3, −2) and of (1, 5) to (6, −2), or by tracing paper. Check: (4, 3) is 4 right and 4 up from the centre; its image (4, −5) is 4 right and 4 down.
Part (d)
reflecting A in $x = 2$: $(x, y) \to (4 - x, y)$, giving at least two of (7, 2), (7, 5), (10, 3)	M1	Use of the vertices of A from part (a). Allow a correct general method such as $x \to -2 - x \to 4 - (-2 - x)$.
translating S by $\begin{pmatrix}6\\0\end{pmatrix}$: (1, 2) → (7, 2), (1, 5) → (7, 5), (4, 3) → (10, 3), or $4 - (-2 - x) = x + 6$	M1	Dependent on the first M1: the translated vertices must be compared with the image after both reflections. Translating S by $\begin{pmatrix}6\\0\end{pmatrix}$ on its own scores M0. Or the general form $x + 6$ with $y$ unchanged.
all three vertices agree (or $x \to x + 6$, $y \to y$), so the result is the translation $\begin{pmatrix}6\\0\end{pmatrix}$	A1	cso. A conclusion must be stated.
Part (e)
reflecting in $x = 2$ first: (3, 2), (3, 5), (0, 3); then in $x = -1$: (−5, 2), (−5, 5), (−2, 3)	M1	Or the general form $x \to 4 - x \to -2 - (4 - x) = x - 6$.
translation $\begin{pmatrix}-6\\0\end{pmatrix}$; not the same position (it is 12 units to the left of the result in (d))	A1	Both the vector and the conclusion needed.
Part (f)(i)
two vertices correct	B1	From the centre (2, 4): (1, 2) is $(-1, -2)$ away, doubled to $(-2, -4)$, giving (0, 0).
(0, 0), (0, 6) and (6, 2)	B1	Multiplying the coordinates of S by 2 gives (2, 4), (2, 10), (8, 6): B0 B0 (the centre is not the origin).
Part (f)(ii)
the angles of E are the same as the angles of S / all lengths are multiplied by the same factor 2, so E is the same shape (similar)	B1	e.g. the vertical side of S is 3 units, the vertical side of E is 6 units.
E is twice as long in every direction, so it is not the same size and cannot be congruent	B1	Congruent shapes must have equal matching sides.
Part (g)
two vertices of F correct	B1	From (2, 2): (0, 0) is $(-2, -2)$ away; $-\frac{1}{2}$ of that is $(1, 1)$, giving (3, 3). Allow ECF from (f)(i).
(3, 3), (3, 0) and (0, 2)	B1
lengths are multiplied by 2 and then by $\frac{1}{2}$, so every side of F equals the matching side of S (e.g. vertical side 3)	B1	Accept a check of the side lengths. (F is S turned through $180^\circ$.)

Answers: (a) A has vertices (−3, 2), (−3, 5) and (−6, 3) · (b) Reflection in the line $y = -x$ · (c) Rotation, $90^\circ$ clockwise, centre $(0, -1)$ · (d) Final image (7, 2), (7, 5), (10, 3) = S translated by $\begin{pmatrix}6\\0\end{pmatrix}$ · (e) Translation $\begin{pmatrix}-6\\0\end{pmatrix}$; no, the final position is different · (f)(i) (0, 0), (0, 6) and (6, 2) · (g) (3, 3), (3, 0) and (0, 2); every length is multiplied by $2 \times \frac{1}{2} = 1$

Common error (b) Writing the mirror line as $y = x$. Test one vertex: reflection in $y = x$ would send (1, 2) to (2, 1), which is not a vertex of T.

Exam tip (f)(i) With a centre that is not the origin, work with the vector from the centre to each vertex, multiply it by the scale factor, then add it back on to the centre.

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A robot vacuum cleaner's path: position vectors and vector geometry

7.2 Vectors, magnitude and vector geometry7.1 Reflections, rotations, translations and enlargements described in words3.1 Gradient, midpoint and length of a line segmentExtended

A robot vacuum cleaner works in a rectangular room OACB. Its charging dock is at the corner O.

Figure 1 shows the room on a grid in which 1 unit represents 1 metre. A is the point $(9, 0)$, C is $(9, 6)$ and B is $(0, 6)$.

$\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$. M is the midpoint of AC.

Figure 1: the room OACB on a 1 m grid, with the dock at O and the line BM shown dashed; the vectors a and b are drawn beside the walls OA and OB

(a)

The robot leaves the dock and makes three straight moves, given in order by the column vectors $\begin{pmatrix}2\\3\end{pmatrix}$, $\begin{pmatrix}3\\-2\end{pmatrix}$ and $\begin{pmatrix}1\\3\end{pmatrix}$. It stops at the point R.

On the grid in Figure 1, draw the robot's path. Write down the position vector $\overrightarrow{OR}$.

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path drawn from O through (2, 3) and (5, 1) to (6, 4)	B1	Three arrows nose to tail.
$\overrightarrow{OR} = \begin{pmatrix}6\\4\end{pmatrix}$	B1	Not the coordinates (6, 4) alone.

Answer: $\overrightarrow{OR} = \begin{pmatrix}6\\4\end{pmatrix}$

(2)

(b)

Calculate the straight-line distance from R back to the dock.

ROUTINENotes

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$\sqrt{6^2 + 4^2}$ or $\sqrt{52}$	M1	Allow ECF from their $\overrightarrow{OR}$.
7.21 m	A1	Accept $2\sqrt{13}$ m or 7.211…

Answer: 7.21 m ($2\sqrt{13}$ m)

(2)

(c)(i)

Find $\overrightarrow{OM}$ in terms of $\mathbf{a}$ and/or $\mathbf{b}$. Give your answer in its simplest form.

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$\overrightarrow{OA} + \overrightarrow{AM}$ with $\overrightarrow{AM} = \frac{1}{2}\mathbf{b}$	M1	$\overrightarrow{AC}$ is equal to $\overrightarrow{OB}$ (opposite sides of a rectangle).
$\mathbf{a} + \frac{1}{2}\mathbf{b}$	A1

Answer: $\overrightarrow{OM} = \mathbf{a} + \dfrac{1}{2}\mathbf{b}$

(2)

(c)(ii)

The point P lies on BM such that $BP : PM = 2 : 1$.

Show that $\overrightarrow{OP} = \dfrac{2}{3}(\mathbf{a} + \mathbf{b})$.

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$\overrightarrow{BM} = -\mathbf{b} + \mathbf{a} + \frac{1}{2}\mathbf{b} = \mathbf{a} - \frac{1}{2}\mathbf{b}$	M1	Allow ECF from (c)(i) for this mark only.
$\overrightarrow{OP} = \mathbf{b} + \frac{2}{3}(\mathbf{a} - \frac{1}{2}\mathbf{b})$	M1	Or $\overrightarrow{OM} + \frac{1}{3}\overrightarrow{MB}$.
$= \frac{2}{3}\mathbf{a} + \mathbf{b} - \frac{1}{3}\mathbf{b} = \frac{2}{3}\mathbf{a} + \frac{2}{3}\mathbf{b} = \frac{2}{3}(\mathbf{a} + \mathbf{b})$	A1	cso. The expansion must be seen.

Answer: $\overrightarrow{OP} = \dfrac{2}{3}(\mathbf{a} + \mathbf{b})$

(3)

(c)(iii)

Explain what your answer to part (c)(ii) tells you about the points O, P and C. Write down the ratio $OP : PC$.

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$\overrightarrow{OC} = \mathbf{a} + \mathbf{b}$, so $\overrightarrow{OP} = \frac{2}{3}\overrightarrow{OC}$: OP and OC are parallel and share the point O, so O, P and C are collinear (P lies on the diagonal OC)	B1	Must mention the common point O, not only “parallel”.
$OP : PC = 2 : 1$	B1

Answer: O, P and C lie on one straight line; $OP : PC = 2 : 1$

(2)

(d)

Use $\mathbf{a} = \begin{pmatrix}9\\0\end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix}0\\6\end{pmatrix}$ to find $\overrightarrow{OP}$ as a column vector. What does this tell you about the point where the robot stopped?

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$\frac{2}{3}\left(\begin{pmatrix}9\\0\end{pmatrix} + \begin{pmatrix}0\\6\end{pmatrix}\right)$ or $\frac{2}{3}\begin{pmatrix}9\\6\end{pmatrix}$	M1	Or $\begin{pmatrix}0\\6\end{pmatrix} + \frac{2}{3}\begin{pmatrix}9\\-3\end{pmatrix}$.
$\begin{pmatrix}6\\4\end{pmatrix}$, the same as $\overrightarrow{OR}$, so the robot stopped at P (on the diagonal OC)	A1	A conclusion about where the robot stopped is needed.

Answer: $\overrightarrow{OP} = \begin{pmatrix}6\\4\end{pmatrix} = \overrightarrow{OR}$: R is the point P

(2)

(e)(i)

The robot's software is updated so that every move is 1.5 times as long, in the same direction. It makes the same three moves again, starting from the dock.

Find the position vector of its new end point, and state which point of the room this is.

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$1.5\begin{pmatrix}6\\4\end{pmatrix}$, or $\begin{pmatrix}3\\4.5\end{pmatrix} + \begin{pmatrix}4.5\\-3\end{pmatrix} + \begin{pmatrix}1.5\\4.5\end{pmatrix}$	M1	Allow ECF from (a).
$\begin{pmatrix}9\\6\end{pmatrix}$, which is the corner C	A1	Accept $\mathbf{a} + \mathbf{b}$.

Answer: $\begin{pmatrix}9\\6\end{pmatrix}$, the corner C

(2)

(e)(ii)

Describe fully the single transformation that maps the robot's original path onto its new path.

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enlargement, centre O (0, 0)	B1	Each point of the new path is 1.5 times as far from O, in the same direction.
scale factor 1.5	B1	Accept $\frac{3}{2}$.

Answer: Enlargement, scale factor 1.5, centre O

(2)

(e)(iii)

Find the new straight-line distance from the end point back to the dock.

ROUTINENotes

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$1.5 \times 7.21$ or $\sqrt{9^2 + 6^2}$	M1	Allow ECF from (b) or (e)(i).
10.8 m	A1	Accept $3\sqrt{13}$ m or 10.81…

Answer: 10.8 m ($3\sqrt{13}$ m)

(2)

Show full solution

Working	Mark	Notes
Part (a)
path drawn from O through (2, 3) and (5, 1) to (6, 4)	B1	Three arrows nose to tail.
$\overrightarrow{OR} = \begin{pmatrix}6\\4\end{pmatrix}$	B1	Not the coordinates (6, 4) alone.
Part (b)
$\sqrt{6^2 + 4^2}$ or $\sqrt{52}$	M1	Allow ECF from their $\overrightarrow{OR}$.
7.21 m	A1	Accept $2\sqrt{13}$ m or 7.211…
Part (c)(i)
$\overrightarrow{OA} + \overrightarrow{AM}$ with $\overrightarrow{AM} = \frac{1}{2}\mathbf{b}$	M1	$\overrightarrow{AC}$ is equal to $\overrightarrow{OB}$ (opposite sides of a rectangle).
$\mathbf{a} + \frac{1}{2}\mathbf{b}$	A1
Part (c)(ii)
$\overrightarrow{BM} = -\mathbf{b} + \mathbf{a} + \frac{1}{2}\mathbf{b} = \mathbf{a} - \frac{1}{2}\mathbf{b}$	M1	Allow ECF from (c)(i) for this mark only.
$\overrightarrow{OP} = \mathbf{b} + \frac{2}{3}(\mathbf{a} - \frac{1}{2}\mathbf{b})$	M1	Or $\overrightarrow{OM} + \frac{1}{3}\overrightarrow{MB}$.
$= \frac{2}{3}\mathbf{a} + \mathbf{b} - \frac{1}{3}\mathbf{b} = \frac{2}{3}\mathbf{a} + \frac{2}{3}\mathbf{b} = \frac{2}{3}(\mathbf{a} + \mathbf{b})$	A1	cso. The expansion must be seen.
Part (c)(iii)
$\overrightarrow{OC} = \mathbf{a} + \mathbf{b}$, so $\overrightarrow{OP} = \frac{2}{3}\overrightarrow{OC}$: OP and OC are parallel and share the point O, so O, P and C are collinear (P lies on the diagonal OC)	B1	Must mention the common point O, not only “parallel”.
$OP : PC = 2 : 1$	B1
Part (d)
$\frac{2}{3}\left(\begin{pmatrix}9\\0\end{pmatrix} + \begin{pmatrix}0\\6\end{pmatrix}\right)$ or $\frac{2}{3}\begin{pmatrix}9\\6\end{pmatrix}$	M1	Or $\begin{pmatrix}0\\6\end{pmatrix} + \frac{2}{3}\begin{pmatrix}9\\-3\end{pmatrix}$.
$\begin{pmatrix}6\\4\end{pmatrix}$, the same as $\overrightarrow{OR}$, so the robot stopped at P (on the diagonal OC)	A1	A conclusion about where the robot stopped is needed.
Part (e)(i)
$1.5\begin{pmatrix}6\\4\end{pmatrix}$, or $\begin{pmatrix}3\\4.5\end{pmatrix} + \begin{pmatrix}4.5\\-3\end{pmatrix} + \begin{pmatrix}1.5\\4.5\end{pmatrix}$	M1	Allow ECF from (a).
$\begin{pmatrix}9\\6\end{pmatrix}$, which is the corner C	A1	Accept $\mathbf{a} + \mathbf{b}$.
Part (e)(ii)
enlargement, centre O (0, 0)	B1	Each point of the new path is 1.5 times as far from O, in the same direction.
scale factor 1.5	B1	Accept $\frac{3}{2}$.
Part (e)(iii)
$1.5 \times 7.21$ or $\sqrt{9^2 + 6^2}$	M1	Allow ECF from (b) or (e)(i).
10.8 m	A1	Accept $3\sqrt{13}$ m or 10.81…

Answers: (a) $\overrightarrow{OR} = \begin{pmatrix}6\\4\end{pmatrix}$ · (b) 7.21 m ($2\sqrt{13}$ m) · (c)(i) $\overrightarrow{OM} = \mathbf{a} + \dfrac{1}{2}\mathbf{b}$ · (c)(ii) $\overrightarrow{OP} = \dfrac{2}{3}(\mathbf{a} + \mathbf{b})$ · (c)(iii) O, P and C lie on one straight line; $OP : PC = 2 : 1$ · (d) $\overrightarrow{OP} = \begin{pmatrix}6\\4\end{pmatrix} = \overrightarrow{OR}$: R is the point P · (e)(i) $\begin{pmatrix}9\\6\end{pmatrix}$, the corner C · (e)(ii) Enlargement, scale factor 1.5, centre O · (e)(iii) 10.8 m ($3\sqrt{13}$ m)

Exam tip (c)(ii) Write the route first, for example $\overrightarrow{OP} = \overrightarrow{OB} + \overrightarrow{BP}$, then replace each step by $\mathbf{a}$ and $\mathbf{b}$. A ratio 2 : 1 means P is $\frac{2}{3}$ of the way from B to M.

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Topic 8 · Probability

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

Setup	8.1	8.2	Other
Training for a penalty shoot-out: relative frequency and expected goals	●	○	9.1
A medical screening test: trees, Venn diagrams and conditional probability	○	●	1.2

8.1 Single events, relative frequency and expected outcomes · 8.2 Combined events, tree diagrams and conditional probabilitymerged group

+ Add to test✓ In your testCC-CIM-81-82-118 marks~20 min

Training for a penalty shoot-out: relative frequency and expected goals

8.1 Single events, relative frequency and expected outcomes8.2 Combined events, tree diagrams and conditional probability9.1 Averages, range and frequency tablesCore

A football striker practises penalties. The table shows the number of penalties he took and the number of goals he scored in each of five training sessions.

Session	1	2	3	4	5
Penalties taken	20	30	40	30	40
Goals scored	14	21	31	23	31

(a)

Find the relative frequency of scoring in session 3.

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$\frac{31}{40}$ or 0.775

Answer: $\frac{31}{40}$ = 0.775

(1)

(b)

Show that the relative frequency of scoring for all five sessions together is 0.75.

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$\frac{14 + 21 + 31 + 23 + 31}{20 + 30 + 40 + 30 + 40}$ or $\frac{120}{160}$	M1	Both totals must be seen.
$\frac{120}{160}$ $= \frac{3}{4} = 0.75$	A1	cso. Averaging the five session relative frequencies is not a correct method: M0 A0.

Answer: $\frac{120}{160}$ = 0.75

(2)

(c)

Figure 1 shows the overall relative frequency of scoring after sessions 1, 2 and 3. For example, after session 2 the striker had scored 35 goals from 50 penalties.

Plot the points for the end of session 4 and the end of session 5. Describe what happens to the overall relative frequency as the number of penalties increases.

Figure 1: the overall relative frequency of scoring after sessions 1, 2 and 3

GRAPHNotes

Show solution

after session 4: $\frac{89}{120}$ $= 0.741…$	B1	Accept 0.74 or 0.742.
points plotted at (120, 0.742) and (160, 0.75)	B1	Allow ±half a small square. Allow ECF from their value for session 4.
it changes less and less and settles (levels off) at about 0.75	B1	Accept “gets closer to 0.75”. Not just “it increases”.

Answer: Points at (120, 0.742) and (160, 0.75); it settles at about 0.75

(3)

(d)

Use the data to find an estimate of the probability that the striker does not score with a penalty.

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$1 - 0.75 = 0.25$ or $\frac{1}{4}$

Accept $\frac{40}{160}$. Allow ECF from (b).

Answer: $\frac{1}{4}$ = 0.25

(1)

(e)

The striker expects to take 28 penalties next season. Find an estimate of the number of these that he will score.

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$28 \times 0.75$ or $28 \times \frac{3}{4}$	M1	Allow ECF from (b).
21	A1	Accept “about 21”.

Answer: 21

(2)

(f)

Find the mean and the median of the number of goals scored per session.

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$\frac{120}{5}$	M1	Or $(14 + 21 + 31 + 23 + 31) \div 5$.
mean = 24	A1
median = 23	B1	From the ordered list 14, 21, 23, 31, 31.

Answer: Mean 24, median 23

(3)

(g)

The goalkeeper in a training match says: “I save one in every three penalties this striker takes.”

Does the data support the goalkeeper's claim? Give a reason, using the numbers.

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compares $\frac{1}{3}$ (≈ 0.33, or about 53 of 160 penalties) with the striker's estimated probability of not scoring, 0.25 (40 of 160)	B1	Any correct numerical comparison.
No: even counting every miss as a save, the striker failed to score only a quarter of the time, and 160 penalties is a large number of trials, so the estimate is reliable	B1	Conclusion with a reason. Allow ECF from (d).

Answer: No: the striker does not score in only about 1 in 4 penalties, fewer than 1 in 3

(2)

(h)(i)

In a match, the striker takes two penalties. Assume that the probability he scores with each penalty is $\dfrac{3}{4}$, independently.

Complete the tree diagram in Figure 2.

Figure 2: tree diagram for two penalties

Extended onlyNo calculatorCONCEPTNotes

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$\frac{3}{4}$ on the first Goal branch and $\frac{1}{4}$ on the first No goal branch	B1	Accept 0.75 and 0.25.
$\frac{3}{4}$ and $\frac{1}{4}$ on both pairs of second branches	B1

Answer: Goal $\frac{3}{4}$, No goal $\frac{1}{4}$ on every pair of branches

(2)

(h)(ii)

Find the probability that he scores with at least one of the two penalties.

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$1 - \frac{1}{4} \times \frac{1}{4}$ or $\frac{3}{4} \times \frac{3}{4} + 2 \times \frac{3}{4} \times \frac{1}{4}$	M1	Allow ECF from their tree.
$\frac{15}{16}$	A1	Accept 0.9375.

Answer: $\frac{15}{16}$

(2)

Show full solution

Working	Mark	Notes
Part (a)
$\frac{31}{40}$ or 0.775	B1
Part (b)
$\frac{14 + 21 + 31 + 23 + 31}{20 + 30 + 40 + 30 + 40}$ or $\frac{120}{160}$	M1	Both totals must be seen.
$\frac{120}{160}$ $= \frac{3}{4} = 0.75$	A1	cso. Averaging the five session relative frequencies is not a correct method: M0 A0.
Part (c)
after session 4: $\frac{89}{120}$ $= 0.741…$	B1	Accept 0.74 or 0.742.
points plotted at (120, 0.742) and (160, 0.75)	B1	Allow ±half a small square. Allow ECF from their value for session 4.
it changes less and less and settles (levels off) at about 0.75	B1	Accept “gets closer to 0.75”. Not just “it increases”.
Part (d)
$1 - 0.75 = 0.25$ or $\frac{1}{4}$	B1	Accept $\frac{40}{160}$. Allow ECF from (b).
Part (e)
$28 \times 0.75$ or $28 \times \frac{3}{4}$	M1	Allow ECF from (b).
21	A1	Accept “about 21”.
Part (f)
$\frac{120}{5}$	M1	Or $(14 + 21 + 31 + 23 + 31) \div 5$.
mean = 24	A1
median = 23	B1	From the ordered list 14, 21, 23, 31, 31.
Part (g)
compares $\frac{1}{3}$ (≈ 0.33, or about 53 of 160 penalties) with the striker's estimated probability of not scoring, 0.25 (40 of 160)	B1	Any correct numerical comparison.
No: even counting every miss as a save, the striker failed to score only a quarter of the time, and 160 penalties is a large number of trials, so the estimate is reliable	B1	Conclusion with a reason. Allow ECF from (d).
Part (h)(i)
$\frac{3}{4}$ on the first Goal branch and $\frac{1}{4}$ on the first No goal branch	B1	Accept 0.75 and 0.25.
$\frac{3}{4}$ and $\frac{1}{4}$ on both pairs of second branches	B1
Part (h)(ii)
$1 - \frac{1}{4} \times \frac{1}{4}$ or $\frac{3}{4} \times \frac{3}{4} + 2 \times \frac{3}{4} \times \frac{1}{4}$	M1	Allow ECF from their tree.
$\frac{15}{16}$	A1	Accept 0.9375.

Answers: (a) $\frac{31}{40}$ = 0.775 · (b) $\frac{120}{160}$ = 0.75 · (c) Points at (120, 0.742) and (160, 0.75); it settles at about 0.75 · (d) $\frac{1}{4}$ = 0.25 · (e) 21 · (f) Mean 24, median 23 · (g) No: the striker does not score in only about 1 in 4 penalties, fewer than 1 in 3 · (h)(i) Goal $\frac{3}{4}$, No goal $\frac{1}{4}$ on every pair of branches · (h)(ii) $\frac{15}{16}$

Common error (b) Taking the mean of the five relative frequencies. The sessions have different numbers of penalties, so add all the goals and divide by all the penalties.

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A medical screening test: trees, Venn diagrams and conditional probability

8.2 Combined events, tree diagrams and conditional probability8.1 Single events, relative frequency and expected outcomes1.2 Fractions, decimals and percentagesExtended

A clinic screens people for a condition that affects 4% of the population. Four versions of the screening test have been trialled. Figure 1 shows, for each version, the percentage of people with the condition who get a positive result.

For every version, 6% of people who do not have the condition also get a positive result.

Figure 1: the percentage of people with the condition who test positive, for four versions of the test

(a)

The clinic uses Version 3 of the test.

Use Figure 1 to write down, as a decimal, the probability that Version 3 gives a positive result for a person who has the condition.

No calculatorGRAPHNotes

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0.9

From the bar height 90%. Accept $\frac{9}{10}$.

Answer: 0.9

(1)

(b)

Complete the tree diagram in Figure 2.

Figure 2: tree diagram for one person taking the test

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0.96 (no condition); 0.9 and 0.1 after Has condition	B1	Allow ECF from (a) for 0.1.
0.06 and 0.94 after No condition	B1

Answer: 0.96; 0.9, 0.1; 0.06, 0.94

(2)

(c)

Find the probability that a person chosen at random tests positive.

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$0.04 \times 0.9 + 0.96 \times 0.06$	M1	Allow ECF from their tree.
0.0936	A1	Accept $\frac{117}{1250}$.

Answer: 0.0936

(2)

(d)

The clinic tests 5000 people. Find the expected number of positive results.

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$5000 \times 0.0936$	M1	Allow ECF from (c).
468	A1

Answer: 468

(2)

(e)

A person tests positive.

Show that the probability that this person has the condition is $\dfrac{5}{13}$.

PROOFNotes

Show solution

P(has condition and positive) $= 0.04 \times 0.9 = 0.036$	M1
$\frac{0.036}{0.0936}$	M1	Their 0.036 divided by 0.0936 (or $\frac{180}{468}$ using 5000 people).
$\frac{0.036}{0.0936} = \frac{360}{936} = \frac{5}{13}$	A1	cso. A decimal 0.3846… alone, without reaching $\frac{5}{13}$, is A0.

Answer: $\dfrac{5}{13}$

(3)

(f)

Give $\dfrac{5}{13}$ as a percentage, correct to 1 decimal place.

ROUTINENotes

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38.5%

Answer: 38.5%

(1)

(g)

A patient says: “The test detects 90% of people with the condition, so if I test positive I am very likely to have it.”

Explain why the patient is wrong.

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most people (96%) do not have the condition, so even a small false-positive rate (6%) gives many positive results from people without it	B1	e.g. 0.0576 of all people are false positives.
these outnumber the true positives (0.0576 > 0.036), so only about 38.5% ($\frac{5}{13}$) of people who test positive have the condition	B1	The comparison or the value from (e) must be used. Allow ECF.

(2)

(h)(i)

The clinic follows up 80 people who tested positive. It records whether each person has fatigue (F) and whether each has a rash (R). The results are shown in the Venn diagram in Figure 3.

Find the value of $x$.

Figure 3: fatigue (F) and rash (R) for the 80 people in the follow-up group

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$x = 14$

$80 - 22 - 8 - 36$.

Answer: $x = 14$

(1)

(h)(ii)

One of these 80 people is chosen at random. Given that this person has fatigue, find the probability that the person also has a rash.

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$\frac{8}{22 + 8}$	M1	Denominator 30, the number in F.
$\frac{4}{15}$	A1	Accept $\frac{8}{30}$ or 0.267.

Answer: $\frac{4}{15}$

(2)

(i)

The clinic changes its procedure. Anyone who tests positive is tested a second time. Assume the two results are independent, with the probabilities from your tree diagram.

A person tests positive twice. Find the probability that this person has the condition.

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P(condition and two positives) $= 0.04 \times 0.9 \times 0.9 = 0.0324$	M1
$\frac{0.0324}{0.0324 + 0.96 \times 0.06 \times 0.06}$ $= \frac{0.0324}{0.035856}$	M1	The denominator must include both paths that give two positives.
$\frac{75}{83}$ or 0.904	A1	Accept 0.9036…, 90.4%.

Answer: $\frac{75}{83}$ ≈ 0.904

(3)

Show full solution

Working	Mark	Notes
Part (a)
0.9	B1	From the bar height 90%. Accept $\frac{9}{10}$.
Part (b)
0.96 (no condition); 0.9 and 0.1 after Has condition	B1	Allow ECF from (a) for 0.1.
0.06 and 0.94 after No condition	B1
Part (c)
$0.04 \times 0.9 + 0.96 \times 0.06$	M1	Allow ECF from their tree.
0.0936	A1	Accept $\frac{117}{1250}$.
Part (d)
$5000 \times 0.0936$	M1	Allow ECF from (c).
468	A1
Part (e)
P(has condition and positive) $= 0.04 \times 0.9 = 0.036$	M1
$\frac{0.036}{0.0936}$	M1	Their 0.036 divided by 0.0936 (or $\frac{180}{468}$ using 5000 people).
$\frac{0.036}{0.0936} = \frac{360}{936} = \frac{5}{13}$	A1	cso. A decimal 0.3846… alone, without reaching $\frac{5}{13}$, is A0.
Part (f)
38.5%	B1
Part (g)
most people (96%) do not have the condition, so even a small false-positive rate (6%) gives many positive results from people without it	B1	e.g. 0.0576 of all people are false positives.
these outnumber the true positives (0.0576 > 0.036), so only about 38.5% ($\frac{5}{13}$) of people who test positive have the condition	B1	The comparison or the value from (e) must be used. Allow ECF.
Part (h)(i)
$x = 14$	B1	$80 - 22 - 8 - 36$.
Part (h)(ii)
$\frac{8}{22 + 8}$	M1	Denominator 30, the number in F.
$\frac{4}{15}$	A1	Accept $\frac{8}{30}$ or 0.267.
Part (i)
P(condition and two positives) $= 0.04 \times 0.9 \times 0.9 = 0.0324$	M1
$\frac{0.0324}{0.0324 + 0.96 \times 0.06 \times 0.06}$ $= \frac{0.0324}{0.035856}$	M1	The denominator must include both paths that give two positives.
$\frac{75}{83}$ or 0.904	A1	Accept 0.9036…, 90.4%.

Answers: (a) 0.9 · (b) 0.96; 0.9, 0.1; 0.06, 0.94 · (c) 0.0936 · (d) 468 · (e) $\dfrac{5}{13}$ · (f) 38.5% · (h)(i) $x = 14$ · (h)(ii) $\frac{4}{15}$ · (i) $\frac{75}{83}$ ≈ 0.904

Exam tip (e) “Given that” tells you the denominator: here it is the probability of testing positive, from both paths of the tree that end in Positive.

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Topic 9 · Statistics

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

Setup	9.1	9.2	Other
City marathon finish times: grouped data, averages and comparisons	●	○	8.1
Screen time and sleep: scatter diagrams, lines of best fit and cumulative frequency	○	●	3.2

9.1 Averages, range and frequency tables · 9.2 Scatter diagrams, correlation, cumulative frequency and histogramsmerged group

+ Add to test✓ In your testCC-CIM-91-92-119 marks~20 min

City marathon finish times: grouped data, averages and comparisons

9.1 Averages, range and frequency tables9.2 Scatter diagrams, correlation, cumulative frequency and histograms8.1 Single events, relative frequency and expected outcomesCore

The organisers of a city marathon take a random sample of 200 runners who finished the race. The table shows their finish times, $t$ minutes.

Finish time $t$ (minutes)	Frequency
$150 \lt t \le 180$	12
$180 \lt t \le 210$	28
$210 \lt t \le 240$	46
$240 \lt t \le 270$	52
$270 \lt t \le 300$	34
$300 \lt t \le 360$	28

(a)

Write down the modal class.

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$240 \lt t \le 270$

Accept 240–270. Do not accept the frequency 52. The last class is twice as wide but has a lower frequency (28), so it cannot be the modal class.

Answer: $240 \lt t \le 270$

(1)

(b)

Find the class that contains the median finish time.

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median is the 100.5th value, or cumulative frequencies 12, 40, 86, 138 seen	M1	Accept the 100th or 101st value.
$240 \lt t \le 270$	A1

Answer: $240 \lt t \le 270$

(2)

(c)

Show that an estimate of the mean finish time is 249.9 minutes.

MULTINotes

Show solution

midpoints 165, 195, 225, 255, 285, 330	M1	At least five correct midpoints used. The last midpoint is 330, not 315.
$\sum fx = 12\times165+28\times195+46\times225+52\times255+34\times285+28\times330\ (=49\,980)$	M1	$\sum fx$ using their midpoints, at least five products correct.
$\frac{49\,980}{200}=249.9$	A1	cso. Division by 6 (the number of classes) scores M0 for this step.

Answer: 249.9 minutes

(3)

(d)

Explain why the value in part (c) is only an estimate of the mean.

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the exact finish times in each class are not known, so the midpoint is used for every runner in a class

Accept equivalent wording. “Because it is rounded” scores B0.

(1)

(e)(i)

One of the 200 runners is chosen at random.

Find the probability that this runner finished in 4 hours or less.

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$12+28+46\ (=86)$ seen, with 4 hours $=240$ minutes	M1
$\frac{86}{200}$ or $\frac{43}{100}$ or 0.43	A1	Accept 43%.

Answer: $\dfrac{43}{100}$

(2)

(e)(ii)

6500 runners finished the marathon.

Use your answer to part (e)(i) to find an estimate of the number of these runners who finished in 4 hours or less.

ROUTINENotes

Show solution

$6500\times\frac{43}{100}$	M1	Allow ECF from (e)(i).
2795	A1	Accept 2790 or 2800 after 2795 seen.

Answer: 2795

(2)

(f)

Figure 1 shows the cumulative frequency curve for the finish times of the 200 runners.

Use Figure 1 to find an estimate of

(1) the median finish time,
(2) the interquartile range of the finish times.

Figure 1: cumulative frequency curve of the finish times of the 200 runners

Extended onlyGRAPHNotes

Show solution

median 248 minutes	B1	Read at cumulative frequency 100. Accept 245 to 251.
$Q_3 \approx 279$ and $Q_1 \approx 218$	M1	Read at cumulative frequencies 150 and 50. Accept 276 to 282 and 215 to 221.
$\text{IQR}\approx 279-218 = 61$ minutes	A1	Accept 56 to 66 from correct readings.

Answer: median ≈ 248 minutes; IQR ≈ 61 minutes

(3)

(g)

For a sample of runners in last year’s marathon, the median finish time was 262 minutes and the interquartile range was 48 minutes.

Make two comparisons between the finish times of the runners last year and this year.

Extended onlyNo calculatorINTERPRETNotes

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this year’s median is lower, so this year’s runners were faster on average (took less time)	B1	Must refer to the median and to time/speed in context. Allow ECF from (f).
this year’s IQR is larger, so this year’s times were more spread out (less consistent)	B1	Must refer to the IQR or spread. “This year was bigger” without saying what is bigger scores B0.

(2)

(h)

Another 20 runners are added to the sample. All 20 of them had finish times in the class $300 \lt t \le 360$.

Calculate the new estimate of the mean finish time, and state whether the class that contains the median changes. Give a reason for your answer.

MULTINotes

Show solution

$\frac{49\,980+20\times330}{220}$ or $\frac{56\,580}{220}$	M1	Allow ECF from (c). Division by 200 scores M0.
257 minutes (257.18…)	A1	Accept 257.2.
no change: the median is now the 110.5th value and the cumulative frequencies 86 and 138 show it is still in $240 \lt t \le 270$	B1	Must give a reason based on the position of the median.

Answer: 257 minutes (3 s.f.); the median class is still $240 \lt t \le 270$

(3)

Show full solution

Working	Mark	Notes
Part (a)
$240 \lt t \le 270$	B1	Accept 240–270. Do not accept the frequency 52. The last class is twice as wide but has a lower frequency (28), so it cannot be the modal class.
Part (b)
median is the 100.5th value, or cumulative frequencies 12, 40, 86, 138 seen	M1	Accept the 100th or 101st value.
$240 \lt t \le 270$	A1
Part (c)
midpoints 165, 195, 225, 255, 285, 330	M1	At least five correct midpoints used. The last midpoint is 330, not 315.
$\sum fx = 12\times165+28\times195+46\times225+52\times255+34\times285+28\times330\ (=49\,980)$	M1	$\sum fx$ using their midpoints, at least five products correct.
$\frac{49\,980}{200}=249.9$	A1	cso. Division by 6 (the number of classes) scores M0 for this step.
Part (d)
the exact finish times in each class are not known, so the midpoint is used for every runner in a class	B1	Accept equivalent wording. “Because it is rounded” scores B0.
Part (e)(i)
$12+28+46\ (=86)$ seen, with 4 hours $=240$ minutes	M1
$\frac{86}{200}$ or $\frac{43}{100}$ or 0.43	A1	Accept 43%.
Part (e)(ii)
$6500\times\frac{43}{100}$	M1	Allow ECF from (e)(i).
2795	A1	Accept 2790 or 2800 after 2795 seen.
Part (f)
median 248 minutes	B1	Read at cumulative frequency 100. Accept 245 to 251.
$Q_3 \approx 279$ and $Q_1 \approx 218$	M1	Read at cumulative frequencies 150 and 50. Accept 276 to 282 and 215 to 221.
$\text{IQR}\approx 279-218 = 61$ minutes	A1	Accept 56 to 66 from correct readings.
Part (g)
this year’s median is lower, so this year’s runners were faster on average (took less time)	B1	Must refer to the median and to time/speed in context. Allow ECF from (f).
this year’s IQR is larger, so this year’s times were more spread out (less consistent)	B1	Must refer to the IQR or spread. “This year was bigger” without saying what is bigger scores B0.
Part (h)
$\frac{49\,980+20\times330}{220}$ or $\frac{56\,580}{220}$	M1	Allow ECF from (c). Division by 200 scores M0.
257 minutes (257.18…)	A1	Accept 257.2.
no change: the median is now the 110.5th value and the cumulative frequencies 86 and 138 show it is still in $240 \lt t \le 270$	B1	Must give a reason based on the position of the median.

Answers: (a) $240 \lt t \le 270$ · (b) $240 \lt t \le 270$ · (c) 249.9 minutes · (e)(i) $\dfrac{43}{100}$ · (e)(ii) 2795 · (f) median ≈ 248 minutes; IQR ≈ 61 minutes · (h) 257 minutes (3 s.f.); the median class is still $240 \lt t \le 270$

Common error (c) The last class, $300 \lt t \le 360$, is twice as wide as the others. Its midpoint is $\dfrac{300+360}{2}=330$, not 315.

Exam tip (f) For a large grouped data set, read the median at $\tfrac12 n = 100$ and the quartiles at $\tfrac14 n = 50$ and $\tfrac34 n = 150$. Draw the reading lines on the graph.

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Screen time and sleep: scatter diagrams, lines of best fit and cumulative frequency

9.2 Scatter diagrams, correlation, cumulative frequency and histograms9.1 Averages, range and frequency tables3.2 Equation of a straight line, parallel and perpendicular linesCore

Twelve students each recorded their average daily screen time, $x$ hours, and their average amount of sleep each night, $y$ hours, over one week.

Student	A	B	C	D	E	F	G	H	I	J	K	L
Screen time $x$ (hours)	3.5	6	1	4.5	8	3	5	1.5	6.5	2.5	4	5.5
Sleep $y$ (hours)	7.5	6.5	9.0	7.2	6.2	8.3	7.4	8.5	6.8	8.0	7.7	6.9

The points for nine of the students are plotted on the scatter diagram in Figure 1.

Figure 1: scatter diagram of nine of the twelve students

(a)

Plot the points for students E, J and L on the scatter diagram.

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two of $(8, 6.2)$, $(2.5, 8.0)$, $(5.5, 6.9)$ plotted correctly	B1	Within half a small square.
all three points plotted correctly	B1

Answer: points at $(8, 6.2)$, $(2.5, 8.0)$ and $(5.5, 6.9)$

(2)

(b)

Describe the correlation shown by the scatter diagram, and what it tells you about these students.

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strong negative (correlation)	B1	Accept “negative”. Do not accept “inverse proportion”.
students with more screen time tend to sleep less	B1	Must be in context. Do not accept “more screen time causes less sleep”.

Answer: strong negative correlation

(2)

(c)

Show that the mean daily screen time of the twelve students is 4.25 hours.

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$1+1.5+2.5+3+3.5+4+4.5+5+5.5+6+6.5+8 = 51$	M1	Total 51 seen, in any order.
$\frac{51}{12}=4.25$	A1	cso

Answer: 4.25 hours

(2)

(d)(i)

Calculate the mean amount of sleep of the twelve students.

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total $=90$ and $\frac{90}{12}$	M1
7.5 hours	A1

Answer: 7.5 hours

(2)

(d)(ii)

A line of best fit is drawn through the mean point and the point $(0.5, 9.0)$.

Find the equation of this line in the form $y = mx + c$.

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$m=\frac{7.5-9.0}{4.25-0.5}=\frac{-1.5}{3.75}\ (=-0.4)$	M1	Allow ECF from (d)(i) with the mean point $(4.25, \text{their } \bar y)$.
$9.0=-0.4(0.5)+c$ or $7.5=-0.4(4.25)+c$	M1	Substituting a point into $y=mx+c$ with their $m$.
$y=-0.4x+9.2$	A1

Answer: $y=-0.4x+9.2$

(3)

(e)

Explain what the gradient of the line tells you in the context of the survey.

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each extra hour of screen time is associated with about 0.4 hours (24 minutes) less sleep

Must have a decrease, the value 0.4 (or 24 minutes) and the context. Allow ECF from (d)(ii).

(1)

(f)

Maya spends 11 hours a day looking at screens.

Use your equation to find an estimate of the amount of sleep Maya gets. Explain why this estimate may not be reliable.

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$-0.4\times11+9.2$	M1	Allow ECF from (d)(ii).
4.8 hours	A1
11 hours is outside the range of the data (1 to 8 hours), so this is extrapolation and the trend may not continue	B1

Answer: 4.8 hours; unreliable because it is extrapolation

(3)

(g)

A newspaper uses this survey to claim that “screens make teenagers sleep less”.

Explain whether the survey shows that this claim is true.

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no: correlation shows a link between screen time and sleep, but does not prove that one causes the other	B1
a valid alternative, e.g. another factor such as stress or late homework could increase screen time and reduce sleep, or students who cannot sleep may use screens more; or only twelve students were surveyed	B1	Any one sensible reason in context.

(2)

(h)

A larger survey records the daily screen time, $t$ hours, of 160 students. The results are summarised in the cumulative frequency table, and the curve drawn from it is shown in Figure 2.

Screen time $t$ (hours)	$t \le 1$	$t \le 2$	$t \le 3$	$t \le 4$	$t \le 5$	$t \le 6$	$t \le 8$
Cumulative frequency	8	30	68	108	136	150	160

Use the table and Figure 2 to find

(1) an estimate of the median daily screen time,
(2) the percentage of these 160 students whose daily screen time is more than 5 hours.

Figure 2: cumulative frequency curve of the daily screen time of 160 students

Extended onlyGRAPHNotes

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median 3.3 hours	B1	Read at cumulative frequency 80. Accept 3.2 to 3.4.
$160-136\ (=24)$	M1
$\frac{24}{160}\times100=15\%$	A1

Answer: median ≈ 3.3 hours; 15%

(3)

Show full solution

Working	Mark	Notes
Part (a)
two of $(8, 6.2)$, $(2.5, 8.0)$, $(5.5, 6.9)$ plotted correctly	B1	Within half a small square.
all three points plotted correctly	B1
Part (b)
strong negative (correlation)	B1	Accept “negative”. Do not accept “inverse proportion”.
students with more screen time tend to sleep less	B1	Must be in context. Do not accept “more screen time causes less sleep”.
Part (c)
$1+1.5+2.5+3+3.5+4+4.5+5+5.5+6+6.5+8 = 51$	M1	Total 51 seen, in any order.
$\frac{51}{12}=4.25$	A1	cso
Part (d)(i)
total $=90$ and $\frac{90}{12}$	M1
7.5 hours	A1
Part (d)(ii)
$m=\frac{7.5-9.0}{4.25-0.5}=\frac{-1.5}{3.75}\ (=-0.4)$	M1	Allow ECF from (d)(i) with the mean point $(4.25, \text{their } \bar y)$.
$9.0=-0.4(0.5)+c$ or $7.5=-0.4(4.25)+c$	M1	Substituting a point into $y=mx+c$ with their $m$.
$y=-0.4x+9.2$	A1
Part (e)
each extra hour of screen time is associated with about 0.4 hours (24 minutes) less sleep	B1	Must have a decrease, the value 0.4 (or 24 minutes) and the context. Allow ECF from (d)(ii).
Part (f)
$-0.4\times11+9.2$	M1	Allow ECF from (d)(ii).
4.8 hours	A1
11 hours is outside the range of the data (1 to 8 hours), so this is extrapolation and the trend may not continue	B1
Part (g)
no: correlation shows a link between screen time and sleep, but does not prove that one causes the other	B1
a valid alternative, e.g. another factor such as stress or late homework could increase screen time and reduce sleep, or students who cannot sleep may use screens more; or only twelve students were surveyed	B1	Any one sensible reason in context.
Part (h)
median 3.3 hours	B1	Read at cumulative frequency 80. Accept 3.2 to 3.4.
$160-136\ (=24)$	M1
$\frac{24}{160}\times100=15\%$	A1

Answers: (a) points at $(8, 6.2)$, $(2.5, 8.0)$ and $(5.5, 6.9)$ · (b) strong negative correlation · (c) 4.25 hours · (d)(i) 7.5 hours · (d)(ii) $y=-0.4x+9.2$ · (f) 4.8 hours; unreliable because it is extrapolation · (h) median ≈ 3.3 hours; 15%

Exam tip (f) Estimates inside the range of the data (here 1 to 8 hours) are interpolation and usually reliable; estimates outside it are extrapolation. Say which one you used.

More 9.2 questions in the question bank

e.g. \(x=0.0444\ldots\), \(100x=4.444\ldots\), \(10x=0.444\ldots\) and \(90x=4\)	M1	Two multiples of \(x\) whose recurring parts line up, subtracted. Or \(\frac{4}{90}\) from a correct method.
\(\frac{2}{45}\)	A1	\(\frac{4}{90}\) not simplified scores M1 A0.

Outer rectangle \((2x+10)\) by \((x+8)\)	B1	Seen or used. Lengths \(2x+8\) and \(x+6\) (adding the width only once) score B0.
\((2x+10)(x+8)-(2x^2+14x+24)\) with \((2x+10)(x+8)=2x^2+26x+80\)	M1	Subtracting their (a) from the expanded outer area. Alternative: strips \(2\times2(2x+6)+2\times2(x+8)\) or equivalent split of the border into rectangles.
\(2x^2+26x+80-2x^2-14x-24=12x+56\)	A1	cso. Every step must be shown, with no sign errors.

\((2x+6+2w)(x+4+2w)-(2x^2+14x+24)\)	M1	Or strips: \(2w(2x+6)+2w(x+4+2w)\), or equivalent.
\(6wx+4w^2+20w\)	A1	Any order of terms.
\(2w(3x+2w+10)\)	A1	Check: \(w=2\) gives \(4(3x+14)\). \(w(6x+4w+20)\) or \(2(3wx+2w^2+10w)\) is not complete and scores A0.

\(3f+20r=38.5\)	B1	Each journey has its own fixed fee, so \(3f\).
\(5f+26r=53.9\)	B1
Correct method to eliminate one letter, e.g. \(15f+100r=192.5\) and \(15f+78r=161.7\), then subtract to get \(22r=30.8\)	M1	Allow one arithmetic slip. Or eliminate \(r\), or use substitution.
\(f=3.5\) and \(r=1.4\)	A1	Fixed fee $3.50, $1.40 per km. Both needed.

a correct first split or division, e.g. \(45 = 5\times9\) or \(45\div3=15\)	M1	A correct factor tree or repeated division with at most one slip.
\(3^2\times5\)	A1	Accept \(3\times3\times5\). Do not accept a list 3, 3, 5 or an answer with + signs.

\(12 = 2^2\times3\) and \(18 = 2\times3^2\)	B1	Both correct. May be seen in a Venn diagram.
LCM \(=2^2\times3^2\times5\) or listing multiples of 45 and testing them against 12 and 18	M1	Uses the highest power of each prime from 12, 18 and 45. Allow ECF from (a).
180	A1	Accept 3 minutes. 3 (the HCF) scores M0 A0.

\(90\times60\div180\) or \(5400\div180\)	M1	Allow ECF from their (b)(i): their \(5400\div\) their LCM, or \(90\div3\).
30	A1	31 (counting the start) scores M1 A0.

\(\frac{1}{4}=\frac{9}{36}\) and \(\frac{2}{9}=\frac{8}{36}\)	M1	Both fractions over a common denominator (36 is the LCM of 4 and 9). Any correct common denominator, e.g. 72, is fine.
\(1-\frac{17}{36}\)	M1	Subtracting the total of the two effects from 1, e.g. \(\frac{36}{36}-\frac{9}{36}-\frac{8}{36}\).
\(\frac{19}{36}\)	A1	cso. Decimal working (e.g. 0.25 + 0.222) scores at most M0 M1 A0.

18, 24, 12 and 6 read from the graph, or a total of 60	B1
\(\frac{24}{60}\)	M1	Their back-rig value over their total.
\(\frac{2}{5}\)	A1	0.4 or 40% alone scores A0. ECF for the correct simplest form of their fraction.

\(24 : 18 : 6\) or a correct partly simplified ratio, e.g. \(12 : 9 : 3\)	M1	Values from their readings in (e)(i).
\(4 : 3 : 1\)	A1	Must be in this order.

\(40 = 2^3\times5\)	B1
LCM \(=2^3\times3^2\times5\)	M1	Highest powers of 2, 3 and 5 from \(2^2\times3\), \(2\times3^2\) and \(2^3\times5\); or listing multiples of 40 tested against 12 and 18.
360	A1	Accept 6 minutes.

\(\frac{26-16}{16}\times100\) or \(\frac{10}{16}\times100\)	M1	Or \(\frac{26}{16}\times100-100\). Dividing by 26 (giving 38.5%) scores M0.
62.5	A1	Accept 62.5% or 62½%.

75 and 120 read from the graph	B1	Both values.
\(\frac{120-75}{75}\times100\)	M1	Their April minus their February, divided by their February.
60	A1	Accept 60%.

\(\frac{120}{630}\)	M1	Total \(60+75+90+120+150+135=630\). Allow ECF from their readings.
\(\frac{4}{21}\)	A1	Simplest form required (divide by the HCF, 30). 0.190… scores A0.

\(1728\div(84+60)\) \((=12)\) or \(84:60=7:5\) and \(1728\div12\) \((=144)\)	M1	Finding the value of one part.
\(12\times84=1008\) or \(7\times144=1008\)	A1	cso

\(28.60\div0.65\) or \(65\% = 28.60\) leading to \(1\% = 0.44\)	M1	\(28.60\times1.35\) or \(28.60\times0.35\) added on scores M0. Checking \(44\times0.65=28.60\) alone is not a method: M0.
44	A1	cso

\(120=2^3\times3\times5\) and \(168=2^3\times3\times7\), or listing factors of both	M1	Shows a method for the HCF.
24 boxes	A1
5 shirts and 7 pairs of socks	B1	Follow through: \(120\div\) their HCF and \(168\div\) their HCF, if both are whole numbers.

\(2\times26 = 52\)	M1	Cost with the new offer.
\(3\times26\times0.65 = 50.70\) or \(78\times0.65\)	M1	Cost of 3 shirts with 35% off.
No, with $52 and $50.70 compared (the 35% reduction was $1.30 cheaper)	A1	Both values must be correct. Alternative: “pay for 2 of 3” is a reduction of \(\frac{1}{3}\), about 33.3%, which is less than 35%, with values seen.

\(26\times1.2\) \((=31.20)\)	M1
\(31.20\times0.8 = 24.96\)	A1	Accept \(26\times0.96\). Must reach $24.96.
the 20% decrease is 20% of the larger amount, $31.20, so it is a bigger decrease ($6.24) than the increase ($5.20)	B1	A comparison of the bases of the two percentages is needed; “percentages don’t cancel” alone is not enough.

\(\frac{6\times10^{10}}{3\times10^{8}}\)	M1	Rounded values must be seen. Allow ECF from (a).
200	A1	Accept \(2\times10^2\).
greater, because the distance was rounded up (5.6 to 6) and the speed was not changed	B1	Needs the reason; “greater” alone scores B0.

\(\frac{5.6\times10^{10}}{3.0\times10^{8}}\)	M1	Allow ECF from (a).
187	A1	186.66… seen and then rounded correctly. 186 scores A0.

\(\frac{4.0\times10^{11}}{3.0\times10^{8}}\) \((=1333.3\ldots\text{ s})\)	M1	Must use \(4.0\times10^{11}\) m (or \(4.0\times10^{8}\times10^{3}\)).
\(1333.3\ldots\div60 = 22.2\ldots\), so 22 minutes	A1	cso. Both 1333.3… s (or 22.2…) and the division by 60 must be seen.

\(\frac{186.66\ldots}{1333.33\ldots}\times100\) or \(\frac{5.6\times10^{7}}{4.0\times10^{8}}\times100\)	M1	Delays are in proportion to distances, so either ratio is fine. Allow ECF from (c) and (d).
14	A1	Accept 14% or 14.0%. Answers from rounded values in the range 13.9 to 14.1 score A1.

\(2^{25}\div2^{15} = 2^{10}\)	M1	Subtracting the indices.
\(2^{10}\) and 1024	A1	Both forms needed.

\(\frac{1}{8}=2^{-3}\) and \(2^{15}\times2^{-3}\), or \(2^{15}\div2^{3}\)	M1
\(2^{12}\)	A1	4096 alone scores M1 A0.

a distance between 160 and 175 million km read from the graph	B1
\(\frac{\text{their reading}\times10^{9}}{3.0\times10^{8}}\div60\), e.g. \(\frac{1.67\times10^{11}}{3.0\times10^{8}}\div60\)	M1	1 million km \(=10^{6}\) km \(=10^{9}\) m. Division by 60 (seconds to minutes) must be seen.
answer in the range 8.9 to 9.7	A1	From a reading in range; e.g. 167 million km gives 9.28 minutes.

\(5.55\times10^{7}\) or \(5.65\times10^{7}\) km, and \(2.95\times10^{8}\) or \(3.05\times10^{8}\) m/s	B1	At least one correct bound for each quantity.
\(\frac{5.65\times10^{10}}{2.95\times10^{8}}\) or \(\frac{5.55\times10^{10}}{3.05\times10^{8}}\)	M1	Upper bound of distance over lower bound of speed, or lower over upper.
182 and 192	A1	Both values (181.967… and 191.525…).

Working	Mark	Notes
Part (a)
\(5.6\times10^{10}\)	B1	\(56\times10^{9}\) is not standard form: B0.
Part (b)
\(\frac{6\times10^{10}}{3\times10^{8}}\)	M1	Rounded values must be seen. Allow ECF from (a).
200	A1	Accept \(2\times10^2\).
greater, because the distance was rounded up (5.6 to 6) and the speed was not changed	B1	Needs the reason; “greater” alone scores B0.
Part (c)
\(\frac{5.6\times10^{10}}{3.0\times10^{8}}\)	M1	Allow ECF from (a).
187	A1	186.66… seen and then rounded correctly. 186 scores A0.
Part (d)
\(\frac{4.0\times10^{11}}{3.0\times10^{8}}\) \((=1333.3\ldots\text{ s})\)	M1	Must use \(4.0\times10^{11}\) m (or \(4.0\times10^{8}\times10^{3}\)).
\(1333.3\ldots\div60 = 22.2\ldots\), so 22 minutes	A1	cso. Both 1333.3… s (or 22.2…) and the division by 60 must be seen.
Part (e)
\(\frac{186.66\ldots}{1333.33\ldots}\times100\) or \(\frac{5.6\times10^{7}}{4.0\times10^{8}}\times100\)	M1	Delays are in proportion to distances, so either ratio is fine. Allow ECF from (c) and (d).
14	A1	Accept 14% or 14.0%. Answers from rounded values in the range 13.9 to 14.1 score A1.
Part (f)(i)
\(2^{25}\div2^{15} = 2^{10}\)	M1	Subtracting the indices.
\(2^{10}\) and 1024	A1	Both forms needed.
Part (f)(ii)
\(\frac{1}{8}=2^{-3}\) and \(2^{15}\times2^{-3}\), or \(2^{15}\div2^{3}\)	M1
\(2^{12}\)	A1	4096 alone scores M1 A0.
Part (g)
a distance between 160 and 175 million km read from the graph	B1
\(\frac{\text{their reading}\times10^{9}}{3.0\times10^{8}}\div60\), e.g. \(\frac{1.67\times10^{11}}{3.0\times10^{8}}\div60\)	M1	1 million km \(=10^{6}\) km \(=10^{9}\) m. Division by 60 (seconds to minutes) must be seen.
answer in the range 8.9 to 9.7	A1	From a reading in range; e.g. 167 million km gives 9.28 minutes.
Part (h)
\(5.55\times10^{7}\) or \(5.65\times10^{7}\) km, and \(2.95\times10^{8}\) or \(3.05\times10^{8}\) m/s	B1	At least one correct bound for each quantity.
\(\frac{5.65\times10^{10}}{2.95\times10^{8}}\) or \(\frac{5.55\times10^{10}}{3.05\times10^{8}}\)	M1	Upper bound of distance over lower bound of speed, or lower over upper.
182 and 192	A1	Both values (181.967… and 191.525…).

\(1728\div144\) or \(720\div60\) \((=12)\)	M1	Pay per hour; or \(720\times\frac{75}{60}\).
900	A1	Accept $900.00.

\(\frac{1000\times3.5\times4}{100}\) \((=140)\)	M1
1140	A1	140 alone (the interest, not the value) scores M1 A0.

\(700\times1.04^{3}\)	M1	Or year-by-year: 728, 757.12, …
\(787.4048\) leading to \(787.40\)	A1	cso. The unrounded value must be seen.

values about $1175 after 5 years and about $1210 after 6 years read from the graph, or the line crossing $1200 between 5 and 6 years	M1	Or \(1000+35n\gt1200\) worked out.
6 years	A1	5.7 or 5.71 scores M1 A0: interest is added only at the end of a year.

\(\frac{700\times1.04^{8}}{700\times1.04^{5}}=1.04^{8-5}=1.04^{3}\)	M1
the multiplier over the first 3 years is also \(\frac{700\times1.04^{3}}{700}=1.04^{3}\), so the percentage increase is the same	A1	A conclusion comparing both multipliers is needed.
12.5	B1	Accept 12.4864% or 12.49%. From \(1.04^{3}=1.124864\).

\(1140\times0.92\) \((=1048.80)\)	M1	Allow ECF from (c).
\(480\times1.08\) \((=518.40)\)	M1
530.40	A1	Money to 2 d.p.: €530.40. Allow ECF from (c): their \(1140\times0.92-518.40\).

\(\frac{1000\times3.5\times2}{100}\) \((=70)\)	M1	Interest for the first 2 years.
\(\frac{1000\times2.5\times3}{100}\) \((=75)\)	M1	Interest for the last 3 years, on the original $1000.
1145	A1	Accept $1145.00. Using 3.5% for 5 years ($1175) scores M0 M0 A0; 145 alone (the interest) scores M1 M1 A0.