Unit Exam 1 — 90 minutes

IGCSE Mathematics 0580 · Section A non-calculator (Q1–6, 40 marks) · Section B calculator allowed (Q7–10, 25 marks) · Money & Finance · Sets · Time

Duration90 minutes

Total Marks65

Questions10

Time Remaining90:00

Syllabus0580

Instructions

Section A (Questions 1–6, 40 marks): no calculator. Every number has been chosen so the arithmetic works by hand — if your working turns into long division of ugly decimals, you have taken a wrong turn.
Section B (Questions 7–10, 25 marks): calculator allowed. Give money to the nearest cent and other answers to 3 significant figures unless the question says otherwise. Suggested split: about 50 minutes on Section A, 40 on Section B.
Answer all questions in the spaces provided. Later parts often use earlier answers.
Show all working. Method marks are awarded even when the final answer is wrong.
This is a challenge paper for the Unit Assessment: every question is at the top of the grade 8–9 range. Expect to have to think.

Question Navigation · Q1–6 Section A (no calculator) · Q7–10 Section B (calculator)

Question 1 — Currency Conversion with Commission

Total: 5 marks · Section A — non-calculator

(a) [3]

Maya is travelling from London to New York. The exchange rate is £1 = $1.25.
She changes £480 into dollars. The bureau first deducts 2% commission from the pounds, then converts what is left into dollars at the exchange rate.
Calculate how many dollars Maya receives.

Model Answer — 1(a)

M1 commission: 2% of £480 = £9.60, leaving 480 − 9.60 = £470.40

M1 convert to dollars: multiply by 1.25 — and 1.25 = 1¼, so ×1.25 is “add a quarter”: 470.40 + 117.60

A1 $588

Check the direction: dollars are worth less than pounds, so she must end up with more dollars than she had pounds ✓

⚠ If you missed marks here: the classic slips are adding the commission (480 + 9.60 = 489.60 → $612) or sliding a decimal point so 2% of 480 becomes 96 instead of 9.60. Commission is money the bureau keeps — it always makes your amount smaller before the conversion happens.

M1 — deducts 2% commission: 480 − 9.60 = £470.40 M1 — multiplies the pounds by 1.25 to convert into dollars A1 — $588

(b) [2]

Back in London, Maya has $164 left. The bureau converts it to pounds at the same rate, £1 = $1.25, and then charges a flat fee of £3, taken from the pounds.
Calculate how many pounds Maya receives.

Model Answer — 1(b)

M1 going dollars → pounds you divide by the rate: 164 ÷ 1.25 = 164 × 4 ÷ 5 = 656 ÷ 5 = £131.20

A1 subtract the fee: 131.20 − 3 = £128.20

⚠ If you missed marks here: the trap is multiplying by 1.25 again ($164 → £205 − 3 = £202), which sails the wrong way across the exchange rate. Sanity-check every currency answer: a dollar is worth less than a pound, so $164 must become fewer than 164 pounds. The tidy non-calculator route for ÷1.25 is ×4 then ÷5.

M1 — divides by 1.25: $164 → £131.20 before the fee A1 — £128.20

Question 2 — Percentages and Reverse Percentages

Total: 7 marks · Section A — non-calculator

(a) [2]

After a price increase of 15%, a concert ticket costs $138.
Calculate the price of the ticket before the increase.

Model Answer — 2(a)

M1 $138 is 115% of the original, so divide by 1.15: 138 ÷ 1.15 = 13800 ÷ 115 = 120

A1 $120

Check forwards: 120 × 1.15 = 120 + 18 = 138 ✓

⚠ If you missed marks here: the famous wrong answer is 138 × 0.85 = $117.30 — taking 15% of the new price. But the 15% was applied to the original, which is a different, smaller number. Whenever the question gives you the AFTER value, you must divide by the multiplier, never multiply by its opposite.

M1 — recognises $138 is 115% and divides by 1.15 A1 — $120

(b) [2]

In a sale, the price of a jacket is reduced by 20%. After the sale, the sale price is increased by 20%.
Kavya says the jacket is now back to its original price. Show that Kavya is wrong, and find the overall percentage change.

Model Answer — 2(b)

M1 combined multiplier: 0.8 × 1.2 = 0.96

A1 0.96 means 96% of the original — an overall 4% decrease, so Kavya is wrong

Why: the 20% increase acts on the smaller sale price, so it puts back less money than the reduction took away

⚠ If you missed marks here: “−20% then +20%” is not zero, because the two percentages are taken of different amounts. Percentage changes multiply, they never simply add. Try it with £100: sale price £80, then +20% of 80 is £16, landing on £96.

M1 — combined multiplier 0.8 × 1.2 = 0.96 A1 — overall 4% decrease (with a reason why Kavya is wrong)

The value of an investment rises by 10% in the first year, then falls by 5% in the second year. At the end of the two years the investment is worth £418.
Calculate the original value of the investment.

Model Answer — 2(c)

M1 combined multiplier over the two years: 1.10 × 0.95 = 1.045

M1 reverse it: original = 418 ÷ 1.045. Without a calculator write 1.045 = 209/200, so 418 × 200 ÷ 209 — and 418 = 2 × 209

A1 £400

Check forwards: 400 → +10% → 440 → −5% (22) → 418 ✓

⚠ If you missed marks here: the trap is calling the overall change “+10 − 5 = +5%” and dividing by 1.05. The 5% fall acts on the already-increased value, so the true multiplier is 1.10 × 0.95 = 1.045, not 1.05. If your division refused to come out neatly, that was the paper telling you the multiplier was wrong — 418 divides by 1.045 exactly.

M1 — combined multiplier 1.10 × 0.95 = 1.045 M1 — divides the final value by 1.045 (e.g. 418 × 200 ÷ 209) A1 — £400

Question 3 — Simple and Compound Interest

Total: 8 marks · Section A — non-calculator

(a) [2]

Priya invests $800 at a rate of 5% per year simple interest.
Calculate the total value of her investment at the end of 3 years.

Model Answer — 3(a)

M1 simple interest = 800 × 5/100 × 3 = $120 — the same $40 each year

A1 total value 800 + 120 = $920

⚠ If you missed marks here: read what the question wants — the total value, not the interest. Writing $120 and stopping loses the answer mark. The other slip is compounding by accident: simple interest is always calculated on the original $800, never on the growing balance.

M1 — interest 800 × 5/100 × 3 = $120 A1 — total value $920

(b) [3]

Rohan invests $800 at 5% per year compound interest.
Calculate the value of Rohan's investment at the end of 3 years, and find how much more it is worth than Priya's.

Model Answer — 3(b)

M1 multiply by 1.05 each year (add 5% of the current balance):

800 → 840 → 882 → 926.10 (adding 40, then 42, then 44.10)

A1 value $926.10

B1 difference from Priya: 926.10 − 920 = $6.10

⚠ If you missed marks here: adding $40 three times is simple interest wearing a disguise. Compound interest grows on the grown balance: year 2 earns 5% of 840, not of 800. Step-by-step doubling of the working (40, 42, 44.10) is faster by hand than computing 1.05³ = 1.157625 and multiplying.

M1 — multiplies by 1.05 each year: 840, 882, … A1 — $926.10 B1 — $6.10 more than Priya's investment

A sum of $5000 is invested at 20% per year compound interest. No money is added or removed.
At the end of which year does the investment first exceed $10 000? You must show the value at the end of each year.

Model Answer — 3(c)

M1 multiply by 1.2 repeatedly (add a fifth each time):

A1 5000 → 6000 → 7200 → 8640 → 10368

8640 has not passed $10 000; 10368 has

A1 first exceeds $10 000 at the end of year 4

⚠ If you missed marks here: two traps live here. Treating 20% as simple interest gives $1000 a year and the answer “year 6” (year 5 lands on exactly $10 000, which does not exceed it). And guessing “20% a year means doubling in 5 years” ignores compounding — the balance doubles during year 4, because each 20% is bigger than the last.

M1 — repeatedly multiplies by 1.2 A1 — year-end values 6000, 7200, 8640, 10368 A1 — end of year 4

Question 4 — Set Notation

Total: 5 marks · Section A — non-calculator

Information for the whole question

The universal set is
ξ = {x : x is an integer and 1 ≤ x ≤ 12}
A = {x : x is a multiple of 3} B = {x : x is a factor of 12}

(a) [1]

List the elements of A ∩ B.

Model Answer — 4(a)

A = {3, 6, 9, 12} and B = {1, 2, 3, 4, 6, 12}

B1 A ∩ B = {3, 6, 12}

⚠ If you missed marks here: list both sets fully before intersecting — the elements everyone forgets are 1 (a factor of every number) and 12 itself (both a multiple of 3 and a factor of 12). ∩ means AND: only elements sitting in both lists survive.

B1 — A ∩ B = {3, 6, 12}

(b) [2]

Find n((A ∪ B)′).

Model Answer — 4(b)

M1 A ∪ B = {1, 2, 3, 4, 6, 9, 12}, so n(A ∪ B) = 7

A1 ξ has 12 elements, so n((A ∪ B)′) = 12 − 7 = 5

(the five are 5, 7, 8, 10 and 11)

⚠ If you missed marks here: n(A ∪ B) is not n(A) + n(B) = 4 + 6 = 10 — that counts 3, 6 and 12 twice. Union means pooling the elements and crossing out duplicates. And the ′ at the end asks for what is left over in ξ, so remember to subtract from 12, not from 10.

M1 — A ∪ B listed without duplicates, n(A ∪ B) = 7 A1 — 5

For each statement, write TRUE or FALSE and give a reason.
(i) 9 ∉ B
(ii) {2, 6} ⊆ A ∩ B

Model Answer — 4(c)

B1 (i) TRUE — 9 is not a factor of 12 (12 ÷ 9 is not a whole number), so 9 is not in B

B1 (ii) FALSE — 2 is not a multiple of 3, so 2 ∉ A, so 2 ∉ A ∩ B; a set containing 2 cannot be a subset of A ∩ B

⚠ If you missed marks here: ⊆ demands that every single element of the small set belongs to the big one — one stray element (here, the 2) sinks it. And read ∉ carefully: statement (i) says 9 is NOT in B, which is true precisely because 9 fails B's membership rule.

B1 — (i) TRUE with reason: 9 is not a factor of 12 B1 — (ii) FALSE with reason: 2 ∉ A, so {2, 6} is not a subset of A ∩ B

Question 5 — Three-Set Venn Diagram

Total: 7 marks · Section A — non-calculator

Information for the whole question

100 students were asked which sports they play: football (F), hockey (H) and tennis (T). The Venn diagram shows the number of students in each region, where x is a number of students.

(a) [3]

Use n(ξ) = 100 to form an equation, and solve it to find x.

Model Answer — 5(a)

M1 sum every region once: 20 + 15 + 12 + 9 + 8 + 12 + 2x + x = 76 + 3x

M1 equate to the total: 76 + 3x = 100

A1 3x = 24, so x = 8

⚠ If you missed marks here: the region everyone drops is the 12 outside all three circles — students who play none of the sports still count in n(ξ) = 100. Leaving them out gives 64 + 3x = 100 and x = 12, which then poisons every later part. Count all eight regions, exactly once each.

M1 — sums all eight regions including the 12 outside: 76 + 3x M1 — forms the equation 76 + 3x = 100 A1 — x = 8

(b) [1]

Find n(F).

Model Answer — 5(b)

B1 all four regions inside the F circle: 20 + 16 + 9 + 8 = 53 (using 2x = 16 and x = 8)

⚠ If you missed marks here: n(F) is the whole circle — footballers who also play other sports are still footballers. The frequent slips: writing 20 (the “F only” region), or substituting x = 8 where the diagram says 2x. The overlap with hockey holds 16 students, not 8.

B1 — n(F) = 20 + 16 + 9 + 8 = 53

Find n((F ∪ H)′).

Model Answer — 5(c)

(F ∪ H)′ is everyone outside both the F and H circles: the tennis-only region and the students who play nothing

B1 12 + 12 = 24

⚠ If you missed marks here: two traps: forgetting the 12 who play no sport (they are firmly inside the complement), and reading (F ∪ H)′ as (F ∩ H)′. Union first, then complement: shade F and H, and take everything else — which includes part of the T circle and the outside.

B1 — n((F ∪ H)′) = 12 + 12 = 24

(d) [2]

One of the students who plays tennis is chosen at random.
Find the probability that this student plays all three sports. Give your answer as a fraction in its simplest form.

Model Answer — 5(d)

M1 the choice is made among tennis players only: n(T) = 12 + 9 + 8 + 8 = 37

A1 P = 8/37 (37 is prime, so nothing cancels)

⚠ If you missed marks here: the words “who plays tennis” shrink the pool before the choice is made — the denominator is n(T) = 37, not 100. Writing 8/100 = 2/25 ignores the condition. This is exactly the conditional-probability phrasing Cambridge loves to hide inside a Venn question.

M1 — denominator n(T) = 37, not 100 A1 — P = 8/37

Question 6 — Timetables, Average Speed and Time Zones

Total: 8 marks · Section A — non-calculator

Information for parts (a) to (c)

The timetable shows two overnight coaches from Avonford to Calder, stopping at Brackley. All times are 24-hour clock; times after midnight are on the next day.

(a) [1]

How long does Coach 2 take for the whole journey from Avonford to Calder, including the stop? Give your answer in hours and minutes.

Model Answer — 6(a)

count through midnight: 22:55 → 00:00 is 1 h 5 min, then 00:00 → 02:00 is 2 h

B1 3 h 10 min

⚠ If you missed marks here: never subtract clock times like decimals — 2.00 − 22.55 is nonsense because an hour has 60 minutes, not 100. Bridge through midnight in two easy hops (to 00:00, then onwards). The other slip is forgetting the journey crosses into the next day entirely.

B1 — 3 h 10 min

(b) [2]

The road distance from Avonford to Brackley is 126 km.
Calculate the average speed of Coach 1 between Avonford and Brackley, in km/h.

Model Answer — 6(b)

M1 21:35 → 23:20 is 1 h 45 min = 1¾ h = 7/4 h, so speed = 126 ÷ 7/4 = 126 × 4/7

126 ÷ 7 = 18, and 18 × 4 = 72

A1 72 km/h

⚠ If you missed marks here: the killer is writing 1 h 45 min as 1.45 h — 45 minutes is ¾ of an hour, so it is 1.75 h. Turning the time into a fraction (7/4) and dividing by flipping it is the clean non-calculator route; decimals invite the 1.45 blunder.

M1 — time 1 h 45 min written as 1.75 h (or 7/4 h) and 126 divided by it A1 — 72 km/h

The road distance from Brackley to Calder is 75 km.
Calculate Coach 1's average driving speed for the whole journey from Avonford to Calder, not counting the 10-minute stop at Brackley.

Model Answer — 6(c)

M1 total distance 126 + 75 = 201 km; driving time 1 h 45 min + 1 h 15 min = 3 h exactly

A1 201 ÷ 3 = 67 km/h

⚠ If you missed marks here: average speed is always total distance ÷ total time. Averaging the two leg speeds — (72 + 60) ÷ 2 = 66 — is wrong because the coach spends longer at the faster speed. And dividing by 3 h 10 min would wrongly include the stop the question told you to exclude.

M1 — total distance 201 km over driving time 3 h (stop excluded) A1 — 67 km/h

(d) [3]

A plane leaves London at 20:40 on Monday, local time. The flight takes 9 hours 45 minutes. The local time in Singapore is 7 hours ahead of London.
Find the local time and day in Singapore when the plane lands.

Model Answer — 6(d)

M1 add the flight time in London time: 20:40 + 9 h 45 min — 20:40 + 9 h = 05:40 Tuesday, + 45 min = 06:25 Tuesday (London time)

M1 Singapore is ahead, so add 7 hours: 06:25 + 7 h

A1 13:25 on Tuesday

⚠ If you missed marks here: do the two steps in order and separately: first fly (add the duration), then change the clock (adjust the zone). “Ahead” means Singapore's clocks show a later time, so you add the 7 hours — subtracting gives 23:25 Monday, a whole half-day out. Dropping the day change at midnight loses the answer mark: the paper asked for time and day.

M1 — 20:40 + 9 h 45 min = 06:25 Tuesday, London time M1 — adds 7 hours (ahead = later, never subtract) A1 — 13:25 on Tuesday

Question 7 — Compound Interest over Many Years

Total: 7 marks · Section B — calculator allowed

(a) [3]

Aditi invests $6500 in an account paying 3.8% per year compound interest. No money is added or removed.
Calculate the value of the investment at the end of 6 years. Give your answer correct to the nearest cent.

Model Answer — 7(a)

M1 the multiplier is 1.038 and it acts once per year, so the 6 goes in the power: value = 6500 × 1.038⁶

M1 1.038⁶ = 1.2508 (4 dp)

A1 $8130.13 (calculator shows 8130.1297… — nearest cent)

Sense check: simple interest would give 6500 + 6 × 247 = $7982, and compound must beat simple ✓

⚠ If you missed marks here: the two classic wrong values are $7982 (simple interest: 6500 × (1 + 0.038 × 6) — the same $247 every year, but compound interest grows on the new balance each year) and $7832.49 (using the power 5 — after 6 complete years the multiplier has acted 6 times, not 5). Keep every calculator digit until the final rounding.

M1 — sets up 6500 × 1.038⁶ (the 6 in the power, not multiplied) M1 — evaluates 1.038⁶ = 1.2508 (4 dp or better) A1 — $8130.13

(b) [2]

Calculate the interest earned during the 6th year only.

Model Answer — 7(b)

M1 value at the end of year 5: 6500 × 1.038⁵ = $7832.49; the 6th year's interest is 8130.13 − 7832.49, or equivalently 0.038 × 7832.49

A1 $297.63 (accept $297.64 if you subtracted the two already-rounded values)

⚠ If you missed marks here: if you got $247, you found 3.8% of the original $6500 — but by year 6 the balance has grown to $7832.49, and the 3.8% acts on that. Interest in a single later year of a compound scheme is always bigger than in year 1: that is the whole point of compounding.

M1 — year-5 value $7832.49 and subtracts (or takes 3.8% of it) A1 — $297.63 (accept $297.64)

Find the number of complete years needed for the investment to first be worth at least double its starting value.

Model Answer — 7(c)

M1 needs the smallest n with 1.038ⁿ ≥ 2: trials give 1.038¹⁸ = 1.9568 (not there yet) and 1.038¹⁹ = 2.0312 ✓

A1 19 years

⚠ If you missed marks here: the classic wrong answer is 18 — that comes from solving n = 18.59 and rounding down, but after 18 years the money has only multiplied by 1.9568, which is not yet double. “At least” and “first exceeds” always round up and then check both neighbours. Solving 1 + 0.038n = 2 to get n ≈ 26.3 is the simple-interest model answering a compound-interest question.

M1 — tests 1.038ⁿ against 2 (trials or logs), showing n = 18 fails A1 — 19 years

Question 8 — Three-Set Venn Diagram with Percentages

Total: 6 marks · Section B — calculator allowed

Information for the whole question

300 travellers were asked which countries they have visited: France (F), Germany (G) and Spain (S). The Venn diagram shows the percentage of the travellers in each region. x% visited all three countries, and the percentage who visited none of the three is 2x%.

(a) [3]

Find the value of x.

Model Answer — 8(a)

M1 the eight regions are percentages of the same whole, so they sum to 100: 18 + 12 + 15 + 9 + 7 + 6 + x + 2x = 67 + 3x

M1 67 + 3x = 100

A1 3x = 33, so x = 11

⚠ If you missed marks here: the region everyone drops is the 2x% outside all three circles — travellers who visited none still count in the 100%. Leaving it out gives 67 + x = 100 and the wrong x = 33 (and 33% visiting all three should have felt absurd). Equating the sum to 300 instead of 100 mixes people with percentages — the diagram is in %, so the total is 100.

M1 — sums all eight regions including the 2x outside: 67 + 3x M1 — forms the equation 67 + 3x = 100 A1 — x = 11

(b) [1]

Calculate the number of travellers who visited exactly two of the three countries.

Model Answer — 8(b)

exactly two = the three pairwise overlaps excluding the centre: (9 + 7 + 6)% = 22%

B1 22% of 300 = 66 travellers

⚠ If you missed marks here: two traps: answering 22 (that is a percentage — the question asks for a number of travellers, so multiply by 300), and including the centre to get 33% → 99 — the x% in the middle visited exactly three countries, not two.

B1 — (9 + 7 + 6)% = 22% of 300 = 66 travellers

One traveller who visited Germany is chosen at random.
Find the probability that this traveller visited all three countries. Give your answer as a fraction in its simplest form.

Model Answer — 8(c)

M1 the pool is Germany only: n(G) = 12 + 9 + 6 + 11 = 38 (%)

A1 P = 11/38 — every region is a percentage of the same 300, so the ratio of percentages is the probability (33/114 people = 11/38 too)

⚠ If you missed marks here: 11/100 ignores the condition “who visited Germany” — the words shrink the pool to the G circle before the choice is made, so the denominator is 38, not 100. You do not need to convert to people first: percentages of the same whole cancel.

M1 — denominator is the whole G circle: 12 + 9 + 6 + 11 = 38 A1 — P = 11/38

Question 9 — Currency Chain with Fees

Total: 6 marks · Section B — calculator allowed

(a) [3]

Mia is travelling from London to Paris. She changes £850 into euros. The bureau first deducts 1.5% commission from the pounds, then converts what is left at £1 = €1.16.
Calculate how many euros Mia receives.

Model Answer — 9(a)

M1 commission: 1.5% of 850 = £12.75, leaving 850 × 0.985 = £837.25

M1 convert: 837.25 × 1.16

A1 €971.21

Direction check: a pound is worth more than a euro, so she must receive more euros than she had pounds ✓

⚠ If you missed marks here: the classic wrong values: €1000.79 comes from adding the commission (850 × 1.015 = 862.75 before converting) — commission is money the bureau keeps; and €721.77 comes from dividing by 1.16 — that is the direction for euros→pounds. Pounds to euros at £1 = €1.16 always multiplies.

M1 — deducts 1.5% commission: 850 → £837.25 M1 — multiplies by 1.16 to convert into euros A1 — €971.21

(b) [3]

In Paris Mia spends €527, leaving her with €444.21. She flies on to New York and converts her remaining euros into dollars. She can choose between two bureaus:
Bureau A: a flat fee of €4 is taken first, then the rest is converted at €1 = $1.08.
Bureau B: no fee, converted at €1 = $1.075.
Which bureau gives Mia more dollars, and how many more? Show your working for both.

Model Answer — 9(b)

M1 Bureau A: (444.21 − 4) × 1.08 = 440.21 × 1.08 = $475.43

M1 Bureau B: 444.21 × 1.075 = $477.53

A1 Bureau B, by $2.10

⚠ If you missed marks here: the instinct “A has the better rate so A wins” is exactly what this question punishes: A's better rate earns only about $2.22 extra across €444.21, but its €4 fee costs $4.32 — so B wins. The other slip is taking the €4 off after converting ($479.75 − 4 = $475.75), which subtracts euros from dollars — a units error. Compute both routes to the end before choosing.

M1 — Bureau A: fee first, (444.21 − 4) × 1.08 = $475.43 M1 — Bureau B: 444.21 × 1.075 = $477.53 A1 — Bureau B, by $2.10

Question 10 — Time Zones and Average Speed

Total: 6 marks · Section B — calculator allowed

(a) [3]

A plane leaves Dubai at 23:45 on Tuesday, local time. The flight to London takes 7 hours 50 minutes. Dubai time is 3 hours ahead of London time.
Find the local time and day in London when the plane lands.

Model Answer — 10(a)

M1 fly first, in Dubai time: 23:45 + 7 h 50 min = 07:35 Wednesday (Dubai time)

M1 Dubai is ahead, so London's clocks show an earlier time: subtract 3 hours

A1 04:35 on Wednesday

⚠ If you missed marks here: 10:35 comes from adding the 3 hours — but “Dubai is ahead” means London is behind, so going Dubai→London you subtract. Do the two steps separately and in order: add the flight time, then adjust the zone, keeping the day attached at every step (23:45 crosses midnight almost immediately).

M1 — 23:45 + 7 h 50 min = 07:35 Wednesday, Dubai time M1 — subtracts 3 hours (Dubai ahead ⇒ London earlier) A1 — 04:35 on Wednesday

(b) [2]

The flight distance is 5470 km.
Calculate the average speed of the plane in km/h, correct to 3 significant figures.

Model Answer — 10(b)

M1 7 h 50 min = 7 + 50/60 = 7.8333… h (on the calculator: 5470 ÷ (7 + 50 ÷ 60))

A1 5470 ÷ 7.8333… = 698.297… = 698 km/h (3 sf)

⚠ If you missed marks here: 729 km/h comes from typing 7 h 50 min as 7.5 — fifty minutes is 50/60 = 0.8333 of an hour, so the time is 7.83 h, not 7.5. Never write minutes after a decimal point; either use a fraction (47/6 h) or the bracket (7 + 50 ÷ 60).

M1 — converts 7 h 50 min to 7.833… h (or 47/6 h) A1 — 698 km/h (3 sf)

The return flight leaves London at 13:20 on Thursday and lands in Dubai at 00:20 on Friday, local time.
Find the flight time.

Model Answer — 10(c)

convert one end into the other zone first: 00:20 Friday in Dubai = 21:20 Thursday in London

B1 13:20 → 21:20 = 8 hours

⚠ If you missed marks here: 11 h comes from subtracting the two clock readings as if they were in the same zone — they are not. Convert the arrival into London time (subtract the 3 hours Dubai is ahead), then subtract. A same-route flight taking 3 hours longer on the way back should have triggered your alarm.

B1 — 8 hours (after converting one end across the 3-hour zone gap)

Self-Assessment

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