Unit Exam 2 — 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]

Leo is travelling from Paris to Geneva. The exchange rate is €1 = 1.40 francs.
He changes €650 into francs. The bureau first deducts 3% commission from the euros, then converts what is left at the exchange rate.
Calculate how many francs Leo receives.

Model Answer — 1(a)

M1 commission: 3% of €650 = €19.50, leaving 650 − 19.50 = €630.50

M1 convert: multiply by 1.4 — by hand, ×1.4 is “the number plus four tenths of it”: 630.50 + 252.20

A1 882.70 francs

Direction check: a euro is worth more than a franc, so he must receive more francs than he had euros ✓

⚠ If you missed marks here: the classic slips: adding the commission (650 + 19.50 = 669.50 → 937.30 francs) — commission is money the bureau keeps; sliding a decimal so 3% of 650 becomes 195; and dividing by 1.4, which sends you the wrong way across the rate (euros → francs multiplies).

M1 — deducts 3% commission: 650 − 19.50 = €630.50 M1 — multiplies the euros by 1.4 to convert into francs A1 — 882.70 francs

(b) [2]

After the trip Leo has 86 francs left. The bureau first takes a flat fee of 2 francs, then converts the remainder into euros at the same rate, €1 = 1.40 francs.
Calculate how many euros Leo receives.

Model Answer — 1(b)

M1 fee first: 86 − 2 = 84 francs; francs → euros divides by the rate: 84 ÷ 1.4 = 840 ÷ 14 = 60

A1 €60

⚠ If you missed marks here: multiplying by 1.4 gives €117.60 — the wrong way across the rate: a franc is worth less than a euro, so 84 francs must become fewer than 84 euros. Converting first and subtracting the fee after gives 86 ÷ 1.4 − 2 = €59.43 — and the ugly decimal 61.4285… on the way was the paper telling you the order was wrong. The tidy route for ÷1.4 is ×10 ÷14.

M1 — fee first, then divides: (86 − 2) ÷ 1.4 A1 — €60

Question 2 — Percentages and Reverse Percentages

Total: 7 marks · Section A — non-calculator

(a) [2]

In a sale every coat is reduced by 12%. A coat costs $132 in the sale.
Calculate the price of the coat before the sale.

Model Answer — 2(a)

M1 $132 is 88% of the original, so divide by 0.88: 132 ÷ 0.88 = 13200 ÷ 88 = 150

A1 $150

Check forwards: 12% of 150 = 18, and 150 − 18 = 132 ✓

⚠ If you missed marks here: the famous wrong answer is 132 × 1.12 = $147.84 — adding 12% of the sale price. The 12% was taken off the original, bigger number, so putting 12% of the smaller number back on cannot restore it. Given the AFTER value, always divide by the multiplier.

M1 — recognises $132 is 88% and divides by 0.88 A1 — $150

(b) [2]

The population of a town rises by 25% during one decade, then falls by 25% during the next.
Amir says the population is back where it started. Show that Amir is wrong, and find the overall percentage change.

Model Answer — 2(b)

M1 combined multiplier: 1.25 × 0.75 = 0.9375 (= 5/4 × 3/4 = 15/16)

A1 0.9375 means 93.75% of the original: an overall 6.25% decrease, so Amir is wrong

Why: the 25% fall acts on the larger, already-grown population, so it removes more people than the rise added

⚠ If you missed marks here: “+25% then −25%” is not zero because the two percentages are taken of different amounts — percentage changes multiply, they never simply add. Try 100: up to 125, then 25% of 125 is 31.25, landing on 93.75. The fraction route 5/4 × 3/4 = 15/16 makes the non-calculator arithmetic instant.

M1 — combined multiplier 1.25 × 0.75 = 0.9375 (or 15/16) A1 — overall 6.25% decrease (with a reason why Amir is wrong)

A price is increased by 20%, and then increased again by p%. The overall increase is 38%.
Find the value of p.

Model Answer — 2(c)

M1 multipliers multiply: 1.2 × m = 1.38 where m is the second multiplier

M1 m = 1.38 ÷ 1.2 = 138/120 = 23/20 = 1.15

A1 p = 15

Check: 1.2 × 1.15 = 1.38 ✓

⚠ If you missed marks here: the trap is 38 − 20 = 18 — that treats the second rise as acting on the original price, but it acts on the already-raised price. Check it: 1.2 × 1.18 = 1.416, an overall 41.6% rise, not 38%. Undo a multiplier by dividing: 1.38 ÷ 1.2, done by hand as 138/120 and cancelling.

M1 — sets up 1.2 × m = 1.38 M1 — m = 1.38 ÷ 1.2 = 1.15 A1 — p = 15

Question 3 — Simple and Compound Interest

Total: 8 marks · Section A — non-calculator

(a) [2]

Dev invests $1200 at a rate of 4% per year simple interest.
Calculate the total value of his investment at the end of 5 years.

Model Answer — 3(a)

M1 simple interest = 1200 × 4/100 × 5 = $240 — the same $48 each year

A1 total value 1200 + 240 = $1440

⚠ If you missed marks here: read what the question wants — the total value, not the interest: writing $240 and stopping loses the answer mark. And simple interest is always calculated on the original $1200; if your yearly amounts grew, you compounded by accident.

M1 — interest 1200 × 4/100 × 5 = $240 A1 — total value $1440

(b) [3]

Zara invests $2000 at 10% per year compound interest.
Calculate the value of Zara's investment at the end of 3 years, and find how much more this is than the same money at 10% per year simple interest.

Model Answer — 3(b)

M1 multiply by 1.1 each year (add 10% of the current balance): 2000 → 2200 → 2420 → 2662 (adding 200, then 220, then 242)

A1 value $2662

B1 simple gives 2000 + 3 × 200 = $2600, so $62 more

⚠ If you missed marks here: adding $200 three times is simple interest wearing a disguise — year 2's 10% is taken of $2200, not of $2000. The growing steps (200, 220, 242) are the signature of compounding, and stepping year by year is faster by hand than computing 1.1³ = 1.331 and multiplying.

M1 — multiplies by 1.1 each year: 2200, 2420, … A1 — $2662 B1 — $62 more than simple interest

A sum of $4000 is invested at 25% per year compound interest. No money is added or removed.
At the end of which year does the investment first exceed $9000? You must show the value at the end of each year.

Model Answer — 3(c)

M1 multiply by 1.25 repeatedly (add a quarter each time):

A1 4000 → 5000 → 6250 → 7812.50 → 9765.63

7812.50 has not passed $9000; 9765.63 has

A1 first exceeds $9000 at the end of year 4

⚠ If you missed marks here: treating 25% as simple interest gives $1000 a year and lands on exactly $9000 at the end of year 5 — which does not exceed it — so that route answers “year 6” and is doubly wrong. Compounding is faster: each 25% is a quarter of a bigger number. And do not stop at 7812.50 — check it against 9000 before writing “year 3”.

M1 — repeatedly multiplies by 1.25 (adds a quarter) A1 — year-end values 5000, 6250, 7812.50, 9765.63 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 2 ≤ x ≤ 14}
A = {x : x is a prime number} B = {x : x is an odd number}

(a) [1]

List the elements of A ∩ B.

Model Answer — 4(a)

A = {2, 3, 5, 7, 11, 13} and B = {3, 5, 7, 9, 11, 13}

B1 A ∩ B = {3, 5, 7, 11, 13}

⚠ If you missed marks here: the element everyone wrongly keeps is 2 — it is prime, but it is not odd, so it fails B and cannot be in the intersection. The element everyone wrongly adds is 9 — odd, but 9 = 3 × 3 is not prime. List both sets in full before intersecting; A ∩ B is not simply “A”.

B1 — A ∩ B = {3, 5, 7, 11, 13} (2 excluded, 9 excluded)

(b) [2]

Find n((A ∪ B)′).

Model Answer — 4(b)

M1 A ∪ B = {2, 3, 5, 7, 9, 11, 13}, so n(A ∪ B) = 7

A1 ξ runs from 2 to 14, so it has 14 − 2 + 1 = 13 elements: n((A ∪ B)′) = 13 − 7 = 6

(the six are 4, 6, 8, 10, 12 and 14 — the even non-primes)

⚠ If you missed marks here: n(A ∪ B) is not n(A) + n(B) = 6 + 6 = 12 — that counts the five odd primes twice. And ξ has 13 members, not 14: a run from 2 to 14 inclusive counts 14 − 2 + 1. Both slips give a wrong final subtraction, so check each stage.

M1 — A ∪ B listed without duplicates, n(A ∪ B) = 7 A1 — 13 − 7 = 6

For each statement, write TRUE or FALSE and give a reason.
(i) 9 ∉ A
(ii) {2, 13} ⊆ A ∩ B

Model Answer — 4(c)

B1 (i) TRUE — 9 = 3 × 3, so 9 is not prime and is not in A

B1 (ii) FALSE — 2 is not odd, so 2 ∉ B, 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 — 13 passing does not save the 2. And a reason must point at the failing element and the failing rule (“2 is not odd”), not just restate the verdict.

B1 — (i) TRUE with reason: 9 = 3 × 3 is not prime B1 — (ii) FALSE with reason: 2 ∉ B, so {2, 13} 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

80 students were asked which instruments they play: piano (P), guitar (G) and drums (D). The Venn diagram shows the number of students in each region, where x is a number of students.

(a) [3]

Use n(ξ) = 80 to form an equation, and solve it to find x.

Model Answer — 5(a)

M1 sum every region once: 11 + 13 + 9 + 4 + 6 + 9 + 3x + x = 52 + 4x

M1 equate to the total: 52 + 4x = 80

A1 4x = 28, so x = 7

⚠ If you missed marks here: the region everyone drops is the 9 outside all three circles — students who play nothing still count in n(ξ) = 80. Leaving them out gives 43 + 4x = 80 and x = 9.25 — and a non-integer x in a counting Venn is always the paper telling you a region went missing. Count all eight regions, exactly once each.

M1 — sums all eight regions including the 9 outside: 52 + 4x M1 — forms the equation 52 + 4x = 80 A1 — x = 7

(b) [1]

Find n(G).

Model Answer — 5(b)

B1 all four regions inside the G circle: 13 + 21 + 6 + 7 = 47 (using 3x = 21 and x = 7)

⚠ If you missed marks here: n(G) is the whole circle — guitarists who also play other instruments are still guitarists. The frequent slips: writing 13 (the “G only” region), or substituting x = 7 where the diagram says 3x — the overlap with piano holds 21 students, not 7.

B1 — n(G) = 13 + 21 + 6 + 7 = 47

Find n((P ∪ G)′).

Model Answer — 5(c)

(P ∪ G)′ is everyone outside both the P and G circles: the drums-only region and the students who play nothing

B1 9 + 9 = 18

⚠ If you missed marks here: two traps: forgetting the 9 who play nothing (they are firmly inside the complement), and reading (P ∪ G)′ as (P ∩ G)′ — that would be 80 − 28 = 52. Union first, then complement: shade P and G, take everything else.

B1 — n((P ∪ G)′) = 9 + 9 = 18

(d) [2]

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

Model Answer — 5(d)

M1 the choice is made among drummers only: n(D) = 9 + 4 + 6 + 7 = 26

A1 P = 7/26 (7 is prime and does not divide 26, so nothing cancels)

⚠ If you missed marks here: the words “who plays drums” shrink the pool before the choice is made — the denominator is n(D) = 26, not 80. Writing 7/80 ignores the condition. This conditional phrasing hides inside a Venn question on almost every recent paper.

M1 — denominator n(D) = 26, not 80 A1 — P = 7/26

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 trains from Dunmore to Fairhaven, stopping at Eastcliff. All times are 24-hour clock; times after midnight are on the next day.

(a) [1]

How long does Train 2 take for the whole journey from Dunmore to Fairhaven, including the stop? Give your answer in hours and minutes.

Model Answer — 6(a)

count through midnight: 23:35 → 00:00 is 25 min, then 00:00 → 02:20 is 2 h 20 min

B1 2 h 45 min

⚠ If you missed marks here: never subtract clock times like decimals — 2.20 − 23.35 is nonsense because an hour has 60 minutes, not 100. Bridge through midnight in two easy hops (to 00:00, then onwards), and remember the arrival is on the next day.

B1 — 2 h 45 min

(b) [2]

The rail distance from Dunmore to Eastcliff is 108 km.
Calculate the average speed of Train 1 between Dunmore and Eastcliff, in km/h.

Model Answer — 6(b)

M1 22:20 → 23:50 is 1 h 30 min = 1½ h = 3/2 h, so speed = 108 ÷ 3/2 = 108 × 2/3

108 × 2 = 216, and 216 ÷ 3 = 72

A1 72 km/h

⚠ If you missed marks here: the killer is writing 1 h 30 min as 1.30 h — thirty minutes is half an hour, so the time is 1.5 h. Turning the time into a fraction (3/2) and dividing by flipping it (×2/3) is the clean non-calculator route.

M1 — time 1 h 30 min written as 1.5 h (or 3/2 h) and 108 divided by it A1 — 72 km/h

The rail distance from Eastcliff to Fairhaven is 63 km.
Calculate Train 1's average driving speed for the whole journey from Dunmore to Fairhaven, not counting the 15-minute stop at Eastcliff.

Model Answer — 6(c)

M1 total distance 108 + 63 = 171 km; driving time 1 h 30 min + 1 h 00 min = 2½ h

A1 171 ÷ 2.5 = 171 × 4 ÷ 10 = 68.4 km/h

⚠ If you missed marks here: average speed is always total distance ÷ total time. Averaging the two leg speeds — (72 + 63) ÷ 2 = 67.5 — is wrong because the train spends longer at the faster speed. Dividing by 2 h 45 min would wrongly include the stop (171 ÷ 2.75 ≈ 62.2). By hand, ÷2.5 is ×4 then ÷10.

M1 — total distance 171 km over driving time 2.5 h (stop excluded) A1 — 68.4 km/h

(d) [3]

A plane leaves Tokyo at 11:50 on Saturday, local time. The flight to Los Angeles takes 12 hours 35 minutes. Tokyo time is 16 hours ahead of Los Angeles time.
Find the local time and day in Los Angeles when the plane lands.

Model Answer — 6(d)

M1 fly first, in Tokyo time: 11:50 + 12 h 35 min = 00:25 Sunday (Tokyo time)

M1 Tokyo is ahead, so Los Angeles shows an earlier time: subtract 16 hours

A1 08:25 on Saturday

⚠ If you missed marks here: yes — the plane lands on the same Saturday it took off, by the calendar: that is what crossing 16 hours' worth of zones westward does, and examiners love it. Adding the 16 hours gives 16:25 Sunday, out by more than a day. And keep the day pinned at every step: 00:25 Sun − 16 h walks back through midnight into Saturday.

M1 — 11:50 + 12 h 35 min = 00:25 Sunday, Tokyo time M1 — subtracts 16 hours (Tokyo ahead ⇒ LA earlier) A1 — 08:25 on Saturday

Question 7 — Compound Interest over Many Years

Total: 7 marks · Section B — calculator allowed

(a) [3]

Rae invests $9200 in an account paying 2.9% per year compound interest. No money is added or removed.
Calculate the value of the investment at the end of 8 years. Give your answer correct to the nearest cent.

Model Answer — 7(a)

M1 the multiplier is 1.029 and it acts once per year, so the 8 goes in the power: value = 9200 × 1.029⁸

M1 1.029⁸ = 1.2570 (4 dp)

A1 $11 564.07 (calculator shows 11564.073…)

⚠ If you missed marks here: the two classic wrong values are $11 334.40 (simple interest: 9200 × (1 + 0.029 × 8) — the same $266.80 every year, but compound interest grows on the new balance each year) and $11 238.17 (power 7 — after 8 complete years the multiplier has acted 8 times). Keep all the digits until the final rounding.

M1 — sets up 9200 × 1.029⁸ (the 8 in the power, not multiplied) M1 — evaluates 1.029⁸ = 1.2570 (4 dp or better) A1 — $11 564.07

(b) [2]

Calculate the interest earned during the 8th year only.

Model Answer — 7(b)

M1 value at the end of year 7: 9200 × 1.029⁷ = $11 238.17; the 8th year's interest is 11564.07 − 11238.17, or equivalently 0.029 × 11238.17

A1 $325.91 (accept $325.90 if you subtracted the two already-rounded values)

⚠ If you missed marks here: if you got $266.80, you found 2.9% of the original $9200 — but by year 8 the balance has grown to $11 238.17, and the 2.9% acts on that. A single later year of compound interest always earns more than year 1.

M1 — year-7 value $11 238.17 and subtracts (or takes 2.9% of it) A1 — $325.91 (accept $325.90)

Find the number of complete years needed for the investment to first exceed $15 000.

Model Answer — 7(c)

M1 needs the smallest n with 9200 × 1.029ⁿ > 15000: trials give year 17 → $14 957.16 (not there yet) and year 18 → $15 390.92 ✓

A1 18 years

⚠ If you missed marks here: the classic wrong answer is 17 — logs (or trial) give n ≈ 17.1 and rounding down leaves you $42.84 short of the target. “First exceeds” always rounds up: check year 17 ($14 957.16 < $15 000) and year 18 ($15 390.92) and state both.

M1 — tests 9200 × 1.029ⁿ against 15000, showing year 17 fails A1 — 18 years

Question 8 — Three-Set Venn Diagram with Percentages

Total: 6 marks · Section B — calculator allowed

Information for the whole question

240 members of a gym were asked which classes they attend: spin (S), yoga (Y) and boxing (B). The Venn diagram shows the percentage of the members in each region. x% attend all three classes, and the percentage who attend none of the three is 3x%.

(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: 16 + 14 + 10 + 8 + 7 + 5 + x + 3x = 60 + 4x

M1 60 + 4x = 100

A1 4x = 40, so x = 10

⚠ If you missed marks here: the region everyone drops is the 3x% outside all three circles — members who attend nothing still count in the 100%. Leaving it out gives 60 + x = 100 and the wrong x = 40 (and 40% attending all three classes should have felt absurd). Equating to 240 instead of 100 mixes members with percentages.

M1 — sums all eight regions including the 3x outside: 60 + 4x M1 — forms the equation 60 + 4x = 100 A1 — x = 10

(b) [1]

Calculate the number of members who attend exactly two of the three classes.

Model Answer — 8(b)

exactly two = the three pairwise overlaps excluding the centre: (8 + 7 + 5)% = 20%

B1 20% of 240 = 48 members

⚠ If you missed marks here: two traps: answering 20 (that is a percentage — the question asks for a number of members, so multiply by 240), and including the centre to get 30% → 72 — the x% in the middle attend exactly three classes, not two.

B1 — (8 + 7 + 5)% = 20% of 240 = 48 members

One member who attends yoga is chosen at random.
Find the probability that this member attends all three classes. Give your answer as a fraction in its simplest form.

Model Answer — 8(c)

M1 the pool is yoga only: n(Y) = 14 + 8 + 5 + 10 = 37 (%)

A1 P = 10/37 — every region is a percentage of the same 240, so the ratio of percentages is the probability (24/88.8… is why you work in %, not people, here)

⚠ If you missed marks here: 10/100 = 1/10 ignores the condition “who attends yoga” — the words shrink the pool to the Y circle before the choice is made, so the denominator is 37, not 100. Percentages of the same whole cancel, so no conversion to people is needed.

M1 — denominator is the whole Y circle: 14 + 8 + 5 + 10 = 37 A1 — P = 10/37

Question 9 — Currency Chain with Fees

Total: 6 marks · Section B — calculator allowed

(a) [3]

Sofia is travelling from New York to Madrid. She changes $1200 into euros. The bureau first deducts 2% commission from the dollars, then converts what is left at $1 = €0.92.
Calculate how many euros Sofia receives.

Model Answer — 9(a)

M1 commission: 2% of 1200 = $24, leaving 1200 × 0.98 = $1176

M1 convert: 1176 × 0.92

A1 €1081.92

Direction check: a dollar is worth less than a euro here, so she must receive fewer euros than she had dollars ✓

⚠ If you missed marks here: the classic wrong values: €1278.26 comes from dividing by 0.92 — but this rate is euros per dollar, so dollars → euros multiplies; and €1126.08 comes from adding the commission (1200 × 1.02 = 1224 before converting) — commission is money the bureau keeps.

M1 — deducts 2% commission: 1200 → $1176 M1 — multiplies by 0.92 to convert into euros A1 — €1081.92

(b) [3]

In Madrid Sofia spends €830, leaving her with €251.92. She flies on to London and converts her remaining euros into pounds. She can choose between two bureaus:
Bureau A: a flat fee of €3.50 is taken first, then the rest is converted at €1 = £0.85.
Bureau B: no fee, converted at €1 = £0.835.
Which bureau gives Sofia more pounds, and how many more? Show your working for both.

Model Answer — 9(b)

M1 Bureau A: (251.92 − 3.50) × 0.85 = 248.42 × 0.85 = £211.16

M1 Bureau B: 251.92 × 0.835 = £210.35

A1 Bureau A, by £0.80 (accept £0.81 from rounded intermediate values)

⚠ If you missed marks here: here “no fee” loses — the mirror image of the usual result, which is why you must compute both to the end instead of guessing from the headline. B's worse rate costs about £3.78 across €251.92, more than A's fee (€3.50 ≈ £2.98). The other slip is taking the €3.50 off after converting (£214.13 − 3.50 = £210.63), which subtracts euros from pounds — a units error.

M1 — Bureau A: fee first, (251.92 − 3.50) × 0.85 = £211.16 M1 — Bureau B: 251.92 × 0.835 = £210.35 A1 — Bureau A, by £0.80

Question 10 — Time Zones and Average Speed

Total: 6 marks · Section B — calculator allowed

(a) [3]

A plane leaves Auckland at 23:30 on Monday, local time. The flight to Santiago takes 11 hours 25 minutes. Auckland time is 16 hours ahead of Santiago time.
Find the local time and day in Santiago when the plane lands.

Model Answer — 10(a)

M1 fly first, in Auckland time: 23:30 + 11 h 25 min = 10:55 Tuesday (Auckland time)

M1 Auckland is ahead, so Santiago shows an earlier time: subtract 16 hours

A1 18:55 on Monday

⚠ If you missed marks here: the plane lands on Monday evening — before the calendar day it left has finished. That is what flying east across 16 hours' worth of zones does, and it is the exact discriminator this question exists for. Adding the 16 hours gives 02:55 Wednesday, out by more than a day; and losing the day at midnight (23:30 + 11:25 crosses it) wrecks everything downstream.

M1 — 23:30 + 11 h 25 min = 10:55 Tuesday, Auckland time M1 — subtracts 16 hours (Auckland ahead ⇒ Santiago earlier) A1 — 18:55 on Monday

(b) [2]

The flight distance is 9670 km.
Calculate the average speed of the plane in km/h, correct to 3 significant figures.

Model Answer — 10(b)

M1 11 h 25 min = 11 + 25/60 = 11.4166… h (on the calculator: 9670 ÷ (11 + 25 ÷ 60))

A1 9670 ÷ 11.4166… = 847.007… = 847 km/h (3 sf)

⚠ If you missed marks here: 860 km/h comes from typing 11 h 25 min as 11.25 — twenty-five minutes is 25/60 = 0.4167 of an hour, so the time is 11.42 h, not 11.25. Never write minutes after a decimal point; use the bracket (11 + 25 ÷ 60) or the degrees-minutes button.

M1 — converts 11 h 25 min to 11.417 h (or 137/12 h) A1 — 847 km/h (3 sf)

The return flight leaves Santiago at 13:10 on Wednesday and lands in Auckland at 17:45 on Thursday, local time.
Find the flight time.

Model Answer — 10(c)

convert one end into the other zone first: 13:10 Wednesday in Santiago = 05:10 Thursday in Auckland

B1 05:10 → 17:45 = 12 h 35 min

⚠ If you missed marks here: 28 h 35 min comes from subtracting the two clock readings as if they were in the same zone — they are not. Convert the departure into Auckland time (add the 16 hours Auckland is ahead), then subtract. A flight nearly 29 hours long on a route flown in 11½ hours the other way should have triggered your alarm.

B1 — 12 h 35 min (after converting one end across the 16-hour zone gap)

Self-Assessment

Tick marks earned, then click Calculate Grade.

Your Score

Out Of

Percentage