IGCSE Mathematics Paper 2 (Extended) — non-calculator
Unit Assessment Mock 4 -- Sets, Percentages, Time, Money -- 40 marks in 45 minutes
45 minutes
40
4
45:00
0580
Instructions
Answer all questions in the spaces provided.
Show all working for calculations.
Give non-exact answers to 3 significant figures unless the question says otherwise.
Your answers will be automatically graded when you submit.
Question Navigation
Question 1 — Sets
Total: 10 marks
Information for parts (a)(i) to (a)(iii)
120 people were asked which of three newspapers they read, so n(ξ) = 120. The three sets are A, B and C.
n(A)
n(B)
n(C)
n(A ∩ B)
n(A ∩ C)
n(B ∩ C)
n(A ∩ B ∩ C)
54
48
40
22
17
15
8
The Venn diagram below has only the middle region filled in. The other seven regions are deliberately blank. Fill them in before you answer — start in the middle and work outwards, because every figure in the table above except the last one has been counted more than once.
(a)(i)[2]
Find n(A ∩ B ∩ C′).
Model Answer — (a)(i)
M1 n(A ∩ B) = 22 counts the middle region as well, so take it off: 22 − 8
A1 n(A ∩ B ∩ C′) = 14
Check: 14 + 8 = 22 = n(A ∩ B) ✓
⚠ If you missed marks here: writing 22 copies a figure straight out of the table, and writing 8 copies the only number on the diagram. Neither is the region asked for. In a three-set diagram the pairwise total n(A ∩ B) always includes the middle, so the "only these two" region is that total minus the middle. Fill the diagram in from the centre outwards and you can read it off.
⚠ If you missed marks here: answering 120 is the most common error on this whole topic — 120 is n(ξ), which is everybody, not the people outside every circle. If instead you got a negative number or something silly, you almost certainly added 54 + 48 + 40 and stopped: that counts each pairwise overlap twice and the middle three times. Add the three singles, subtract the three pairs, add the middle back.
(a)(iii)[1]
Find n(A′).
Model Answer — (a)(iii)
B1 120 − 54 = 66
The long way agrees: 19 + 16 + 7 (the parts of B and C outside A) + 24 outside everything = 66 ✓
⚠ If you missed marks here: answering 23 gives the part of A on its own, which is the opposite of what a dash means, and answering 42 (19 + 16 + 7) forgets the 24 people outside all three circles. A′ is everything not in A, the outside region included. One subtraction from n(ξ) does it in a single line.
Information for parts (b)(i) to (b)(iii)
At a sports centre, 24 members play judo (J) and 18 members play fencing (F). Every member plays at least one of the two, and x members play both.
(b)(i)[2]
Write down expressions, in terms of x, for the number of members who play judo only and for the number who play fencing only.
Model Answer — (b)(i)
B1 judo only = 24 − x
B1 fencing only = 18 − x
Answer: 24 − x and 18 − x
⚠ If you missed marks here: writing 24 for judo only is the classic slip: 24 is the whole judo circle, and it already contains the x members who also fence. The overlap always has to come off the total before you have an "only" region. Writing x − 24 is the sign error to watch — you are taking the small part off the big one, not the other way round.
(b)(ii)[2]
The number of members who play judo only is three times the number who play fencing only.
Form an equation in x and solve it.
Model Answer — (b)(ii)
M1 24 − x = 3(18 − x), so 24 − x = 54 − 3x and 2x = 30
A1x = 15
Check: judo only = 9, fencing only = 3, and 9 = 3 × 3 ✓
⚠ If you missed marks here: the bracket is where this is won or lost. 3(18 − x) is 54 − 3x, not 54 − x, and forgetting to multiply the second term is the single most common algebra slip in the whole paper. Putting the 3 on the wrong side — 3(24 − x) = 18 − x — gives a negative x, which cannot be a number of people: that is your signal to swap it round.
(b)(iii)[1]
Find n(ξ).
Model Answer — (b)(iii)
B1 9 + 15 + 3 = 27 members
Or 24 + 18 − 15 = 27, taking the overlap off once ✓
⚠ If you missed marks here: 42 comes from adding 24 and 18 and leaving the overlap counted twice. The sentence "every member plays at least one" is what tells you the outside region is 0, so the three inner regions are the whole universal set — if you had ignored that sentence you would have had no way to answer this at all.
Question 2 — Percentages
Total: 10 marks
(a)[3]
The price of a phone, including sales tax at 20%, is $540.
Calculate the amount of sales tax paid.
Model Answer — (a)
M1 the $540 is 120% of the pre-tax price, so pre-tax price = 540 ÷ 1.2
M1 540 ÷ 1.2 = 5400 ÷ 12 = $450
A1 tax = 540 − 450 = $90
Check forwards: 20% of 450 = 90, and 450 + 90 = 540 ✓
⚠ If you missed marks here: $108 is the trap, and it is a reasonable-looking one — it is 20% of $540. But the tax was worked out on the price before tax, not on the price you were given, so 540 is 120% and not 100%. The word including is the signal to divide. The other loss is stopping at $450, which is the pre-tax price, not the tax.
(b)[3]
Two shops sell the same suitcase.
Shop P raises its price of $80 by 25%. Shop Q reduces its price of $125 by 20%.
Determine which shop now sells the suitcase more cheaply. You must show your working.
Model Answer — (b)
M1 Shop P: 80 × 1.25 = 80 + 20
M1 Shop Q: 125 × 0.8 = 125 − 25
A1 both come to $100, so neither shop is cheaper — the prices are the same
⚠ If you missed marks here: comparing 25% with 20% and naming Shop Q decides the question without doing it — those percentages are percentages of different starting prices, so they cannot be compared directly. Work both prices out in full. And do not talk yourself out of a tie: "they are the same" is a perfectly good answer, and here it is the right one.
Information for parts (c)(i) and (c)(ii)
A printing machine is bought for $20 000. Its value falls by 20% of its value at the start of each year.
(c)(i)[3]
Calculate the value of the machine at the end of 3 years.
Model Answer — (c)(i)
M1 the multiplier for a 20% fall is 0.8, applied once for each year
M1 20 000 → 16 000 → 12 800 (or 20 000 × 0.8³)
A1 $10240
⚠ If you missed marks here: $8000 comes from taking 60% off in one go — that is simple, not compound, and it is wrong because each year’s 20% is 20% of a smaller amount than the year before. On a non-calculator paper do not reach for the power key in your head: halve and halve again. 10% of 20 000 is 2000, so 20% is 4000; then repeat on 16 000, then on 12 800.
(c)(ii)[1]
Calculate the total percentage fall in the value of the machine over the 3 years.
Model Answer — (c)(ii)
B1 fall = 20 000 − 10 240 = 9760, and 9760 ÷ 20 000 = 0.488, so 48.8%
Or straight from the multiplier: 0.8³ = 0.512, and 1 − 0.512 = 0.488 ✓
⚠ If you missed marks here: 60% is the answer you get by adding three 20% falls together, and it is always too big — compound percentages never simply add. Also watch which number you divide by: the percentage change is measured against the original $20 000, not against the $10 240 you finished with.
Question 3 — Time
Total: 10 marks
Information for parts (a)(i) and (a)(ii)
A cricket match starts at 10:48. Play is stopped for rain at 13:06.
(a)(i)[2]
Find the playing time before the stoppage, in hours and minutes.
Model Answer — (a)(i)
M1 10:48 → 11:00 is 12 minutes, 11:00 → 13:00 is 2 hours, 13:00 → 13:06 is 6 minutes
A1 12 + 6 = 18 minutes, so 2 hours 18 minutes
Check: 10:48 + 2 h = 12:48, and 12:48 + 18 min = 13:06 ✓
⚠ If you missed marks here: 3 hours 18 minutes, or 2 hours 42 minutes, both come from lining the times up in columns and subtracting: 13 − 10, then 06 − 48. Minutes do not borrow in tens, they borrow in sixties, so column subtraction quietly gives the wrong answer here every time. Count up through the o’clock instead; it never fails.
(a)(ii)[2]
Play restarts after a delay of 1 hour 35 minutes, and the match then lasts a further 2 hours 47 minutes.
Find the time at which the match ends.
Model Answer — (a)(ii)
M1 restart time = 13:06 + 1 h 35 min = 14:41
A1 14:41 + 2 h = 16:41, and 41 + 47 = 88 minutes = 1 h 28 min, so the match ends at 17:28
Check in minutes: 1 h 35 min + 2 h 47 min = 262 minutes = 4 h 22 min, and 13:06 + 4 h 22 min = 17:28 ✓
⚠ If you missed marks here: 16:88 is not a time — once the minutes pass 60 you must carry an hour. The safest method when two durations are added is to convert both to minutes, add, and only convert back at the very end: 95 + 167 = 262 minutes, and 262 = 4 × 60 + 22.
Information for parts (b)(i) and (b)(ii)
The timetable shows four trains from Westhill to Redbridge on one morning. A dash means that the train does not stop at that station.
T1
T2
T3
T4
Westhill (depart)
06:52
08:14
09:06
10:38
Marlow
07:19
—
09:33
11:05
Denton
07:44
08:51
09:58
11:30
Redbridge (arrive)
08:26
09:28
10:40
12:12
(b)(i)[2]
Omar joins his train at Marlow and must arrive at Redbridge before 10:45. He wants to travel as late as possible.
Write down the time at which he leaves Marlow.
Model Answer — (b)(i)
M1 T4 reaches Redbridge at 12:12, which is after 10:45, and T2 does not stop at Marlow at all, so the latest train he can use is T3
A1 he leaves Marlow at 09:33
⚠ If you missed marks here: 09:06 is the Westhill time in the same column — right train, wrong row. 10:40 is the Redbridge time in the same column — right train, wrong row again. And if you tried to use T2 you missed the dash, which means that train runs straight past Marlow. Find the column first, then run your finger along the row the question names, and say the row name out loud before you write the time down.
(b)(ii)[2]
Hana says that T4 takes longer than T1 for the whole journey from Westhill to Redbridge.
Show that Hana is wrong.
Model Answer — (b)(ii)
M1 T1: 06:52 → 07:00 is 8 min, 07:00 → 08:00 is 1 h, 08:00 → 08:26 is 26 min
A1 T1 takes 1 h 34 min and T4 takes 10:38 → 12:12 = 1 hour 34 minutes as well, so the two journeys take exactly the same time and Hana is wrong
⚠ If you missed marks here: both of these times are column-subtraction traps — 26 − 52 and 12 − 38 both need a borrow of 60, not of 10. If you got 2 h 34 for one of them and 1 h 34 for the other you would have "proved" Hana right on a false arithmetic. When a question says show that, the two workings have to be written out in full; an answer with no method earns nothing here.
(c)[2]
A runner completes 12 km in 1 hour 20 minutes.
Calculate her average speed in kilometres per hour.
⚠ If you missed marks here: dividing by 1.20 gives 10 km/h and is wrong, because 1 h 20 min is 1.333... hours, not 1.20 hours. The unit in the answer tells you the unit you need in the sum: km per hour means the time must be in hours before you divide. Twenty minutes is a third of an hour, and a third is 0.333..., never 0.20.
Question 4 — Money and finance
Total: 10 marks
Information for parts (a)(i) and (a)(ii)
A taxi company charges $3.20 for the first kilometre of a journey, and then $1.40 for each further kilometre.
⚠ If you missed marks here: $18.60 charges 11 kilometres at $1.40 and adds the $3.20 on top, which double-charges the first kilometre. The first kilometre is already paid for by the $3.20, so only 10 kilometres are left to price. Write the journey out as "1 km + 10 km" before you multiply anything.
(a)(ii)[2]
A different journey costs $31.20.
Calculate the length of this journey, in kilometres.
Model Answer — (a)(ii)
M1 take the first kilometre off first: 31.20 − 3.20 = $28.00, then 28 ÷ 1.40 = 280 ÷ 14 = 20 further kilometres
⚠ If you missed marks here: 20 km is the near miss that costs the A mark — the division gives the number of further kilometres, and the first one still has to be added back. Undo the charging in reverse order: the $3.20 went on first, so it comes off first, and the +1 goes back on last.
(b)[4]
A shopkeeper pays $48 for a watch and puts a marked price on it.
In a sale he takes 25% off the marked price, and on that sale price he still makes a profit of 20% of what he paid.
Calculate the marked price of the watch.
Model Answer — (b)
M1 a 20% profit on $48 means the sale price is 48 × 1.2
M1 sale price = 48 + 9.60 = $57.60
M1 the sale price is 75% of the marked price, so marked price = 57.60 ÷ 0.75 = 57.60 × 4/3
A1 marked price = $76.80
Check forwards: 25% of 76.80 is 19.20, so the sale price is 57.60, and 57.60 − 48 = 9.60, which is 20% of 48 ✓
⚠ If you missed marks here: $72 comes from multiplying by 1.25 to undo the 25% discount. Undoing a 25% cut is a division by 0.75, never a multiplication by 1.25 — the two are not inverses, and this is the same trap as a price that goes down 20% and back up 20%. Work the chain forwards on paper first (cost → marked → sale), then undo it one step at a time from the end.
(c)[2]
The same book costs £18 in London and €22 in Paris. The exchange rate is £1 = €1.25.
Nadia says the book is cheaper in Paris. Determine whether she is correct, and find the difference in price. Give your answer in pounds.
Model Answer — (c)
M1 to compare, put both prices in the same currency: €22 ÷ 1.25 = 22 × 0.8 = £17.60
A1 Nadia is correct, and the book is cheaper in Paris by 18 − 17.60 = £0.40
⚠ If you missed marks here: €22 looks bigger than £18, but the two numbers are in different currencies and cannot be compared until one of them is converted. Multiplying by 1.25 instead of dividing gives £27.50 and turns the answer upside down — the rate reads "one pound buys 1.25 euros", so going from euros back to pounds must make the number smaller. Check the direction of every conversion against that sentence before you press on.
Self-Assessment
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0
40
0%
A* : 70%+
A : 60-69%
B : 50-59%
C : 40-49%
D : 30-39%
E : 20-29%
U : <20%
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