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IGCSE Mathematics Paper 2 (Extended) — non-calculator

Unit Assessment Mock 5 -- Sets, Percentages, Time, Money -- 40 marks in 45 minutes

Duration45 minutes

Total Marks40

Questions4

Time Remaining45:00

Syllabus0580

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 the whole question

A sports centre has 90 members.
54 of the members use the pool ( P ).
48 of the members use the gym ( G ).
12 of the members use neither the pool nor the gym.

The Venn diagram below is empty — not one of the four regions has been filled in. Work out the number in each region for yourself before you answer any part.

(a)(i) [2]

Find n( P ∩ G ).

Model Answer — (a)(i)

n( P ∪ G ) = 90 − 12 = 78 M1 (the members who use at least one)

n( P ∩ G ) = 54 + 48 − 78 M1 for the inclusion–exclusion step

so n( P ∩ G ) = 24 A1

⚠ If you missed marks here: the overlap is not given to you anywhere, and it is not 12. Adding 54 and 48 counts every member of the overlap twice, so 54 + 48 = 102 is 78 real people plus one extra count of the overlap. That is why the overlap is 102 − 78.

M1 — found n( P ∪ G ) = 78 from 90 − 12 A1 — overlap 24, by 54 + 48 − 78

(a)(ii) [2]

Find n( P ∩ G ′).

Model Answer — (a)(ii)

P ∩ G ′ is the part of P that lies outside G M1

54 − 24 = 30 A1

⚠ If you missed marks here: you almost certainly wrote 54. 54 is n( P ) — the whole of the P circle, overlap included. P ∩ G ′ is only the crescent, so the overlap has to come off. Never answer a region question with a number you were handed in the stem.

M1 — identified the region as n( P ) minus the overlap A1 — 30

(a)(iii) [2]

Find n( G ′).

Model Answer — (a)(iii)

G ′ is everything outside G: the pool-only members and the members who use neither M1 (30 + 12, or 90 − 48)

n( G ′) = 42 A1

⚠ If you missed marks here: the two answers people give here are 12 and 30, and both are one region short. G ′ means everything that is not in G — that is the pool-only crescent and the region outside both circles. The quick check is 90 − 48.

M1 — complement taken from n(ξ), or both outside regions added A1 — 42

Information for the whole question

60 students were asked which of three clubs they attend: Art ( A ), Music ( M ) and Drama ( D ).
The Venn diagram shows the number of students in each of the eight regions.

(b)(i) [2]

Write down, in set notation, the set of students who attend Drama but attend neither Art nor Music.

Model Answer — (b)(i)

The region is inside D and outside both of the others M1

D ∩ A ′ ∩ M ′ A1 (D ∩ ( A ∪ M )′ is equally correct)

⚠ If you missed marks here: writing 7 is answering a different question. This part asks for the name of the region in set notation, not how many people are in it. "Not Art and not Music" is A ′ ∩ M ′, which is the same set as ( A ∪ M )′.

M1 — D intersected with a correct complement A1 — D ∩ A ′ ∩ M ′ or D ∩ ( A ∪ M )′

(b)(ii) [2]

Find n( A ∪ ( M ∩ D )).

Model Answer — (b)(ii)

M ∩ D is the two regions 4 and 3, so n( M ∩ D ) = 7 M1

n( A ) = 11 + 6 + 5 + 3 = 25; the 3 is already inside A, so only the 4 is new

25 + 4 = 29 A1

⚠ If you missed marks here: the usual slip is 25 + 7 = 32, which counts the centre region twice. A union is not an addition sum — shade A, then shade M ∩ D on top of it, and count each region once. Brackets first, every time.

M1 — n( M ∩ D ) = 7, or the correct regions shaded A1 — 29

Question 2 — Percentages

Total: 10 marks

(a) [3]

A machine loses 10% of its value during each year that it is owned.
At the end of 2 years the machine is worth $32 400.
Calculate the value of the machine when it was new.

Model Answer — (a)

Each year multiplies the value by 0.9, so 2 years multiply it by 0.9 × 0.9 = 0.81 M1

new value = 32 400 ÷ 0.81 M1

32 400 ÷ 81 = 400, so the value is $40 000 A1

Check forwards: 40 000 × 0.9 = 36 000, and 36 000 × 0.9 = 32 400. ✓

⚠ If you missed marks here: if you got $39 204 you multiplied by 1.21 instead of dividing by 0.81, and if you got $38 880 you added 20% back on. This is a reverse question: 32 400 is 81% of what you want, so you divide. Always check a reverse answer by running it forwards again.

M1 — two-year multiplier 0.81 (or 0.9 used twice) M1 — divided 32 400 by that multiplier A1 — $40 000

(b)(i) [2]

The length of each side of a square photograph is increased by 20%.
Calculate the percentage increase in the area of the photograph.

Model Answer — (b)(i)

Each side is multiplied by 1.2, so the area is multiplied by 1.2 × 1.2 = 1.44 M1

1.44 is an increase of 0.44, so the area rises by 44% A1

Check with numbers: 10 × 10 = 100 becomes 12 × 12 = 144. ✓

⚠ If you missed marks here: the answer is not 20% and it is not 40%. Area is a length times a length, so both lengths grow — the multiplier has to be squared. If you are unsure, invent a square of side 10 and do it with real numbers; that check takes ten seconds.

M1 — area multiplier 1.2² = 1.44 (or a worked numerical example) A1 — 44%

(b)(ii) [2]

The area of a different square is increased by 69%.
Calculate the percentage increase in the length of each side.

Model Answer — (b)(ii)

area multiplier = 1.69, so the length multiplier is √1.69 M1

13² = 169, so √1.69 = 1.3 and each side rises by 30% A1

Check: 1.3 × 1.3 = 1.69. ✓

⚠ If you missed marks here: halving 69 to get 34.5% is the trap. Going from length to area you square the multiplier, so going from area to length you take the square root of it — and 1.69 has an exact root because 13² = 169. Non-calculator questions are built around numbers like that; look for the perfect square.

M1 — took the square root of the area multiplier 1.69 A1 — 30%

In a survey, 35% of the people chose tea and the rest chose coffee.
120 more people chose coffee than chose tea.
Calculate the number of people in the survey.

Model Answer — (c)

coffee = 100% − 35% = 65% B1

the difference is 65% − 35% = 30% of the survey, and that is 120 people M1

30% → 120, so 10% → 40 and 100% → 400 people A1

Check: 35% of 400 = 140 and 65% of 400 = 260, and 260 − 140 = 120. ✓

⚠ If you missed marks here: the common wrong answers are 343 (from 120 ÷ 0.35) and 185 (from 120 ÷ 0.65). The 120 is not the tea group and it is not the coffee group — it is the gap between them, which is 30% of the whole survey. Name the percentage your known number belongs to before you divide by anything.

B1 — coffee is 65% M1 — identified 120 as 30% of the total A1 — 400

Question 3 — Time

Total: 10 marks

(a)(i) [2]

An orchestra rehearsal begins at 16:38 and ends at 19:12 on the same day.
Find the length of the rehearsal, in hours and minutes.

Model Answer — (a)(i)

16:38 → 17:00 is 22 min, then 17:00 → 19:00 is 2 h, then 19:00 → 19:12 is 12 min M1

22 + 12 = 34, so the rehearsal lasts 2 hours 34 minutes A1

In minutes: 19:12 = 1152 and 16:38 = 998, and 1152 − 998 = 154 = 2 h 34 min. ✓

⚠ If you missed marks here: if you wrote 3 h 26 you subtracted the columns as if a clock ran in tens: 19 − 16 = 3 and then 38 − 12 = 26. A clock is base 60 and the minutes here go up, from 38 to 12, so there is a borrow. Never column-subtract a time. Count up through the o’clock, or turn both times into minutes since midnight.

M1 — counting up through the hour, or both times converted to minutes A1 — 2 hours 34 minutes

(a)(ii) [3]

The next rehearsal lasts 48 minutes longer than the one in part (a)(i).
It must end at 20:05.
Find the time at which the next rehearsal must begin.

Model Answer — (a)(ii)

new length = 2 h 34 min + 48 min = 2 h 82 min = 3 h 22 min M1

20:05 − 3 h = 17:05, then 17:05 − 22 min M1

start time = 16:43 A1

In minutes: 20:05 = 1205 and 154 + 48 = 202, and 1205 − 202 = 1003 = 16:43. ✓

⚠ If you missed marks here: two separate base-60 traps sit in this part. First, 34 + 48 = 82 minutes, which is 1 h 22 — not "3 h 82". Second, taking 22 minutes off 17:05 crosses the hour, so it is 16:43, not 17:83 and not 16:83. Carry and borrow in sixties, or work entirely in minutes since midnight and convert once at the end.

M1 — new length 3 h 22 min M1 — subtracted the length from 20:05 A1 — 16:43

Information for the whole question

The timetable shows four buses from Hartwell to Thornbury on one morning.
All times are on the same day and are given using the 24-hour clock.

	Bus 1	Bus 2	Bus 3	Bus 4
Hartwell (depart)	07:26	08:49	09:37	11:14
Oakmere	07:58	09:28	10:09	11:52
Thornbury (arrive)	08:21	10:02	10:44	12:28

(b)(i) [2]

Find the journey time from Hartwell to Thornbury of the bus that leaves Hartwell at 09:37.
Give your answer in hours and minutes.

Model Answer — (b)(i)

09:37 is Bus 3, which arrives at Thornbury at 10:44 B1 (correct column used)

09:37 → 10:00 is 23 min, 10:00 → 10:44 is 44 min, 23 + 44 = 67 min

1 hour 7 minutes A1

⚠ If you missed marks here: if you got 1 h 13 you used Bus 2’s arrival time of 10:02, and if you got 32 minutes you stopped at Oakmere. A timetable is read down a single column: find the column whose top cell matches the departure time in the question, then use only that column. Put a finger on it.

B1 — used Bus 3 (09:37 and 10:44) A1 — 1 hour 7 minutes (67 minutes)

(b)(ii) [3]

Find the difference between the longest and the shortest of the four journey times from Hartwell to Thornbury.
You must show the time taken by each bus.

Model Answer — (b)(ii)

Bus 1: 07:26 → 08:21 = 55 min · Bus 2: 08:49 → 10:02 = 73 min M1

Bus 3: 09:37 → 10:44 = 67 min · Bus 4: 11:14 → 12:28 = 74 min M1

longest 74 min (Bus 4), shortest 55 min (Bus 1), difference = 19 minutes A1

⚠ If you missed marks here: you cannot see which bus is slowest by looking — Bus 4 leaves last and is also the slowest, and Bus 2 looks similar to Bus 3 but takes six minutes longer. Every one of the four has to be worked out. Column-subtracting even one of them poisons the comparison: 08:49 → 10:02 is 73 minutes, not "2 h 43".

M1 — at least two journey times correct M1 — all four journey times correct A1 — 19 minutes

Question 4 — Money and finance

Total: 10 marks

Information for the whole question

A shop sells porridge oats in three packs.

    Pack A: 500 g for £2.00
    Pack B: 800 g for £2.50, with 25% extra free
    Pack C: 1.4 kg for £4.00, or a 4.5 kg sack for £12.00

(a) [3]

Work out which pack gives the best value for money.
You must show your working.

Model Answer — (a)

Pack A: 500 ÷ 2.00 = 250 g per £1 M1 (one correct rate)

Pack B: 25% extra free gives 800 × 1.25 = 1000 g, still for £2.50, so 1000 ÷ 2.50 = 400 g per £1 M1

Pack C: the sack is the better of its two prices, 4500 ÷ 12 = 375 g per £1 (the 1.4 kg pack is only 1400 ÷ 4 = 350 g per £1)

Pack B is the best value (400 g per £1) A1

⚠ If you missed marks here: if you skipped the words 25% extra free you would have scored Pack B at 800 ÷ 2.50 = 320 g per £1 and chosen Pack C — the wrong pack. Extra free means more oats for the same money: the £2.50 does not move, the 800 g does. Pack C also has two prices and you must compare them before you compare packs. Work in grams per £1 for all three and pick the biggest, or cost per 100 g and pick the smallest — but never mix the two directions in one answer.

M1 — a correct rate for one pack M1 — the extra-free handled (1000 g for £2.50) and the sack rate found A1 — Pack B, with all three rates shown

(b) [3]

Ishan invests £2500 at a rate of r % per year simple interest.
At the end of 4 years his investment is worth £2950.
Find the value of r .

Model Answer — (b)

total interest = 2950 − 2500 = £450 M1

simple interest is the same every year, so one year earns 450 ÷ 4 = £112.50 M1

112.50 as a percentage of 2500 = 112.5 ÷ 2500 = 0.045, so r = 4.5 A1

Check: 4.5% of 2500 = £112.50, and 4 × 112.50 = £450. ✓

⚠ If you missed marks here: two slips are common. The first is dividing £2950 by something — the £2500 is not interest, it is the money he started with, so it must come off first. The second is finding the yearly interest and then forgetting to turn it into a percentage of the original 2500. With simple interest every year earns the same amount, which is what makes dividing by 4 legal here; you could not do that with compound interest.

M1 — interest = £450 M1 — one year’s interest = £112.50 A1 — r = 4.5

Information for the whole question

The cash price of a television is £800.
Farida does not pay cash. Instead she pays a deposit of 15% of the cash price, followed by 12 equal monthly payments of £62.

(c)(i) [2]

Calculate the total amount Farida pays for the television.

Model Answer — (c)(i)

deposit = 15% of 800 = £120 M1

payments = 12 × 62 = £744, total = 120 + 744 = £864 A1

⚠ If you missed marks here: the deposit is part of what she pays, not a discount off it — answering £744 leaves it out. And 15% of £800 is £120, not £12: 10% is £80 and 5% is £40.

M1 — deposit £120 or 12 × 62 = £744 A1 — £864

(c)(ii) [2]

Calculate the extra amount Farida pays, as a percentage of the cash price.

Model Answer — (c)(ii)

extra = 864 − 800 = £64 M1

64 ÷ 800 = 0.08, so she pays 8% more A1

Check: 8% of 800 = £64, and 800 + 64 = £864. ✓

⚠ If you missed marks here: a percentage increase is always measured against the original, which here is the £800 cash price, not the £864 she ended up paying. Dividing by 864 gives 7.4% and is the standard error. Write the fraction as extra ÷ original before you touch the numbers.

M1 — extra = £64 A1 — 8%

Self-Assessment

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