2026 O-Level E-Maths Paper 1 Answers (4052/01) – Worked Solutions + Free PDF
Suggested answers and full worked solutions for all 28 questions from Acer Academy's Maths team. Answers are to 3 s.f. and angles to 1 d.p., as the E-Math paper instructs.
Working is in red; the final answer sits on the answer line, with the marks in brackets.
1
(a)
Simplify 6a − 5(2a − 3).
= 6a − 10a + 15
= −4a + 15
Answer−4a + 15[1]
(b)
Factorise cd − c + d − 1.
c(d − 1) + 1(d − 1)
= (d − 1)(c + 1)
Answer(d − 1)(c + 1)[2]
2
A = p × q² × r⁴ and B = p² × q × r³, where p, q and r are prime factors with p > q > r.
(a)
Express 120 as a product of prime factors, in index form.
2
120
2
60
2
30
3
15
5
5
1
Answer2³ × 3 × 5[1]
(b)(i)
Find the HCF of A and B in terms of p, q and r.
A = p × q² × r⁴
B = p² × q × r³
HCF = lowest power of each common prime = p × q × r³
Answerp × q × r³[1]
(ii)
Given that the HCF is 120, find A.
p × q × r³ = 120 = 2³ × 3 × 5
so r = 2
p > q, so p = 5 and q = 3
A = p × q² × r⁴ = 5 × 3² × 2⁴ = 720
Answer A =720[2]
(iii)
Find the LCM of A and B as a product of prime factors.
LCM = highest power of each prime = p² × q² × r⁴ = 5² × 3² × 2⁴
Answer5² × 3² × 2⁴[1]
3
A semicircle has centre O, with A, B and C on the arc. ∠AOB = 44° and ∠OAC = 16°.
(a)
Find angle BOC, giving a reason.
△OCA is isosceles (OC = OA are radii)
∠AOC = 180° − 16° − 16° = 148°
∠BOC = 148° − 44° = 104°
Angle BOC =104°because
△OCA is isosceles, so ∠AOC = 148°. Hence ∠BOC = 148° − 44° = 104°.
[2]
(b)
Find angle BAC, giving a reason.
△OBA is isosceles (OB = OA are radii)
∠OAB = (180° − 44°)/2 = 68°
∠BAC = 68° − 16° = 52°
Angle BAC =52°because
△OBA is isosceles, so ∠OAB = 68°. Hence ∠BAC = 68° − 16° = 52°.
[2]
Also correct: Another valid reason: the angle at the centre is twice the angle at the circumference, so ∠BAC = 1/2 × ∠BOC = 1/2 × 104° = 52°.
4
Show that (5x − 3y + 16x + 17y)/(5x − 3y + x + 7y) has the same value for all positive x and y.
Answer
= (21x + 14y)/(6x + 4y)
= 7(3x + 2y)/(2(3x + 2y))
= 7/2
The result is a constant, so the values of x and y do not affect the expression.
Answer7/2, a constant[2]
5
The list is 8, 8, 9, 10, 11, 12. Two more integers are added, after which the median is 9 and the range is 7.
Find the two integers that were added.
There are now 8 integers, so the median is the mean of the 4th and 5th values.
Median = 9 = (9 + 9)/2, so the other new integer is 9.
Range = 7 = 12 − x, so x = 5.
Answer5and9[2]
6
Omar cycles from village A to village B at 18 km/h, then from B to village C at 12 km/h. AB and BC are both x km.
(a)(i)
Write an expression in x for the total time Omar takes to cycle from A to C.
Total time = x/18 + x/12 = 5x/36
Answer5x/36hours[1]
(ii)
Hence find Omar's average speed from A to C.
Average speed = (total distance)/(total time)
= 2x × 36/(5x)
= 72/5 = 14.4
Answer14.4km/h[2]
(b)
Convert 12 km/h to metres per second.
12 km/h = (12 × 1000 m)/(60 × 60 s) = 10/3 m/s
Answer10/3m/s[1]
7
x° is obtuse and sin x° = 0.9867. Find x.
sin−1(0.9867) = 80.6°
x is obtuse, so x = 180° − 80.6° = 99.4°
Answer x =99.4[1]
Watch out: The calculator gives the acute angle, 80.6°. x is obtuse, so take 180° − 80.6°.
8
72 adults were asked whether they visited a theatre (T) or a cinema (C) last month. 12 visited neither. Theatre only was 4 fewer than the number who visited both; cinema only was 10 more than the number who visited both.
(a)
Find x, the number of adults who visited both.
n(T ∪ C) = 72 − 12 = 60
Theatre only = x − 4, cinema only = x + 10
60 = (x − 4) + x + (x + 10)
60 = 3x + 6
x = 54/3 = 18
Answer x =18[2]
(b)
Find n((T ∩ C) ∪ (T ∪ C)′).
n(T ∩ C) = x = 18, n((T ∪ C)′) = 12
18 + 12 = 30
Answer30[1]
(c)
One adult who visited a theatre is picked at random. Find the probability that they also visited a cinema.
Visited a theatre = x + (x − 4) = 32
Also visited a cinema = x = 18
P = 18/32 = 9/16
Answer9/16[1]
Watch out: Divide by the 32 adults who visited a theatre, not by all 72 adults.
9
(a)
Write 0.0000123 in standard form.
Answer1.23 × 10−5[1]
(b)
A = 1.5 × 10n and B = 4.8 × 10n+1. Find 5B/A in standard form.
5B/A = (24 × 10n+1)/(1.5 × 10n)
= 16 × 101
= 1.6 × 102
Answer1.6 × 102[2]
10
25 students' puzzle times, t minutes: 10 < t ≤ 20: 1 student; 20 < t ≤ 25: 2; 25 < t ≤ 30: 8; 30 < t ≤ 40: 7; 40 < t ≤ 70: 7.
(a)
Calculate an estimate of the mean time.
Mid-values: 15, 22.5, 27.5, 35, 55
Mean = (15×1 + 22.5×2 + 27.5×8 + 35×7 + 55×7)/25
= 910/25 = 36.4
Answer36.4minutes[1]
(b)
Lihui then takes 39 minutes and her time is added. Without calculating, will the estimated mean increase or decrease? Give a reason.
decreasebecause
When we add Lihui's 39 minutes, she fits into the 30 < t ≤ 40 class.
We take her time to be the mid-value, 35, for the calculation.
35 is below the estimated mean of 36.4, so the estimated mean decreases.
[1]
Watch out: The estimate uses each class's mid-value, not the actual time. Lihui counts as 35, not 39, so the mean goes down.
11
Brian : Amal = 3 : 4 and Brian : Priya = 4 : 7. Priya receives $150 more than Amal.
Find how much Brian receives.
B : A = 3 : 4 = 12 : 16
B : P = 4 : 7 = 12 : 21
P − A = 21 − 16 = 5 units
5 units → $150, 1 unit → $30
Brian = 12 units = $360
Answer $360[3]
12
Box-and-whisker plots of reaction times (milliseconds) for a group of 20-year-olds and a group of 50-year-olds.
Make two comparisons of the two groups' reaction times, using figures.
1On average, the 20-year-olds reacted faster, since their median (384 ms) is lower than the 50-year-olds' median (412 ms).
2The 20-year-olds are more consistent, since their IQR is smaller than the 50-year-olds' (72 < 84).
Answer20-year-olds are faster and more consistent[3]
13
Find all the integers x such that 21 < 5 − 4x ≤ 47.
21 < 5 − 4x → 4x < −16 → x < −4
5 − 4x ≤ 47 → −4x ≤ 42 → x ≥ −10.5
So −10.5 ≤ x < −4
Answer−10, −9, −8, −7, −6, −5[3]
Watch out: Dividing or multiplying by a negative number flips the inequality sign.
14
F is (−3, −4), G is (5, −14) and GH = −820.
(a)
Find the coordinates of H.
OH = OG + GH = 5−14 + −820 = 5 − 8−14 + 20 = −36
Answer (−3,6)[1]
Watch out: Add GH to the position of G, not to the origin: OH = OG + GH.
(b)
Find the equation of line GH.
Gradient = rise/run = 20/−8 = −5/2
y = −5/2x + c, through (5, −14):
−14 = −5/2(5) + c, so c = −1.5
Answery = −5/2x − 3/2[3]
(c)
Calculate the length of FG.
FG = √[(5 − (−3))² + (−14 − (−4))²]
= √(8² + 10²) = √164 = 12.8
Answer12.8[2]
15
2x + 6y = −13 and 3x − 8y = 6.
Solve the simultaneous equations algebraically.
Equation 1 × 3: 6x + 18y = −39
Equation 2 × 2: 6x − 16y = 12
Subtract: 34y = −51, y = −3/2
Substitute: 2x + 6(−3/2) = −13
2x = −4, x = −2
Answer x =−2y =−3/2[3]
16
Write 3c/(2c − 3)² + 4/(2c − 3) as a single fraction in its simplest form.
= 3c/(2c − 3)² + 4(2c − 3)/(2c − 3)²
= (3c + 8c − 12)/(2c − 3)²
= (11c − 12)/(2c − 3)²
Answer(11c − 12)/(2c − 3)²[2]
17
Rani spends £86 on her credit card in the UK. Her bank charges $141.56, which includes a 2.75% currency conversion fee.
Calculate the exchange rate the bank used, in pounds per Singapore dollar.
x × 102.75% = $141.56
x = $141.56/1.0275 = $137.77
Rate = £86/$137.77 = £0.624
Answer $1 = £0.624[3]
Watch out: $141.56 already includes the fee, so divide by 1.0275. Taking 2.75% off $141.56 gives the wrong amount.
18
Two circles have centre O. AB = 19 cm is a chord of the larger circle and a tangent to the smaller circle. The shaded region is the ring between the two circles.
Calculate the shaded area.
The tangent is perpendicular to the radius and bisects the chord:
half of AB = 19/2 = 9.5 cm
By Pythagoras, R² − r² = 9.5² = 90.25
Shaded area = πR² − πr² = π(R² − r²) = 90.25π
Answer90.25π (or 284)cm²[2]
Also correct: 90.25π cm² is exact; give 284 cm² (3 s.f.) as well to be safe.
19
(a)
A box holds blue, yellow and pink marbles. P(pink) = 0.37. Find the probability of picking a blue or yellow marble.
1 − 0.37 = 0.63
Answer0.63[1]
(b)
A bag holds 20 counters: 10 red, 6 white and 4 green. Three are picked at random. Find the probability that exactly two are red.
(R, R, other) or (other, R, R) or (R, other, R)
= 10/20 × 9/19 × 10/18 × 3
= 15/38
Answer15/38[2]
Watch out: The counters are not replaced, so the denominators go 20, 19, 18. Multiply by 3 for the three possible orders.
20
P is a port and lighthouse L is due east of P. Ship B is 18 km due north of P, on a bearing of 304° from L. Ship C is due west of B, on a bearing of 331° from P.
(a)
Write down the bearing of L from B.
180° − 56° = 124°
Answer124°[1]
Watch out: The bearing of L from B is the reverse of B from L: 304° − 180° = 124°.
Raise every part of the bracket to the power −4/3:
27: 27−4/3 = 1/(∛27)4 = 1/34 = 1/81
x: power 6 × (−4/3) = −8, giving x−8
y: power −3 × (−4/3) = 4, giving y4
z: power 9 × (−4/3) = −12, giving 1/z−12 = z12
= 1/81 x−8 y4 z12
Answer1/81 x−8 y4 z12[3]
24
Vertical posts AB = 21 m, CD = 15 m and EF stand on level ground, with B, F and D in a straight line. AED and BEC are straight lines.
(a)
Show that triangles ABE and DCE are similar.
∠AEB = ∠DEC (vertically opposite angles)
AB and CD are vertical posts, so AB ∥ CD
∠ABE = ∠DCE (alternate angles)
By AA, triangles ABE and DCE are similar.
Answer△ABE ~ △DCE (AA)[2]
Also correct: Any two correct pairs of equal angles, each with a reason, prove similarity, e.g. ∠BAE = ∠CDE (alternate angles, AB ∥ CD).
(b)
Calculate the height of post EF.
△EFD ~ △ABD: EF/21 = FD/BD
△EFB ~ △CDB: EF/15 = BF/BD
Adding: EF/21 + EF/15 = (FD + BF)/BD = 1
EF ( 1/21 + 1/15 ) = 1
EF = 35/4 = 8.75
Answer EF =8.75m[3]
25
OPQ is a major sector with centre O. Its angle is 216° and its major arc is 18π cm long.
(a)
Find the radius of the major sector.
216/360 × 2πr = 18π
r = 15
Answer15cm[2]
(b)
The sector is made into a cone by joining OP and OQ. Calculate the volume of the cone.
Base circumference = arc length: 2πr = 18π, r = 9
Slant height = 15
h = √(15² − 9²) = 12
V = 1/3 π (9)² (12) = 324π
Answer324π (or 1020)cm³[4]
Also correct: 324π cm³ is exact; give 1020 cm³ (3 s.f.) as well to be safe.
26
A set of 25 numbers has mean 30 and standard deviation 10.
Explain how each number must change so the mean is 75 and the s.d. is 20.
Step 1: multiply every number by 2 → mean = 2 × 30 = 60, s.d. = 2 × 10 = 20
Step 2: add 15 to every number → mean = 60 + 15 = 75, s.d. stays at 20
Answermultiply by 2, then add 15[1]
Also correct: add 7.5 to every number, then multiply by 2 (mean (30 + 7.5) × 2 = 75, s.d. 10 × 2 = 20).
27
The frequency, f Hz, of a note from a string is proportional to √T, where T newtons is the tension. The tension is increased so that the frequency increases by 100%.
Calculate the percentage increase in the tension.
f = k√T, so T = f²/k²
A 100% increase in frequency gives 2f:
2f = k√T′, so T′ = 4f²/k² = 4T
T becomes 4 times: 100% → 400%
Answer300%[3]
Watch out: A 100% increase means the frequency doubles. The tension becomes 4 times as large: an increase of 300%, not 400%.
28
The graph of y = x² + bx + c, where b and c are integers, has a minimum point at (3, −11).
Find the value of b and the value of c.
Minimum point (3, −11):
y = (x − 3)² − 11
= x² − 6x + 9 − 11
= x² − 6x − 2
Answer b =−6c =−2[3]
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