2026 O-Level E-Maths Paper 2 Answers (4052/02) – Worked Solutions + Free PDF
Suggested answers and full worked solutions for all 9 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.
Also correct: y = (4x² − 7)/(8 − 4x²) is the same fraction, with top and bottom multiplied by −1.
4
A tank is a prism with a rectangular base. Its cross-section ABCDEF is rectangle ABEF on top of trapezium BCDE. AB = 1.5 m, CD = 3 m, GH = 5 m (so AF = BE = 5 m) and FG = 6 m (the length of the tank). The tank's vertical height is 4 m.
(a)(i)
Work out the volume of the tank.
Rectangle ABEF: 5 × 1.5 = 7.5 m²
Trapezium BCDE: height = 4 − 1.5 = 2.5 m, area = 1/2(5 + 3)(2.5) = 10 m²
Cross-section = 7.5 + 10 = 17.5 m²
Volume = 17.5 × 6 = 105 m³
Answer105m³[3]
(a)(ii)
A pipe fills the empty tank at 5 m³ every 1 minute 30 seconds. Find the time to fill the tank, in minutes.
1 minute 30 seconds = 1.5 minutes
Time = 105/5 × 1.5 = 21 × 1.5 = 31.5 minutes
Answer31.5minutes[3]
(b)(i)
The circumference of circle P is 10% greater than that of circle Q, and the circumference of Q is 20% smaller than that of circle R. The area of P is d% smaller than the area of R. Find d.
Let the circumference of R be x.
Q = 0.8x, P = 1.1 × 0.8x = 0.88x
Ratio of lengths P : R = 0.88 : 1
Ratio of areas P : R = 0.88² : 1² = 0.7744 : 1
d = (1 − 0.7744) × 100 = 22.56
Answer d =22.56[3]
Watch out: Areas scale with the square of lengths: square 0.88. Don't just take 10% and 20% off.
(b)(ii)(a)
Solid cylinders CP and CR have bases P and R and the same height. Find volume of CP : volume of CR, in the form n : 1.
The cylinders have the same height, so volume ratio = base area ratio
= 0.7744 : 1
Answer0.7744: 1[1]
(b)(ii)(b)
Find curved surface area of CP : curved surface area of CR, in the form n : 1.
Curved surface area = 2πrh, and h is the same for both
so the ratio = rP : rR = 0.88 : 1
Answer0.88: 1[1]
Watch out: Curved surface area depends on r, not r², so the ratio is 0.88 : 1, not 0.7744 : 1.
5
Part (b): quadrilateral BCDE has a point A inside it, with AD ∥ BC and AC ∥ ED. ∠ABC = ∠AED and AB = AE.
(a)
Make an accurate construction of quadrilateral PQRS, where PQ = 10 cm, ∠QPS = 62°, PS = 6.4 cm, ∠PQR = 80° and QR = 5.2 cm.
Draw PQ = 10 cm.
At P, measure 62° with a protractor and mark S on that line, 6.4 cm from P.
At Q, measure 80° and mark R on that line, 5.2 cm from Q.
Join S to R.
Answersee construction[2]
Check: Measure your finished drawing: RS should be about 6.1 cm.
(b)(i)
Show that triangle ABC is congruent to triangle AED, giving a reason for each statement.
AB = AE (given) (S)
∠ABC = ∠AED (given) (A)
∠ACB = ∠CAD (alternate angles, AD ∥ BC)
∠CAD = ∠ADE (alternate angles, AC ∥ ED)
so ∠ACB = ∠ADE (A)
By AAS, triangle ABC is congruent to triangle AED.
Answer△ABC ≡ △AED (AAS)[3]
Also correct: SAA and AAS are the same congruence test.
(b)(ii)
∠ABC = x° and ∠DAE = y°. Show that ∠ACD = ((x + y)/2)°, giving a reason for each statement.
∠BAC = ∠DAE = y° (corresponding angles of congruent triangles)
Explain what the sum of the elements of T represents.
368.40 + 693.55 = $1061.95 is the total profit Robert made
from all the cakes he sold over the two days.
Answertotal profit over the two days[1]
(c)(i)
A spinner lands on blue with probability 1/4, otherwise red. Xu spins it 5 times. Find the probability that it lands on blue all 5 times.
(1/4)⁵ = 1/1024
Answer1/1024[1]
(c)(ii)
Hence find the probability that it lands on red at least once and on blue at least once.
P(red) = 1 − 1/4 = 3/4
P(at least one red and at least one blue) = 1 − P(all blue) − P(all red)
= 1 − 1/1024 − 243/1024 = 780/1024 = 195/256
Answer195/256[1]
Watch out: Take away both 'all blue' and 'all red'. Taking away only 'all blue' gives P(at least one red) instead.
7
Triangle ABC has AB = (2x + 1) cm, AC = (8x − 9) cm and ∠BAC = 30°. Its area is 3 cm².
(a)
Show that 16x² − 10x − 21 = 0.
Area = 1/2 × AB × AC × sin 30°
3 = 1/2(2x + 1)(8x − 9) × 1/2
12 = (2x + 1)(8x − 9) = 16x² − 10x − 9
16x² − 10x − 21 = 0
Answer16x² − 10x − 21 = 0[2]
(b)
Solve 16x² − 10x − 21 = 0 by factorisation.
(2x − 3)(8x + 7) = 0
Check: 16x² + 14x − 24x − 21 = 16x² − 10x − 21 ✓
2x − 3 = 0 gives x = 3/2; 8x + 7 = 0 gives x = −7/8
Answer x =3/2or x =−7/8[3]
(c)
State, with a reason, which solution is not valid here.
x = −7/8because
it gives AB = 2(−7/8) + 1 = −0.75 cm, and a length cannot be negative.
[1]
(d)
Hence find the obtuse angle ACB.
x = 1.5, so AB = 2(1.5) + 1 = 4 cm and AC = 8(1.5) − 9 = 3 cm
BC² = 4² + 3² − 2(4)(3) cos 30° = 4.2154…
BC = 2.0531… cm
BC/(sin 30°) = AB/(sin ∠ACB)
sin ∠ACB = (4 sin 30°)/2.0531… = 0.9741…
The calculator gives 76.9°, but ∠ACB is obtuse:
∠ACB = 180° − 76.9° = 103.1°
Answer ∠ACB =103.1°[5]
Watch out: sin⁻¹ on a calculator always gives the acute angle. ∠ACB is obtuse, so subtract from 180°.
Check: Check with the cosine rule: cos ∠ACB = (3² + 2.0531² − 4²)/(2(3)(2.0531)) = −0.226, which gives 103.1° directly.
8
Part (b): in triangle OAB, OA = 3a + b and OB = a + 2b. AB is extended to C so that AB : BC = 5 : 2. M is the midpoint of OA, and N lies on OB with ON : OB = 7 : 9.
(a)(i)
P has position vector 4−3 and Q has position vector 72. Find PQ.
PQ = OQ − OP = 72 − 4−3 = 35
Answer PQ =35[1]
(a)(ii)
R lies on PQ with PR = 3/7PQ. Find |PR|.
|PQ| = √(3² + 5²) = √34
|PR| = 3/7 × √34 = 2.4989…
Answer2.50[2]
(b)(i)
Express AB in terms of a and b, in its simplest form.
AB = OB − OA = (a + 2b) − (3a + b)
Answer AB =−2a + b[1]
(b)(ii)
Show that MNC is a straight line.
OM = 1/2OA = 1.5a + 0.5b
ON = 7/9OB = 7/9a + 14/9b
AC = 7/5AB = −2.8a + 1.4b
MN = ON − OM = −13/18a + 19/18b = 1/18(−13a + 19b)
MC = MO + OA + AC = −1.5a − 0.5b + 3a + b − 2.8a + 1.4b = 1/10(−13a + 19b)
So MC = 18/10MN: MN and MC are parallel and share the point M,
so M, N and C lie on a straight line.
AnswerMNC is a straight line[3]
(b)(iii)
Hence find the ratio MN : NC.
MN : MC = 1/18 : 1/10 = 10 : 18 = 5 : 9
NC = MC − MN, so MN : NC = 5 : (9 − 5) = 5 : 4
Answer5:4[1]
Watch out: MN : MC is 5 : 9, but the question asks for MN : NC, which is 5 : 4.
9
A city is planning for its 2036 population. A line graph shows the population, in thousands, from 2001 to 2026. New housing: each unit costs $160 000 to build and houses 2 people; the office pays 1/5 of the cost. Public transport: the cost is proportional to the percentage population increase from 2026 to 2036, and a 10% increase costs $8 million; the office pays all of it. Essential services: the city has 2 water treatment plants (each serves 7500 people, $500 000 per extra plant) and 4 electricity substations (each serves 2700 people, $300 000 per extra substation); the office pays all of it.
(a)
How much will the city planning office pay for each new housing unit?
1/5 × $160 000 = $32 000
Answer $32 000[1]
(b)
Use the line graph to estimate the city's population in 2036.
From 2021 to 2026 the graph rises from 10.75 to 10.80 thousand.
Extending this line to 2036 gives 10.9 thousand.
Answer10 900[2]
Watch out: The axis is in thousands: 10.9 means 10 900 people.
(c)
The total cost is paid with a loan at 1.5% compound interest per year for 4 years. The office claims the interest will not exceed $1 million. Is the claim correct? Show your calculations.
Population: 10 800 in 2026 and about 10 900 in 2036, an increase of 100 people
