1
Algebraic manipulation
3 marks
a
Simplify. 6a − 5(2a − 3)
[1]
Show working
6a − 5(2a − 3) = 6a − 10a + 15
−4a + 15
b
Factorise. cd − c + d − 1
[2]
Show working
Concept: Factorisation by grouping
cd − c + d − 1 = c(d − 1) + 1(d − 1)
(c + 1)(d − 1)
2
Prime factorisation, HCF & LCM
5 marks
a
Write 120 as a product of its prime factors. Give your answer in index form.
[1]
Show working
Two integers, A and B, are written as products of prime factors. A = p × q² × r⁴, B = p² × q × r³, where p > q > r.
b(i)
Find, in terms of p, q and r, the highest common factor (HCF) of A and B.
[1]
Show working
Concept: HCF — lowest power of each common factor
HCF = pmin(1,2) × qmin(2,1) × rmin(4,3)
p × q × r³
b(ii)
The highest common factor (HCF) of A and B is 120. Find the value of A.
[2]
Show working
p × q × r³ = 120 = 2³ × 3 × 5, and since p > q > r: p = 5, q = 3, r = 2
A = p × q² × r⁴ = 5 × 3² × 2⁴ = 5 × 9 × 16
A = 720
b(iii)
Find the lowest common multiple (LCM) of A and B. Give your answer as a product of its prime factors.
[1]
Show working
Concept: LCM — highest power of each factor
LCM = p² × q² × r⁴ = 5² × 3² × 2⁴
2⁴ × 3² × 5²
3
Circle theorems
4 marks
The diagram shows a semicircle, centre O. A, B and C are points on the semicircle. Angle AOB = 44° and angle OAC = 16°.
a
Find angle BOC. Give a reason for your answer.
[2]
Show working
Concept: Isosceles triangle (radii) + angle sum of a triangle
OA = OC (radii of the same circle), so triangle OAC is isosceles → angle OCA = angle OAC = 16°
Angle AOC = 180° − 16° − 16° = 148° (angle sum of triangle OAC, base angles of isosceles triangle)
Angle BOC = angle AOC − angle AOB = 148° − 44°
Angle BOC = 104°
b
Find angle BAC. Give a reason for your answer.
[2]
Show working
Concept: Angle at centre = 2 × angle at circumference
Angle BOC and angle BAC are both subtended by the same arc BC
Angle BAC = angle BOC ÷ 2 = 104° ÷ 2 (angle at centre is twice the angle at circumference, same arc BC)
Angle BAC = 52°
4
Algebraic fractions
2 marks
In the expression (5x − 3y + 16x + 17y) / (5x − 3y + x + 7y), x and y are positive numbers. Show that the value of the expression is the same for all values of x and y.
[2]
Show working
Concept: Simplifying algebraic fractions
Numerator = 5x − 3y + 16x + 17y = 21x + 14y = 7(3x + 2y)
Denominator = 5x − 3y + x + 7y = 6x + 4y = 2(3x + 2y)
Expression = 7(3x+2y) / 2(3x+2y) = 7/2, a constant independent of x and y (shown)
5
Median & range
2 marks
Here is a list of numbers: 8, 8, 9, 10, 11, 12. Two more integers are added to the list of numbers. The median is now 9 and the range is now 7. Find the two integers.
[2]
Show working
Concept: Median & range with an enlarged data set
The range increases from 12 − 8 = 4 to 7, so a new smallest value must be added: 12 − 7 = 5
With 8 numbers in order, the median is the mean of the 4th and 5th values; inserting 5 and 9 gives 5, 8, 8, 9, 9, 10, 11, 12 — median = (9+9)/2 = 9
The two integers are 5 and 9
6
Speed, distance & time
4 marks
Omar cycles from village A to village B at an average speed of 18 km/h. He then cycles from village B to village C at an average speed of 12 km/h. The distance from village A to village B is the same as the distance from village B to village C.
a(i)
Let x kilometres represent the distance between village A and village B. Write down an expression in terms of x for the total time taken for Omar to cycle from village A to village C.
[1]
Show working
Total time = x/18 + x/12 = (2x + 3x)/36
5x/36 hours
a(ii)
Hence, find Omar's average speed for the journey from village A to village C.
[2]
Show working
Concept: Average speed = total distance ÷ total time
Total distance = 2x km
Average speed = 2x ÷ (5x/36) = 2x × 36/5x
14.4 km/h
b
Convert 12 km/h to metres per second.
[1]
Show working
12 km/h = 12 ÷ 3.6 m/s
10/3 m/s = 3.33 m/s (3 s.f.)
7
Trigonometry — obtuse angles
1 mark
The angle x° is an obtuse angle and sin x° = 0.9867. Find the value of x.
[1]
Show working
x = 180° − sin⁻¹(0.9867)
x = 99.4 (1 d.p.)
8
Venn diagrams & probability
4 marks
72 adults were asked whether they visited a theatre (T) or a cinema (C) in the previous month.
• 12 did not visit a theatre or a cinema.
• The number who visited a theatre only was 4 less than the number who visited both.
• The number who visited a cinema only was 10 more than the number who visited both.
a
Let x represent the number of adults who visited both. Find the value of x. You may use the Venn diagram to help you.
[2]
Show working
Theatre only = x − 4, cinema only = x + 10
(x − 4) + x + (x + 10) + 12 = 72 → 3x + 18 = 72 → 3x = 54
x = 18
b
Find the value of n((T∩C)∪(T∪C)′).
[1]
Show working
Concept: Set notation — "both" and "neither"
n(T∩C) = x = 18 (both); n((T∪C)′) = 12 (neither); the two regions do not overlap, so add them
18 + 12 = 30
c
One of the adults that visited a theatre is picked at random. Find the probability that the adult also visited a cinema.
[1]
Show working
Total who visited a theatre = (x − 4) + x = 2(18) − 4 = 32
Visited both = 18
P = 18/32 = 9/16
9
Standard form
3 marks
a
Write 0.000 012 3 in standard form.
[1]
Show working
b
In standard form, A = 1.5 × 10n and B = 4.8 × 10n+1. Find 5B/A. Give your answer in standard form.
[2]
Show working
5B/A = 5(4.8 × 10n+1) / (1.5 × 10n) = 24 × 10n+1 / 1.5 × 10n = 16 × 10¹
1.6 × 10²
10
Mean from grouped data
2 marks
A group of 25 students each complete a puzzle. The table shows information about the times, in minutes, the students took to complete the puzzle.
| time (t minutes) | 10 < t ≤ 20 | 20 < t ≤ 25 | 25 < t ≤ 30 | 30 < t ≤ 40 | 40 < t ≤ 70 |
|---|
| frequency | 1 | 2 | 8 | 7 | 7 |
|---|
a
Calculate an estimate for the mean time.
[1]
Show working
Concept: Estimated mean — use class midpoints
Midpoints: 15, 22.5, 27.5, 35, 55
Σfx = 1(15)+2(22.5)+8(27.5)+7(35)+7(55) = 15+45+220+245+385 = 910
Mean = 910 ÷ 25
36.4 minutes
b
Lihui attempted the puzzle later than the group. She took 39 minutes to complete the puzzle. Her time is added to the table. Without doing any further calculations, decide whether the estimated mean will increase or decrease. Give a reason for your answer.
[1]
Show working
Decrease — 39 minutes falls in the 30 < t ≤ 40 class, so it is represented by the midpoint 35 in the estimate, which is lower than the current estimated mean of 36.4 minutes, pulling the mean down.
11
Ratio
3 marks
Brian, Amal and Priya share some money. The ratio Brian's share : Amal's share is 3 : 4. The ratio Brian's share : Priya's share is 4 : 7. Priya receives $150 more than Amal. Find how much Brian receives.
[3]
Show working
Concept: Combining ratios using a common term
Brian:Amal = 3:4 → Amal:Brian = 4:3; Brian:Priya = 4:7
Common term is Brian (LCM of 3 and 4 is 12): Amal:Brian = 16:12, Brian:Priya = 12:21 → Amal:Brian:Priya = 16:12:21
Priya − Amal = 21 − 16 = 5 parts = $150 → 1 part = $30
Brian receives 12 × $30 = $360
12
Box-and-whisker plots
3 marks
A group of 20-year-olds and a group of 50-year-olds had their reaction times, in milliseconds, tested. The box-and-whisker plots show the distributions of their times.
Make two comments comparing the reaction times of the two groups. Use figures to support your answers.
[3]
Show working
Concept: Comparing medians and interquartile ranges
20-year-olds: median = 384 ms, IQR = 418 − 340 = 78 ms
50-year-olds: median = 418 ms, IQR = 444 − 360 = 84 ms
1. The 20-year-olds generally have a faster reaction time, as their median (384 ms) is lower than the 50-year-olds' median (418 ms).
2. The 20-year-olds have more consistent reaction times, as their interquartile range (78 ms) is smaller than the 50-year-olds' (84 ms).
13
Inequalities
3 marks
Find all the integers that satisfy the inequalities 21 < 5 − 4x ≤ 47.
[3]
Show working
21 < 5 − 4x → 4x < −16 → x < −4
5 − 4x ≤ 47 → −4x ≤ 42 → x ≥ −10.5
Combined: −10.5 ≤ x < −4
x = −10, −9, −8, −7, −6, −5
14
Vectors & coordinate geometry
6 marks
F is the point (−3, −4), G is the point (5, −14) and vector GH = (−8, 20).
a
Find the coordinates of H.
[1]
Show working
OH = OG + GH = (5, −14) + (−8, 20)
H = (−3, 6)
b
Find the equation of the line GH.
[3]
Show working
Gradient of GH = 20 ÷ (−8) = −5/2
Using G(5, −14): −14 = −5/2(5) + c → −14 = −25/2 + c → c = −3/2
y = −5/2 x − 3/2 (i.e. 2y = −5x − 3)
c
Calculate the length of the line FG.
[2]
Show working
FG = √[(−3−5)² + (−4−(−14))²] = √[(−8)² + 10²] = √(64+100) = √164
12.8 (3 s.f.)
15
Simultaneous equations
3 marks
Use an algebraic method to solve these simultaneous equations. 2x + 6y = −13, 3x − 8y = 6. You must show your working.
[3]
Show working
×3 the first equation: 6x + 18y = −39 …(1)
×2 the second equation: 6x − 16y = 12 …(2)
(1) − (2): 34y = −51 → y = −3/2
Sub into 2x + 6y = −13: 2x − 9 = −13 → 2x = −4
x = −2, y = −3/2
16
Algebraic fractions
2 marks
Write as a single fraction in its simplest form: 3c/(2c−3)² + 4/(2c−3).
[2]
Show working
Common denominator (2c−3)²: [3c + 4(2c−3)] / (2c−3)² = (3c + 8c − 12) / (2c−3)²
(11c − 12) / (2c − 3)²
17
Currency conversion
3 marks
Rani is on holiday in the UK. She spends £86 on her credit card. The bank charges her account $141.56 which includes a 2.75% fee for the currency conversion. Calculate the exchange rate used by the bank. Give your answer in UK pounds per Singapore dollar.
[3]
Show working
Concept: Percentages & currency conversion
86 × (100+2.75)/100 = 88.365 (the $ amount equivalent to £86 before the conversion fee is added on top)
Exchange rate = 88.365 ÷ 141.56
$1 = £0.624 (3 s.f.)
18
Circles — chord & tangent
2 marks
The diagram shows two circles centred at O. AB is a chord of the larger circle and a tangent to the smaller circle. The length of AB is 19 cm.
Calculate the shaded area.
[2]
Show working
Concept: Tangent–radius right angle + Pythagoras
Let r = radius of small circle, R = radius of large circle. Since AB is tangent to the small circle at the midpoint M of AB, OM ⊥ AB, so AM = 9.5 cm
By Pythagoras: R² = AM² + r² = 9.5² + r²
Shaded area = πR² − πr² = π(9.5² + r²) − πr² = π(9.5²) = 90.25π
284 cm² (3 s.f.)
19
Probability
3 marks
a
A box contains blue marbles, yellow marbles and pink marbles. The probability of picking a pink marble at random is 0.37. Find the probability of picking a blue or yellow marble at random.
[1]
Show working
P(blue or yellow) = 1 − P(pink) = 1 − 0.37
0.63
b
A bag contains 20 counters. 10 of the counters are red, 6 are white and 4 are green. Three counters are picked at random from the bag. Find the probability that exactly two of the counters are red.
[2]
Show working
Concept: Probability without replacement
P(2 red, 1 not-red in some order) = (10/20 × 9/19 × 10/18) × 3, since there are 3 positions the non-red counter can occupy
15/38
20
Bearings
5 marks
The diagram shows the positions of a port, P, a lighthouse, L, and two ships, B and C. B is 18 km due north of the port, and B is on a bearing of 304° from the lighthouse. C is due west of B, and C is on a bearing of 331° from the port. The lighthouse is due east of the port.
a
Write down the bearing of L from B.
[1]
Show working
Bearing of B from L is 304°, so the bearing of L from B is the reverse bearing: 304° − 180°
124°
b
Calculate the distance PL.
[2]
Show working
Concept: Alternate angles with parallel north lines
Angle PBL = 56° (alternate angles, the north lines at B and L are parallel)
In right triangle PBL (right angle at P): tan56° = PL/18 → PL = 18 tan56°
PL = 26.7 km (3 s.f.)
c
Calculate the distance PC.
[2]
Show working
Concept: Right angle between a north line and an east-west line
Angle BPC = 29° (given); angle PBC = 90° (BP is a north-south line, BC is due east-west, so they are perpendicular)
In right triangle PBC: cos29° = BP/PC = 18/PC
PC = 18/cos29° = 20.6 km (3 s.f.)
21
Algebraic fractions — factorisation
3 marks
Simplify (3x−2)² − (x−3)² all over 4x² − 8x − 5.
[3]
Show working
Concept: Difference of two squares
Numerator = [(3x−2)+(x−3)][(3x−2)−(x−3)] = (4x−5)(2x+1)
Denominator = 4x²−8x−5 = (2x−5)(2x+1)
(4x − 5) / (2x − 5)
22
Algebraic equations
3 marks
Solve. (2x+5)/3 − (2x²−1)/(3x−1) = 1
[3]
Show working
Multiply through by 3(3x−1): (2x+5)(3x−1) − 3(2x²−1) = 3(3x−1)
Expand: (6x²+13x−5) − (6x²−3) = 9x−3 → 13x − 2 = 9x − 3
4x = −1
x = −1/4
23
Indices
3 marks
Write (27x⁶y⁻³/z⁹)−4/3 in the form axpyqzr where a, p, q and r are constants.
[3]
Show working
Invert for the negative power: = (z⁹/(27x⁶y⁻³))4/3 = (z⁹y³/(27x⁶))4/3
Raise each factor to the power 4/3: z9×4/3 = z¹², y3×4/3 = y⁴, 274/3 = 81, x6×4/3 = x⁸
(1/81) x⁻⁸ y⁴ z¹²
24
Similar triangles
5 marks
The diagram shows three vertical posts, AB, CD and EF, which are built in a straight line with points B, D and F on horizontal ground. The height of AB is 21 m and the height of CD is 15 m. AED and BEC are straight lines.
a
Show that triangle ABE and triangle DCE are similar. Give a reason for each statement you make.
[2]
Show working
Concept: AA similarity test
Angle AEB = angle DEC (vertically opposite angles)
AB is parallel to CD (both vertical posts) → angle BAE = angle CDE (alternate angles, AB ∥ CD)
By the AA similarity test, triangle ABE and triangle DCE are similar (shown)
b
Calculate the height of EF.
[3]
Show working
Concept: Ratio of corresponding sides of similar triangles
AB/DC = 21/15 = 7/5, so AE/DE = 7/5, giving DE/AD = 5/12
Angle ABD = angle EFD = 90° (both vertical lines); angle ADB = angle EDF (common angle) → triangle ABD is similar to triangle EFD (AA similarity test)
EF/AB = ED/AD = 5/12 → EF = 5/12 × 21
EF = 8.75 m
25
Sectors & cones
6 marks
OPQ is a major sector of a circle, centre O. The angle of the major sector is 216° and the length of the major arc is 18π cm.
a
Find the radius of the major sector.
[2]
Show working
Arc length = (216/360) × 2πr → 18π = 0.6 × 2πr → 18 = 1.2r
r = 15 cm
The major sector is formed into a cone by joining the two radii, OP and OQ, together.
Calculate the volume of the cone.
[4]
Show working
Concept: Arc length of sector = circumference of cone's base
Base circumference = major arc length: 2πrcone = 18π → rcone = 9 cm
Slant height l = radius of sector = 15 cm (radius of sector becomes slant height of cone)
Height h = √(15² − 9²) = √(225−81) = √144 = 12 cm
Volume = 1/3 × π × r² × h = 1/3 × π × 9² × 12 = 324π
1020 cm³ (3 s.f.)
26
Mean & standard deviation
1 mark
A set of 25 numbers has a mean of 30 and a standard deviation of 10. Explain how every number in the set must be changed so that the mean becomes 75 and the standard deviation becomes 20.
[1]
Show working
Concept: Linear transformation of mean & standard deviation
Doubling every number scales both the mean and standard deviation by 2: mean 30 → 60, s.d. 10 → 20
Multiply each number by 2, then add 15 (adding a constant shifts the mean from 60 to 75 without changing the standard deviation)
27
Proportion
3 marks
The frequency, f Hz, of a musical note produced by a string is proportional to the square root of the tension, T newtons, in the string. The tension in a string is increased so that there is a 100% increase in the frequency. Calculate the percentage increase in the tension.
[3]
Show working
Concept: Proportion with a square root
f = k√T. A 100% increase in frequency means the new frequency is 2f
2f = k√Tnew → dividing by f = k√T: 2 = √(Tnew/T) → Tnew = 4T
Percentage increase = (4T − T)/T × 100% = 300%
28
Completing the square
3 marks
The diagram shows the graph of y = x² + bx + c where b and c are integers.
The curve has a minimum point at (3, −11). Find the value of b and the value of c.
[3]
Show working
Concept: Completed-square form y = (x−h)²+k
Minimum point (3, −11) → y = (x−3)² − 11
Expand: y = x² − 6x + 9 − 11 = x² − 6x − 2
b = −6, c = −2
FAQ
Common questions about this paper
What topics are tested in the GCE O-Level E-Math Paper 1 (Oct/Nov 2026)?
This paper covers algebraic manipulation and factorisation, prime factorisation and HCF/LCM, circle theorems, ratio, speed-distance-time, trigonometry, Venn diagrams and probability, standard form, statistics (mean from grouped data, box-and-whisker plots), coordinate geometry and vectors, simultaneous equations, bearings, similar triangles, sectors and cones, and proportion — effectively a full sweep of the Sec 3/4 Elementary Mathematics (4052) syllabus.
How do you find an unknown angle at the centre of a circle using the angle properties of a circle?
Use the fact that the angle at the centre is twice the angle at the circumference subtended by the same arc, together with the fact that the angle sum of a triangle is 180°. In this paper's Q3, angle AOC is found using the isosceles triangle formed by two radii, angle BOC follows by subtraction, and angle BAC then follows from the centre/circumference relationship.
How do you use a Venn diagram to solve set notation problems like n((T∩C)∪(T∪C)′)?
Represent each given description (only theatre, only cinema, both, neither) as a region in the Venn diagram, form an equation from the total to find the unknown, then read off the required regions directly — n(T∩C) is the "both" region and n((T∪C)′) is "neither", and since the two regions don't overlap, the answer is simply their sum.
How do you prove two triangles are similar using the AA similarity test?
Identify two pairs of equal angles, each with a reason such as vertically opposite angles, alternate angles on parallel lines, or a shared/common angle, then state that the two triangles are similar by the AA (angle-angle) similarity test. Once proven similar, corresponding sides are in the same ratio, which lets you solve for an unknown length such as EF in Q24.
How do you find the volume of a cone formed by folding a sector of a circle?
The arc length of the sector becomes the circumference of the cone's circular base, which gives the base radius. The sector's own radius becomes the cone's slant length. Pythagoras' theorem then gives the cone's vertical height from the slant length and base radius, and volume = ⅓ × π × radius² × height.
