SPS EM 2026 4G3 P2 Solutions

St. Patrick’s School – Preliminary Examination 2026 Mathematics 4052 Paper 2 – Worked Solutions Qn Answer Mark 1(a) 5 x − 2 − 3 2x + 1 = 5(2x + 1) − 3(x − 2) (x − 2)(2x + 1) = 10x + 5 − 3x + 6 (x − 2)(2x + 1) = 7x + 11 (x − 2)(2x + 1) [2] 1(b) V r2 = h(R2 + r2) V r2 − hr2 = hR2 r2(V − h) = hR2 r2 = hR2 V − h r = R r h V − h [2] 1(c) 2x2 − xy − 6x + 3y 9 − x2 = x(2x − y) − 3(2x − y) (3 − x)(3 + x) = (2x − y)(x − 3) −(x − 3)(x + 3) = y − 2x x + 3 [3] 1(d) (2x3y−2)4 ÷ (8x5y−3) = 16x12y−8 8x5y−3 = 2x7y−5 = 2x7 y5 [2] 2(a)(i) Deposit = 0.20 × 2400 = 480 Instalments = 24 × 92 = 2208 Total HP price = 480 + 2208 = $2688 [2] 2(a)(ii) Difference = 2688 − 2400 = 288 Percentage = 288 2400 × 100% = 12% [2] 1

Qn Answer Mark 2(b) 85% of price = 1734 Original price = 1734 0.85 = $2040 [2] 2(c)(i) Value = 6666 × 1.082 = 6666 × 1.1664 = $7775.2224 = $7775.22 (nearest cent) [2] 2(c)(ii) Continue year by year from the year 3 value: End of year 3: 6666 × 1.083 = 8397.24 ( < 8888) End of year 4: 6666 × 1.084 = 9069.02 ( > 8888) The value first exceeds $ 8888 after 4 complete years. Profit = 9069.02 − 6666 = $2403.02 [3] 3(a) AC2 = 112 + 82 − 2(11)(8) cos 58◦ = 185 − 176 cos 58◦ = 91.73 AC = 9.58 cm [3] 3(b) sin ∠ADC 9.578 = sin 42◦ 7 sin ∠ADC = 9.578 sin 42◦ 7 = 0.9155 ∠ADC = 113.7◦ (obtuse) ∠CAD = 180◦ − 42◦ − 113.7◦ = 24.3◦ [3] 3(c) CD sin ∠CAD = 7 sin 42◦ CD = 7 sin 24.3◦ sin 42◦ = 4.30 cm [2] 3(d) Area ABC = 1 2 (11)(8) sin 58◦ = 37.31 Area ACD = 1 2 (9.578)(7) sin 24.3◦ = 13.79 Area ABCD = 37.31 + 13.79 = 51.1 cm2 [2] 4(a)(i) Median ≈ 4.1 kg (read at cumulative frequency = 60) [1] 2

Qn Answer Mark 4(a)(ii) Q1 = 3.5 kg (CF = 30), Q 3 = 5.0 kg (CF = 90) IQR = Q3 − Q1 = 5.0 − 3.5 = 1.5 kg [2] 4(b) Below 3 kg: CF at 3.0 kg ≈ 13 Above 6.5 kg: 120 − CF(6.5) = 120 − 116 = 4 Discounted = 13 + 4 = 17 Percentage = 17 120 × 100% = 14.2% [3] 4(c) Five-number summary for shipment 2: minimum 3, lower quartile 4, median 5, upper quartile 6.5, maximum 7.5 (range 4.5). Box drawn from 4 to 6.5 with median line at 5; whiskers to 3 and 7.5. [3] 4(d) Shipment 1: within 3–6.5 kg ≈ 103/120 ≈ 86%, so ≈14% discounted. Shipment 2: Q 3 = 6.5 kg, so 75% lie within 3–6.5 kg; 25% (above 6.5 kg) discounted. Shipment 1 is more desirable: a larger proportion ( ≈86% vs 75%) falls within the acceptable 3–6.5 kg range, so fewer melons are discounted. [2] 5(a) d = 0.1(3)3 − 1.2(3)2 + 3.3(3) + 4 = 5.8 [1] 5(b) Plot the nine points from the table. Join them with a single smooth curve. [3] 5(c)(i) Draw a tangent to the curve at t = 5. Gradient ≈ −1.2 (accept −1.0 to −1.4) [2] 5(c)(ii) Minimum depth ≈ 2.2 m (at t ≈ 6.2) [1] 5(d) Depth is decreasing for 1.8 < t < 6.2 (approx). [1] 5(e) Draw the line d = 0.2t + 2 on the grid. It meets the curve on the falling part at t = 5. t = 5 is 5 hours after 6 am, so the ship must leave by 11 am. [2] 6(a)(i) 1 2 1 3 πr2h

= 48π 1 2 1 3 π(6)2h

= 48π 6πh = 48π h = 8 cm [2] 3

Qn Answer Mark 6(a)(ii) l = p 62 + 82 = 10 cm Half curved surface = 1 2 πrl = 1 2 π(6)(10) = 30π Half base = 1 2 πr2 = 1 2 π(6)2 = 18π Triangular face = 1 2 (12)(8) = 48 Total surface area = 30π + 18π + 48 = 199 cm2 [4] 6(b) Detergent = 3.20 × 3 2 3 = 3.20 × 27 8 = $10.80 Material = 0.80 × 3 2 2 = 0.80 × 9 4 = $1.80 Total = 10.80 + 1.80 + 0.15 = $12.75 [3] 7(a) ∠BCA = 32 ◦ (AC bisects ∠BCD) ∠BOA = 2 × 32◦ = 64◦ (∠ at centre = 2 × ∠ at circumference) [2] 7(b) O and M are the midpoints of AC and DC, so OM ∥ AD (midpoint theorem). ∠OMC = ∠ADC = 90 ◦ (corr. ∠s, OM ∥ AD; ∠ in semicircle) ∠OCM = ∠ACB = 32 ◦ (AC bisects ∠BCD) ∴ △COM is similar to △CAB (2 pairs of equal angles) [2] 7(c) In △ABC and △ADC: ∠ABC = ∠ADC = 90 ◦ (∠ in semicircle) ∠BCA = ∠DCA = 32 ◦ (AC bisects ∠BCD) AC is common. ∴ △ABC ∼= △ADC (AAS) ∴ BC = DC [2] 7(d) BC = 18.4 cos 32◦ = 15.60 cm The perpendicular from O bisects BC, so the half-chord = 7.802 cm. d = p 9.22 − 7.8022 = 4.88 cm (3 s.f.) [2] 4

Qn Answer Mark 8(a)(i) − − →AB = 2b − 3a [1] 8(a)(ii) − − →BP = − − →BA + − →AP = (3a − 2b) + (1.5a + b) = 4.5a − b [1] 8(b) − − →OQ = − →OA + − →AP + − − →P Q = 3a + (1.5a + b) + kb = 4.5a + (1 + k)b − − →BQ = − − →OQ − − − →OB = 4.5a + (k − 1)b BR = h a is parallel to a, and B, R, Q are collinear, so BQ is parallel to a. ∴ coefficient of b = 0: k − 1 = 0 k = 1 [2] 8(c) − − →OP = 3a + (1.5a + b) = 4.5a + b − − →OR = − − →OB + − − →BR = ha + 2b − →P R= − − →OR − − − →OP = (h − 4.5)a + b Since PR ∥ AB, PR = λ AB: (h − 4.5)a + b = λ(2b − 3a) Comparing coefficients of b: 1 = 2λ λ = 1 2 Comparing coefficients of a: h − 4.5 = −3λ = −1.5 h = 3 [3] 8(d) With k = 1: − − →RQ = − − →BQ − − − →BR = 4.5a − 3a = 1.5a BR : RQ = 3 : 1.5 = 2 : 1 R lies on BQ, two-thirds of the way from B to Q (BR is twice RQ). [2] 5

Qn Answer Mark 8(e) − →AR = − − →AB + − − →BR = (2b − 3a) + 3a = 2b Since AR = 2b = OB and BR = 3a = OA, OARB is a parallelogram. Diagonal AB divides it into two equal triangles, △AOB and △ABR. Area of △ABR = Area of △AOB = 60 units2 [2] 9(a) x = 1: tickets sold = 2300; x = 2: tickets sold = 2200. Row x: new price = 8 + x, tickets sold = 2400 − 100x, revenue R = (8 + x)(2400 − 100x). [1] 9(b) (8 + x)(2400 − 100x) = 23100 19200 + 1600x − 100x2 = 23100 100x2 − 1600x + 3900 = 0 x2 − 16x + 39 = 0 (x − 3)(x − 13) = 0 x = 3 or x = 13 x = 3: price = $ 11, tickets = 2100 ( ≥ 2000 ✓) x = 13: price = $ 21, tickets = 1100 ( < 2000 ×) ∴ ticket price = $ 11. [4] 9(c) R = (8 + x)(2400 − 100x) = 0 when x = −8 or x = 24. Line of symmetry: x = −8 + 24 2 = 8 Maximum revenue is at x = 8, so price = 8 + 8 = $ 16. Rmax = (16)(2400 − 800) = $25 600 [2] 9(d) Sponsor A ( $ 15 000 if ≥ 1900 tickets): 2400 − 100x ≥ 1900, so x ≤ 5. Best R = (13)(1900) = 24700 Total (A) = 24700 + 15000 = $39 700 Sponsor B (50c per $ 1 of sales = 1.5R): maximise R at x = 8. Total (B) = 1.5 × 25600 = $38 400 $ 39 700 > $ 38 400, so accept Sponsor A and set the price at $ 13. [4] 6