SPS EM 2026 4G3 P1 Solutions

St. Patrick’s School – Preliminary Examination 2026 Mathematics 4052 Paper 1 – Worked Solutions Qn Answer Mark 1 3 r 15.62 − 9.4 0.028 = 3√ −92.35 = −4.52 [1] 2(a) 2p4 × 6p−5 = 12p−1 = 12 p [1] 2(b) 4(3x − 2) − 5 = 12x − 8 − 5 = 12x − 13 [1] 3(a) 0.00047 = 4.7 × 10−4 [1] 3(b) 1 km = 1 ×103 m 0.4 nm = 4 ×10−10 m Percentage = 1 × 103 4 × 10−10 × 100% = 2.5 × 1014 % [1] 4 F = k/d 2. Halving d multiplies the force by 4. New force = 4 × 0.8 = 3.2 N Increase = 3.2 − 0.8 = 2.4 N [2] 5 2x = y + 10 and 4x + 3y = 220 4x + 3(2x − 10) = 220 10x − 30 = 220 x = 25 y = 2(25) − 10 = 40 [3] 6 Smallest cube side = LCM(24, 16, 10). Side = 240 mm Number of blocks = 2403 24 × 16 × 10 = 3600 [3] 1

Qn Answer Mark 7 Make the number of boys equal (LCM of 5 and 7 = 35): Class A → 35 : 14, Class B → 35 : 20 Total girls = 14 + 20 = 34 Total children = 70 + 34 = 104 Fraction = 34 104 = 17 52 [3] 8 x = 7 ± p (−7)2 − 4(3)(1) 2(3) = 7 ± √ 37 6 x = 2.18 or x = 0.15 [3] 9 In △AMB and △CMD: AM = CM (M is the midpoint of AC) BM = DM (M is the midpoint of BD) ∠AMB = ∠CMD (vertically opposite angles) ∴ △AMB ∼= △CMD (SAS) ∴ ∠BAM = ∠DCM (corresponding angles of congruent triangles) These are equal alternate angles, so AB ∥ CD. [3] 10(a)(i) Midpoints: 35, 45, 55, 65 Mean = 6(35) + 10(45) + 16(55) + 8(65) 40 = 2060 40 = 51.5 min [1] 10(a)(ii) s.d. = r 109800 40 − 51.52 = √ 92.75 = 9.63 min [1] 10(b) Adding 5 to every value increases the mean by 5, to 56.5 min. The standard deviation is unchanged (9.63 min): adding a constant does not change the spread. [2] 11(a) P(even score) = 12/16 = 3/4, P(odd score) = 4/16 = 1/4. 3/4 ̸= 1/4, so Aisha is more likely to win — the game is unfair. [1] 11(b) P(Aisha) = 3/4, P(Ben) = 1/4. 3/4 = 3 × 1/4, so Aisha is three times as likely to win, not twice. ∴ Aisha is incorrect. [2] 2

Qn Answer Mark 12 1 2 (9.8)(15) sin∠P QR= 67.67 sin ∠P QR= 0.9207 ∠P QR= 67.0◦ or 113.0◦ [3] 13 Let the sphere radius be r. Then cylinder radius = r and height = 4r. Cylinder = πr2(4r) = 4πr3 Two spheres = 2 × 4 3 πr3 = 8 3 πr3 Fraction empty = 4πr3 − 8 3 πr3 4πr3 = 1 3 [3] 14(a) ∠ABC = 90 ◦ (angle in semicircle) ∠DBC = 180 ◦ − 90◦ = 90◦ (angles on straight line ABD) ∠DCB = 180◦ − 90◦ − 60◦ = 30◦ (angle sum of △BDC) [2] 14(b) If DC were a tangent at C, then ∠DCB would equal ∠BAC = 28◦ (tangent–chord, alternate segment). But ∠DCB = 30 ◦ ̸= 28◦, so DC is not a tangent. [1] 15(a) Figure 4: shaded = 16, unshaded = 20. [1] 15(b) Unshaded in Figure n = (n+2) 2 − n2 = 4n + 4. [1] 15(c) 4n + 4 = 96 n = 23 Shaded = 232 = 529 [2] 16 Let OD = r, so OA = 2r. Perimeter = 2r(1.2) + r(1.2) + r + r = 5.6r 5.6r = 44.8 r = 8 Area = 1 2 (1.2)(162) − 1 2 (1.2)(82) = 153.6 − 38.4 = 115.2 cm2 [4] 3

Qn Answer Mark 17(a) Write each side as a power of 3: 3k × 36 = 3−4k k + 6 = −4k 5k = −6 k = − 6 5 [2] 17(b) 4725 = 3 3 × 52 × 7 LCM(A, B) = pˆ(r+1) × q2 × 7 q = 5 p = 3 r + 1 = 3 r = 2 [2] 18 (i) C (depth falls at a constant rate — straight line decreasing) (ii) D (area = πr2, increasing and concave up) (iii) B (distance increases at a decreasing rate, then levels off) [3] 19(a) Interior angles: A = 110 ◦, B = 30 ◦+70◦ = 100◦, E = 40 ◦+80◦ = 120◦, D = 30 ◦+80◦ = 110◦. 110 + 100 + 120 + 110 +x = 540 x = 100 [2] 19(b) ∠ABE = ∠ADE = 30 ◦, so A, B, D, E lie on a circle (equal angles subtended by AE, angles in the same segment). In △AED: ∠DAE = 180 ◦ − 120◦ − 30◦ = 30◦, so ∠DAB = 110 ◦ − 30◦ = 80◦. For ABCD: ∠DAB + ∠BCD = 80◦ + 100◦ = 180◦, so ABCD is a cyclic quadrilat- eral. Hence all five points A, B, C, D, E lie on the same circle. [3] 20(a) −15 < 3x − 3 ≤ 15 −12 < 3x ≤ 18 −4 < x≤ 6 [2] 20(b) Open circle at −4, filled circle at 6, joined by a line. [1] 20(c) Smallest y2 − x2: take smallest y 2 (y = 2) and largest x 2 (x = 6). y2 − x2 = 22 − 62 = −32 [1] 21(a) From A, measure a bearing of 062 ◦ and mark F at 6.4 cm (12.8 m). [1] 21(b) P is due south of F with AP = 8 cm (16 m). Mark P and draw PA, PF. [1] 4

Qn Answer Mark 21(c) PF runs due north–south, so the shortest distance from A to PF is the easterly distance: d = 12.8 sin 62◦ = 11.30 m Greatest angle = tan−1 5.3 11.30 = 25.1◦ [2] 22(a) 1 2 (56)(v) = 672 v = 24 m/s [1] 22(b) Deceleration = 24 56 − 24 = 0.75 m/s2 [1] 22(c) 0 to 24 s: curve from (0, 0), increasing gradient (concave up), reaching 288 m. 24 to 56 s: curve continues up with decreasing gradient (concave down), reaching 672 m at 56 s. [2] 23(a) B = (0 − 3, 7 + 5) = (−3, 12) [1] 23(b) |p| = p (−3)2 + 52 = √ 34 = 5.83 [2] 23(c) Gradient = 12 − 7 −3 − 0 = − 5 3 y = − 5 3 x + 7 [2] 24(a) A ∪ B = {1,2,3,4,6,9,12} so (A ∪ B)’ = {5,7,8,10,11} n(A ∪ B)’ = 5. [1] 24(b) A’ ∩ B = {1, 2, 4 }. [1] 24(c) True. 9 ∈ A (multiple of 3) and 9 /∈ B (not a factor of 12), so 9 ∈ B’; hence 9 ∈ A ∩ B’. [1] 24(d) B = {1,2,3,4,6,12}, C = {7,11}, so B ∩ C = ∅. The events cannot occur together, so X and Y are mutually exclusive. [1] 25(a) Tuesday: 40(t + 6) litres Wednesday: 60(t − 4) litres [1] 5

Qn Answer Mark 25(b) 40(t + 6) = 60(t − 4) 40t + 240 = 60t − 240 20t = 480 t = 24 min [3] 25(c) Capacity = 40(24 + 6) = 1200 litres [1] 26(a) P = 4.50 6.00 ! [1] 26(b) QP = 120 80 150 60 ! 4.50 6.00 ! = 1020 1035 ! [1] 26(c) Each element is the total takings from meal-set sales that day: $ 1020 on Monday and $ 1035 on Tuesday. [1] 26(d)(i) R = 0.3 0 0 0 .4 ! [1] 26(d)(ii) RP = 1.35 2.40 ! Q(RP) = 120 80 150 60 ! 1.35 2.40 ! = 354 346.50 ! Profit: Monday $ 354, Tuesday $ 346.50. [2] 6