OPSS_Prelims_2024_4E_Math_P1_Marking_Scheme

Paper 1 Qn No. Solutions 1a 7.97× 1022 1b Sum must be ≤ 349 349 3 = 116.333 3 even numbers are : 114, 116, 118 2a 2𝑦 − 1 < 11𝑦 4 𝑎𝑛𝑑 11𝑦 4 < 1 4 8𝑦 − 4 < 11𝑦 𝑎𝑛𝑑 11𝑦 < 1 −3𝑦 < 4 𝑎𝑛𝑑 𝑦 < 1 11 𝑦 > −1 1 3 𝑎𝑛𝑑 𝑦 < 1 11 −1 1 3 < 𝑦 < 1 11 2b −25 + (−5)2 = 0 3a 756 = 22 × 33 × 7 3b 756 p = 22 × 33 × 7 × 𝑝 495 = 32 × 5 × 11 Smallest p = 5 × 11 = 55 Or LCM = 22 × 33 × 5 × 7 × 11 756𝑝 = 22 × 33 × 7 × 𝑝 𝑝 = 5 × 11 = 55 3c 756 × 𝑎 𝑏 = perfect cube 22 × 33 × 7 × 𝑎 𝑏 = perfect cube a = 2, b = 7 4a 𝑎2−3𝑎+2 9𝑎2−1 ÷ 2𝑎−2 6𝑎−2 = (𝑎−2)(𝑎−1) (3𝑎−1)(3𝑎+1) × 2(3𝑎−1) 2(𝑎−1) = 𝑎−2 3𝑎+1 5a (2𝑥3𝑦2)−4 (10𝑥−2𝑦3)2 ÷ √27𝑥−3𝑦63

= 2−4𝑥−12𝑦−8 100𝑥−4𝑦6 × 1 (27𝑥−3𝑦6) 1 3 = 1 1600𝑥8𝑦14 × 1 3𝑥−1𝑦2 = 1 4800𝑥7𝑦16 5b 3𝑥+2 × 3(35) = 1 3𝑥+2 × 31(35) = 1 3𝑥+2 × 36 = 30 𝑥 + 2 + 6 = 0 𝑥 = −8 6a 𝑎3 − 2𝑎2𝑏 − 4𝑎 + 8𝑏 = 𝑎2(𝑎 − 2𝑏) − 4(𝑎 − 2𝑏) = (𝑎2 − 4)(𝑎 − 2𝑏) = (𝑎 − 2)(𝑎 + 2)(𝑎 − 2𝑏) 6b 3𝑥+2 9𝑥 = 1 7𝑥−3 (3𝑥 + 2)(7𝑥 − 3) = 9𝑥 21𝑥2 − 9𝑥 + 14𝑥 − 6 = 9𝑥 21𝑥2 − 4𝑥 − 6 = 0 Using general formula 𝑥 = 0.638 𝑜𝑟 𝑥 = −0.448 7 Bank A: Total amount = = 20000 (1 + 3 100)5 = $23,185.48 Bank B: Total amount = 20000 + 3.2 100 × 20000 × 5 =$23,200 Bank C: Total amount = $23,000 He should invest in Bank B as the interest he get is the highest. 8a Probability of all good apple = 17 20 × 16 19 × 15 18 n rP      + 1001

= 0.59649 Probability of at least 1 bad = 1- 0.59649 = 0.404 or 23 57 8b No. 17 20 × 16 19 × 3 18 refers to the probability of (Good, Good, Bad). It can also be (Good, Bad, Good) or (Bad, Good, Good). Probability should be 17 20 × 16 19 × 3 18 + 17 20 × 3 19 × 16 18 + 3 20 × 17 19 × 6 18 = 3 ( 17 20 × 16 19 × 3 18) 9 19 28 30 32 48 50 70 72 Median = 32+48 2 = 40 Lower quartile = 28+30 2 = 29 Upper quartile = 50+70 2 = 60 10a 10b 10c 4 𝑥 + 2𝑥3 = 0 4 𝑥 = −2𝑥3 no intersection for 𝑦 = 4 𝑥 𝑎𝑛𝑑 𝑦 = −2𝑥3 19 29 40 60 72 x 0 x 0

0 solution 11a −(𝑥 + 2)(𝑥 − 5) = −(𝑥2 − 5𝑥 + 2𝑥 − 10) = −(𝑥2 − 3𝑥 − 10) = −[(𝑥 − 1.5)2 − (−1.5)2 − 10] = −[(𝑥 − 1.5)2 − 12.25] = −(𝑥 − 1.5)2 + 12.25 𝑝 = −1.5. 𝑘 = 12.25 11b 11c (1.5, 12.25) 12a Angle ACD = 2 angle AOD = 67o÷ 2 = 33.5o (angle at centre = 2 angle at circumference) Angle BAO = 67o (alt angles, AB // DO) Angle BCA = 180o −67o−90o = 23o (right angle in semicircle) Angle BCD = 23o + 33.5o = 56.5o 12b Angle DAB and angle DCB are angles in opposite segments in a circle and adds up to 180o 12c Angle DBA = angle DCA = 33.5o (angles in same segment) Angle DBC = 90o − 33.5o = 56.5o (right angle in semicircle) 13 √3𝑝3 + 6𝑟2 = 5𝑟 2 3𝑝3 + 6𝑟2 = 25𝑟2 4 12𝑝3 + 24𝑟2 = 25𝑟2 x y 10 5 -2

12𝑝3 = 𝑟2 𝑟 = ±√12𝑝3 14a 𝐴1 𝐴2 = ( 𝑙1 𝑙2 ) 2 9 49 = ( 𝑙1 𝑙2 ) 2 𝑙1 𝑙2 = 3 7 𝑉1 𝑉2 = ( 𝑙1 𝑙2