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 ) 3 𝑉1 𝑉2 = 27 343 V1 = 1200 ÷ 343 × 27 =94.5 cm3 14b Radius 3 refers to that of water though qn may be a bit vague Volume of water = 94.5 × 80% = 75.6 1 3 𝜋(3)2ℎ = 75.6 h = 8.02 cm Accept 3 as radius of cone also 𝑉1 𝑉2 = 4 5 𝑟1 𝑟2 = √ 4 5 3 𝑟1 = √ 4 5 3 × 3 = 2.78495 1 3 𝜋(2.78495)2ℎ = 75.6 H = 9.31 cm 15a 𝐼 = 𝑘 𝑑2 8 = 𝑘 32 k = 72 𝐼 = 72 𝑑2 15b 𝐼 = 72 (0.25𝑑)2 𝐼 = 72 0.0625𝑑2 𝐼 = 1152 𝑑2 1152 72 = 16 𝑡𝑖𝑚𝑒𝑠 % change = 1152−72 72 × 100% = 1500%

16ai Angle FGA= (7−2)180 7 = 128.57° Angle OGF = 128.57° ÷ 2 = 64.3° 16aii Angle GFH = 180 − 64.3 = 115.7° Angle EFH = 128.57 − 115.7 = 12.9° 16b Sum of exterior angle = 360o 18 + 22 + 32 + 4(17) + (𝑛 − 7)(20) = 360 140 + 20𝑛 − 140 = 360 𝑛 = 18 17a Square is a type of rectangle Square and rectangle is a type of parallelogram 17b 18a 𝐴𝐵 = √(6 − 0)2 + (3 − (−11))2 = √36 + 196 =15.2 units 18b Gradient = 6−0 3−(−11) = 3 7 y = mx +C At (3,6), 6 = 3 7 (3) + 𝐶 C =4 5 7 y = 3 7x +4 5 7 18c y = mx +4 5 7 at x = 0 y = 4 5 7 ξ q p r s ξ q p r s kite

E = (0, 4 5 7) 18d 1 2 (𝐵𝐶)(6) = 22.5 BC = 7.5 C = (−3.5, 0) 18e Coordinate A to C x shifted -6.5 units y shifted -6 units Coordinate D = (−11 − 6.5 , 0 − 6) = (−17.5 , −6) 18f 7𝑦 − 3𝑥 + 15 = 0 7𝑦 = 3𝑥 − 15 𝑦 = 3 7 𝑥 − 15 7 As line l has the same gradient 3 7 as AB, the two lines are parallel and will not meet 19ai 19aii 86o 19bi See diagram above 19bii See diagram above 19c Draw perpendicular bisector of WX It is not possible to so, as the perpendicular bisector of WX, the perpendicular bisector of XY, and the angle bisector of WXY do not meet at a single point

20a 20b a = 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑠𝑝𝑒𝑒𝑑 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑡𝑖𝑚𝑒 = 20 15 = 1.33 m/s2 20c Total distance = (0.5)(20)(15) + (20)(20) + (0.5)(10)(20) = 650 m Average speed = 650 45 =14.4 m/s