ADSS Prelim 2025 P1 [Answers]

1 Marking Scheme for Sec 4E/5N EMath Prelim P1 (2025)125142xx2528xxM110x > –3310x A12Total amount = nrP1001 21262040011001.84281.84xx M1A1315 = 3 x 545 = 3 x 3 x 5M1Thus, the smallest value of x = 9A14aBy observation, median mark is 55.A14bThe lower quartile mark is 40 marks, representing the bottom 25% of students. Thus, Rachel is wrong.A15aAnswer A : The size of the worker symbol distorts the graph. A1(Only accept this answer)5bAnswer A : The area is not proportional to the actual decrease in unemployed workers. The number of unemployed workers in 2020 is half of that in 2010. However, the size of worker symbol is less than half the size of worker symbol for 2010. Or Answer B : It exaggerates the differences between the number of unemployed workers.A16asinMBC4041 A16bcosBCN4041 A1

2 7Q : P : R =125 :100 : 87.5 = 10 : 8 : 7 M1P : R = 8 : 7A1813816bb3(1)4(3)22bbM1By comparing, 3(1)4(3)bb 33412157bbb M1157bA1922222222(21)122(21)212(21)(21)42(21)4(21)aaaaaaaaaaaaaaaaaaa M1M1A110a1 cm : 200 000 cm1 cm : 2 km8 cm : 16 kmA110b1 cm : 2 km1 cm2 : 4 km2 M1150 cm2 : 600 km2 A111a204abb= 4b (5a – 1)A111b326xyabayxb=3(2)(2)xbyabyM1=(2)(3)byxaA112a22khch

3 22(2.81)(8)(53.673.67)5.8215.c A112b22khchhc = k2 + h2 22khch M12khchA113asum of interior angles = (n – 2)(180o) 285o + fourth angle = ((4 – 2)(180o) Fourth angle = 360o ‒ 285o = 75o A113bGiven: interior angle of a regular polygon is 165o 1 ext angle = 180o – 165o = 15o M1 1 ext angle 360on 3602415oon Number of sides is 24. A114aThe elements are {5, 6, 7, 8, 9, 11, 12, 13, 14} A114bAB {4,5,7,11}AC 8C 9B A215422222228126()92(3)(9)(9)2(3)(3)(3)(9)2(3)(3)(9)2xxxxxxxxxxxxx M1M1A116a

4 M1A116bThe maximum point is (–2.5, –10.25).A117a318x 318x 6xB1a = 11 ‒ 6 = 5 b = 11 + 6 = 17 c = 17 + 6 = 23 A117b61n A117c6n is always an even number regardless the value of n. 61n will always be an odd number. Therefore, the value of every term must be odd for all values of n. B118a2.4 x 10–4 A118b5.4246 x 104 megametres = 5.4246 x 104 x 106 m = 5.4246 x 1010 m A118c5049(5.210)(2.010)= (1049)(5.2 x 10 ‒ 2.0) M1 = (1049)(52 ‒ 2.0) = (1049)(50) = 5.0 x 1050 A119aPerimeter of shaded region =60602(4)442(8)360360ooooM1= 20.566 cm = 20.6 cmA119bArea of shaded region = 226060(8)(4)360360oooo 226060(8)(4)360360ooooM1= 25.133 cm2 = 25.1 cm2 A120a22163.549areaofraisedstageareaofcentralstage B11616491633areaofraisedstageareaofwaterstage A120b

5 B1A121aBy Pythagoras’ Thm, 222ACBCAB BC2 = 1102 ‒ 702 = 7200 BC = 84.8528 = 84.9 km A121bAngle BAC = 170cos110 = 50.479o M1The bearing of B from A = 90o ‒ Angle BAC = 039.5o M1A122Gradient of the line AB2121yyxx 5136(2)4 M1B1Equation of AB is 34yxq ………………. (1)Sub (6, ‒5) into (1)35(6)4q 12q B1Thus, equation of AB is3142yx When y = 0, 31042p 23p B1The y-coordinate of E127236Thus, y-intercept of CD is 76 Since CD is parallel to AB, they have the same gradient. Thus, the equation of CD is 3746yx A123aInterquartile range = Q3 – Q1 = 56 ‒ 38M1 = 18 A1

6 23bP(obtained a distinction grade) 200180200 M1 20120010 A123cThe shape of the cumulative frequency curve of the second group of students will similar with the one of the first group of students in steepness in the middle 50% of the data , but shifted to the right.A124aangle ODC 18082492oo (base s of isos. )A124bangle CED 824122ooCOD (angle at the centre = 2 x angle at circumference)B1angle AEC = 180o – angle ABD – angle CED (s in opp. segments) = 180o – 55o – 41o = 84oA124cLet angle BDC be y.angle DBC + xo + angle OCD + y = 180o ( sum of )41o + xo + 49o + y = 180o ………………(1)angle BAE + xo + angle CDE ‒ y = 180o (s in opp. segments) 70o + xo + 49o ‒ y = 180o ……………….. (2) M1(1)+ (2), 209o + 2xo = 360o M1 xo = 75.5o A125a3331022091371412.7788.973288.9731.99522xxx M1M1A125bStandard Deviation = $7.74A125c1.The mean expense of Paul is more than Susan’s, implying that he has spent more than Susan.A12.The standard deviation of Susan’s expenses is lower than Paul’s, indicating a more consistent in her spending.A126a

7 Coordinates of P is (–11, 3)A126b22222((5))(24)148(5)(2)148(5)1484144(5)12127()17aaaaorarejectedor Thus, a 17 M1M1A126cGiven 183RT→17512624212RQOQOR→→→ 186313RT→ By observation, 23RQRT→→ M11.Since RTkRQ→→ and R is the common point of both vectors, R, Q and T lie on the same straight line. A12.The magnitude of RT is 2 times of the magnitude of RQ.A1