Broadrick EM 2024 Maths Sec Prelim - P1 MS

Broadrick Secondary School4E5N Preliminary Examination 2024Paper 1 Marking Scheme1-0.17898 = -0.1790 (4sf)B12aMean = = 2.28 sB1b = 0.901 s (3sf)B13 B1B14HCF = LCM = X = = 36Y = = 162B1B15a1 : 50 0001cm: 0.5 kmActual dist = =3.75 kmB1b1cm: 0.5 km1 cm2: 0.25 km2Area on map = =9 cm2 M1A16Largest density = largest mass/smallest vol= = 21 g/ cm3 M1A17 or 5 (any odd integer more than 1) B1B18a Obtuse =98.3° (1dp)M1A1

bAngle ABC = =50.6562Area = =42.9 cm2 (3sf)M1 (formula to find area with their angle)A19In Singapore:Amount of Yen = In JapanAmount of Yen = Her claim is not true. She will receive (237.21 Yen) more if she changes in Japan.M1M1A110aThe vertical axis did not start from 0.(Optional: The scores in 2022 looked like it had increased to 6 times but the increase was from 55 to 80 (which is slightly less than double.)B1bI disagree.Although the increase in the height of the bar looks the same from 2022 to 2024 for both classes, the scale of the two graphs are different. It exaggerates the increase in test scores of Class B.B111 -- perfect cube (powers are multiples of 3) B1B112a = B1b M1(change either to power of 3)M1 (combine to or to a single power of 3)A1

13aA={2,3,4,6,8,12}B1bSince , B={…,11, 13, 17, 19}C={…,9, 16} B1cA={2,3,4,6,8,12}C={4,9,16}If , smallest p = 5B114X = = 40%M1A115In triangle OAP and BPA, * (alt angle, AB // OP)Since OA = AP = PB (same radius). * (base angles of isos triangle)*AP = PA (common side)Therefore (AAS)Hence AB = OPB2 (any 2 of the 3 stmts)A1 (conclude AAS and equal sides)16aAB: PQ: The gradient of PQ should be steeper than that of AB.Hence she is not correct.M1 (rearrange to find gradient of either line)A1 (compare both gradient and conclude)

bi M1A1biiMin point (-1.25, -7.5625)B1, B117 M1A118ai B1aii M1 (grouping)A1b M1A119a =21490.79038= $21490.79 (2dp)M1A1b r = 0.622%M1A120

M1, M1A121a = 6.32 units (3sf)M1A1bR is (1, −3)By sketching and counting, From Q to P: horizontally -6 and vertically -2Since y=x is reflection line, From Q to R: horizontally -2 and vertically -6B122Int angle of pentagon Int angle of square = 90°Interior angle of regular polygon = Exterior angle = Since n is a positive integer, it is possible to form a regular polygon, hence a closed loop.M1M1 (int or ext of polygon)A123a M1, M1A1b M1

A124a When n is odd, 5n is odd and 5n-1 is evenSince product of odd and even number is always even, is even.When n is even, 5n is even and 5n-1 is odd.Since product of odd and even number is always even, is even.Hence the sum of the first n terms in the sequence is always even.M1A1b1st term: 2nd term: 3rd term: Thereforenth term = M1A125a B1b = B2 (B1 for any 2 correct)cIt represents the amount of money Kevin and Molly spent respectively at Bookstore A.B1d B1

26a1.5 km/min = =25 m/sB1bArea = dist travelled M1A1c = 19m/sOR M1A1dArea = dist travelled T is 0915M1A127a3BD = 2DC M1A1bi

Since AC is parallel to BR and M1M1 A1bii A1