CGS 2025 EMATH PRELIM P2 MS

Name: Solution Register No.: Class: CRESCENT GIRLS’ SCHOOL SECONDARY FOUR PRELIMINARY EXAMINATION 2025 MATHEMATICS 4052/02 Paper 2 26 August 2025 2 hours 15 minutes Candidates answer on the Question Paper. READ THESE INSTRUCTIONS FIRST Write your name, register number and class on all the work you hand in. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Answer all the questions. The number of marks is given in brackets [ ] at the end of each question or part question. If working is needed for any question it must be shown with the answer. Omission of essential working will result in loss of marks. The total of the marks for this paper is 90. The use of an approved scientific calculator is expected, where appropriate. If the degree of accuracy is not specified in the question, and if the answer is not exact, give the answer to three significant figures. Give answers in degrees to one decimal place. For 𝜋𝜋, use either your calculator value or 3.142. Question 1 2 3 4 5 6 7 8 9 10 Marks 10 8 7 9 9 8 9 10 10 10 Table of Penalties Question No. Presentation –1 Accuracy/ Units –1 Parent’s / Guardian’s Signature This document consists of 24 printed pages For Examiner’s Use 90

2 Crescent Girls’ School 2025 Prelim S4 Math P2 Mathematical Formulae Compound Interest Total amount = 1 100 n rP + Mensuration Curved surface area of a cone = rlπ Surface area of a sphere = 24 rπ Volume of a cone = 21 3 rhπ Volume of a sphere = 34 3 rπ Area of triangle ABC = 1 sin2 ab C Arc length = rθ , where 𝜃𝜃 is in radians Sector area = 21 2 r θ , where 𝜃𝜃 is in radians Trigonometry sin sin sin abc ABC= = 2 22 2 cosa b c bc A=+− Statistics Mean = fx f ∑ ∑ Standard deviation = 22fx fx ff ∑∑ − ∑∑ 

3 Crescent Girls’ School 2025 Prelim S4 Math P2 [Turn Over 1 (a) Solve the inequality 34 532 2 x xx + <− −< . Solution: 7 2 and 5 3 4 234 34 1 06 0 34 3 4 22 10 xx xx xx x x xx −< < − −< ++ + +< − >> 10 7x > Answer …………………… [3] (b) It is given that 2 )(7 2 3 ma xF a −= , where 0m ≠ and 0a ≠ . (i) Find, in terms of a, the value of F when xa= − and 3m = . Solution: 2 )3(7 2( ) 3 72 aaF a a −−= = − Answer F = ………………… [1] (ii) Express x in terms of F, a and m. Solution: 2 2 2 ) 7 2 7 2 7 (7 2 3 372 2 3 3 3 2 2 a a am aF m ma xF a aFax m aFx m aFx m −= −= = = ± −± − = − Answer x = ………………… [2]

4 Crescent Girls’ School 2025 Prelim S4 Math P2 (c) Solve the equation 23 752 xx xx + +=+− . Solution: 22 2 22 2 2 23 752 ( 2)( 2) 3 ( 5) 7( 5)( 2) 15 7( 3 10) 4 15 4 7 21 70 3 6 66 0 2 22 0 43 xx xx x x xx x x x xxx xx xx xx xx x + +=+− + −+ += + − + = +− += −+ − +− +−= +−= 22 2 4(1)( 22) 2(1) 2 92 2 3.80 or 5.80 (3sf) x −± − −= −±= = − Answer x = ………………… or ………………… [4]

5 Crescent Girls’ School 2025 Prelim S4 Math P2 [Turn Over Solution: 3 probability 0.7 0.7 0.7 0.34 ×× = = Answer …………………… [2] (b) Find the probability that Emily does not log in on Wednesday. Solution: 0.3) (0.3 0.8) 0.45 probability (0.7 += ×× = Answer …………………… [2] (c) Each time the user logs in, she will earn 10 reward points. If the user does not log in, no reward points will be given. (i) Find the probability that Emily earns exactly 20 points by Wednesday. Solution: 0.3) (0.3 0.2) 0.27 probability (0.7 += ×× = Answer …………………… [2] (ii) Find the probability that Emily earns more than 10 points by Wednesday. Solution: 6 p (ro .bab 0 it 1 03 0l .iy 8) .7 − = = × or 0.27 (0.7 0.7) 0.76 probability = +× = Answer …………………… [2] 2 A company conducted a survey in a girl’s school and found that a user’s likelihood of logging into a particular app each day depends on whether she logged in the previous day. If the user logs in on a particular day, the probability that she logs in again the following day is 0.7. If the user does not log in on a particular day, the probability that she does not log in the next day is 0.8. It is given that Emily logs in on Monday. (a) Find the probability that Emily logs in on Tuesday, Wednesday and Thursday.

6 Crescent Girls’ School 2025 Prelim S4 Math P2 3 A, B, C, D and E are points on a circle. AE = AB, AE is parallel to BD, angle 22BEC = ° and angle 80EAC = ° . (a) Find angle AEB . Solution: 22 (angles in the same segment) 22 80 102 180 102 (base angles of isos. triangle)2 39 BAC BAE AEB ∠= ° ∠ = °+ ° = ° °− °∠= = ° Answer Angle AEB = …………………… [2] (b) Find angle DEC . Solution: 180 102 (angles in opp segment) 78 180 78 (int angles, / / ) 102 102 39 22 41 BDE AED AE BD CED ∠ = °− ° = ° ∠ = °− ° = ° ∠ = °− °− ° = ° 80° 22° E A B C D

7 Crescent Girls’ School 2025 Prelim S4 Math P2 [Turn Over Alternative Solution: (alt angles, / / ) 39 180 39 3 22 (angles in opp segment) 41 DBE BEA AE BD DEC ∠= ∠ = ° ∠ = °− °× − ° = ° Answer Angle DEC = …………………… [3] (c) Angle 78AFE = ° . Determine the position of point F. Justify your answer. Solution: 180 39 22 80 (sum of angles in triangle) 39 ACE∠ = °− °− °− ° = ° 2AFE ACE∠= ∠ Point F lies on the centre of the circle. [2]

8 Crescent Girls’ School 2025 Prelim S4 Math P2 4 In the diagram, BD is perpendicular to AD. It is given that AC = BC and the gradient of BC is 3 4− . (a) Find the equation of line BC. Solution: 3 ( 5)4 3 35 44 5y x yx −= − − = −+ Answer …………………… [2] (b) Find the value of n. Solution: 3 35(13)44 1 n = −+ = − Alternative Solution: 53 5 13 4 1 n n − = −− = − Answer n = ………………… [1] A (m, n) B (13, n) C (5, 5) D y x O

9 Crescent Girls’ School 2025 Prelim S4 Math P2 [Turn Over (c) Show that 3m = − . Solution: ABC∆ is isosceles. Points A and B are symmetrical about the vertical through C. 13 52 m + = 3m = − [1] (d) Find the area of triangle ABC. Solution: 2 1area of triangle 16 6 2 48 units ABC = ×× = Answer ………………… units2 [2] (e) Hence, find the length of BD. Solution: 22(5 ( 3)) 1s (5 ( u 1 0 ni )) t AC = +−= −− − 1 10 482 9.6 units BD BD ×× = = Answer ………………… units [3]

10 Crescent Girls’ School 2025 Prelim S4 Math P2 5 A, B and C are points on the horizontal ground. B is at a bearing of 290° and 800 m away from A. C is due north of A and at a bearing of 60° from B. (a) Find BC. Solution: (alternate angles)60BCA °∠= 800 sin 70 sin 60 800 sin 70sin 60 868.05 868 m (3sf) BC BC =°° = °° × = = Answer ……………………. m [2] (b) Show that AC is approximately 707.64 m. Answer Solution: (360 290 ) (sum of angles in a triangle) 5 180 60 0 CBA °− °− °− ° = ° ∠= 800 sin 50 sin 60 800 sin 50sin 60 707.64 m (5sf) (shown) AC AC × =°° = °° = C NB NA A B 800 m 290° 60°