2025 3E SPS EOY P2 ANS

NAME FORM CLASS ACAD CLASS INDEX NO. Answers ST. PATRICK’S SCHOOL END-OF-YEAR EXAMINATION 2025 SUBJECT : Mathematics DATE : 6 October 2025 (4052/02) LEVEL : Secondary 3 Express DURATION : 2 hours 15 minutes Candidates answer on the Question Paper. READ THESE INSTRUCTIONS FIRST Write your Name, Form Class, Acad Class and Index No. in the spaces at the top of the page. 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 questions. 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 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  , use either your calculator value of 3.142, unless the equation requires the answer in terms of  . The number of marks is given in brackets [ ] at the end of each question or part question. The total of the marks for this paper is 90 . For Examiner’s Use Type of Penalties Question Number 90 Presentation 1 2 This paper consists of 19 printed pages including this cover page.

2 Mathematical Formulae Compound Interest Total amount = 1 100 n rP    Mensuration Curved surface area of a cone = rl Surface area of a sphere = 24 r Volume of a cone = 21 3 r h Volume of a sphere = 34 3 r Area of triangle ABC = 1 sin2 ab C Arc length = r , where  is in radians Sector area = 21 2 r  , where  is in radians Trigonometry sin sin sin a b c A B C  2 2 2 2 cosa b c bc A   Statistics Mean = fx f   Standard deviation = 22fx fx f f          

3 Answer all questions. 1 (a) Solve the equation 2( 3) 16x  . 3 4 or 3 4 7 1 x x x x       Answer x = ...………. or ………… [ 2] (b) Simplify 2 4 6 2 9 9 x x x    . 2(2 3) (2 3)( 3) 2 3 x x x x      Answer ...….……………………. [ 2] (c) Rearrange the formula 1px x p   to make p the subject of the formula. 2 2 1 1 ( 1)( 1) (1 ) ( 1)( 1) ( 1) 1 x xp p x p xp x x p x x xp x p x                Answer p = …..……………….……. [ 3]

4 2 (a) Simplify 2 3 2 1 5 (4 ) 2 a b a b , express your answer in positive index notation. 4 6 5 5 16 2 8 a b a b a b  Answer .…………….…..… ..…….…. [2] (b) The mass of a dust particle is 7.5×10−6 kg. Find the total mass of 2×108 such particles, giving your answer in standard form. 8 6 3 2 10 7.5 10 1.5 10      Answer .……………………… …. kg [1] (c) Solve the equation 5x+1 = 252. 1 4 5 5 1 4 3 x x x      Answer .……………………… …. …. [2]

5 3 The diagram shows a sector OAB of a circle with centre O and radius OA. A semi-circle with centre C and radius 12 cm is drawn inside the sector such that OA is the tangent to the semicircle at D, where angle ODC = 2  radians. Angle AOB = 6  radians and CD = 12 cm. (a) Show that the length of OC = 24 cm. Answer 12sin 6 24 cm OC OC    [1] (b) Find the perimeter of the shaded region 36 6 6 cm AB r           12 6 2 8 cm BD            2 236 24 12 = 12 3 cm AD   Perimeter = 6 12 3 8 64.76690684 64.8 cm      Answer .……………………... cm [ 4] 12 cm rad

6 (c) Find the area of the shaded region.   2 2 2 2 2 2 1Area of sector = (36) 2 6 108 cm 1Area 24 12 12 2 72 3 cm 1Area of sector (12) 2 6 2 48 c AOB OCD BCD                         2 2 m Area = 108 72 3 48 = 63.7879 63.8 cm     Answer .………………...…… cm 2 [4]

7 4 (a) A laptop is sold in a shop for $1600. During a promotion, the shop offers a 25% discount on the laptop. If the shop originally bought the laptop from the supplier at $1000, calculate the percentage profit made by the shop. 75Discounted price = 1600100 $1200 1200 1000Percentage profit = 100%1000 20%      Answer ..……………………..… % [3] (b) Sam is travelling to Japan. He goes to a money changer to convert his Singapore dollars (SGD) into Japanese yen (JPY). The exchange rate is 1 SGD =112 JPY. The money changer also charges a 2% commission fee on the amount of SGD exchanged. (i) Sam prepares to exchange 1500 SGD inclusive of commission fee. Calculate the commission fee he needs to pay. 102% -------- $1500 2% -------- 29.41 Answer ..…………...……… … SGD [2] (ii) Find the amount of money (in JPY) Sam receives after deducting the commission fee. Amount = (1500 29.41) 112 164706.08 JPY    Answer .…………...……...… JPY [2]

8 (iii) During his trip, Sam spends 90 000 JPY. When he returns, he converts the remaining yen back to SGD at the new rate 1 SGD = 110 JPY. How much money (in SGD) will he get back? 164706.08 90000Amount = 110 679.14618 679.15 SGD    Answer …………...………... SGD [2]

9 5 (a) Construct triangle ABC on the given line AB, where angle BAC = 45° and angle ABC = 75°. Answer [2] (b) Measure and state the length of AC. Answer AC = ..………………. cm [1] (c) Construct the perpendicular bisector of AB. [1] (d) Construct the angle bisector of angle ABC. [1] (e) Label the point of intersection as P. Measure and state the length of AP. Answer AP = …………..……. cm [1] A B

10 6 A tank has a capacity of 600 litres. Tap A can fill the tank completely in a minutes. Tap B can fill the tank completely in (a – 5) minutes. A drain pipe can empty the full tank in 30 minutes. (a) Write down the rate of Tap A to fill the tank. Answer …… 600 a ..………… l/min [1] (b) Write down the rate of Tap B to fill the tank. Answer ……… 600 5a …….…. l/min [1] (c) Both taps are turned on together with the drain pipe. The tank is filled in 20 minutes. Form an equation in terms of a. Show that it reduces to a² – 29a + 60 = 0. Answer [3] 2 2 600 600 600 600 5 30 20 1 1 1 1 5 20 30 5 1 ( 5) 12 12(2 5) ( 5) 24 60 5 0 29 60 (shown) a a a a a a a a a a a a a a a a                  