ACSI 2021 Y3 Core Mathematics Paper 2 Solution

FINAL EXAMINATION 2021 YEAR 3 INTEGRATED PROGRAMME CORE MATHEMATICS PAPER 2 XXXXXX xxth October 2021 1 hour 30 minutes ADDITIONAL MATERIALS: Answer Paper (7 sheets) Graph Paper ( 1 sheet) INSTRUCTIONS TO STUDENTS Do not open this examination paper until instructed to do so. A calculator is required for this paper. Answer all the questions on the answer sheets provided. At the end of the examination, fasten the answer sheets together. Unless otherwise stated in the question, all numerical answers must be given exactly or correct to three significant figures. Answers in degrees are to be given to one decimal place. INFORMATION FOR STUDENTS The maximum mark for this paper is 80. _____________________________________________________ This question paper consists of 4 printed pages. [Turn over

ACS(Independent)MathDept/Y3IPCoreMathPaper2/2021/FinalExamination 2 Full marks are not necessarily awarded for a correct answer with no working. Answers must be supported by working and/or explanations. Where an answer is incorrect, some marks may be given for correct method, provided this is shown by written working. You are therefore advised to show all working. Answer all the questions on the answer sheets provided. Begin each question on a new page. 1. [Maximum mark: 12] (a) Evaluate 3 3 ln(3.256) 1.325 (0.25 2.38) e+ − , leaving your answer correct to 3 significant figures. [2] 5.4623 −9.6636 = −0.565 (b) Simplify ( ) 3 2 1 1 2 4 82 4 3 (3 ) 3 x y yx −   , expressing your answer in positive indices. [3] (3𝑥 𝑦2) 3 9𝑦 (1 3 𝑥2) = 27𝑥3 𝑦6 3𝑦𝑥2 = 27𝑥3 3𝑦7𝑥2 = 9𝑥 𝑦7 Most students just use calculator to find the answer, a small group of students still give the wrong answer. Careless in expanding the number 3 for both cube and square. Some students did not leave it in positive indices.

ACS(Independent)MathDept/Y3IPCoreMathPaper2/2021/FinalExamination 3 (c) Simplify 13 21 27 2(3 ) 39 yy yy + +− + . [3] (33)𝑦+1 + 2(33𝑦) 3𝑦+2(32)𝑦−1 = 33𝑦+3 + 2(33𝑦) 3𝑦+232𝑦−2 = 27(33𝑦) + 2(33𝑦) 33𝑦 Let 𝑥 = 33𝑦 = 27𝑥 + 2𝑥 𝑥 = 29𝑥 𝑥 = 29 (d) Solve for x if 2 651xx+− = . [4] 5𝑥2+𝑥−6 = 50 𝑥2 + 𝑥 − 6 = 0 (𝑥 − 2)(𝑥 + 3) = 0 𝑥 = −3 or 𝑥 = 2 Careless in factorising the 3 or expanding the power-power rule wrongly. Some students see the question as 3y instead of 33y Easy question to solve.

ACS(Independent)MathDept/Y3IPCoreMathPaper2/2021/FinalExamination 4 2 [Maximum mark: 11] (a) Expand and simplify ( )( ) 2 2 2 2 1 n p n m p mm n n m p  − + −+−  − . [3] (𝑚2 − 𝑛2) (𝑛2(𝑝 − 1) + 𝑚2(𝑝 − 1) 𝑝 − 1 ) = (𝑚2 − 𝑛2) ((𝑛2 + 𝑚2)(𝑝 − 1) 𝑝 − 1 ) = (𝑚2 − 𝑛2)(𝑛2 + 𝑚2) = 𝑛4 − 𝑚4 (b) Find the sum of 35 x x− and 2 6 10 9 25 x x +− − , expressing your answer as a single fraction in its simplest form. [3] 𝑥 3𝑥 − 5 − 6𝑥 + 10 9𝑥2 − 25 = 𝑥 3𝑥 − 5 − 2(3𝑥 + 5) (3𝑥 − 5)(3𝑥 + 5) = 𝑥 (3𝑥 − 5) − 2 (3𝑥 − 5) = 𝑥 − 2 (3𝑥 − 5) [Common mistake] Did not simplify into the 2 terms answers. [Reminder] Always give the answer in simplest form. Simplest means shortest unless the question ask Factorise completely! Did not minus but sum up the 2 terms. Did not factorise and express in simplest form by giving 3x2-x-10 as the numerator.

ACS(Independent)MathDept/Y3IPCoreMathPaper2/2021/FinalExamination 5 (c) Solve the equation ( )( ) 23 5 4 04 2 4 xx x x x + +=− + − , leaving your answers in 2 decimal places. [5] 3𝑥2 + 5 (𝑥 − 4)(𝑥 + 2) − 4𝑥 𝑥 − 4 = 0 3𝑥2 + 5 − 4𝑥(𝑥 + 2) = 0 3𝑥2 + 5 − 4𝑥2 − 8𝑥 = 0 −𝑥2 − 8𝑥 + 5 = 0 𝑥 = −(−8) ± √(−8)2 − 4(−1)(5) 2(−1) 𝑥 = −8.58 or 0.58 3 [Maximum mark: 9] A circular cylinder container of base radius xe cm, and height 2xe cm is fully filled with water. (a) Given that the volume of water in the container is 12900 cm3, find the value of x . [3] Volume of container = 𝜋(𝑒𝑥)2(𝑒2𝑥) = 12900 𝑒4𝑥 = 12900 𝜋 ln 𝑒4𝑥 = ln 12900 𝜋 4𝑥 = 8.32025 𝑥 = 2.08 Many students see the question wrongly. (4x became 4) Did not leave the final answers in 2 d.p. Wrong formula to find the Volume of container.

ACS(Independent)MathDept/Y3IPCoreMathPaper2/2021/FinalExamination 6 The water is poured into a rectangular tank of base area 850 cm2 and height 60 cm. (a) Find the depth of the water in the rectangular tank. [2] Let h be the depth of the water in the rectangular tank, 850(ℎ) = 12900 ℎ = 15.2 cm (c) If a solid metal sphere of radius 14 cm is then put into the rectangular tank and the sphere is totally immersed in the water, will the water overflow? Explain your answer. [4] Volume of sphere = 4 3 𝜋(14)3 Volume of sphere = 11494 cm3 Total volume of water = 11494 + 12900 = 24394 cm3 Total volume of vessel = (850)(60) = 51000 cm3 Remaining space in the tank = 51000 – 24394 = 26606 cm3 Since volume of water is less than total volume of vessel, water will not overflow. OR Volume of sphere = 4 3 𝜋(14)3 Volume of sphere = 11494 cm3 Height increased = 11494/850 = 13.5 cm New height of water in the tank = 13.5 + 15.2 = 28.7 cm Since new height of water in the tank is less than the height of the tank, water will not overflow. Did not give the final answer in 3 s.f. as instructed on the cover page. Almost all students know how to explain their findings, but minority use the wrong formula to find the Volume of Sphere. Almost all students know how to explain their findings, but minority use the wrong formula to find the Volume of Sphere.

ACS(Independent)MathDept/Y3IPCoreMathPaper2/2021/FinalExamination 7 4 [Maximum mark: 7] (a) Solve 2 391 log (1 2 ) log (5 )xx− − = − . [4] 1 = log3(5 − 𝑥) + log3(1 − 2𝑥) log3[(5 − 𝑥)(1 − 2𝑥)] = 1 (5 − 𝑥)(1 − 2𝑥) = 3 5 − 10𝑥 − 𝑥 + 2𝑥2 = 3 2𝑥2 − 11𝑥 + 2 = 0 𝑥 = 0.188 or 𝑥 = 5.31 (rejected) (b) Evaluate 2 3 4 5 63log 3 log 4 log 5 log 6 ... log (64)     , find the value of n . [3] log 3 log 2 × log 4 log 3 × log 5 log 4 × log 6 log 5 × … × log(64) log 63 = log 64 log 2 = log 26 log 2 = 6 Almost all students did it correctly, but did not check the final answer which need to reject the answer (x = 5.31) Only minority applied the law of logarithms wrongly in combining the terms. Did not check the equation properly end up 5 – 3 = – 2 Some students did not try at all, but some students applied the wrong law of logarithms as below. ( ) ( ) 23log (2 1) log (3 1)...... 1 0 1 0 .... 1 +  + = +  + =

ACS(Independent)MathDept/Y3IPCoreMathPaper2/2021/FinalExamination 8 5 [Maximum mark: 15] O, A, B and C are four points on ground level. C is on a bearing of 50o from O. B is due east of A and south of C. OB = OC = 60 m and 90oAOB= . A vertical flag pole, TC, 25 m high, is located at C. Find (a) Bearing of O from B, [2] ∡𝑂𝐶𝐵 = 50𝑜 ∡𝑂𝐵𝐶 = 50𝑜 Bearing of O from B = 360𝑜 − 50𝑜 = 310𝑜 (b) the distance OA, [3] ∡𝑂𝐵𝐴 = 90 − 50𝑜 = 40𝑜 tan 40𝑜 = 𝑂𝐴 60 𝑂𝐴 = 60 × tan 40𝑜 𝑂𝐴 = 50.3 m m m Well attempted Some students used the wrong formula for tangent. Note that: 𝑡𝑎𝑛𝑔𝑒𝑛𝑡 = 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡

ACS(Independent)MathDept/Y3IPCoreMathPaper2/2021/FinalExamination 9 (c) the distance BC, [3] ∡𝐶𝑂𝐵 = 180 − 100 = 80𝑜 Using Cosine Rule, 𝐵𝐶2 = 602 + 602 − 2(60)(60) cos 80𝑜 𝐵𝐶2 = 5949 𝐵𝐶 = 77.1 OR ∡𝐶𝑂𝐵 = 180 − 100 = 80𝑜 Using Sine Rule, 𝐵𝐶 sin 80𝑜 =