ACSI 2018 Y3 Core Mathematics Paper 2 Solution

FINAL EXAMINATION 2018 YEAR 3 INTEGRATED PROGRAMME CORE MATHEMATICS PAPER 2 MONDAY 9th October 2017 1 hour 30 minutes 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 5 printed pages. [Turn over

ACS(Independent)MathDept/Y3IPCoreMathPaper2/2018/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: 6] (i) Express 𝑦−𝑥 𝑥2−𝑥𝑦−2𝑦2 + 2 3(𝑥+𝑦) as a single fraction in its simplest form. [3] 𝑦 − 𝑥 (𝑥 − 2𝑦)(𝑥 + 𝑦) + 2 3(𝑥 + 𝑦) = 3𝑦 − 3𝑥 + 2(𝑥 − 2𝑦) 3(𝑥 − 2𝑦)(𝑥 + 𝑦) = 3𝑦 − 3𝑥 + 2𝑥 − 4𝑦 3(𝑥 − 2𝑦)(𝑥 + 𝑦) = −𝑦 − 𝑥 3(𝑥 − 2𝑦)(𝑥 + 𝑦) = 1 3(2𝑦 − 𝑥) (ii) Hence or otherwise, find the value of x when w = 1.25 and y = 0.03 if 22 2 2 3( ) yx wx xy y x y − +=− − + . [3] 1 = 3𝑤(2𝑦 − 𝑥) 1 3𝑤 = 2𝑦 − 𝑥 𝑥 = 2𝑦 − 1 3𝑤 𝑥 = 2(0.03) − 1 3(1.25) 𝑥 = −0.207 Students failed to factorize the denominator. They should OBSERVE that the other fraction has 𝑥 + 𝑦 as denominator. Hence one of the factor must be 𝑥 + 𝑦

ACS(Independent)MathDept/Y3IPCoreMathPaper2/2018/FinalExamination 3 2 [Maximum mark: 13] (a) Expand and simplify 3( ) 4[2 2(3 4 )]y x x y x y− + − − − . [3] 3𝑦 − 3𝑥 + 4[2𝑥 − 𝑦 − 6𝑥 + 8𝑦] = 3𝑦 − 3𝑥 + 4(−4𝑥 + 7𝑦) = 3𝑦 − 3𝑥 − 16𝑥 + 28𝑦 = 31𝑦 − 19𝑥 (a) Solve the equation √11𝑥2 + 45 = 4𝑥. [3] 11𝑥2 + 45 = 16𝑥2 5𝑥2 = 45 𝑥2 = 9 𝑥 = ±3 (reject −3) (b) The solution of 𝑥√5 = 𝑥√3 + √48 is 𝑎 + 𝑏√15. Find the values of the integers 𝑎 and 𝑏. [4] 𝑥√5 − 𝑥√3 = 4√3 𝑥(√5 − √3) = 4√3 𝑥 = 4√3 √5−√3 𝑥 = 4√3 √5−√3 × √5+√3 √5+√3 𝑥 = 4√15+12 2 𝑥 = 6 + 2√15 Students need to learn how to check which answer should be rejected Very poorly attempted. • Students need to realise that √48 can be simplified to a simpler surd first - 4√3 • Common practise to shift all 𝑥 terms to one side of the equation

ACS(Independent)MathDept/Y3IPCoreMathPaper2/2018/FinalExami