ACSI 2022 Y3 Core Mathematics Paper 2 Solution

Anglo - Chinese School (Independent) FINAL EXAMINATION 2022 YEAR 3 INTEGRATED PROGRAMME CORE MATHEMATICS PAPER 2 Thursday 6th October 2022 1 hour 30 minutes ADDITIONAL MATERIALS: Answer Paper (6 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/Y3IPCoreMathPaper2/2022/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 t herefore 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] (a) Simplify 22 22 2a ab b b a ab − + +− − . [3] (b) Subtract 2 4 4x − from 11 22 xx−−+ , expressing your answer as a single fraction in its simplest form. [3] 22 22 2 22 2 22 2 () () () ( )[( ) 1] ( )( ) ( )1 () a ab b b a ab ab ba ab ab ab ab ab ab a ba b ab ab − + +−= − − +−= − − −−= − − −−= −+ −−= + 2 2 2 22 22 2 11 4 2 24 11 4 22 4 11 4 22 4 224 44 24 44 2 4 2( 2) 2 4 ( 2)( 2) 2 xx x xxx xx x xx xx x xx xx x xx x −−−+ − = −−−−+ − = −+− −+ − ++−=−− −− = −− −− −− − + −= = =− −+ − Students are very careless and many are weak in algebraic manipulation. Many students made the following mistake: 2() ( )( ) 0 0 ab ba a ba b abba ab ab − +− −+ −+−= + = + = Students were careless with the negative signs. Some students have problem understanding which term to subtract.

ACS(Independent)MathDept/Y3IPCoreMathPaper2/2022/FinalExamination 3 2. [Maximum mark: 8] Joe bought x number of books, each at the same price, for a total cost of $336. (a) Write down an expression for the cost of each book in terms of x . [1] Joe sold 20 of them for $480, and the rest at a loss of $4 per book. (b) Write down an expression for the total amount, in dollars, he received for all the books. [2] (c) Given that Joe made a profit of $184 altogether, form an equation in x and show that it reduces to 2 94 1680 0xx−+ = . [2] 336 x Selling price of the remaining books = 336 4x  − Number of books left = 20x− Total mount collected = ( ) 336480 20 4x x +− −  ( ) ( ) 2 2 336480 20 4 336 184 33620 4 40 6720336 4 80 40 67204 376 0 4 376 6720 0 94 1680 0 x x x x x x x x xx xx +− −− = − −=  −− += −− + = −+ − = −+ = Students don’t understand the term “sold 20 of them for $480” They think each book is being sold for $480. Very poorly done. Many students associated the answer in (b) to be $184 which is the profit only. Hence they have difficult showing the expression.

ACS(Independent)MathDept/Y3IPCoreMathPaper2/2022/FinalExamination 4 (d) Hence, solve the equation 2 94 1680 0xx−+ = and state the cost price of each book. [3] 3. [Maximum mark: 12] (a) Evaluate 2 3 2 2.15log 0.25 e −  +  , leaving your answer correct to 2 significant figures. [3] (b) Given that ( ) ( ) 31 43 23144 216 2 3 xy zpp p −−÷= , evaluate x , y and z . [4] 2 2 94 1680 0 ( 94) ( 94) 4(1)(1680) 2 24 70 xx x x x −+ = −− ± − −= = = Price of each book = $336 70 or $336 24 Price of each book = $4.80 or $14 2 3 2 3 2.15log 0.25 log (0.596191) lg 0.596191 lg 3 0.47 e −  +  = = =− ( ) ( ) 31 43 23 13 42 4 33 3 32 6 36 111 6 ( 1) 3 ( 1) 6 1 74 5 144 216 (2 3 ) (2 3 ) (2 3 ) (2 3 ) 23 23 pp pp pp p p −− −− −− −− −− − ÷ = ÷ = ÷ = = Hence, 7 4 5 x y z = = = There is no reason why students should reject either of the answer. Many students rejected the answer $4.80 because it is not a whole number. There is no need to apply log rules to this question, except change of base. The whole expression in the bracket should be keyed into the calculator to obtain 0.596191. But students manipulated and simplified because they are stuck with e, which is a constant Very poorly attempted. Students very confused with laws of indices and they did not change the numbers to base 2 and base 3 even though the right hand side of the equation has already given a clue that the numbers must be in base 2 and 3 so that they can compare the powers.

ACS(Independent)MathDept/Y3IPCoreMathPaper2/2022/FinalExamination 5 (c) Simplify 22 1 6 2(3 ) 48 ww w+ + + . [3] (d) Find the range of values of x if 22 5277xx−− < . [3] 22 1 22 2 22 22 2 22 2 2 6 2(3 ) 48 2 3 2(3 ) 28 3 (2 2) 2 (4) 8 3 (2 2) 4(2 2) 3 4 ww w ww w w ww w ww w w + + + + += + += + += + = 22 52 2 2 77 2 5 21 2 5 30 ( 3)(2 1) 0 1 32 xx xx xx xx x −− < − −< − −< − +< −<< Many misconceptions with the law of indices. For example: 226 23ww= × 226 66ww= × These are very basic and fundamental concepts of indices, students should not make these mistakes. It is recommended that students draw the curve and shade the necessary region to visualise the values needed. Some students give the wrong inequalities because they did not draw the graph above.

ACS(Independent)MathDept/Y3IPCoreMathPaper2/2022/FinalExamination 6 4. [Maximum mark: 8] (a) Solve the equation 134yy+ = . [4] (b) Solve the equation ( ) 2 33 9log log ( 3) log (9 )xx x+ −= . [4] 1 1 34 lg 3 lg 4 ( 1) lg 3 lg 4 lg 3 lg 3 lg 4 lg 4 lg 3 lg 3 (lg 4 lg 3) lg 3 lg 3 lg 4 lg 3 3.82 yy yy yy yy yy y y y + + = = += += −= −= = − = 3 (3) 4 31 43 31 43 31lg lg43 1lg 3 3lg 4 3.82 yy y y y y y y = =  =  =    =  = ( ) ( ) ( ) 2 33 9 33 3 33 2 2 log log ( 3) log (9 ) log log ( 3) log (3 ) log [ ( 3)] log (3 ) 33 60 ( 6) 0 6 0 xx x xx x xx x x xx xx xx x x + −= + −= −= −= −= −= = =

ACS(Independent)MathDept/Y3IPCoreMathPaper2/2022/FinalExamination 7 5. [Maximum mark: 12] In the figure, A, B and C are three points on a horizontal field. A is due west of B, the bearing of B from C is 125o, AB = 430 m and BC = 460 m. (a) Find (i) the distance between A and C, [3] (ii) ACB , [2] A B C 125o 460 m 430 m Angle ∡𝐶𝐶𝐶𝐶𝐶𝐶 = 90 (180 125 )o oo−− Angle ∡𝐶𝐶𝐶𝐶𝐶𝐶 = 35o 2 22 2 430 460 2(430)(460) cos35 72443.45 269 oAC AC AC =+− = = sin35 sin 269.15 430 sin 0.91636 66.4o ACB ACB ACB = = = Generally quite well done. Most students have good conceptual understanding of this topic.

ACS(Independent)MathDept/Y3IPCoreMathPaper2/2022/FinalExamination 8 (iii) the bearing of C from A, [2] (iv) the area of ABC∆ . [2] At a certain instant, a hot air balloon is at a point which is directly above C. (b) Given that the angle of elevation of the hot air balloon from B is 5.2 o, find the angle of elevation of the hot air balloon fro