MGS 4048 02 AS

This question paper consists of 25 printed pages and 3 blank pages. Class Index Number Name : ____Solutions______________________ METHODIST GIRLS’ SCHOOL Founded in 1887 PRELIMINARY EXAMINATION 2022 Secondary 4 Wednesday MATHEMATICS 4048/02 3 August 2022 Paper 2 2 h 30 min Candidates answer on the Question Paper. INSTRUCTIONS TO CANDIDATES Write your name, class and index number in the spaces at the top of this 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 the 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. Forp, use either your calculator value or 3.142, unless the question requires the answer in terms of p 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 100. 100

2 Methodist Girls’ School Mathematics Paper 2 Sec 4 Preliminary Examination 2022 Mathematical Formulae Compound Interest Total amount =  100 r 1 n P Mensuration Curved surface area of a cone = rlp Surface area of a sphere = 42rp Volume of a cone = hr2 3 1p Volume of a sphere = 3 3 4 rp Area of a triangle ABC = 2 1absinC Arc length = r , where  is in radians Sector area = 21 2r, where  is in radians Trigonometry C c B b A a sin sin sin  a 2 = b 2 + c 2 – 2bc cos A Statistics Mean =   f fx Standard deviation = 22        f fx f fx

3 Methodist Girls’ School Mathematics Paper 2 Sec 4 Preliminary Examination 2022 1 (a) Simplify 2 4 5 3 (3 )x x  as a single fraction in is lowest term. 2 2 4 5 3 (3 ) 4 5 3 ( 3)(3 ) 4(3 ) 5 ( 3)(3 ) 4 7 ( 3) x x x x x x x x x x             or 2 2 2 2 4 5 3 (3 ) 5 4 (3 ) 3 5 4(3 ) (3 ) 4 7 (3 ) x x x x x x x x           or 2 2 2 4 5 3 ( 3) 4( 3) 5 ( 3) 4 7 ( 3) x x x x x x        Answer ......…………………. [2] (b) (c) Simplify 2 6 3 3 64x y     , leaving your answer in positive index. 2 6 3 3 4 2 4 2 64 16 16 x y x y x y         or   2 3 6 3 2 4 2 4 234 4 16x y x y x y  Answer ………………………. [2] Solve these simultaneous equations. 2 1 4 3 6 3 33 164 x y x y     2 1 4 (1)3 6 3 33 16 (2)4 3(2) 2, 6 32 (3)2 3(1) 9, 6 12 (4)2 x y x y x y x y               (3) (4), 3 20 20 263 3 20sub into (2),3 3 11 11 233 3 y y y x x         Answer x = ................................. y = ................................. [3]

4 Methodist Girls’ School Mathematics Paper 2 Sec 4 Preliminary Examination 2022 (d) (i) Express 25 4 7x x  in the form 2( )a x b c  . 2 2 2 2 2 2 5 4 7 45( ) 75 4 45 7 10 10 2 45 5 75 25 2 45 7 5 5 x x x x x x x                                          Answer ………………………. [2] (ii) Hence, solve 25 4 9x x  . 2 2 2 5 4 7 2 2 45 7 25 5 2 49 5 25 2 7 5 5 9 or 15 x x x x x x x                   Answer ………………………. [3]

5 Methodist Girls’ School Mathematics Paper 2 Sec 4 Preliminary Examination 2022 2 (a) A fruit basket contains 2 apples and 2 oranges. A second bag contains 1 apple, 1 orange and 2 kiwis. Ben picks one fruit from each bag. (i) Draw a possibility diagram to show all possible outcome. Answer 2nd Bag 1st Bag Apple Orange Kiwi Kiwi Apple AA AO AK AK Apple AA AO AK AK Orange OA OO OK OK Orange OA OO OK OK [2] (ii) Find the probability that Ben picks (a) two apples, P(two apples) = ଵ ଼ Answer ………………………. [1] (b) different fruits, P(different kind) = ଷ ସ Answer ………………………. [1] (c) at least one orange. P(at least one orange) = ହ ଼ Answer ………………………. [1]

6 Methodist Girls’ School Mathematics Paper 2 Sec 4 Preliminary Examination 2022 (b) Dream Renovation Company has three different packages. The table below shows the quantity of each type of work included in the different packages. Package Plumbing Work Window Carpentry Work A 6 5 6 B 4 2 7 C 5 3 9 This information is represented by the matrix Q = 6 5 6 4 2 7 5 3 9       . The unit cost price of each work is as follows. Plumbing Work $180 Window Work $420 Carpentry Work $390 (i) (a) Write down the 3 × 1 matrix P to represent the unit cost of each work. 180 420 390 P       Answer P = ……….………….[1] (b) Evaluate QP. 6 5 6 180 4 2 7 420 5 3 9 390 5520 4290 5670 QP                  Answer QP = …….………….. [1]

7 Methodist Girls’ School Mathematics Paper 2 Sec 4 Preliminary Examination 2022 (c) State what the elements of QP represent. Answer The elements of QP represent the total cost of the three types of work in package ……………………………………………………………………………………………….. A, B and C respectively. …………………………………………………………………………………………….[1] (ii) A condominium’s manager, Amy, decided to engage Dream Renovation Company to do renovation works. She ordered 85 package A, 90 Package B and 85 Package C. A matrix Y when multiplied by QP, will give the total cost of this renovation. (a) State matrix Y. 85 90 85 Answer Y= ………………………. [1] (b) Using matrix multiplication, find the total cost of this renovation.   Total cost 5520 85 90 85 4290 5670 (1337250)         Therefore, cost is $1337250 Answer $ ………………………. [2]

8 Methodist Girls’ School Mathematics Paper 2 Sec 4 Preliminary Examination 2022 3 In the diagram, the points A, B, C, D lie on circle I. The points A, E, D, F lie on circle II. AEC is a straight line. DF and AF are tangents to circle I. CD and AB are extended to meet at point G. ∠ECD = 67°, ∠CBD = 23° and ∠AED = 134°. (a) Stating your reason clearly, find (i) ∠AFD, 180 134 46 (angles in the opp segment)AFD     Answer ∠AFD = ……………….. (ii) ∠CAD, 23 (angles in the same segment)CAD   Answer ∠CAD = ……………….. (iii) a line that has the same length as DF. AF (tangents from ext point) Answer line ………….......… [1] [1] [2] 67° 134° A B C D E F G Circle I Circle II 23°

9 Methodist Girls’ School Mathematics Paper 2 Sec 4 Preliminary Examination 2022 (b) Explain why the centre of circle I lie on the circumference of circle II. 134 2 67 = 2 AED ACD        Hence, since angle at centre is twice angle at circumference, E is the centre of circle I. Since E lies on circle II, the centre of circle I lies on circle II. [3] (c) Show that triangle BCG is similar to triangle DAG. State your working and reasons clearly. Answer 90 (angles in a semicircle) 180 90 90 (adj angles on a str line) 90 (angles in a semicircle) (commo