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              