[Presbyterian High] [2023] 4E EM 4052 Prelims P1 MS

4052/01/4E Preliminary Examination/2023 PRESBYTERIAN HIGH SCHOOL MATHEMATICS 4052/01 PAPER 1 14 August 2023 Monday 2 hours 15 minutes MARK SCHEME

2 Presbyterian High School 4052/01/4E Prelim/2023 Answer all the questions. 1 Solve 7 18 3 .x x  7 18 3 4 18 9 --- B12 x x x x     2 (a) Calculate 3 2 26.18 4.52 0.4 . Write your answer correct to 5 significant figures. 8593.4 --- B1 (b) Write your answer to part (a) in standard form. 38.5934 10 --- B1 3 (a) Express 784 as a product of prime factors. 2 784 2 392 2 196 2 98 7 49 7 7 1 (b) It is given that a and b are prime numbers. Find the smallest values of a and b such that 784 a b is a perfect cube.   4 2 7784 2 7 7 and 2 --- B1, B12 a a bb       4 Expand and simplify   5 1w w  .    2 2 5 1 5 5 --- M1: at least 2 correctly expand ed terms 4 5 --- A1 w w w w w w w          4 2784 2 7 --- B1 

3 Presbyterian High School 4052/01/4E Prelim/2023 5 The bar graph below shows the results of a survey conducted on the service quality of a hotel. Service quality of a hotel (a) Find the percentage of respondents who answered ‘Strongly Satisfied’ and ‘Satisfied’. 18 27 100%18 27 20 10 5 56.25% --- B1       (b) Suggest the use of another statistical diagram to represent the results of the survey conducted, that can show the relative size of a part in relation to the whole. Pie Chart --- B1 6 Find the largest integer that satisfies 2 3 4.y  2 3 4 2 7 7 B1: seen this answer2 y y y     The largest integer is 3. ---- B1 Number of respondents 0 10 20 30 Satisfied Strongly Satisfied Somewhat Satisfied Unsatisfied Strongly Unsatisfied Responses

4 Presbyterian High School 4052/01/4E Prelim/2023 7 P is directly proportional to 3.Q When 2, 64Q P  . When the value of Q is halved, the value of P changes by a factor of m. Find the value of m.   3 3 3 3 3 3 When 2 and 64 64 2 ---- M1: attempt to fin d the proportionality constant by substitution 8 8 18 82 8new P k Q Q P k k P Q QP Q Q                    Hence the factor m is 1 .8 --- A1 8 The diagram shows a quadrilateral playground ABCD. A circular fence is constructed around the playground such that the vertices, A, B, C and D of the playground touch the circumference of the fence. (a) Construct the perpendicular bisector of AB. [1] (b) Construct the bisector of angle ADC. [1] (c) A sand pit is to be constructed inside the circular fence but outside the quadrilateral playground. The sand pit is nearer to AD than CD and nearer to B than A. Shade the region for the sand pit to be constructed. [1] A B C D (a) B1 (b) B1 (c) B1

5 Presbyterian High School 4052/01/4E Prelim/2023 9 The diagram below shows the graph of   2 3 4y x h   . (a) Find the value of h.     2 Substitute 0, 8 : 8 3 0 4 --- M1: shows subsitution 2 or 2 (reject) 2 --- A1 h h h        (b) Explain why the graph of   23 1y x h   does not cut the x-axis. Either one 1. The minimum point of the graph   23 1y x h   is  , 1h or  2, 1 . 2. The equation   2 1 has no solution for .3x h x  [1] 10 A group of six students took a Mathematics quiz and the marks were recorded below. 8 10 9 13 10 9 (a) Calculate the standard deviation. 1.57 (b) Two other students also took the quiz, and their marks were recorded. Given that the mean mark obtained by the eight students was 10 and the mode was also 10, find the marks of these two students. The two marks are 10 and 11. --- B1, B1  0, 8 O x y 

6 Presbyterian High School 4052/01/4E Prelim/2023 11 The distance-time graph shows the journey Tan took to run from town A to B. (a) Find the distance Tan ran in the first two hours. 8 km --- B1 (b) Calculate the average speed, in m/s, for the whole journey Tan ran. Average speed 16000 --- M1: attempt to convert km to m or h to sec5 60 60   8 m/sec or 0.889 m/sec --- A19 Time 10 am 11 am 12 noon 1 pm 2 pm 3 pm 4 8 12 16 0 B A Distance from A (km)

7 Presbyterian High School 4052/01/4E Prelim/2023 12 Simplify 2 2 2 3 4 9 y y y    .       2 2 2 3 4 9 2 3 1 --- M1, M1: factorise numerator and denominator 2 3 2 3 1 --- A12 3 y y y y y y y y y          13 In the diagram, ABC is a straight line and triangles ABE and BCD are equilateral triangles. Show that triangle ABD and triangle EBC are congruent. Give a reason for each statement you make. Answer 1) (sides of an equilateral triangle / given) 2) (sides of an equilateral triangle / given) --- B1: at least one statement with reason provided 3) 180 60 (adj. on a st. line) AB EB BD BC ABD       = --- B1: show equivalent angles with explanation 120 triangle is congruent to triangle (S AS) --- B1: with name EBC ABD EBC     of test [3] A B C D E

8 Presbyterian High School 4052/01/4E Prelim/2023 14 The first three terms in a sequence of numbers, 1 2 3, , , ... T T T are given below. 1 2 3 11 2 1 1 2 3 1 1 3 4 T T T       . . . (a) Write down 4.T 4 1 1 --- B14 5T   (b) Show that the total sum of 1 2 3, , , ... , nT T T T in the above sequence is 11 . 1n  1 1 1 1 11 ...2 2 3 1 n n                       --- M1: seen either the formation or 1 1 1n n      11 --- [AG1: shown]1n   15 A, B and C are points 1, 0 , 3, 8 and 2, 1 respectively. (a) Find the length of AB. Length of AB     2 2 1 3 0 8     --- M1: correct application of length formula 8.94 units (3s.f.) --- B1 (b) Find the equation of the line that passes through B and has the same gradient as AC. 1 0 1 --- M12 ( 1) 3 1Equation of line passing through B has the gradient 3 mAC    same The equation of the line:  1 1 or 8 3 (No marks without simplificat ion)3 3y x c y x     1 7 --- A13y x 

9 Presbyterian High School 4052/01/4E Prelim/2023 16 (a) Find the interior angle of a regular 18-sided polygon.  18 2 180 --- M118 160 --- A1      (b) An n-sided polygon has two of its exterior angles at 45 and 75 .  If the remaining exterior angles are each 20, calculate the value of n.  45 75 2 (20) 360 --- M1 14 --- A1 n n      17 (a) Simplify 1 6 3 9 a b       and leave your answer in positive index notation. 2 1 6 3 9 2 3 3 --- M1: applied indices law with at most one error 1= --- A1 a b a b a b          (b) Given that 4 32 2 2x x  , find x. 4 3 1 4 3 1 3 3 2 2 2 2 2 2 --- M1: able to convert to appropri ate index form 2 2 13