2026 CCH MAIN P1 MS

2026 Preliminary Exam/CCHMS/Secondary 4/Mathematics/4052/01 [Turn over Name: Class: Class Register Number: PRELIMINARY EXAMINATION 2026 SECONDARY 4 MATHEMATICS 4052/01 Paper 1 Thursday 20 August 2026 Candidates answer on the Question Paper. 2 hours 15 minutes READ THESE INSTRUCTIONS FIRST Write your name, class and index number on all the work you hand in. 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. The number of marks is given in brackets [ ] at the end of each question or part question. 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 total of the marks for this paper is 90. 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. For , use either your calculator value or 3.142. For Examiner’s Use Total / 90 This document consists of 21 printed pages and 1 blank page. Marking Scheme

2 2026 Preliminary Exam/CCHMS/Secondary 4/Mathematics/4052/01 Mathematical Formulae Compound interest Total amount = Mensuration Curved surface area of a cone = Surface area of a sphere = Volume of a cone = Volume of a sphere = Area of triangle ABC = Arc length = , where is in radians Sector area = , where is in radians Trigonometry Statistics Mean = Standard deviation = nrP       1001 lr  2 4 r hr 3 1 2 3 3 4 r Cba sin 2 1 r  2 2 1 r  C c B b A a sin sin sin  Acbcba cos 2 222    f xf 22           f xf f xf

3 2026 Preliminary Exam/CCHMS/Secondary 4/Mathematics/4052/01 [Turn over Answer all the questions. 1 (a) Factorise 24 2 8p q pq p   .            2 24 2 8 4 8 2 4 2 2 B1 4 1 2 B1 p q pq p p pq p q p p q p q p p q               Alternative            2 24 2 8 4 2 8 4 1 2 1 4 B1 4 1 2 B1 p q pq p p p q pq p p q p p p q               Answer ……………………… [2] (b) Simplify   22 3 3 2 xyx y z z  .       322 3 2 3 2 3 1 2 2 2 2 3 1 2 2 3 2 2 2 M1,M1 Any 2 Laws of Indices A1 xyx y x y z x yz z z x y y yz             Answer ……………………… [3] SPACE SPACE   4 1 2p p q  yz

4 2026 Preliminary Exam/CCHMS/Secondary 4/Mathematics/4052/01 2 (a) Solve the inequalities 7 4 2 3 8.x x x         7 4 2 and 4 2 3 8 9 3 10 B1 Any corr ect inequality 3 3 10 A1 x x x x x x x x              Answer ……………………… [2] (b) Illustrate your answer to part (a) in the number line below. [1] 3 Solve the equation 2 1 4 363 x x   .     2 2 1 1 363 4 1 1 36 M1 Change 3 into 99 4 1 36 M1 Comparing power with same base36 1 A1 x x x x x x x x            Alternative     2 2 2 2 2 2 2 1 1 36 3 4 1 1 6 M1 Change 4 into 23 2 1 1 M1 Comparing power with same base6 6 2 2 1 A1 x x x x x x x x x              Answer x ………………… [3] SPACE x 0 1 2 3 4 5 6 7 8 9 10 3 10x  1

5 2026 Preliminary Exam/CCHMS/Secondary 4/Mathematics/4052/01 [Turn over 4 Expand   4 3 1x x  .        2 2 4 3 1 4 4 3 3 4 3 B2 All 3 terms correct B1 Any one wrong term x x x x x x x            Answer …… ………………… [2] SPACE 5 Light takes approximately 8 minutes to travel 150 million kilometres. (a) Find the average speed of light, in metres per second. Give your answer in standard form. Answer ………………… m/s [2] (b) The star Proxima Centauri is 4×1013 km from Earth. Find the time, in years, it takes for light to travel from Proxima Centauri to Earth. [Assume 1 year = 365 days]                  13 3 8 4 10 10 Time 3.125 10 60 60 24 365 M1 Time formula with 60 60 24 365 4.06 years 3 sig. fig. A1 m         Answer ………… …… .. years [2] SPACE     6 3 8 150 10 10Speed M1 Speed formula, correct 150 and 8 608 60 3.125 10 m/s A1 m s       150 million kilometres Converting minutes into seconds Convert into metres 24 3x x  83.125 10 4.06     13 3 8 8 Distance (m) secondsSpeed (m/s) 4 10 10= 3.125 10 / 1.28 10 seconds 60 2133 333.333 minutes 35 555.55 hours 1481.481days 4.058...years 4.06 years 3 Time m m s sf            

6 2026 Preliminary Exam/CCHMS/Secondary 4/Mathematics/4052/01 6 (a) Sketch the graph of   2 2 4y x   on the axes below. Indicate clearly the turning point and the points where the graph crosses the axes. [2] (b) On the same graph in (a), draw and label the line of symmetry of the graph [1] 7 In a regular n-sided polygon, the interior angle is 3 times that of the exterior angle. Find the value of n. Answer n = ………………… [2] y x O Part (a) G1 – Correct shape passing through O G1 – Correct x-, y- intercepts and maximum point (can use dotted lines from x– and y–axes. Part (b) B1 – Correct line of symmetry with label (equation) (2, 4) x = 2 (Must label) | 4  The line of symmetry is required as a solution. Therefore, the line should be ‘solid’ Method 1     Let exterior angle be 3 180 M1 4 180 45 360 45 8 A1 x x x x x n           3x x Method 2     2 2 2 360180 3 [M1] 2 2 3 2 6 2 6 8 0 8 0 8 or 0 N.A. [A1] n n n n n n n n n n n n n n n n n n                          8

7 2026 Preliminary Exam/CCHMS/Secondary 4/Mathematics/4052/01 [Turn over SPACE SP 8 Solve the following equations. (a) 4 3 3 4 x x    Answer x = ……… or ……… [2] (b) 3 1 37 2 x x   Answer x = ……………… …. [3]         3 1 37 2 2 3 7 1 3 B1 Common denominator (equivalent fractions)14 2 6 7 7 42 M1 Simplifying 9 1 42 9 43 7 434 A1 Accept , not 3 s.f.9 9 x x x x x x x x x                              2 2 4 3 3 4 4 4 9 B1 Multiplying both side s by 3 4 16 9 25 5 or 5 B1 x x x x x x x x                  5 5 74 9

8 2026 Preliminary Exam/CCHMS/Secondary 4/Mathematics/4052/01 9 Solve these simultaneous equations. 2 5 2 0 x y x y     Answer x = …………………. y = …………………. [3] SP Method 1                   2 5 1 2 0 2 1 2 : 2 4 10 3 2 3 : 5 10 M1 Attempting to make coeff icient of or equal 2 A1 1 A1 x y x y x y y x y y x                Method 2                 5 2 1 2 0 2 Subst 1 into 2 : 2 5 2 0 M1 Attempting to make o r the subject 10 4 2 0 5 10 2 A1