E Math Paper 1 (Solution)

00062720080 1 4052/01/O/N/26 [Turn over SECONDARY 4 PRELIMARY EXAMINATION in preparation for the General Certificate of Education Ordinary Level CANDIDATE NAME SOLUTIONS CENTRE NUMBER S . INDEX NUMBER MATHEMATICS 4052/01 Paper 1 October/November 2026 2 hours 15 minutes Candidates answer on the Question Paper. READ THESE INSTRUCTIONS FIRST Write your Centre number, index number and name 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. DO NOT WRITE ON ANY BARCODES. 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. Gives answers in degrees to one decimal place. For π, use either your calculator value or 3.142. This document consists of 21 printed pages and 1 blank page. 0006270

00062720080 2 4052/01/O/N/26 Mathematical Formulae Compound interest Total amount = P(1+ r 100) n Mensuration Curved surface area of a cone = πrl Surface area of a sphere = 4πr2 V olume of a cone = 1 3 πr2h V olume of a sphere = 4 3 πr3 Area of a triangle ABC = 1 2 ab sin C Arc length = rθ, where θ is in radians Sector area = 1 2 r2θ, where θ is in radians Trigonometry a sin A = b sin B = c sin C a2 = b2 + c2 – 2bc cos A Statistics Mean = Σ fx Σ f Standard deviation = √Σ fx 2 Σ f − (Σ fx Σ f ) 2

00062720080 3 4052/01/O/N/26 [Turn over Answer all questions. 1 sin x° = 0.7954 Give two possible values of x in the range 0 ⩽ x ⩽ 180. Answer x = ………….. or …………… [2] __________________________________________________________________________________ 2 The resistance of a wire, R, in ohms is inversely proportional to the square of the current, I, in ampere. The current of a wire is increased by 60%. Calculate the percentage reduction in the resistance of a wire. Answer ..…………………………… % [2] __________________________________________________________________________________ x° = sin−1 (0.7954) = 52.7° or 180° − 52.7° = 127.3° 52.7 127.3 R = k I2 When I = 1, R = k. When I = 1.6, R = k 2.56. Percentage reduction = [(k − k 2.56) × 1 k] ×100% = [(2.56k − k 2.56 ) × 1 k] ×100% = (1.56k 2.56 × 1 k) ×100% = 1.56 2.56 ×100% = 60.9% (3 s.f.) 60.9

00062720080 4 4052/01/O/N/26 3 The line graph shows the annual mean temperature over the six years. (a) (i) Tabita says that there is one outliner in the graph. Identify the outliner. ……………………………………….…………………………………………….. [1] (ii) Explain how this outliner affects the rest of the data in the graph. ……………………………………….…………………………………………….. ……………………………………….…………………………………………….. [1] (b) (i) State one misleading feature of the graph. ……………………………………….…………………………………………….. ……………………………………….…………………………………………….. [1] (ii) Explain how this feature affects the reader’s interpretation of the graph. ……………………………………….…………………………………………….. ……………………………………….…………………………………………….. [1] __________________________________________________________________________________ 25.0 2019 2020 2021 2022 2023 2024 Year Temperature (oC) Annual mean temperature 25.5 26.0 26.5 27.0 27.5 28.0 28.5 29.0 2023 The mean temperature in 2023 is much lower than the other years. This pulls the mean down, making it seems like the temperature is cooler overall than it actually was in most years. The vertical axis, representing temperature, does not start from 0 oC. The reader may think that the difference in temperature from 2023 to 2024 increases by about 3 oC but in the graph, it looks like there is a huge increase.

00062720080 5 4052/01/O/N/26 [Turn over 4 A = (a b c) B = ( u v w x y z ) C = BA Explain why matrix C is not possible. ……………….……………………………………………………….……………….…… ……………….……………………………………………………….……………….…… ……………….……………………………………………………….……………….…… ……………….……………………………………………………….……………….…… [2] __________________________________________________________________________________ 5 Jamie invested $3700 at a simple interest rate of 1.4% per year for 3 years. Work out the total value of his investment at the end of 3 years. Answer $ .……………………………. [2] __________________________________________________________________________________ The order of matrix A is 1 by 3 and the order of matrix B is 3 by 2. Since the number of columns of matrix B is not equal to the number of rows of matrix A, hence, matrix C is not possible. Simple interest, I = PRT 100 = (3700)(1.4)(3) 100 = $155.40 Total amount = $3700 + $155.40 = $3855.40 3855.40

00062720080 6 4052/01/O/N/26 6 The diagram shows the speed-time graph for part of a car journey. (a) Calculate the acceleration of the car in the first 10 seconds of the journey. Answer .………………………. m/s2 [1] (b) The car decelerates at 2.5 m/s2 after T seconds. Calculate the total distance travelled by car. Answer ………………………… m [2] __________________________________________________________________________________ Speed (m/s) Time (seconds) 10 20 T 0 32 Acceleration = 20 − 0 10 − 0 = 1 2 = 0.5 m/s2 0.5 0 − 20 32 − T = −2.5 T = 24 seconds Area under graph = area of trapezium 1 2(14 + 32)(20) = 460 m 460

00062720080 7 4052/01/O/N/26 [Turn over 7 Solve the equation x 5 − 2x − 3 4 = −3. Answer x = …………………………. [3] __________________________________________________________________________________ 8 Written as a product of its prime factors, 72 = 23 × 32. (a) Write 540 as a product of its prime factors. Answer ………………………………. [1] (b) Find the lowest common multiple (LCM) of 72 and 540. Answer ………………………………. [1] (a) The number 540p q is a perfect square, where p and q are prime numbers and p > q. Find the values of p and q. Answer p = ………………….……………. q = ………………….……………. [2] __________________________________________________________________________________ x 5 − 2x − 3 4 = −3 4x − 5(2x − 3) = −60 4x − 10x + 15 = −60 6x = 75 x = 12.5 12.5 540 = 22 × 33 × 5 72 = 23 × 32 540 = 22 × 33 × 5 LCM = 23 × 33 × 5 = 1080 ___________________ 1080 √540p q = √22×33×5×p q = √22×33×5×5 3 = √22× 32×52 = 2 × 3 × 5 = 30 5 3

00062720080 8 4052/01/O/N/26 9 ξ = {integers x : 1 ⩽ x ⩽ 15} The Venn diagram shows the elements of ξ and three sets P, Q and R. Use one of the symbols below to complete each statements. ∅ ⊂ ⊆ ∉ ∈ ξ (a) {1, 10} ………… P [1] (b) 7 ………… Q [1] (c) P ∩ R = ………… [1] __________________________________________________________________________________ 10 A salesman earns a 3% commission on the sales he makes. He earned $35 after selling a laptop that was on a 25% discount. Find the original selling price of the laptop. Answer $ .……………………………. [2] __________________________________________________________________________________ ξ P Q R 1 2 3 4 5 6 7 8 9 11 10 12 13 14 15 ⊂ ∈ 12 discounted selling price = $35 ÷ 3% =