AISS 2021 Sec 3 Express EOY Paper 1 Marking Scheme

_____________________________________________________________________________________ This document consists of 20 printed pages. AISS EOY / 3E / P1 / 2021 [Turn over AHMAD IBRAHIM SECONDARY SCHOOL END-OF-YEAR EXAMINATION 2021 SECONDARY 3 EXPRESS Name: Class: Register No.: MARKING SCHEME MATHEMATICS Paper 1 Candidates answer on the Question Paper. 4048/01 29 September 2021 2 hours 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 questions. Give non-exact numerical answers to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. The use of an approved scientific calculator is expected, where appropriate. You are reminded of the need for clear presentation in your answers. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. The total number of marks for this paper is 80. For Examiner’s Use /80

2 AISS EOY/3E/P1/2021 Mathematical Formulae Compound Interest Total amount = n rP     1001 Mensuration Curved surface area of a cone = πrl Surface area of a sphere = 4πr2 Volume of a cone = 3 1 πr2h Volume of a sphere = 3 4 πr3 Area of triangle ABC = 2 1 ab sin C Arc length = rθ, where θ is in radians Sector area = 2 1 r2θ, where θ is in radians Trigonometry C c B b A a sinsinsin  a2 = b2 + c2 − 2bc cos A Statistics Mean = f fx   Standard deviation = 22          f fx f fx

3 AISS EOY/3E/P1/2021 [Turn over Answer all the questions. 1 (a) Calculate   23 56.8 18.22 1.4 . Write down the first five digits on your calculator display. Answer …………………………….. [1] (b) Write your answer to part (a) correct to 3 significant figures. Answer …………………………….. [1] 2 A pond contains 65.62 10 drops of water and 92.91 10 micro-organisms. Let N be the average number of micro-organism per drop of water. Find the value of N. Give your answer in standard form. 9 6 2 2.91 10 5.62 10 517.79 5.18 10 N     Answer N =…………...……………..[2] 3 A 4-litre orange juice mixture (water and orange juice syrup) has 16% of orange juice syrup. How much water must be added to reduce the concentration of orange juice syrup to 8%? Amount of orange juice syrup =0.16 4 0.64 let the amount of water to be added be litres, 0.64 100% 8%4 0.08 0.32 0.64 4 4 of water must be added. l l x x x x l      37.459 B1 37.5 B1 M1 A1 M1 A1

4 AISS EOY / 3E / P1 / 2021 Alternative Method 8% represents 0.64 M1 1% represents 0.08 92% represents 7.36 Amount of water to be added = 7.36 – 3.36 = 4 A1 Answer ……………...…………….. l [2] 4 Without using a calculator, show that 2021 202077  is a multiple of 3. Answer 2021 2020 2020 2020 2020 77 7 (7 1) 6(7 ) 3(2 7 )     Since 3 is a factor of 202027 , 2021 202077  is a multiple of 3. [2] M1 A1

5 AISS EOY/3E/P1/2021 [Turn over 5 The graph shows the number of Mathematics Olympiad Prize wineners over a number of years. (a) State one misleading feature of the graph. ……………………………………………………………………………………… ……………………………………………………………………………………….. [1] (b) Explain how this feature affects the reader’s interpretation of the graph. [1] 6 Simplify 1 12 4 8 625 .x y     1 12 4 8 1 8 4 12 2 3 625 625 5 x y y x y x        Answer …………………………….. [2] 15 20 25 30 35 2013 2014 2015 2016 Great increase in the number of Mathematics Olympiad Prize Winners The title of the graph is biased. OR The vertical axis does not start with 0. B1 It does not allow the reader to make his own judgement. OR It exaggerates the difference in the number of winners over the years. B1 Year Number of prize winners M1 A1

6 AISS EOY / 3E / P1 / 2021 7 Simplify 22 2 82 12 6 xy x xy   22 2 22 82 12 6 2(4 ) 6 (2 ) 2(2 )(2 ) 6 (2 ) 2(2 ) 6 2 3 xy x xy xy x x y x y x y x x y xy x xy x         Answer …….……………………….. [3] 8 Factorise completely 2 4 2a a b a b   . 2 4 2 2 2 2 22 2 (1 ) (1 ) ( )(1 ) ( )(1 )(1 ) a a b a b a a b a a b a a b a a               Answer …….……………………….. [3] 9 A bank g uarantees an interest of $ x at the end of 2 yea rs with a compound interest of 4% per annum, compounded half yearly, for a principal amount of $5000. Calculate the amount of guaranteed interest earned at the end of 2 years. 4 4 2Total amount = $5000 (1+ )100 Interest 2= $5000 (1+ ) $5000100 $412.16   Answer $…….…………………… [3] A1 M1 M1 A1 M1 M1 M2 Factorise numerator and denominator A1

7 AISS EOY/3E/P1/2021 [Turn over 10 5 men are hired to paint a house. If an additional man is hired, the painting can be completed 4 days earlier. Calculate the number of additional men to be hired if the painting is to be completed 18 days earlier. 12 4 456 6 5 120 120 18 120 120 185 1206 20 original new kD M DD kk kk k DD M M M          Additional no. of men = 15 Answer …………………………….. [3] 11 It is given that 2 3 2 12 4 rp qr   . Express r in terms of p and q. 2 3 2 2 3 2 3 3 2 2 3 2 2 3 2 3 3 3 2 3 3 3 12 4 12 4 4 1 2 2 1 4 ( 2) 1 4 14 2 14 2 rp qr rp qr p q p r r p r r p q r p p q pqr p pqr p                  Answer r =………………………….. [3] M1 M1 A1 M1 M1 A1 Alternative Method: 5 men – x days 6 men – ( 4)x days 5 6( 4) 24 xx x   5 men – 24 days 20 men –6 days Additional no. of men = 15

8 AISS EOY / 3E / P1 / 2021 12 Solve the equation 2 4 1353 x x  . 2 2 2 2 2 4 1 353 2(3) 3(3)( 5) (4 1)( 5) 6 9 45 4 20 5 4 10 56 0 2 5 28 0 ( 5) ( 5) 4(2)( 28) 2(2) 5.19(3 ) 2.69(3 ) x x x x x x x x x xx xx x x sf or x sf                           Answer x = ………… or ………….. [3] 13 (a) The sketch shows the graph of .ny kx Write down a possible value of n. Answer n=………………………….. [1] (b) State the range of values of k. Answer ….………………………….. [1] y x M1 M1 A1, both answers B1 2 0k  B1 O

9 AISS EOY/3E/P1/2021 [Turn over (c) Sketch the graph of 3xy Answer [1] 14 Two geometrically similar figurines are made of the same material. The base areas of the smaller and larger figurines are 210 cm2 and 680.4 cm2 respectively. (a) Find the ratio of the height of the smaller figurine : the height of the larger figurine. height of smaller figurine 210 height of larger figurine 680.4 25 81 5 9 ratio= 5:9    Answer …………… : …………….. [1] B1, curve must pass through (0, 1 ) B1 y x 1 O

10 AISS EOY / 3E / P1 / 2021 (b) The mass of the smaller figurine is 750 g. Find the mass of the larger figurine. 3 3 750 5 mass of larger figurine 9 750mass of larger figurine 5 9 4374g     Answer ………………………… g [2] 15 (a) Solve the inequalities 32 4 1 1634 xx x     . 32 4 1 1634 32 413 3 2 12 3 19 1 9 4 1 16 4 16 4 64 17 68 4 1 49 xx x x x xx x x and xx xx x x x           