P45864A 2016 Pearson Education Ltd. 1/1/1/1/ Write your name here Surname Pearson Edexcel International GCSE Mathematics A Paper 3HR Thursday 26 May 2016 Morning Time: 2 hours Centre Number Other names *P45864A0120* Candidate Number Higher Tier Paper Reference You must have: Ruler graduated in centimetres and millimetres, protractor, compasses, pen, HB pencil, eraser, calculator. Tracing paper may be used. Instructions Use black ink or ball-point pen. Fill in the boxes at the top of this page with your name, centre number and candidate number. Answer all questions. Without sufficient working, correct answers may be awarded no marks. Answer the questions in the spaces provided there may be more space than you need. Calculators may be used. You must NOT write anything on the formulae page. Anything you write on the formulae page will gain NO credit. Information The total mark for this paper is 100. The marks for each question are shown in brackets use this as a guide as to how much time to spend on each question. Advice Read each question carefully before you start to answer it. Check your answers if you have time at the end. 4MA0/3HR Total Marks Turn over
Pythagoras Volume of cone = Theorem 2 c a a 2 + b 2 = c 2 hyp cross section adj r r b opp length International GCSE MATHEMATICS FORMULAE SHEET HIGHER TIER Curved surface area of cone = adj = hyp opp = hyp opp = adj or sin cos tan cos sin tan opp hyp adj hyp opp adj 1 3 r 2 h Volume of prism = area of cross section h Circumference of circle = 2 Area of circle = r 2 Volume of cylinder = r 2 h Curved surface area of cylinder = 2 rh l r r a Sine rule: sin A *P45864A0220* A b 2 Cosine rule: a b 2 2 + c 2bc cos A length Area of triangle c C b sin B Area of a trapezium = a h b a B c sin C ab sin C 1 (a+b)h 2 The Quadratic Equation 2 The solutions of ax + bx + c 0, where a 0, are given by x Volume of sphere = b+ 2 b 4ac 2a 1 2 r 4 3 3 rl Surface area of sphere = 4 h In any triangle ABC r r 2
Answer ALL TWENTY TWO questions. Write your answers in the spaces provided. You must write down all the stages in your working. 1 Rafael and Roger played tennis against each other 30 times. Each of the times they played, either Rafael won or Roger won. The ratio of the number of times Rafael won to the number of times Roger won is 7 : 3 (a) Work out the number of times Rafael won. Total = 3+7 Rafael wins 7 out of every 10 plays 7/10 is probability he wins 30 trials 30 x 7/10 = 21 In a school, there are 75 girls in the tennis squad. The ratio of the number of boys in the tennis squad to the number of girls in the tennis squad is 4 : 3 (b) Work out the number of boys in the tennis squad. 75 girls correspondes to the ratio value of 3 25 students per unit 25 x 4 = 100 21 100 (Total for Question 1 is 4 marks) *P45864A0320* 3 Turn over
2 (a) Factorise fully 2x 2 4x A = 2p + 3q (b) Work out the value of p when A = 32 and q = 7 32 = 2p+ 3(7) 32-21 = 2p P = 5.5 3 There are 50 marbles in a bag. 35 of the marbles are brown. Otti takes at random a marble from the bag. He records the colour of the marble and puts the marble back in the bag. He does this 300 times. Work out an estimate for the number of brown marbles he takes. Probability he takes a brown marble = 35/50 300 trials 35/50 x 300 = 210 2x( x -2) 5.5 p = (Total for Question 2 is 5 marks) 210 (Total for Question 3 is 2 marks) 4 *P45864A0420*
4 Work out the size of an exterior angle of a regular polygon with 8 sides. 5 In a sale, normal prices are reduced by 8% (a) The normal price of a jacket is 28 Work out the price of the jacket in the sale. (b) In the sale, the price of a shirt decreases by 3 Work out the normal price of the shirt. Sum of exterior angles = 360 8(x) = 360 360/8 = x = 45 Reduction by 8% = multiple of.92 28 x 0.92 = 25.76 8% of the initial cost = 3 3 / 0.08 = original cost = 37.5 *P45864A0520* 45 (Total for Question 4 is 2 marks) 25.76 37.5 (Total for Question 5 is 6 marks) 5 Turn over
6 (a) Solve the inequalities 4 < 3x + 5 11 (b) Write down the integer values of x which satisfy 4 < 3x + 5 11 6-3< x < 2 Not -3, 2 is included -2,-1,0,1,2-9 < 3x < 6-3< x < 2 7 Write 792 as a product of its prime factors. Show your working clearly. Divide by prime numbers until left with a prime. 2x 396 2x2x198 2x2x2x99 2x2x2x3x33 2x2x2x3x3x11 *P45864A0620* -3< x < 2-2,-1,0,1,2 (Total for Question 6 is 5 marks) 2x2x2x3x3x11... (Total for Question 7 is 3 marks)
8 (a) Describe fully the single transformation that maps shape P onto shape Q. Translation by 5...... (b) Rotate shape Q 90 clockwise about (1,0) Label the new shape R. P y 5 4 3 2 1 4 3 2 1 O 1 2 3 4 5 1 2 3 4 5 *P45864A0720* Q (Total for Question 8 is 4 marks) x 7 Turn over
9 Li throws a 6-sided biased dice once. The table shows the probability that the dice will land on 1, 2, 3, 5 or 6 8 Number 1 2 3 4 5 6 Probability 0.15 0.1 0.05 0.2 0.15 (a) Work out the probability that the dice will land on 4 Probalities must sum to 1 1-(0.15 + 0.1 + 0.05 + 0.2 + 0.15) = x = 0.35 (b) Work out the probability that the dice will land on an odd number. Can land on 1,3,5 0.15 + 0.05 + 0.2 =0.4 *P45864A0820* 0.35 0.4 (Total for Question 9 is 4 marks)
10 Julie asked 50 children how many exercise sessions they each took part in last month. The table shows information about her results. Number of exercise sessions Frequency 0 to 6 13 7 to 13 10 14 to 20 16 21 to 27 7 28 to 34 4 Calculate an estimate for the total number of exercise sessions the children took part in last month. Sum of Midpoint x class frequency 3x13 + 10x10 + 17x16 + 24x 7 + 31x4 = 703 *P45864A0920* (Total for Question 10 is 3 marks) 11 The line L passes through the point (3,1) and is parallel to the line with equation y = 7 2 2x. Find an equation for the line L. Gradient is -2 General equation of a line: Y -y1 = m(x-x1) Y -1 = -2(X -3) Y = -2x + 7 703 Y = -2x + 7 (Total for Question 11 is 3 marks) 9 Turn over
12 (a) Simplify fully 11 a a a 2 5 (b) Make p the subject of p + 4q = 3p + 5 (c) Expand and simplify (2y + 3)(4y 1) 1 (d) Simplify (8a 6 b 3 3 ) p -3p = 5-4q -2p = 5-4q 2p = 4q -5 p = 2q -5/2... (Total for Question 12 is 8 marks) 10 *P45864A01020*
13 Here is the quadrilateral ABCD. 3.6cm A D 9.8cm Angle BAD = 90 and angle BCD = 90 AB = 9.8cm AD = 3.6cm BC = 8.4cm Calculate the length of DC. Using Pythagoras : DB = 9.8 + 3.6 = 109 DC = DB -BC DC = 109-8.4 DC = 6.2 C B 8.4cm Diagram NOT accurately drawn 6.2 cm (Total for Question 13 is 4 marks) *P45864A01120* 11 Turn over
14 Linford and Alan race against each other in a competition. If one of them wins a race, he wins the competition. If the race is a draw, they run another race. They run a maximum of three races. Each time they race, the probability that Linford wins is 0.35 Each time they race, the probability that there is a draw is 0.05 (a) Complete the probability tree diagram. 12 0.35 0.05 0.6... race 1 race 2 race 3 Linford wins draw Alan wins......... Linford wins draw Alan wins (b) Calculate the probability that Linford wins the competition. Possible combinations : linford wins Draw, linford wins Draw, draw, linford wins 0.35 + 0.05 x 0.35 + 0.05 x 0.05 x 0.35 = 0.368375 0.35 0.35 *P45864A01220*... 0.05 0.05... 0.6 0.6... Linford wins draw Alan wins 0.368375... (Total for Question 14 is 5 marks)
15 y = x 3 9 2 x2 54x + 10 (a) Find dy dx The curve with equation y = x 3 9 2 x2 54x + 10 has two turning points. (b) Find the x coordinate of each of these two points. Turning point when differential = zero Factorise:... (Total for Question 15 is 5 marks) *P45864A01320* 13 Turn over
16 The incomplete histogram shows information about the heights of a group of children. 14 Frequency density 110 120 130 140 150 Height of children (cm) There were 10 children with heights between 130cm and 135cm. (a) How many children had heights between 110cm and 130cm? 1 small square = 1/2 20(0.5) + 20(0.5) + 15(0.5) = 35 There were 6 children with heights between 135cm and 145cm. (b) Show this information on the histogram. 12 squares *P45864A01420* 35 (1) (Total for Question 16 is 4 marks)
17 D Diagram NOT accurately drawn 9cm C 4.5cm Triangle ABE is similar to triangle ACD. AED and ABC are straight lines. EB and DC are parallel. AE = 5cm, BC = 4.5cm, BE = 4cm, CD = 9cm (a) Calculate the length of AD. Scale factor = 9/4 9/4 x 5 = 11.25 (b) Calculate the length of AB. 4/9(4.5 + AB ) = AB 18+4AB = 9AB 5AB = 18 AB = 3.6 The area of quadrilateral BCDE is xcm 2 The area of triangle ABE is ycm 2 (c) Find an expression for y in terms of x. Give your answer as simply as possible. 4cm B E 5cm Area scale factor = length scale factor squared =81/16 BCDE + ABE = Asf x ABE X + Y = 81/16 Y 16X = 65Y A 11.25 cm 3.6 cm 16X/65 y = (Total for Question 17 is 7 marks) *P45864A01520* 15 Turn over
18 f is the function such that (a) Find f(0.5) (b) Find ff( 1) x f(x) = 3x + 1 (c) Find the value of x that cannot be included in any domain of f (d) Express the inverse function f 1 in the form f 1 (x) =... Show clear algebraic working. Make x the subject Swap x&y (1) (1) f 1 (x) = (Total for Question 18 is 7 marks) 16 *P45864A01620*
19 A, B and C are points on a circle, centre O. PA and PC are tangents to the circle. Angle ABC = 100 Calculate the size of angle APC. O A 100 Reflex angle AOC = 200 as angle at the center is twice angle at the circumference Therefore obtuse angle = 180 Where radius meets a tangent the angle is 90 360 = 160 + 90 + 90 + APC Therefore APC = 20 C B 20 P Diagram NOT accurately drawn (Total for Question 19 is 3 marks) *P45864A01720* 17 Turn over
2 50x 8 20 (a) Simplify fully 10x 4 Show clear algebraic working. Difference of two squares (b) Given that a is a positive integer, show that is always a multiple of 3 Cancel common factors ( ) 3a 12a + a 3a Is just an integer, so it is equivalent too 3(n) Therefore must be a multiple of 3 (Total for Question 20 is 6 marks) 18 *P45864A01820*
21 Solve 3 4 2k+8 = 24 Show your working clearly. k = (Total for Question 21 is 4 marks) *P45864A01920* 19 Turn over
22 P Q The diagram shows a circle, centre C. PR is a chord of the circle. The area of the shaded region is 100 cm 2 Angle PCR = 30 Calculate the length of the arc PQR. Give your answer correct to 3 significant figures. C 30 R Diagram NOT accurately drawn Area of triangle = 0.5(r)(r)sin(c) Area of whole sector = πr (θ/360) Shaded area = sector - triangle = πr (θ/360) - 0.5(r)(r)sin(c) = 100 r (π 30/360-0.5sin30) = 100 r = 100 / (π -3) Arc length = θ/360 x 2π r 30/360 x 2 π ( 100 / (π -3) ) = 48.2 48.2 cm (Total for Question 22 is 6 marks) TOTAL FOR PAPER IS 100 MARKS 20 *P45864A02020*