Cambridge IGCSE Mathematics 0580 — 2017 Feb/March Paper 4 · Variant 2

1 The Smith family paid $5635 for a holiday in India. The total cost was divided in the ratio travel : accommodation : entertainment = 10 : 17 : 8. (a) Calculate the percentage of the total cost spent on entertainment. ..........................................% [2] (b) Show that the amount spent on accommodation was $2737. [2] (c) The $5635 was the total amount Mr Smith received from an investment he made 5 years ago. Compound interest at a rate of 2.42% per year was paid on this investment. Calculate the amount he invested 5 years ago. $ ........................................... [3] (d) Mr Smith, his wife and their three children visit a theme park. The tickets cost 2500 Rupees for an adult and 1650 Rupees for a child. Calculate the total cost of the tickets. ................................. Rupees [2] (e) One day the youngest child spent 130 Rupees on sweets. On this day the exchange rate was 1 Rupee = $0.0152 . Calculate the value of the sweets in dollars, correct to the nearest cent. $ ........................................... [2]

Mark scheme: Question Answer Marks Part Marks 1 (a) 22.9 or 22.85 to 22.86 2 8 M1 for [× 100] oe 10 + 17 + 8 17 5635 (b) 5635 × or better [= 2737] 2 M1 for 10 + 17 + 8 (10 + 17 + 8)  2.42  5 (c) 5000 3 M2 for 5635 = k 1 + oe    100   2.42  or B1 for  1 +   100  (d) 9950 2 M1 for 2 × 2500 or 3 × 1650 (e) 1.98 final answer 2 B1 for 1.976 or 1.98 not final answer or M1 for 130 × 0.0152

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2 y 12 10 P 8 6 4 2 S Q x –10 –8 –6 –4 –2 0 2 4 6 8 10 –2 –4 R –6 –8 (a) Describe fully the single transformation that maps (i) shape P onto shape Q, ...................................................................................................................................................... ...................................................................................................................................................... [3] (ii) shape P onto shape R, ...................................................................................................................................................... ...................................................................................................................................................... [2] (iii) shape P onto shape S. ...................................................................................................................................................... ...................................................................................................................................................... [3] (b) (i) Draw the reflection of shape S in the line y = x . [2] (ii) Write down the matrix that represents the transformation in part (b)(i). [2] f p

Mark scheme: 2 (a) (i) Rotation 1 90° [anticlockwise] oe 1 (9, 5) 1 (ii) Translation 1  −8    oe 1  −14  (iii) Enlargement 1 1 [sf] 1 3 (− 8, − 2) 1 (b) (i) Image at (1, − 3) (2, − 3) (2, − 5) 2 M1 for triangle correct size and orientation, wrong position or SC1 for correct reflection in y = –x  0 1  (ii)   2 B1 for 1 correct column or row  1 0 

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3 The table shows some values for y = 1.5 x - 1. x –2 –1 0 1 2 3 4 5 y – 0.56 – 0.33 2.38 4.06 6.59 (a) Complete the table. [3] (b) Draw the graph of y = 1.5 x - 1 for - 2 G x G 5 . y 7 6 5 4 3 2 1 x –2 –1 0 1 2 3 4 5 –1 [4] (c) Use your graph to solve the equation 1.5 x - 1 = 3. 5 . x = ................................................ [2] (d) By drawing a suitable straight line, solve the equation 1.5 x - x - 2 = 0 . x = ............................ or x = ............................[3] (e) (i) On the grid, plot the point A at (5, 5). [1] (ii) Draw the tangent to the graph of y = 1.5 x - 1 that passes through the point A. [1] (iii) Work out the gradient of this tangent. .............................................. [2]

Mark scheme: 3 (a) 0 0.5 oe 1.25 oe 1, 1, 1 (b) Fully correct smooth curve 4 B3 FT for 7 or 8 points or B2 FT for 5 or 6 points or B1 FT for 3 or 4 points (c) 3.6 to 3.8 2 M1 for y = 3.5 soi (d) line y = x + 1 ruled M1 −1.55 to −1.40 4.55 to 4.8 A1 A1 If 0 scored SC1 for y = x + 1 stated or implied or for 2 correct values given (e) (i) Point plotted at (5, 5) 1 (ii) Tangent ruled from A 1 (iii) 1.2 to 1.4 B2 B2 and M1 dep on reasonable attempt at tangent from (5, 5) M1 for change in y/ change in x of their ruled line 1 1  1 1 1 

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4 Ravi spins a biased 5-sided spinner, numbered 1 to 5. The probability of each number is shown in the table. Number 1 2 3 4 5 1 1 1 Probability x x 6 4 3 (a) Find the value of x. x = ................................................ [3] (b) Ravi spins the spinner once. Find the probability that the number is 2 or 3. .............................................. [2] (c) Ravi spins the spinner twice. Find the probability that (i) the number is 2 both times, .............................................. [2] (ii) the sum of the numbers is 3. .............................................. [3] (d) Ravi spins the spinner 72 times. Calculate how many times he expects the number 1. .............................................. [1]

Mark scheme: 1 1  1 1 1  4 (a) oe 3 M2 for  1 − − −  oe 8 2  6 4 3  1 1 1 or M1 for + + seen oe or idea that 6 4 3 all sum to 1 7 1 1 (b) oe 2 M1 for + oe 12 3 4 1 1 1 (c) (i) oe 2 M1 for × oe 16 4 4 2 1 1 (ii) oe 3 M2 for 2 × × oe 24 6 4 1 1 or M1 for × oe 6 4 (d) 12 1

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5 (a) (i) Factorise 3x 2 + 11x - 4 . .............................................. [2] (ii) Solve the equation 3x 2 + 11x - 4 = 0 . x = ............................ or x = ............................[1] 2 1 1 2 (b) (i) Show that - = simplifies to 2x + 3x - 6 = 0 . 2x + 11 x - 4 2 [4] (ii) Solve the equation 2x 2 + 3x - 6 = 0 . You must show all your working and give your answers correct to 2 decimal places. x = ............................ or x = ............................[4]

Mark scheme: 5 (a) (i) ( 3 x − 1)( x + 4 ) 2 M1 for ( 3 x + b )( x + c ) with bc = − 4 or 3c + b = 11 or for 3x(x + 4) – 1(x + 4) or for x(3x – 1) + 4(3x – 1) 1 (ii) oe and −4 1 3 (b) (i) 2 × 2 ( x − 4 ) − 2 ( 2 x + 11) = ( 2 x + 11)( x − 4 ) M2 M1 for common denom or better 2( 2 x + 11)( x − 4 ) seen or attempt to multiply through by denoms 2( x − 4) − (2 x + 11)  1  or for =   (2 x + 11)( x − 4)  2  2 x 2 + 11x − 8 x − 44 or better B1 or for other correct relevant 2 bracket expansion if alt method used 4 x − 16 − 4 x − 22 = 2 x 2 − 8 x + 11x − 44 2 A1 correct solution reached with all brackets 2 x + 3 x − 6 = 0 expanded and no errors or omissions seen −±3 ( 3 ) 2 − 4 ( 2 )( − 6 ) 2 (ii) 2 B1 for ( 3 ) − 4 ( 2) ( −6 ) ) or better 2 × 2  3  2 or  x +  oe  4  −+3 q −−3 q and B1 for or or better 2(2) 2(2) 3 57 3 57 or − + oe or − − oe 4 16 4 16 −2.64 and 1.14 final ans cao B1B1 SC1 for −2.6 or –2.637... and 1.1 or 1.137... or −2.64 and 1.14 seen in working or 2.64 and −1.14 as final answers

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6 (a) G H x° Z NOT TO SCALE E 27° F In the diagram, EH is parallel to FG. The straight lines EG and FH intersect at Z. Angle ZFG = 27°. (i) Find the value of x. x = ................................................ [1] (ii) EH = 5 cm, FG = 9 cm and ZG = 7 cm. Calculate EZ. EZ = .......................................... cm [2] (b) The diagram shows points A, B, C and D on the circumference of a circle, centre O. AD is a straight line, AB = BC and angle OAB = 52°. D NOT TO SCALE O C A 52° B Find angle ADC. Angle ADC = ................................................ [3] (c) The diagram shows points P, Q, R and S on the circumference of a circle, centre O. VT is the tangent to the circle at Q. S 27° NOT TO P SCALE O R 63° V T Q Complete the statements. (i) Angle QPS = angle QRS = ................ ° because ........................................................................ ...................................................................................................................................................... [2] (ii) Angle SQP = ................ ° because .............................................................................................. ...................................................................................................................................................... [2] (iii) Part (c)(i) and part (c)(ii) show that the cyclic quadrilateral PQRS is a .................................................... [1]

Mark scheme: 6 (a) (i) 27 1 7 9 (ii) 3.89 or 3.888 to 3.889 2 M1 for = oe EZ 5 (b) 76 cao 3 B2 for ABC = 104 or AOC = 152 or COD = 28 or OBA = 52 and OBC = 52 or BCD = 128 and OCB = 52 or B1 for any one of OBA,OBC, OCB = 52 or BCD = 128

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7 The table shows information about the time taken by 400 people to complete a race. Time taken 45 1 m G 50 50 1 m G 60 60 1 m G 70 70 1 m G 90 90 1 m G 100 100 1 m G 120 (m minutes) Frequency 23 64 122 136 26 29 (a) Calculate an estimate of the mean time taken. ...................................... min [4] (b) (i) Complete the cumulative frequency table. Time taken m G 50 m G 60 m G 70 m G 90 m G 100 m G 120 (m minutes) Cumulative 23 400 frequency [2] (ii) On the grid, draw a cumulative frequency diagram to show this information. 400 300 Cumulative 200frequency 100 0 m 40 50 60 70 80 90 100 110 120 Time taken (minutes) [3] (iii) Use your diagram to estimate (a) the median, ...................................... min [1] (b) the inter-quartile range, ...................................... min [2] (c) the 60th percentile. ...................................... min [2]

Mark scheme: 7 (a) 72.7 or 72.70 to 72.71 nfww 4 M1 for midpoints soi (condone 1 error or omission) (47.5, 55, 65, 80, 95, 110) M1 for use of ∑fx with x in correct interval including both boundaries (condone 1 further error or omission) (1092.5, 3520, 7930, 10880, 2470, 3190) M1 (dep on 2nd M1) for ∑fx ÷ 400 (b) (i) [23] 87 209 345 371 [400] 2 B1 for 2 or 3 correct (ii) Correct graph 3 B1FT their (b)(i) for 6 correct heights B1 for 6 points at upper ends of intervals on correct vertical line B1FT (dep on at least B1) for increasing curve or polygon through 6 points After 0 scored, SC1FT their (b)(i) for 5 correct points plotted (iii) (a) 69 to 70 1 (b) 20 to 23 2FT FT their cumulative freq curve M1 for correct UQ or LQ for their cumulative freq curve (c) 72 to 75 2 M1 for 240 soi

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8 (a) In triangle TXZ, TX = 12.5 cm and angle TZX = 37°. Y is a point on the line XZ such that TY = 9.9 cm, angle XTY = 23° and angle TYZ = 72°. T 23° NOT TO SCALE 12.5 cm 9.9 cm 72° 37° X Y Z (i) Calculate XY. XY = .......................................... cm [4] (ii) Calculate TZ. TZ = .......................................... cm [3] (b) The diagram shows a shape made up of three identical sectors of a circle, each with sector angle 65°. The perimeter of the shape is 20.5 cm. 65° 65° NOT TO SCALE 65° Calculate the radius of the circle. ........................................ cm [4]

Mark scheme: 8 (a) (i) 5.14 or 5.135 to 5.142 nfww 4 M2 for [ XY 2 = ] 12.52 + 9.92 − 2 × 12.5 × 9.9 × cos 23 or M1 for implicit version A1 for 26.4 to 26.5 OR B1 for [XYT = ] 108 or [TXY = ] 49 12.5sin 23 M2 for oe sin(180 − 72) sin(180 − 72) sin23 or M1 for = oe 12.5 XY

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9 Bernie buys x packets of seeds and y plants for his garden. He wants to buy more packets of seeds than plants. The inequality x 2 y shows this information. He also wants to buy • less than 10 packets of seeds • at least 2 plants. (a) Write down two more inequalities in x or y to show this information. .............................................. .............................................. [2] (b) Each packet of seeds costs $1 and each plant costs $3. The maximum amount Bernie can spend is $21. Write down another inequality in x and y to show this information. .............................................. [1] (c) The line x = y is drawn on the grid. Draw three more lines to show your inequalities and shade the unwanted regions. y 10 9 8 7 6 5 4 3 2 1 x 0 2 4 6 8 10 12 14 16 18 20 22 [5] (d) Bernie buys 8 packets of seeds. (i) Find the maximum number of plants he can buy. ............................................. [1] (ii) Find the total cost of these packets of seeds and plants. $ ................................................. [1]

Mark scheme: 9 (a) x < 10 oe 1 Accept x - 9 y ⩾ 2 oe 1 Accept y > 1 (b) x + 3 y - 21 oe 1 Mark answer line isw (c) ruled broken line x =10 B1 or ruled line x = 9 ruled line y = 2 B1 or ruled broken line y = 1 ruled line from (0, 7) to (21, 0) B2 SC1 for line with negative gradient correct only at (0, 7) or (21, 0) correct region indicated cao 1 (d) (i) 4 1 (ii) 20 1

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10 (a) The diagram shows a regular F E hexagon ABCDEF of side 10 cm. NOT TO SCALE A D B C (i) Show that angle BAF = 120°. [2] (ii) The vertices of a rectangle PQRS F E touch the sides FA, AB, CD and DE. P S PS is parallel to FE and AP = x cm. NOT TO SCALE A D Q R B C Use trigonometry to find the length of PQ in terms of x. PQ = .......................................... cm [3] (iii) PF = (10 – x) cm. Show that PS = (20 − x) cm. [3] (b) F E K N NOT TO SCALE A D L M B C The diagram shows the vertices of a square KLMN touching the sides of the same hexagon ABCDEF, with KN parallel to FE. Use your results from part (a)(ii) and part (a)(iii) to find the length of a side of the square. ........................................ cm [4]

Mark scheme: 10 (a) (i) (6 − 2) × 180 or (2 × 6 – 4) × 90 M1 or (360 ÷ 6) (6 − 2) × 180 ÷ 6 or (2 × 6 – 4) × 90 ÷ 6 M1dep dep on previous M1 or 180 − (360 ÷ 6) (ii) 1.73x or x 3 oe 3 M2 for 2 x sin60 or 2 x cos30 oe or for x 2 + x 2 − 2 × x × x × cos120 or M1 for x sin60 or x cos30 oe or for x 2 + x 2 −×2 x × x × cos120

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11 On Monday, Ankuri sent this text message to two friends. Today is Day Number 1. Tomorrow, please add 1 to the Day Number and send this text message to two friends. All the friends who receive a text message follow the instructions. (a) Complete the table. Day Monday Tuesday Wednesday Thursday Friday Saturday Sunday Day Number 1 2 3 Number of text messages sent 2 4 today [4] (b) Write down an expression for the number of text messages sent on Day Number n. .............................................. [1] (c) Ankuri thinks that, by the end of Day Number 3, the total number of text messages that have been sent is 2 4 - 2 . (i) Show that she is correct. [2] (ii) Complete the statement. The total number of text messages sent by the end of Day Number 5 is ........................ which is equal to 2k – 2 where k = ........................ . [2] (iii) Write down an expression for the total number of text messages sent by the end of Day Number n. .............................................. [1] (iv) Find the Day Number when the total number of text messages sent by the end of the day is 1022. .............................................. [1]

Mark scheme: 11 (a) 4 5 6 7 1 8 16 32 64 128 3 B2 for 3 or 4 correct or B1 for first 2 correct If 0 scored, SC1 for 4 values correctly doubled FT one error (b) 2n oe 1 (c) (i) 2 + 4 + 8 = 14 1 16 − 2 = 14 1 or for 14 + 2 = 16 = 24 (ii) 62 2 B1 for each and 6 (iii) 2n + 1 − 2 oe 1 (iv) 9 1

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