Cambridge IGCSE Mathematics 0580 — 2013 Oct/Nov Paper 3 · Variant 3

0580/33/O/N/13 · 9 questions · 104 marks · ≈117 min

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Mark scheme5 pages

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Questions as text

Q1 · Adam owns a farm

1 Adam owns a farm. For Examiner′s Use (a) He plans to keep twenty hens. He works out what he thinks this will cost. Complete the following table. Item Cost ($) Equipment 500 20 hens costing $12 each 3 years supply of feed costing $25 per month TOTAL [3] (b) The equipment actually costs $600. The ratio of costs is equipment : hens : feed = 5 : 3 : 9 . (i) Show that the total cost is now $2040. Answer(b)(i) [2] (ii) Adam actually buys more than 20 hens, each costing $12. How many hens does he buy? Answer(b)(ii) ............................................... [2] (c) Adam makes $2920 from selling his hens’ eggs. For Examiner′s Use Calculate his percentage profi t on the $2040. Answer(c) ........................................... % [2] (d) Adam borrows $1500 for 3 years at a rate of 5.5% per year compound interest. Calculate the interest he will pay, correct to the nearest cent. Answer(d) $ ................................................ [3] _____________________________________________________________________________________

Mark scheme: Qu. Part Answers Mark Part Marks 1 (a) 240 900 1,1 [Total] 1640 1FT 500 + their 2 costs (b) (i) 600 ÷ 5 × 17 M2 M1 for 600 ÷ 5 or 17 ÷ 5 (ii) 30 2 M1 for 2040 ÷ 17 × 3 Or 120 × 3, soi by 360 2920 − 2040 (c) 43.1 2 M1 for × 100 oe 2040 2920 or ( − )1 × 100 oe 2040 2920 or × 100 − 100 oe 2040 (d) 261.36 cao 3 M1 for 1500 × 1.0553 oe M1FT for their 1761.36 – 1500 If only 1 scored SC1 for correctly rounding to 2 decimal places from at least 3 decimal places SC2 if only 1761.36 seen

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Q2 · The diagram shows four quadrilaterals drawn on a 1 cm2 grid

2 The diagram shows four quadrilaterals drawn on a 1 cm2 grid. For Examiner′s Use y 8 7 X 6 5 A 4 3 2 1 x –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 –1 –2 –3 B C –4 –5 –6 –7 –8 (a) Write down the mathematical name of the quadrilateral X. Answer(a) ............................................... [1] (b) Describe fully the single transformation that maps quadrilateral X onto quadrilateral For Examiner′s Use (i) A, Answer(b)(i) ...................................................................................................................... ........................................................................................................................................... [3] (ii) B, Answer(b)(ii) ..................................................................................................................... ........................................................................................................................................... [2] (iii) C. Answer(b)(iii) .................................................................................................................... ........................................................................................................................................... [3] (c) (i) Calculate the length of the longest side of quadrilateral X. Show that your answer rounds to 3.16 cm, correct to 3 signifi cant fi gures. Answer(c)(i) [2] (ii) Calculate the perimeter of quadrilateral X. Answer(c)(ii) ......................................... cm [3] (iii) Find the perimeter of quadrilateral C. Answer(c)(iii) ......................................... cm [1] _____________________________________________________________________________________

Mark scheme: 2 (a) Kite 1 (b) (i) Rotation 1 90° clockwise (or 270° anti- 1 clockwise) oe [centre] origin oe 1 (ii) Translation 1  − 2  Accept 2 left and 10 down oe   1  − 10  IGCSE – October/November 2013 0580 33 (iii) Enlargement 1 [Scale Factor] –3 1 [centre] (–3, 4) 1 (c) (i) [x2 =] 32 + 12 M1 M1 for 32 + 12 or better [x =] 3 2 + 1 2 or [x = 9 + 1 Needs a value to 3 or more decimal places M1dep or 10 and = 3.162… (ii) 9.15 3 B1 for 2 or 1.41 or better seen M1 for 2 x 3.16 + 2 x their 1.41... soi by 9.14 If zero scored SC1 if answer in range 8.6 to 9.6 (iii) 27.45 to 27.5 1FT their (c)(ii) ×3

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Q3 · Using only the integers from 1 to 50, fi nd For Examiner′s Use (i) a multiple of both 4…

3 (a) Using only the integers from 1 to 50, fi nd For Examiner′s Use (i) a multiple of both 4 and 7, Answer(a)(i) ............................................... [1] (ii) a square number that is odd, Answer(a)(ii) ............................................... [1] (iii) an even prime number, Answer(a)(iii) ............................................... [1] (iv) a prime number which is one less than a multiple of 5. Answer(a)(iv) ............................................... [1] (b) Find the value of (i) ^ 5 h 2 , Answer(b)(i) ............................................... [1] (ii) 2–3 × 63. Answer(b)(ii) ............................................... [2] _____________________________________________________________________________________

Mark scheme: 3 (a) (i) 28 1 (ii) 25 or 49 or 9 or 1 1 (iii) 2 1 (iv) 19 or 29 1 1 (b) (i) 5 1 B1 for or 216 seen 8 (ii) 27 2

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Q4 · A regular polygon has 9 sides

4 (a) A regular polygon has 9 sides. For Examiner′s For this polygon, calculate Use (i) the size of one exterior angle, Answer(a)(i) ............................................... [2] (ii) the size of one interior angle. Answer(a)(ii) ............................................... [1] (b) C w° B 24° y° NOT TO SCALE D O E x° z° A F In the diagram, A, B, C and D are points on the circumference of a circle, centre O. AB is the diameter and EF is a tangent to the circle at A. AB is parallel to DC and angle ACD = 24°. Find (i) w, Answer(b)(i) w = ............................................... [1] (ii) x, Answer(b)(ii) x = ............................................... [1] (iii) y. Answer(b)(iii) y = ............................................... [1] (c) Complete the statement. z = ............................... because ................................................................................................ ................................................................................................................................................... [2] _____________________________________________________________________________________

Mark scheme: 4 (a) (i) 40 2 M1 for 360 ÷ 9 (ii) 140 1FT 180 – their (a)(i) (b) (i) [w =] 90 1 (ii) [x =] 24 1 (iii) [y =] 66 1FT 180 – (their w + their x) (c) [z =] 66 1FT (90 – their x) or their y [Angle between] tangent [and] 1 diameter/radius [=] 90°

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Q5 · Complete the table for y = 5 + 3x – x2

5 (a) (i) Complete the table for y = 5 + 3x – x2. For Examiner′s Use x –2 –1 0 1 2 3 4 5 y –5 5 7 5 –5 [3] (ii) On the grid, draw the graph of y = 5 + 3x – x2 for –2 Y x Y 5. y 8 7 6 5 4 3 2 1 x –2 –1 0 1 2 3 4 5 –1 –2 –3 –4 –5 –6 [4] (b) Use your graph to solve the equation 5 + 3x – x2 = 0 . Answer(b) x = ..................... or x = ..................... [2] (c) (i) On the grid, draw the line of symmetry of y = 5 + 3x – x2. [1] For Examiner′s Use (ii) Write down the equation of this line of symmetry. Answer(c)(ii) ............................................... [1] (d) (i) On the grid, draw a straight line from (–1, 1) to (3, 5). [1] (ii) Work out the gradient of this line. Answer(d)(ii) ............................................... [2] (iii) Write down the equation of this line in the form y = mx + c. Answer(d)(iii) y = ............................................... [1] _____________________________________________________________________________________

Mark scheme: 5 (a) (i) 1, 7, 1 1, 1, 1 (ii) 8 points correctly plotted P3FT P2FT for 6 or 7 correct P1FT for 4 or 5 correct Correct smooth curve through all 8 C1 correct points IGCSE – October/November 2013 0580 33 (b) –1.1 to –1.3 and 4.1 to 4.3 1FT, 1FT (c) (i) Line x = 1.5 drawn 1 (ii) x = 1.5 oe 1FT Equation of their line in (c)(i) (d) (i) Ruled continuous line drawn 1 rise (ii) 1 2 M1 for for their line run (iii) [y =] x + 2 1FT their (d)(ii) + their 2

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Q6 · Alison scored the following number of runs in 15 cricket matches

6 Alison scored the following number of runs in 15 cricket matches. For Examiner′s Use 12 3 27 35 0 7 52 4 18 30 18 7 94 61 7 (a) For these scores, (i) work out the median, Answer(a)(i) ............................................... [2] (ii) write down the mode, Answer(a)(ii) ............................................... [1] (iii) calculate the mean. Answer(a)(iii) ............................................... [2] (b) These are the averages for the number of runs scored by Bethan in the 15 matches. Median = 21 Mode = 13 Mean = 20 Alison says that her scores are better than Bethan’s scores. Bethan says that her scores are better than Alison’s scores. Explain how they could both be correct. Answer(b) ................................................................................................................................. ................................................................................................................................................... ................................................................................................................................................... [2] (c) Alison puts her 15 scores into 4 groups and shows them in a pie chart. For Examiner′s Use (i) Complete the table. Score Frequency Sector Angle 0 to 25 9 216° 26 to 50 51 to 75 76 to 100 [3] (ii) Complete the pie chart and label the sectors. 0 to 25 [3] (d) Estimate the probability that in the next match Alison will score more than 25 runs. Give your answer as a fraction in its simplest form. Answer(d) ............................................... [2] _____________________________________________________________________________________

Mark scheme: 6 (a) (i) 18 2 M1 for evidence of ordering (ii) 7 1 (iii) 25 2 M1 for sum of 15 items ÷ 15 soi (b) Alison with reference to [higher] mean 1FT Strict FT and Bethan with reference to [higher] median 1FT Strict FT (c) (i) [Frequencies] 3, 2, 1 1 [Angles] 72°, 48°, 24° 2 B1 for 1 correct or M1 for one frequency ÷ 15 × 360 or × 24 (ii) Two correct sectors on pie chart 2FT B1FT for 1 correct sector Only ft if (c)(i) angles total 144 3 ‘correct’ labels 1 Independent 2 6 (d) 2 B1 for 0.4 or 40% or or any equivalent 5 15 fraction

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Q7 · For Examiner′s D Use E C NOT TO SCALE 2.25 m 1.5 m A B 1.0 m The diagram shows a…

7 For Examiner′s D Use E C NOT TO SCALE 2.25 m 1.5 m A B 1.0 m The diagram shows a trapezium ABCD. AB = 1.0 m, AD = 2.25 m, BC = 1.5 m and angle DEC = 90°. (a) Using trigonometry, calculate angle DCE. Answer(a) Angle DCE = ............................................... [3] (b) Calculate the area of the trapezium ABCD. Answer(b) .......................................... m2 [2] (c) ABCD is the cross-section of a box. The box is 2 m long. Calculate the volume of the box. D C 2 m B A Answer(c) .......................................... m3 [1] (d) On the grid, complete the net of the box. For Examiner′s The base and one face of the box have been drawn for you. Use The scale is 2 cm to 1 m. [4] _____________________________________________________________________________________

Mark scheme: 7 (a) [Angle DCE =] 36.9 or 36.8699 to 36.9 3 B1 for [DE =] 0.75 soi their DE M1 for than DCE = 0.1 (b) 1.875 or 1.88 2 M1 for 0.5 × (1.5 + 2.25) × 1.0 oe (c) 3.75 1FT their (b) × 2 IGCSE – October/November 2013 0580 33 (d) 3 rectangles and 1 trapezium correctly 4 B1 for rectangle to right 6 by 8 squares placed on the grid with correct scale and B1 for an accurate and correctly placed size. trapezium B1 for a rectangle to left 9 by 8 squares B1 for rectangle 5 by 8 squares and further to the left

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Q8 · Here is a sequence of patterns made using identical polygons

8 Here is a sequence of patterns made using identical polygons. For Examiner′s Use Pattern 1 Pattern 2 Pattern 3 (a) Write down the mathematical name of the polygon in Pattern 1. Answer(a) ............................................... [1] (b) Complete the table for the number of vertices (corners) and the number of lines in Pattern 3, Pattern 4 and Pattern 7. Pattern 1 2 3 4 7 Number of vertices 8 14 Number of lines 8 15 [5] (c) (i) Find an expression for the number of vertices in Pattern n. Answer(c)(i) ............................................... [2] (ii) Work out the number of vertices in Pattern 23. Answer(c)(ii) ............................................... [1] (d) Find an expression for the number of lines in Pattern n. For Examiner′s Use Answer(d) ............................................... [2] (e) Work out an expression, in its simplest form, for (number of lines in Pattern n) – (number of vertices in Pattern n). Answer(e) ............................................... [2] _____________________________________________________________________________________ Question 9 is printed on the next page.

Mark scheme: 8 (a) Octagon 1 (b) [Pattern 3] 20 and 22 1 [Pattern 4] 26, 29 1, 1 [Pattern 7] 44, 50 1, 1 (c) (i) 6n + 2 oe final answer 2 B1 for 6n + a or bn + 2 b ≠ 0 (ii) 140 oe 1FT ft linear expression in (c)(i) (d) 7n + 1 oe final answer 2 B1 for 7n + c or dn + 1 d ≠ 0 (e) n – 1 final answer 2FT B1FT for n + j or kn 1 k ≠ 0 3V 2 3V 3V

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Q9 · For 9 (a) The formula for the volume, V, of a cone with radius r, and height h, is V = 3…

1 For 9 (a) The formula for the volume, V, of a cone with radius r, and height h, is V = 3 πr2h . Examiner′s Use (i) To make r the subject of this formula, the fi rst step is 3V = πr2h. Show the remaining steps to make r the subject of this formula. Answer(a)(i) r = ............................................... [2] (ii) An ice-cream cone has a volume of 141 cm3 and height 15 cm. Show that the radius of the cone is 3 cm, correct to the nearest whole number. Answer(a)(ii) [2] (b) The open end of an ice-cream cone is a circle of radius 3 cm. Calculate the circumference of this circle. Answer(b) ......................................... cm [2] (c) The volume of a ball of ice-cream is 113 cm3. The ball of ice-cream costs $2.15 . Calculate the cost of 1 cm3 of the ice-cream. Give your answer in cents, correct to 1 decimal place. Answer(c) ...................................... cents [3]

Mark scheme: 3V 3V 3V 9 (a) (i) [r =] 2 B1 for [r2 =] or seen or better πh π h 3 x141 (ii) [r =] M1FT their formula πx15 [r =] 2.99… A1 (b) 18.9 or 18.8 or 18.849 to 18.852 2 M1 for 2 × π × 3 oe (c) 1.9 [cents] cao 3 M1 for 2,15 (or 215) ÷ 113 A1 for 0.019 (0…) or 1.9 (0…) soi

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