Cambridge IGCSE Mathematics 0580 — 2011 May/June Paper 4 · Variant 1

4 (a) For Examiner's 12 cm Use H G NOT TO 6 cm SCALE 14 cm F The diagram shows triangle FGH, with FG = 14 cm, GH = 12 cm and FH = 6 cm. (i) Calculate the size of angle HFG. Answer(a)(i) Angle HFG = [4] (ii) Calculate the area of triangle FGH. Answer(a)(ii) cm2 [2] (b) For Q Examiner's 18 cm Use R NOT TO SCALE 12 cm 117° P The diagram shows triangle PQR, with RP = 12 cm, RQ = 18 cm and angle RPQ = 117°. Calculate the size of angle RQP. Answer(b) Angle RQP = [3]

Mark scheme: 6 + 14 − 12 4 (a) (i) (cos (HFG)) = M2 M1 for implicit form 2 × 6 × 14 58.4 (58.41…) A2 A1 for 0.5238… (ii) 0.5 × 6 × 14 × sin (their 58.4) oe M1 ft their (i) 35.8 or 35.77 to 35.78 A1ft Correct or ft their (i) sin(117 ) × 12 (b) (sin (RQP)) = M2 M1 for implicit form 18 36.4 or 36.44... A1  3  k 

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5 For y Examiner's Use 4 A B 3 2 1 x –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 –1 C –2 –3 –4 –5 –6 (a) On the grid above, draw the image of  − 3  (i) shape A after translation by the vector   , [2]  − 2  (ii) shape A after reflection in the line x = −1 . [2] (b) Describe fully the single transformation which maps (i) shape A onto shape B, Answer(b)(i) [3] (ii) shape A onto shape C. Answer(b)(ii) [3] (c) Find the matrix representing the transformation which maps shape A onto shape B.   Answer(c)   [2]    − 1 0  (d) Describe fully the single transformation represented by the matrix   .  0 − 1  Answer(d) [3]

Mark scheme: − 3   k  5 (a) (i) Correct translation (see diagram) 2 SC1 for translation by or by      k   −2  (ii) Correct reflection (see diagram) 2 SC1 for reflection in y = −1 (b) (i) Stretch, 1 (factor) 3, 1 y-axis or x = 0 invariant 1 (ii) Rotation 1 90° clockwise 1 Accept −90° (1, − 1) 1  3 0   1 0   k 0  (c) (i) ft from (b)(i) 2 ft SC1 for (ft from (b)(i)) or        0 1   0 3   0 1  with k algebraic or numeric but ≠ 1 or 0 (ii) Rotation, 1 180° 1 Origin 1 Accept O or (0,0)

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6 For F Examiner's Use NOT TO SCALE C D E 14 cm 36 cm A 19 cm B In the diagram, ABCDEF is a prism of length 36 cm. The cross-section ABC is a right-angled triangle. AB = 19 cm and AC = 14 cm. Calculate (a) the length BC, Answer(a) BC = cm [2] (b) the total surface area of the prism, Answer(b) cm2 [4] (c) the volume of the prism, Answer(c) cm3 [2] (d) the length CE, Answer(d) CE = cm [2] (e) the angle between the line CE and the base ABED. Answer(e) [3]

Mark scheme: 6 (a) 23.6 (23.60…) 2 M1 for 142 + 192 (b) 2300 or 2303 to 2304 cao 4 M3 for 2 × ½ × 14 × 19 + 14 × 36 + 19 × 36 + their BC × 36 M2 for 4 of these added M1 for ½ × 14 × 19 (c) 4788 or 4790 cao 2 M1 their triangle area × 36 (d) 43(.0) or 43.04 to 43.05 cao 2 M1 for (their (a))2 + 362 or 362 + 192 + 142  14  (e) 18.9° to 19.02° cao 3 M2 for inv sin   or  their CE   14  inv tan  2 2  or  19 + 36   2 2  19 + 36   inv cos or complete longer  their CE    methods (M1 for clearly identifying angle CEA) IGCSE – May/June 2011 0580 41

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4 For 7 (a) Complete the table of values for the equation y = , x ≠ 0. Examiner's x 2 Use x O4 O3 O2 O1 O0.6 0.6 1 2 3 4 y 0.25 0.44 11.11 4.00 0.44 [3] 4 (b) On the grid, draw the graph of y = for O4 Y x Y O0.6 and 0.6 Y x Y 4 . x 2 y 12 11 10 9 8 7 6 5 4 3 2 1 x –4 –3 –2 –1 0 1 2 3 4 –1 –2 [5] 4 For (c) Use your graph to solve the equation = 6 . Examiner's x 2 Use Answer(c)x = or x = [2] (d) By drawing a suitable tangent, estimate the gradient of the graph where x = 1.5. Answer(d) [3] 4 (e) (i) The equation O x + 2 = 0 can be solved by finding the intersection of the graph x 2 4 of y = and a straight line. x 2 Write down the equation of this straight line. Answer(e)(i) [1] (ii) On the grid, draw the straight line from your answer to part (e)(i). [2] 4 (iii) Use your graphs to solve the equation O x + 2 = 0. x 2 Answer(e)(iii) x = [1]

Mark scheme: 7 (a) 1(.00) 4(.00) 11.1(1) 1(.00) 0.25 3 B2 for 4 correct, B1 for 3 correct (b) 10 points plotted P3 ft B2 for 8 or 9 points correct ft B1 for 6 or 7 points correct ft Correct shaped curve through 10 points C1 ft ft their points if shape correct – ignore anything (condone 2 points slightly missed) between – 0.6 and 0.6 2 separate curves not crossing x-axis and B1 Independent not touching or crossing y-axis (c) −0.85 to – 0.75 cao 1 0.75 to 0.85 cao 1 (d) Tangent drawn (ruled) at x = 1.5 T1 Allow slight daylight – 3 to −2 2 Dep on T1 M1 evidence rise/run dependent on tangent SC1 for answer in range 2 to 3 Answer implies M but not the T mark (e) (i) y = x − 2 oe 1 (ii) line ruled to cross curve 2 ft Dependent on (i) in form y = mx + c, m ≠ 0, c ≠ 0 B1 for gradient ft or y intercept ft but again to cross curve at all possible points (iii) 2.5 to 2.7 cao 1 Dependent on (e)(i) correct

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8 The table below shows the marks scored by a group of students in a test. For Examiner's Use Mark 11 12 13 14 15 16 17 18 Frequency 10 8 16 11 7 8 6 9 (a) Find the mean, median and mode. Answer(a) mean = median = mode = [6] (b) The table below shows the time (t minutes) taken by the students to complete the test. Time (t) 0 I=t Y=10 10 I=t Y=20 20 I=t Y=30 30 I=t Y=40 40 I=t Y=50 50 I=t Y=60 Frequency 2 19 16 14 15 9 (i) Cara rearranges this information into a new table. Complete her table. Time (t) 0 I=t Y=20 20 I=t Y=40 40 I=t Y=50 50 I=t Y=60 Frequency 9 [2] (ii) Cara wants to draw a histogram to show the information in part (b)(i). Complete the table below to show the interval widths and the frequency densities. 0 I=t Y=20 20 I=t Y=40 40 I=t Y=50 50 I=t Y=60 Interval 10 width Frequency 0.9 density [3] (c) Some of the students were asked how much time they spent revising for the test. For Examiner's 10 students revised for 2.5 hours, 12 students revised for 3 hours and n students revised for Use 4 hours. The mean time that these students spent revising was 3.1 hours. Find n. Show all your working. Answer(c) n = [4]

Mark scheme: 8 14.2 3 M1 for Σ fx (10 × 11 + 8 × 12 + 16 × 13 + 11 × 14 + 7 × 15 + 8 × 16 + 6 × 17 + 9 × 18 ) (1065) (allow one error or omission) M1dep for ÷ Σf (10 + 8 + 16 + 11 + 7 + 8 + 6 + 9) (75) (allow one further error or omission) 14 2 M1 for 37th, 37.5th or 38th seen 13 1 (b) (i) 21, 30, 15 2 B1 for 2 correct (ii) 20 20 10 (10) 3 1, 1, 1 for each correct vertical pair 1.05 1.5 1.5 (0.9) (c) 10 × 5.2 + 12 × 3 + 4 n ( = )1.3 M2 M1 for either numerator or denominator seen 10 + 12 + n multiplying across and collecting terms M1 dep on linear numerator and denominator their (68.2 − 25 − 36) = their (4 − 3.1) × n (n =) 8 www 4 A1 IGCSE – May/June 2011 0580 41

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9 Peter wants to plant x plum trees and y apple trees. For Examiner's He wants at least 3 plum trees and at least 2 apple trees. Use (a) Write down one inequality in x and one inequality in y to represent these conditions. Answer(a) , [2] (b) There is space on his land for no more than 9 trees. Write down an inequality in x and y to represent this condition. Answer(b) [1] (c) Plum trees cost $6 and apple trees cost $14. Peter wants to spend no more than $84. Write down an inequality in x and y, and show that it simplifies to 3x + 7y Y 42. Answer(c) [1] (d) On the grid, draw four lines to show the four inequalities and shade the unwanted regions. For Examiner's y Use 12 11 10 9 8 7 6 5 4 3 2 1 x 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 [7] (e) Calculate the smallest cost when Peter buys a total of 9 trees. Answer(e) $ [2] Question 10 is printed on the next page.

Mark scheme: 9 (a) x [=3 y [=2 1, 1 (b) x + y Y 9 1 (c) 6x + 14y Y 84 1 (d) x = 3 y = 2 1, 1 Accept clear and freehand lines long enough to define the correct quadrilateral x + y = 9 2 SC1 for line through (0, 9) or (9, 0) Line from (0, 6) to (14, 0) 2 B1 for through (0, 6) or (14, 0) Correct quadrilateral unshaded or clearly 1 indicated (e) $ 70 2 B1 for considering (7, 2)

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10 The first and the nth terms of sequences A, B and C are shown in the table below. For Examiner's (a) Complete the table for each sequence. Use 1st term 2nd term 3rd term 4th term 5th term nth term Sequence A 1 n3 Sequence B 4 4n Sequence C 4 (n + 1)2 [5] (b) Find (i) the 8th term of sequence A, Answer(b)(i) [1] (ii) the 12th term of sequence C. Answer(b)(ii) [1] (c) (i) Which term in sequence A is equal to 15 625? Answer(c)(i) [1] (ii) Which term in sequence C is equal to 10 000? Answer(c)(ii) [1] (d) The first four terms of sequences D and E are shown in the table below. Use the results from part (a) to find the 5th and the nth terms of the sequences D and E. 1st term 2nd term 3rd term 4th term 5th term nth term Sequence D 5 16 39 80 Sequence E 0 1 4 9 [4]

Mark scheme: 10 (a) (A 1) 8 27 64 125 2 B1 for 3 correct (B 4) 8 12 16 20 1 (C 4) 9 16 25 36 2 B1 for 3 correct (b) 512 1 169 1 (c) 25 1 99 1 (d) 145 n3 + 4n oe 1, 1 16 (n + 1)2 – 4n oe but isw 1, 1 Likely oe is (n – 1)2

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