Cambridge IGCSE Mathematics 0580 — 2011 Oct/Nov Paper 4 · Variant 3

1 For 0.8 m Examiner's Use 0.5 m NOT TO SCALE 1.2 m 1.2 m d 0.4 m A rectangular tank measures 1.2 m by 0.8 m by 0.5 m. (a) Water flows from the full tank into a cylinder at a rate of 0.3 m3/min. Calculate the time it takes for the full tank to empty. Give your answer in minutes and seconds. Answer(a) min s [3] (b) The radius of the cylinder is 0.4 m. For Examiner's Calculate the depth of water, d, when all the water from the rectangular tank is in the cylinder. Use Answer(b) d = m [3] (c) The cylinder has a height of 1.2 m and is open at the top. The inside surface is painted at a cost of $2.30 per m2. Calculate the cost of painting the inside surface. Answer(c) $ [4]

Mark scheme: Qu. Answers Mark Part Marks 1 (a) 1 min 36 s www 3 M1 for 1.2 × 0.8 × 0.5 (= 0.48) A1 1.6 or 96 If A0, B1 for correctly converting to min and sec Dep on M1 (b) 0.954 to 0.956 www 3 their 0.48 M2 for or M1 for π × 0.42 × d = ‘0.48’ π × 0.4 2 (c) 8.09 to 8.10 www 4 M1 for π × 0.42 (0.503) condone × 2 and M1 for π × 0.8 × 1.2 (3.02) M1 for their area × 2.3 (dep M1 M1)

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2 (a) Complete the table of values for y = 2x. For Examiner's Use x –2 –1 0 1 2 3 y 0.25 1 2 8 [2] (b) On the grid, draw the graph of y = 2x for O2 Y x Y 3. y 10 9 8 7 6 5 4 3 2 1 x –2 –1 0 1 2 3 –1 [3] For Examiner's (c) (i) On the grid, draw the straight line which passes through the points (0, 2) and (3, 8). [1] Use (ii) The equation of this line is y = mx + 2. Show that the value of m is 2. Answer(c)(ii) [1] (iii) One answer to the equation 2x =2x + 2 is x = 3. Use your graph to find the other answer. Answer(c)(iii) x = [1] (d) Draw the tangent to the curve at the point where x = 1. Use this tangent to calculate an estimate of the gradient of y = 2x when x = 1. Answer(d) [3]

Mark scheme: 2 (a) 0.5, 4 1+1 (b) 6 points plotted ft P2 P1 for 5 points Correct shaped curve through 6 points C1 Ignore to left of x = −2 (exponential) (c) (i) Correct ruled line reaching both L1 points (ii) 6 ÷ 3 oe 1 Allow ‘test’ with a coordinate on the line (not 0, 2) (iii) –0.8 to –0.6 1 Dep on L1 (d) Tangent drawn at (1, 2) T1 Not chord, allow up to 1 mm daylight Rise/run attempt using correct scales M1 Dep on T1 1.2 to 1.6 cao A1

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3 (a) For D Examiner's NOT TO Use SCALE C 40° 30° A B E ABCD is a quadrilateral with angle BAD = 40°. AB is extended to E and angle EBC = 30°. AB = AD and BD = BC. (i) Calculate angle BCD. Answer(a)(i) Angle BCD = [3] (ii) Give a reason why DC is not parallel to AE. Answer(a)(ii) [1] (b) A regular polygon has n sides. 5n Each exterior angle is degrees. 2 Find the value of n. Answer(b) n = [3] (c) For Examiner's Use NOT TO SCALE O C 25° A B The diagram shows a circle centre O. A, B and C are points on the circumference. OC is parallel to AB. Angle OCA = 25°. Calculate angle OBC. Answer(c)Angle OBC = [3]

Mark scheme: 3 (a) (i) 50 www3 3 B1 for angle ADB or ABD = 70 B1 for angle DBC = 80 (ii) Angle DCB ≠ angle CBE oe 1 Accept angle CDB ≠ angle ABD 5n 360 (b) 12 B3 M2 for = oe 2 n or M1 for 360 soi (c) 65 www 3 OAC = 25, CAB = 25, OBA = 50, BOC = 50, AOB = 80, AOC = 130 B1 each, max 2 IGCSE – October/November 2011 0580 43

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4 For y Examiner's Use 6 4 P 2 W x –6 –4 –2 0 2 4 6 –2 –4 –6 (a) Draw the reflection of shape P in the line y = x. [2] − 2  (b) Draw the translation of shape P by the vector   . [2]  1  (c) (i) Describe fully the single transformation that maps shape P onto shape W. Answer(c)(i) [3] (ii) Find the 2 by 2 matrix which represents this transformation.     Answer(c)(ii)   [2]      1 0  (d) Describe fully the single transformation represented by the matrix   . 0 2   Answer(d) [3]

Mark scheme: 4 (a) Image (1, –1), (1, –2), (4, –2), (3, –1) 2 B1 if vertices plotted only or reflects in y = –x − 2   k  (b) Image (–3, 2), (–4, 2), (–4, 5), (–3, 4) 2 B1 for translation by   or    k   1  (c) (i) Rotation only, 1 Spoilt if extras 90 clockwise oe, 1 (Centre) (0, 0) oe 1  0 1  (ii)   2 B1 for one row or one column correct  − 1 0  (d) Stretch only, 1 Spoilt if extras (Factor) 2, 1 x-axis oe invariant 1

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5 (a) The cost of a bottle of juice is 5 cents more than the cost of a bottle of water. For Mohini buys 3 bottles of water and 6 bottles of juice. Examiner's The total cost is $5.25. Use Find the cost of a bottle of water. Give your answer in cents. Answer(a) cents [4] (b) The cost of a biscuit is x cents. The cost of a cake is (x + 3) cents. The number of biscuits Roshni can buy for 72 cents is 2 more than the number of cakes she can buy for 72 cents. (i) Show that x2 + 3x O 108 = 0. Answer(b)(i) [3] (ii) Solve the equation x2 + 3x O 108 = 0. Answer(b)(ii) x = or x = [3] (iii) Find the total cost of 2 biscuits and 1 cake. Answer(b)(iii) cents [1]

Mark scheme: 5 (a) 55 www B4 M3 for 3w + 6(w + 5) = 525 oe in $ or (3j – 5) + 6j = 525 oe in $ or M2 for j = w + figs5 oe and 3w + 6j = figs525 or M1 for w and w + figs5 or j and j – figs5 72 72 72 72 (b) (i) = 2 oe M2 M1 for or x −x + 3 x x + 3 72(x + 3) – 72x = 2x(x + 3) oe M1 Dep on 3 terms above Fractions removed, isw − 3 ± 441 (ii) –12, 9 www 3 M2 for (x + 12)(x – 9) or 2 or SC1 for ( x + a )( x + b ) where ab = –108 − 3 ± 3 2 − 4 × 1 × − 108 or a + b = 3 or 2 (iii) 30 1 ft 3 × a positive root + 3

More questions on Equations

6 For Q P Examiner's Use NOT TO SCALE 3 cm D C 4 cm A B 12 cm The diagram shows a triangular prism of length 12 cm. The rectangle ABCD is horizontal and the rectangle DCPQ is vertical. The cross-section is triangle PBC in which angle BCP = 90°, BC = 4 cm and CP = 3 cm. (a) (i) Calculate the length of AP. Answer(a)(i) AP = cm [3] (ii) Calculate the angle of elevation of P from A. Answer(a)(ii) [2] (b) (i) Calculate angle PBC. For Examiner's Use Answer(b)(i) Angle PBC = [2] (ii) X is on BP so that angle BXC = 120°. Calculate the length of XC. Answer(b)(ii) XC = cm [3]

Mark scheme: 6 (a) (i) 13 or 13.0 www 3 M1 for 32 + 42 oe Equiv if find AC first and M1 for 12 2 + their (3 2 + 4 2 ) 3 3 (ii) 13.32 to 13.35 or 13.3 2 M1 for sin = or tan = oe their AP their AC 3 (b) (i) 36.86 to 36.87 or 36.9 2 M1 for tan (PBC) = oe 4 4 sin their (b )( i ) (ii) 2.770 to 2.774 or 2.77 3 M2 for or M1 for correct sin 120 implicit eqn IGCSE – October/November 2011 0580 43

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8 f(x) = x2 + x O1 g(x) = 1 O 2x h(x) = 3x Examiner's Use (a) Find the value of hg(–2). Answer(a) [2] (b) Find g –1(x). Answer(b) g O1(x) = [2] (c) Solve the equation f(x) = 0. Show all your working and give your answers correct to 2 decimal places. Answer(c) x = or x = [4] (d) Find fg(x). Give your answer in its simplest form. Answer(d) fg(x) = [3] (e) Solve the equation h – 1(x) = 2. Answer(e) x = [1]

Mark scheme: 8 (a) 243 2 B1 for (g(–2) =) 5 seen or 3(1–2x) 1 − x x − 1 2 M1 for x = 1 − 2y or x = (1 − y)/2 (b) or final ans 2 − 2 − 1 ± 12 − 41()( −)1 B2 B1 for 12 − 4 1()( −)1 or better ( 5 ) seen (c) 2)1( anywhere p + q p − q If in form or r r B1 for p = –1 and r = 2(1) –1.62, 0.62 B1B1 SC1 for –1.62 and 0.62 seen or –1.6 or –1.618.. and 0.6 or 0.618… (d) 4x2 – 6x + 1 final ans www3 3 M1 for (1 – 2x)2 + (1 – 2x) – 1 or better and B1 for (1 – 2x)2 = 1 – 2x – 2x + 4x2 or better (e) 9 1 IGCSE – October/November 2011 0580 43

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9 For Examiner's Set A Use S U M S Set B M I N U S The diagram shows two sets of cards. (a) One card is chosen at random from Set A and replaced. (i) Write down the probability that the card chosen shows the letter M. Answer(a)(i) [1] (ii) If this is carried out 100 times, write down the expected number of times the card chosen shows the letter M. Answer(a)(ii) [1] (b) Two cards are chosen at random, without replacement, from Set A. Find the probability that both cards show the letter S. Answer(b) [2] (c) One card is chosen at random from Set A and one card is chosen at random from Set B. Find the probability that exactly one of the two cards shows the letter U. Answer(c) [3] (d) A card is chosen at random, without replacement, from Set B until the letter shown is either I or U. Find the probability that this does not happen until the 4th card is chosen. Answer(d) [2]

Mark scheme: 9 Accept fraction, %, dec equivalents (3sf or better when not exact) throughout but not ratio or words isw incorrect cancelling/conversion to other forms 1 (a) (i) oe 1 4 1 (ii) 25 cao 1ft ft their × 100 to 3sf or better or rounding or 4 truncating to integer Not 25/100 2 2 1 (b) oe cao 2 M1 for × 0.167, 16.7% 12 4 3 7 1 4 3 1 (c) oe cao 3 M2 for × + × 20 4 5 4 5 1 4 3 1 or M1 for × or × 4 5 4 5 After 0, SC1 for 7 correct in list (condone UU in addition) 6 3 2 1  2  (d) oe cao 2 M1 for × × ×   60 5 4 3  2 

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10 Hassan stores books in large boxes and small boxes. For Each large box holds 20 books and each small box holds 10 books. Examiner's He has x large boxes and y small boxes. Use (a) Hassan must store at least 200 books. Show that 2x + y [ 20. Answer(a) [1] (b) Hassan must not use more than 15 boxes. He must use at least 3 small boxes. The number of small boxes must be less than or equal to the number of large boxes. Write down three inequalities to show this information. Answer(b) [3] (c) On the grid, show the information in part (a) and part (b) by drawing four straight lines and shading the unwanted regions. y 20 18 16 14 12 10 8 6 4 2 x 0 2 4 6 8 10 12 14 16 18 20 [6] (d) A large box costs $5 and a small box costs $2. For Examiner's (i) Find the least possible total cost of the boxes. Use Answer(d)(i) $ [1] (ii) Find the number of large boxes and the number of small boxes which give this least possible cost. Answer(d)(ii) Number of large boxes = Number of small boxes = [2] Question 11 is printed on the next page.

Mark scheme: 10 (a) 20x + 10y ≥ 200 1 In (a), (b) −1 once for wrong symbol (b) x + y ≤ 15, y ≥ 3, y ≤ x 3 B1 for each (c) All lines long enough to make full boundary of region, accept dashed or solid lines, 2 mm acc at intercepts 2x + y = 20 ruled B2 B1 for ruled line through (10, 0) or (0, 20) x + y = 15 ruled B1 y = x ruled B1 y = 3 ruled B1 −1 once, freehand Quadrilateral identified R1 Allow if slight inaccuracy(s) in diagonal lines Allow any clear indication of region (d) (i) 47 cao 1 (ii) 7, 6 cao 2 M1 for any 5x + 2y in their region evaluated to equal their 47 IGCSE – October/November 2011 0580 43 8 

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11 (a) For y Examiner's Use 5 4 3 Q 2 P 1 x –3 –2 –1 0 1 2 3 4 5 The points P and Q have co-ordinates (–3, 1) and (5, 2). (i) Write as a column vector.     Answer(a)(i) =   [1]     − 1  (ii) = 2    1  Mark the point R on the grid. [1] (iii) Write down the position vector of the point P.     Answer(a)(iii)   [1]     (b) For U Examiner's Use L NOT TO u SCALE M K O V v In the diagram, = u and = v. 2 3 K is on UV so that = and L is on OU so that = . 3 4 M is the midpoint of KL. Find the following in terms of u and v, giving your answers in their simplest form. (i) Answer(b)(i) = [4] (ii) Answer(b)(ii) = [2] Question 12 is printed on the next page.

Mark scheme:  8  11 (a) (i)   1  1  (ii) Point (3, 4) indicated 1 − 3  (iii)   1  1  5 2 (b) (i) − u + v oe 2 terms 4 M1 for any correct route L to K 12 3 e.g. LU + UK and B1 for LU = u/4 oe or OL = ¾u oe and B1 for UK = 2/3 (v – u) oe or VK = 1/3(u – v) oe all Bs are soi 13 1 (ii) u + v oe 2 terms 2 M1 for correct route from O to M e.g. OL + LM 24 3 (can be in terms of u, v)

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12 (a) The nth term of a sequence is n(n+1). For Examiner's Use (i) Write the two missing terms in the spaces. 2, 6, , 20, [2] (ii) Write down an expression in terms of n for the (n + 1)th term. Answer(a)(ii) [1] (iii) The difference between the nth term and the (n + 1)th term is pn + q. Find the values of p and q. Answer(a)(iii) p = q = [2] (iv) Find the positions of the two consecutive terms which have a difference of 140. Answer(a)(iv) and [2] (b) A sequence u1, u2, u3, u4, …………. is given by the following rules. u1 = 2, u2 = 3 and un = 2u n − 2 + u n −1 for n [ 3. For example, the third term is u3 and u3 = 2u1 + u2 = 2 × 2 + 3 = 7. So, the sequence is 2, 3, 7, u4, u5, ….. (i) Show that u4 = 13. Answer(b)(i) [1] (ii) Find the value of u5 . Answer(b)(ii) u5 = [1] (iii) Two consecutive terms of the sequence are 3413 and 6827 . Find the term before and the term after these two given terms. Answer(b)(iii) , 3413, 6827, [2]

Mark scheme: 12 (a) (i) 12, …, 30 2 B1 each (ii) (n + 1)(n + 2) oe 1 isw if expand incorrectly (iii) p = 2 1 q = 2 1 (iv) 69(th), 70(th) 2 M1 for their 2n + 2 = 140 soi (b) (i) 2 × 3 + 7 1 Accept 2 × 3 + 2 × 2 + 3 (ii) 27 1 (iii) 1707, …, 13 653 1,1

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