Cambridge IGCSE Mathematics 0580 — 2021 May/June Paper 4 · Variant 3

1 (a) (i) Yasmin and Zak share an amount of money in the ratio 21 : 19. Yasmin receives $6 more than Zak. Calculate the total amount of money shared by Yasmin and Zak. $ ................................................. [2] (ii) In a sale, all prices are reduced by 15%. (a) Yasmin buys a blouse with an original price of $40. Calculate the sale price of the blouse. $ ................................................. [2] (b) Zak buys a shirt with a sale price of $29.75 . Calculate the original price of the shirt. $ ................................................. [2] (b) Xavier’s salary increases by 2% each year. In 2010, his salary was $40 100. (i) Calculate his salary in 2015. Give your answer correct to the nearest dollar. $ ................................................. [3] (ii) In which year is Xavier’s salary first greater than $47 500? ................................................. [3] (c) In January 2020, the population of a town was 5% more than its population in January 2018. In January 2021, the population of this town was 2% less than its population in January 2020. Calculate the overall percentage increase in the population from January 2018 to January 2021. ............................................. % [2]

Mark scheme: Question Answer Marks Partial Marks 1(a)(i) 120 2 M1 for 6 ÷ (21 – 19) oe soi 2 x or for = 6 40 1(a)(ii)(a) 34 2 15 M1 for 40 – × 40 oe or better 100 or B1 for 6 1(a)(ii)(b) 35 2  15  M1 for  1 −  × p = 29.75 or better  100  1(b)(i) 44 274 cao 3 B2 for 44273 to 44274 or 44270  2  5 or M1 for 40100 ×  1 +  oe  100  1(b)(ii) 2019 nfww 3 M2 for one correct trial of n = 8 or n = 9 either to find a salary or, if working with 1.02n and 47 500÷ 40 100 [= 1.1845], to find a value of 1.02n or B2 for final answer 9 or 4 nfww or M1 for  2  n their 44 274 ×  1 +  = 47 500 oe  100   2  n or 40 100 ×  1 +  = 47 500 oe  100  or for at least one trial giving a value greater than their 44 274 1(c) 2.9 [increase] 2  5  2  M1 for  1 + × 1 −  oe  100  100  implied by 1.029 or 102.9[%]

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2 (a) y = px 2 + t (i) Find the value of y when p = 3, x = 2 and t = -13. y = ................................................. [2] (ii) Rearrange the formula to write x in terms of p, t and y. x = ................................................. [3] (b) (i) Factorise. 15x 2 - 2x - 8 ................................................. [2] (ii) Solve the equation. 15x 2 - 2x - 8 = 0 x = ..................... or x = ........................ [1] (c) Factorise completely. x 3 - 16xy 2 ................................................. [3] (d) Simplify. 2x - 1 - 4ax + 2a 2x 2 - x ................................................. [4]

Mark scheme: 2(a)(i) –1 2 M1 for 3 × 22 – 13 oe 2(a)(ii) y − t 3 M1 for correct rearrangement to isolate x2 [±] oe final answer term p M1 for correct division by p M1 for correct square root Incorrect answer scores a maximum of M2 If 0 scored, SC1 for a correctly rearranged formula with p = 3 and t = – 13 substituted 2(b)(i) (5 x − 4)(3 x + 2) oe final answer 2 B1 for ( ax + b )( cx + d ) where either ac = 15 and bd = –8 or ad + bc = –2 or 5x(3x + 2) – 4(3x + 2) or 3x(5x – 4) + 2(5x – 4) or correct factors seen and spoiled 2(b)(ii) 4 2 1 FT a factorised quadratic oe and − oe 5 3 2(c) x ( x + 4 y )( x − 4 y ) final answer 3 B2 for ( x 2 + 4 xy )( x − 4 y ) or ( x + 4 y )( x 2 − 4 xy ) or answer in the form x(a + b)(a – b) or correct answer seen and spoiled or B1 for x ( x 2 − 16 y 2 ) oe or ( x + 4 y )( x − 4 y ) 2(d) 1 −a2 4 B2 for (2x – 1)(1 – 2a) oe oe final answer or B1 for 2x – 1 – 2a(2x – 1) x or 2x(1 – 2a) – (1 – 2a) B1 for x(2x – 1)

More questions on Algebraic manipulation

3 (a) Zoe’s test scores last term were 6 7 7 7 8 9 9 10 10. Find (i) the range, ................................................. [1] (ii) the mode, ................................................. [1] (iii) the median. ................................................. [1] (b) The cumulative frequency diagram shows information about the time taken by each of 200 students to solve a problem. 200 180 160 140 120 Cumulative 100 frequency 80 60 40 20 0 0 2 4 6 8 10 12 14 16 18 20 Time (minutes) Use the diagram to find an estimate of (i) the median, .......................................... min [1] (ii) the interquartile range. .......................................... min [2] (c) The test scores of 200 students are shown in the table. Score 5 6 7 8 9 10 Frequency 3 10 43 75 48 21 Calculate the mean. ................................................. [3] (d) The height, in cm, of each of 200 plants is measured. The histogram shows the results. 4 3 Frequency 2 density 1 0 50 60 70 80 90 100 110 120 130 140 150 160 Height (cm) Calculate an estimate of the mean height. You must show all your working. ............................................ cm [6]

Mark scheme: 3(a)(i) 4 1 3(a)(ii) 7 1 3(a)(iii) 8 1 3(b)(i) 14 1 3(b)(ii) 4 2 B1 for [ l.q. =] 11 or [u.q =] 15 3(c) 8.09 3 M1 for 5 × 3 + 10 × 6 + 43 × 7 + 75 × 8 + 48 × 9 + 21 × 10 M1 dep ÷ 200 3(d) 30, 70, 40, 36, 24 seen B2 B1 for 3 or 4 correct or M1 for 1 × (80 – 50), 3.5 × (100 – 80), 4 × (110 – 100), 3.6 × (120 – 110) and 0.6 × (160 – 120) oe (their 30 × 65 + their 70 × 90 M3 M1 for midpoints soi + their 40 × 105 + their 36 × 115 M1 for Σfx , x in interval or boundary of + their 24 × 140) ÷ 200 interval M1 dep on second M1 for ÷ 200 99.75 A1

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4 (a) A is the point (1, 5) and B is the point (3, 9). M is the midpoint of AB. (i) Find the coordinates of M. (...................... , ...................... ) [2] (ii) Find the equation of the line that is perpendicular to AB and passes through M. Give your answer in the form y = mx + c . y = ................................................. [4] - 2 - 2 (b) The position vector of P is and the position vector of Q is e 3o e 5o. (i) Find the vector PQ. [2] f p (ii) R is the point such that PR = 3PQ . Find the position vector of R. [2] f p (c) U NOT TO SCALE u Y T O t OT = t , OU = u and UY = 2YT. (i) Find OY in terms of t and u. Give your answer in its simplest form. OY = ................................................. [2] (ii) Z is on OT and YZ is parallel to UO. Find OZ in terms of t and/or u. Give your answer in its simplest form. OZ = ................................................. [1]

Mark scheme: 4(a)(i) (2, 7) 2 B1 for each coordinate 4(a)(ii) 1 4 Correct equivalent in different form − x + 8 oe scores 3 marks. 2 9 − 5 4 M1 for gradient of AB = or or 2 3 − 1 2 M1 dep for gradient 1 p = −their grad of AB M1 (dep on previous M1) for substitution of their midpoint into y = (their p)x + c oe where their p ≠ 0 4(b)(i) 0 2 0 k  B1 for  or  2 k 2  4(b)(ii)  − 2  2 FT their PQ    9  0 B1FT for  6 4(c)(i) 2 1 1 2  2 t + u or (2t + u) final answer M1 for UY = ( t –u) oe 3 3 3 3  1 or TY = (u – t) oe 3 or correct route soi 4(c)(ii) 2 1 t cao 3

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5 Solve the simultaneous equations. (a) x + 2y = 13 x + 5y = 22 x = ................................................. y = ................................................. [2] (b) y = 2 - x y = x 2 + 2x + 2 x = .................... y = .................... x = .................... y = .................... [4]

Mark scheme: 5(a) [x =] 7 2 B1 for each [y =] 3 5(b) [x =] 0, [y =] 2 4 B3 for x = 0 and x = –3 or B2 for x2 + 3x = 0 [x =] –3, [y =] 5 or M1 for 2 – x = x2 + 2x + 2 If 0 scored award B1 for x = 0, y = 2 or x = –3, y = 5 from no/incorrect working ALTERNATIVE B3 for y = 2 and y = 5 or B2 for y2 – 7y + 10 = 0 or M1 for y = (2 – y)2 + 2(2 – y) + 2 If 0 scored award B1 for x = 0, y = 2 or x = –3, y = 5 from no/incorrect working

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6 In a class of 24 students, 18 students like homework (H ), 15 students like tests (T ) and 1 student does not like homework and does not like tests. (a) Complete the Venn diagram to show this information. H T ...... ...... ...... 1 ...... [2] (b) Write down the number of students who like both homework and tests. ................................................. [1] (c) Find n (H l + T ) . ................................................. [1] (d) A student is picked at random from the class. Write down the probability that this student likes tests but does not like homework. ................................................. [1] (e) Two students are picked at random from the class. Find the probability that both students do not like homework and do not like tests. ................................................. [1] (f) Two of the students who like homework are picked at random. Find the probability that both students also like tests. ................................................. [3]

Mark scheme: 6(a) 2 i.e. 8, 10 and 5 correctly placed T H B1 for 10 correctly placed or M1 for 18 – x , x and 15 – x correctly 8 10 5 placed on diagram and x = 10 seen 1 6(b) 10 1 FT their Venn diagram 6(c) 5 1 FT their Venn diagram 6(d) 5 1 FT their 5 on the Venn diagram oe 24 6(e) 0 1 6(f) 5 3 their10 their 9 oe M2 for × 17 18 17 their10 their 9 or B1FT for or seen 18 17 25 After 0 scored, SC1 for answer oe 81

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7 (a) x –2 1 Write down the inequality in x shown by the number line. ................................................. [2] (b) (i) Write x 2 + 4x + 1 in the form ( x + p) 2 + q . ................................................. [2] (ii) Use your answer to part (b)(i) to solve the equation x 2 + 4x + 1 = 0 . x = .......................... or x = .......................... [2] (iii) Use your answer to part (b)(i) to write down the coordinates of the minimum point on the graph of y = x 2 + 4x + 1. (...................... , ...................... ) [2] (iv) On the diagram, sketch the graph of y = x 2 + 4x + 1. y O x [2]

Mark scheme: 7(a) –2 < x ⩽ 1 2 B1 for –2 < x or x ⩽ 1 7(b)(i) ( x + 2 ) 2 − 3 2 M1 for ( x + 2 ) 2 + k 7(b)(ii) ( x + 2 ) 2 = 3 M1 FTdep their (b)(i) for k < 0 –3.73 or –3.732... and B1 –0.268 or –0.2679... 7(b)(iii) (–2, –3) 2 2 FT their ( x + 2 ) − 3 B1 for each coordinate 7(b)(iv) Correct sketch 2 Parabola with minimum point in correct 2222 quadrant and both x-intercepts negative and positive y-intercept 1111 4444 -2-2-2-2 0000 0000 2222 4444 B1 for parabola with minimum point. -1-1-1-1 -2-2-2-2 -3-3-3-3 -4-4-4-4

More questions on Algebraic manipulation

8 (a) A solid cuboid measures 20 cm by 12 cm by 5 cm. (i) Calculate the volume of the cuboid. .......................................... cm3 [1] (ii) (a) Calculate the total surface area of the cuboid. .......................................... cm2 [3] (b) The surface of the cuboid is painted. The cost of the paint used is $1.52 . Find the cost to paint 1 cm 2 of the cuboid. Give your answer in cents. ........................................ cents [1] 9x (b) A solid metal cylinder with radius x and height is melted. 2 All the metal is used to make a sphere with radius r. Find r in terms of x. 4 3 [The volume, V, of a sphere with radius r is V = rr .] 3 r = ................................................. [3] (c) NOT TO SCALE 20 cm 5 cm 150 cm The diagram shows a cylinder of length 150 cm on horizontal ground. The cylinder has radius 20 cm. The cylinder contains water to a depth of 5 cm, as shown in the diagram. Calculate the volume of water in the cylinder. Give your answer in litres. ........................................ litres [7]

Mark scheme: 8(a)(i) 1200 1 8(a)(ii)(a) 800 3 M2 for [2 ×] (20 × 12 + 20 × 5 + 12 × 5) or M1 for 20 × 12 or 20 × 5 or 12 × 5 8(a)(ii)(b) 0.19 1 FT 152 ÷ their 800 8(b) 3 x 3 3 27 x 3 [π] or 1.5x B2 for r = or better 2 8[π] 4 3 2 9 x or M1 for πr = πx × 3 2 8(c) 13.6 or 13.59 to 13.61 7 If chord is AB and O is centre of the cross section −1  20 − 5  M2 for 2 × cos   oe  20  20 − 5 or M1 for cos = oe 20 theirAOB 2 M1 for × π × 20 360 1  82.8π  or (20)2   2  180  1 M1 for × 202 × sin(their AOB) oe 2 M1 for their area × 150 M1 for their volume ÷ 1000

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9 (a) C NOT TO SCALE D 60° 14 cm 45° 35° B A Calculate the perimeter of the quadrilateral ABCD. ............................................ cm [7] (b) B NOT TO SCALE A The diagram shows a cube. The length of the diagonal AB is 8.5 cm. (i) Calculate the length of an edge of the cube. ............................................ cm [3] (ii) Calculate the angle between AB and the base of the cube. ................................................. [3]

Mark scheme: 9(a) 42.3 or 42.28 to 42.30... 7 AB M1 for = cos35 oe 14 AD M1 for = sin35 oe 14 B1 for [C =] 75 14sin60 M3 for [BC =] oe sin their 75 14sin45 and [DC] oe sin their 75 14sin60 14sin45 or M2 for or oe sin their 75 sin their 75 sin their 75 sin60 or M1 for = oe 14 BC sin their 75 sin45 or = oe 14 CD 9(b)(i) 4.91 or 4.907... 3 B2 for [l2 =] 24.1 or 24.08... 8.5 2 or M2 for √3 l = 8.5 or [l =] oe 3 or M1 for l 2 + l 2 + l 2 = 8.5 2 oe 9(b)(ii) 35.3 or 35.26 to 35.3 nfww 3 their (b)(i) M2dep for sin (angle) = oe 8.5 or M1 for clear recognition of correct angle

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10 f ( )x = 3 x - 2 g ( )x = 5 x - 7 h ( )x = x 2 + x j ( )x = 3 x (a) Find (i) f(2), ................................................. [1] (ii) g(2), ................................................. [1] (iii) gf(2). ................................................. [1] (b) Find f -1 ( )x . f -1 ( )x = ................................................. [2] (c) Find hf(x), giving your answer in the form ax 2 + bx + c . ................................................. [3] (d) Find the derivative of h(x). ................................................. [1] (e) (i) Find x when j -1 ( )x = 4 . x = ................................................. [1] (ii) Simplify j -1 j ( )x . ................................................. [1]

Mark scheme: 10(a)(i) 4 1 10(a)(ii) 3 1 10(a)(iii) 13 1 FT 5 × their (a)(i) – 7 10(b) x + 2 2 y 2 final answer M1 for y + 2 = 3x or for = x − 3 3 3 or for x = 3y – 2 10(c) 9x2 – 9x + 2 final answer 3 2 M1 for ( 3 x − 2 ) + 3 x − 2 2 2 B1 for ( 3 x − 2 ) = 9 x − 6 x − 6 x + 4 10(d) 2x + 1 1 10(e)(i) 81 1 10(e)(ii) x 1 Not y = x

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11 (a) These are the first four terms of a sequence. 11 7 3 -1 (i) Write down the next term. ................................................. [1] (ii) Write down the term to term rule for this sequence. ............................................................................................................................................. [1] (iii) Find the nth term of this sequence. ................................................. [2] 2n (b) The nth term of a different sequence is . n + 1 (i) Find the difference between the 5th term and the 6th term of this sequence. Give your answer as a fraction. ................................................. [2] 3 (ii) Is a term in this sequence? 4 Show how you decide. [3]

Mark scheme: 11(a)(i) –5 1 11(a)(ii) Subtract 4 oe 1 11(a)(iii) 15 – 4n oe final answer 2 B1 for k – 4n or 15 – jn j ≠ 0 11(b)(i) 1 2 12 10 or equivalent fraction B1 for and 21 7 6 11(b)(ii) 3 M2 3 2 n n = oe M1 for = oe 5 4 n + 1 or or 2n ⩾ n + 1 but 3 < 4. M1 for 2n > n + 1 but 3 < 4 No, n is not an integer oe A1 or 3 No, is less than 1, oe 4

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