Cambridge IGCSE Mathematics 0580 — 2022 May/June Paper 4 · Variant 1

1 (a) The list shows 15 midday temperatures, in degrees Celsius, in Suntown. 17 21 21 18 23 22 25 19 21 17 19 18 21 24 23 (i) Complete the stem-and-leaf diagram to show this information. 1 7 2 Key: 1|7 represents 17 °C [2] (ii) Find the median. ............................................. °C [1] (iii) Find the upper quartile. ............................................. °C [1] (iv) Rahul draws a pie chart to show this information. Calculate the sector angle for the number of days the temperature is 18 °C. ................................................. [2] (b) 0 50 100 150 200 Mass (grams) The box-and-whisker plot shows information about the masses, in grams, of some apples. (i) Find the median. ............................................... g [1] (ii) Find the range. ............................................... g [1] (iii) Find the interquartile range. ............................................... g [1] (c) (i) The time, t minutes, spent on homework in one week by each of 200 students is recorded. The table shows the results. Time (t minutes) 40 1 t G 60 60 1 t G 80 80 1 t G 90 90 1 t G 100 100 1 t G 150 Frequency 6 10 70 84 30 Calculate an estimate of the mean. .......................................... min [4] (ii) A new table with different class intervals is completed. Time (t minutes) 40 1 t G 90 90 1 t G 150 Frequency 86 114 On a histogram the height of the bar for the 40 1 t G 90 interval is 17.2 cm. Calculate the height of the bar for the 90 1 t G 150 interval. ............................................ cm [2]

Mark scheme: 1(a)(i) 1 7 7 8 8 9 9 2 1 1 1 1 2 3 3 4 5 2 B1 for one row correctly ordered or for fully correct unordered stem-and-leaf diagram or for a correct diagram with one error or omission 1(a)(ii) 21 1 1(a)(iii) 23 1 1(a)(iv) 48 2 M1 for   2 360 15  or   360 2 15  1(b)(i) 120 1 1(b)(ii) 130 1 1(b)(iii) 60 1 1(c)(i) 93.4 4 M1 for mid-values soi M1 for fx M1 dep on second M for 200   fx 1(c)(ii) 19 2 M1 for 86 50 or 114 60

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2 (a) Alex, Bobbie and Chris share strawberries in the ratio Alex : Bobbie : Chris = 3 : 2 : 2. Chris receives 12 strawberries. Calculate the total number of strawberries shared. ................................................. [2] (b) In a sale, a shop reduces all prices by 12%. (i) Dina buys a book which has an original price of $6.50 . Calculate how much Dina pays for the book. $ ................................................. [2] (ii) Elu pays $11 for a toy. Calculate the original price of the toy. $ ................................................. [2] (c) Feri invests some money. The rate of interest for the first year is 2.5%. At the end of the second year the overall percentage increase of Feri’s investment is 6.6%. Find the rate of interest for the second year. ..............................................% [2] (d) A radioactive substance decays at an exponential rate of 2% per day. The initial mass is 80 g. (i) Find the mass at the end of 5 days. ............................................... g [2] (ii) Find how many more whole days, after day 5, it takes for the mass to reduce to less than 67 g. ................................................. [3]

Mark scheme: 2(a) 42 2 M1 for 12 ÷ 2 or better 2(b)(i) 5.72 2 M1 for 100 12 6.50 100   oe or B1 for 0.88 oe 2(b)(ii) 12.5[0] 2 M1 for 100 12 11 100    x or better oe Question Answer Marks Partial Marks 2(c) 4 2 M1 for 100 2.5 100 6.6 [...] 100 100     oe 2(d)(i) 72.3 or 72.31... 2 M1 for 5 100 2 80 100        oe 2(d)(ii) 4 nfww 3 B2 for answer 9 nfww or M2 for correct trials with values giving either side of 67 or M1 for 100 2 80 100        n = 67 or  100 2 67 100          k their i or an evaluated trial with n ⩾ 6 or k ⩾ 1

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3 (a) Geeta buys x apples, ( x + 7) oranges and ( 2x - 1) bananas. The total number of pieces of fruit Geeta buys is 30. (i) Find the number of apples Geeta buys. ................................................. [3] (ii) The cost of one apple is 15 cents. The cost of one orange is 18 cents. The total cost of all the fruit is $5.55 . Find the cost, in cents, of one banana. ........................................ cents [3] (b) (i) Solve. 3w 1 - 1 = 16 2 w = ................................................. [2] 3 ( 2 - y ) 1 (ii) - 1 = 16 2 Find the value of y. y = ................................................. [2] (c) (i) Solve the simultaneous equations. 2p + q = 2 1 p - q =- 2 p = ................................................. q = ................................................. [2] (ii) Hence, for 0° G u G 360° and 0° G v G 360° , solve the simultaneous equations. 2 sin u + cos v = 2 1 sin u - cos v =- 2 u = ................... or u = ................... v = ................... or v = ................... [4]

Mark scheme: 3(a)(i) 6 3 B2 for 4x + 6 = 30 or better or M1 for x + x + 7 + 2x – 1 [ = 30] 3(a)(ii) 21 3 M2 for (555 – their x  15 – their (x + 7) × 18) ÷ their (2x – 1) or M1 for their x  15 or their (x + 7) × 18 3(b)(i) 8 2 M1 for isolating the term in w or correctly removing all fractions e.g. 3 1 1 16 2  w or better or 3w – 16 = 8 3(b)(ii) 3  2 M1 for 2 8  y or 1 2 8  y or 2  y their w or better Question Answer Marks Partial Marks 3(c)(i) [p =] 1 2 oe [q =] 1 2 B1 for each If zero scored, SC1 for 2 values satisfying one of the original equations 3(c)(ii) [u =] 30 and 150 [v =] 0 and 360 4 B1 for each OR SC1 for sin u = their p and cos v = their q SC1 if their two different angles for u sum to 180 or if their different two angles for v sum to 360

More questions on Equations

14 f ( x) = 2 x - 1 g ( x) = 3 x - 2 h ( x) = , x ! 0 j ( x) = 5x x (a) Find (i) f ( 2) , ................................................. [1] (ii) gf ( 2) . ................................................. [1] (b) Find g -1 ( x) . g -1 ( x) = ................................................. [2] (c) Find x when h ( x) = j (- 2) . x = ................................................. [2] (d) Write f ( x) - h ( x) as a single fraction. ................................................. [2] (e) Find the value of jj ( 2 ) . ................................................. [1] (f) Find x when j -1 ( x) = 4 . x = ................................................. [2]

Mark scheme: 4(a)(i) 3 1 4(a)(ii) 7 1 FT their (i) 3×their (i) 2  4(b) 2 3  x oe final answer 2 M1 for y + 2 = 3x or 2 3 3   y x or x = 3y – 2 4(c) 25 2 M1 for 2 1 5  x oe 4(d) 2 2 1   x x x final answer 2 M1 for 2x – 1 – 1 x 4(e) 2.98 × 1017 or 2.980... × 1017 1 4(f) 625 2 M1 for x = j(4)

More questions on Functions

5 (a) ABCDEFGH is a regular octagon with sides of length 6 cm. The diagram shows part of the octagon. O is the centre of the octagon and M is the midpoint of AB. A M B NOT TO SCALE O (i) (a) Show that angle OAM is 67.5°. [2] (b) Calculate the area of the octagon. .......................................... cm2 [4] (ii) Find the area of the circle that passes through the vertices of the octagon. .......................................... cm2 [3] (b) NOT TO SCALE 4 m 0.45 m The diagram shows a horizontal container for water with a uniform cross-section. The cross-section is a semicircle. The radius of the semicircle is 0.45 m and the length of the container is 4 m. (i) Calculate the volume of the container. ........................................... m3 [2] (ii) NOT TO SCALE 0.3 m The greatest depth of the water in the container is 0.3 m. The diagram shows the cross-section. Calculate the number of litres of water in the container. Give your answer correct to the nearest integer.

Mark scheme: 5(a)(i)(a) 8 2 180 8 2    oe M2 8 2 180 8   or 360 2 8 4 90 8   5(a)(i)(b) 174 or 173.8.... 4 M3 for 1 6 2 OM oe or   2 1 2   OA sin45 oe or 1 6 67.5 2   OA sin oe where OA and OM are as in the M2 or M2 for 3 tan67.5  OM oe or for 3 67.5       OA cos or 6 67.5 45 sin sin oe or M1 for tan67.5 3  OM oe or for 3 67.5 cos OA oe or for 45 67.5 6  sin sin OA oe 5(a)(ii) 193 or 193.0 to 193.1 3 M2 for 2 3 67.5        cos oe or M1 for 3 67.5 cos r or 45 67.5 6  sin sin r Question Answer Marks Partial Marks 5(b)(i) 1.27 or 1.272 to 1.273 2 M1 for 2 1 0.45 4 2           or   2 1 0.45 4 2     5(b)(ii) 742 or 743 6 M5 for a method leading to the volume of water e.g. 2 0.15 cos 0.45 4 {2 0.45 360            inv 2 1 0.15 0.45 sin 2 cos 2 0.45                inv } oe OR M2   2 0.15 cos 0.45 2 0.45 360           inv oe or   2 0.15 90 cos 0.45 2 0.45 360            inv oe or M1 for use of 2 0.45 360     oe M2 for 2 1 0.15 0.45 sin 2 cos 2 0.45               inv oe or   1 0.15 0.15 0.45 sin cos 2 2 0.45                 inv oe Question Answer Marks Partial Marks 5(b)(ii) or M1 for use of 2 1 0.45 2  × sinθ oe or  1 2 0.15 0.45 sinβ 2     oe If 0 scored, SC1 for invcos 0.15 0.45       or invsin 0.15 0.45       or 2 2 0.45 0.15  soi

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6 (a) y 12 11 10 9 8 7 6 5 4 3 2 1 – 1 0 1 2 3 4 5 x – 1 – 2 – 3 – 4 – 5 – 6 – 7 – 8 – 9 – 10 The diagram shows the graph of y = f ( x) for - 1.5 G x G 5 . (i) Find f ( 2) . ................................................. [1] (ii) Solve the equation f ( x) = 0 for - 1.5 G x G 5 . x = .................. or x = .................. or x = .................. [3] (iii) f ( x) = k has three solutions for - 1.5 G x G 5 where k is an integer. Find the smallest possible value of k. k = ................................................. [1] (iv) On the grid, draw a line y = mx so that f ( x) = mx has exactly one solution for - 1.5 G x G 5 . [2] (b) y = 3 x 2 - 12x + 7 dy (i) Find the value of when x = 5 . dx ................................................. [3] (ii) Find the coordinates of the point on the graph of y = 3x 2 - 12x + 7 where the gradient is 0. ( ...................... , ...................... ) [2] p 2 dy 6(c) When y = 2x + qx , = 14x + 6x . dx Find the value of p and the value of q. p = ................................................. q = ................................................. [2]

Mark scheme: 6(a)(i) –3 1 6(a)(ii) –1 1.55 to 1.6 4.4 to 4.45 3 B1 for each 6(a)(iii) –8 1 6(a)(iv) Ruled line through origin intersecting curve once 2 B1 for ruled line through origin 6(b)(i) 18 3 B2 for 6x – 12 or B1 for 6x or –12 6(b)(ii) (2, –5) 2 B1 for each. If 0 scored, M1 for their 6x – 12 = 0 or states 0  dy dx 6(c) [p = ] 7 [q = ] 3 2 B1 for each

More questions on Differentiation

7 D NOT TO SCALE 12 km 9 km 14 km A C 25° 32° 123° B (a) Calculate angle ACD. Angle ACD = ................................................. [4] (b) Show that BC = 7.05 km , correct to 2 decimal places. [3] (c) Calculate the shortest distance from B to AC. ............................................ km [3] (d) Calculate the length of the straight line BD. BD = ............................................ km [4] (e) C is due east of A. Find the bearing of D from C. ................................................. [2]

Mark scheme: 7(a) 39.6 or 39.57.... 4 M2 for [cos =] 2 2 2 14 12 9 2 14 12     or M1 for 92 = 142 + 122 – 2 × 14 × 12 × cos ACD A1 for 0.7708... or 0.771 or 37 48 oe 7(b) 14sin25 sin123 M2 M1 for sin123 sin 25 14 BC  oe 7.054… A1 7(c) 3.74 or 3.735 to 3.739 3 M2 for 7.05 × sin 32 or M1 for recognition that the line from B is perpendicular to AC 7(d) 11.8 or 11.83 to 11.85 4 M1 for 32 + their(a) soi M2 for  2 2 12 7.05 2 12 7.05 cos( 32)      their a or M1 for    2 2 2 12 7.05 cos 32 2 12 7.05       BD their a 7(e) 309.6 or 309.57... 2 FT 270 + their(a) M1 for 270 + their(a) oe

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8 (a) (i) Use set notation to describe the shaded region in the Venn diagram. B A ................................................. [1] (ii) Shade the correct region in each Venn diagram. K L Q P M Q , P ' ( K , L ) + M ' [2] (b) V E N N D I A G R A M The diagram shows 11 cards. (i) One of these cards is chosen at random. Write down the probability that the letter on the card is not A. ................................................. [1] (ii) A card is chosen at random from these 11 cards and then replaced. A second card is then chosen at random. Find the probability that exactly one card has the letter N. ................................................. [3] (c) E M ............. .......... ............. ............. 50 students are asked if they like English (E) and if they like mathematics (M). 3 say they do not like English and do not like mathematics. 33 say they like English. 42 say they like mathematics. (i) Complete the Venn diagram. [2] (ii) A student is chosen at random. Find the probability that this student likes English and likes mathematics. ................................................. [1] (iii) Two students are chosen at random. Find the probability that they both like mathematics. ................................................. [2] (iv) Two students who like English are chosen at random. Find the probability that they both also like mathematics. ................................................. [2]

Mark scheme: 8(a)(i) A B 8(a)(ii) 2 B1 for each 8(b)(i) 9 11 1 8(b)(ii) 36 121 oe 3 M2 for 2 9 2 11 11   oe or M1 for 2 9 11 11  oe If 0 scored SC1 for 36 110 8(c)(i) 3, 5, 28, 14 correctly placed 2 B1 for 28 in the intersection 8(c)(ii) 28 50 oe 1 FT their 28 where their 28 50  8(c)(iii) 123 175 oe 2 M1 for 42 41 50 49  8(c)(iv) 63 88 oe 2 FT their 28 M1 for 28 28 1 33 32   their their

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9 (a) (x – 1) cm NOT TO x cm SCALE (2x + 1) cm x cm The area of the rectangle is 29cm2 greater than the area of the square. The difference between the perimeters of the two shapes is k cm. Find the value of k. You must show all your working. k = ................................................. [6] (b) NOT TO SCALE (y + 1) cm y cm The volume of the larger cube is 5cm3 greater than the volume of the smaller cube. (i) Show that 3y 2 + 3y - 4 = 0 . [4] (ii) Find the volume of the smaller cube. Show all your working and give your answer correct to 2 decimal places. .......................................... cm3 [4]

Mark scheme: 9(a) 2 30 0    x x B3 M1 for 2 (2 1)( 1) 29     x x x oe B1 for 2 (2 1)( 1) 2 2 1       x x x x x oe soi    6 5   x x oe M1 or correct factors for their 3 term quadratic equation or for correct substitution into quadratic formula or correctly completing the square for their 3 term quadratic equation 6  x cao B1 12 or 2 × their x evaluated or 2  k x stated B1 FT 9(b)(i)   3 3 1 5    y y oe M1   3 3 2 1 3 3 1      y y y y soi B2 B1 for   2 2 1 1      y y y y oe soi Completion to 3y2 + 3y – 4 = 0 A1 With no errors or omissions 9(b)(ii) 2 3 3 4(3)( 4) 2 3     B2 or B1 for 2 3 4(3)( 4)   or for 3 ... 2 3   or 3 ... 2 3   0.44 B2 B1 for 0.758 or 0.7583...

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