Cambridge IGCSE Mathematics (US) 0444 — 2022 May/June Paper 4 · Variant 1

1 (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]

Mark scheme: Question Answer Marks Partial Marks 1(a)(i) 6 3 B2 for 4x + 6 = 30 or better or M1 for x + x + 7 + 2x – 1 [ = 30] 1(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 1(b)(i) 8 2 M1 for isolating the term in w or correctly 3w 1 removing all fractions eg 1 or better 16 2 or 3w – 16 = 8 1(b)(ii) 3 2 y y 1 y M1 for 2  8 or 2  or 2  their w or 8 better

2 (a) Write down an example of continuous data. ................................................. [1] (b) A class of 24 students takes a test. The table shows their marks. Mark 6 7 8 9 10 Frequency 1 3 8 3 9 (i) Find (a) the range, ................................................. [1] (b) the mode, ................................................. [1] (c) the median. ................................................. [1] (ii) A pie chart is drawn to show the information in the table. Calculate the sector angle for the number of students who scored 10 marks. ................................................. [2] (c) 0 50 100 150 200 Mass (grams) The box 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] (d) (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: 2(a) e.g. of continuous data 1 2(b)(i)(a) 4 1 2(b)(i)(b) 10 1 2(b)(i)(c) 8.5 1 2(b)(ii) 135 2 9 M1 for [ 360] or 360[ 9] 24 24 2(c)(i) 120 1 2(c)(ii) 130 1 2(c)(iii) 60 1 2(d)(i) 93.4 4 M1 for mid-values soi M1 for fx M1 dep on second M for fx  200 2(d)(ii) 19 2 86 114 M1 for or 50 60

3 (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) Each day the mass of a radioactive substance decays at a rate of 2% of its mass on the previous 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: 3(a) 42 2 M1 for 12 ÷ 2 or better 3(b)(i) 5.72 2 100  12 M1 for  6.50 oe 100 or B1 for 0.88 oe 3(b)(ii) 12.5[0] 2 100  12 M1 for  x  11 or better oe 100 3(c) 4 2 100  2.5 100  6.6 M1 for  [...]  oe 100 100 3(d)(i) 72.3 or 72.31... 2 5  100  2  M1 for 80   oe  100  3(d)(ii) 4 nfww 3 B2 for answer 9 nfww or M2 for correct trials with values giving either side of 67  100  2  n or M1 for 80   = 67  100   100  2  k 67 or their i      100  or an evaluated trial with n  6 or k  1

4 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) x  2 2 y 2 oe final answer M1 for y + 2 = 3x or  x  or x = 3y – 2 3 3 3 4(c) 25 2 1 2 M1 for  5 oe x 4(d) 2 x 2  x  1 2 1 final answer M1 for 2x – 1 – x x 4(e) 2.98 × 1017 or 2.980... × 1017 1 4(f) 625 2 M1 for x = j(4)

5 (a) ABCDEFGH is a regular octagon with sides of length 6 cm. The diagram shows part of the octagon. O is the center 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 liters of water in the container. Give your answer correct to the nearest integer. ........................................ liters [6]

Mark scheme: 5(a)(i)(a)  8  2   180 M2  8  2   180 360 oe M1 for or or 8  2 8 8  2 8 4   90 8 5(a)(i)(b) 174 or 173.8.... 4 1 M3 for 6 OM oe 2 1 2 or   OA  sin45 oe 2 1 or 6 OA  sin 67.5 oe 2 where OA and OM are as in the M2 or M2 for OM 3 tan67.5 oe  3  6  sin 67.5 or for OA   or oe  cos 67.5  sin 45 OM or M1 for  tan67.5 oe 3 3 or for  cos 67.5 oe OA sin 45 sin 67.5 or for  oe 6 OA 5(a)(ii) 193 or 193.0 to 193.1 3 2  3  M2 for   oe  cos67.5  3 sin 45 sin67.5 or M1 for  cos67.5 or  r 6 r 5(b)(i) 1.27 or 1.272 to 1.273 2  1  2 M1 for  π  0.45  4    2  1 2 or  π  0.45  4  2 5(b)(ii) 742 or 743 6 M5 for a method leading to the volume of water  0.15  invcos    0.45  2 e.g. 4  {2   π  0.45 360 1 2   0.15     0.45  sin  2 invcos    } oe 2   0.45   OR  0.15  invcos    0.45  2 M2  2    π  0.45 oe 360  0.15  90  invcos    0.45  2 or  2    π  0.45 oe 360 2 oe or M1 for use of  π  0.45 360 1 2   0.15   M2 for  0.45  sin  2 invcos    oe 2   0.45   1   0.15    2  oe or  0.15  0.45  sin  invcos     2   0.45   1 2 or M1 for use of  0.45 × sinθ oe 2 or  2  1  0.15  0.45  sinβ oe 2 If 0 scored,  0.15   0.15  SC1 for invcos   or invsin    0.45   0.45  or 0.45 2  0.15 2 soi

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) By drawing a suitable straight line solve the equation f ( )x = 10 - 2 x . x = ................................................ [3] (v) 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) Line L passes through the point ( 4, - 1) and is perpendicular to the line y = 2x + 5 . Work out the equation of line L, giving your answer in the form y = mx + b . y = ................................................ [4]

Mark scheme: 6(a)(i) –3 1 6(a)(ii) –1 3 B1 for each 1.55 to 1.6 4.4 to 4.45 6(a)(iii) –8 1 6(a)(iv) Correct ruled line B2 At least through (4, 2) and (5, 0) B1 for ruled line with gradient – 2 or through (0, 10) 4.45 to 4.55 B1 6(a)(v) Ruled line through origin intersecting 2 B1 for ruled line through origin curve once 6(b) 1 4 M1 for grad = 2 y  x  1 final answer 2 1 M1 for grad of perpendicular =  their 2 M1 for –1 = (their m)  4+b oe

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 14 2  12 2  9 2 M2 for [cos =] 2  14  12 or M1 for 92 = 142 + 122 – 2 × 14 × 12 × cos ACD 37 A1 for 0.7708... or 0.771 or oe 48 7(b) 14sin 25 M2 sin123 sin 25 M1 for  oe sin123 14 BC 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 12 2  7.05 2 2 12  7.05  cos(their a  32) or M1 for 12 2  7.05 2  BD 2 cos  their a  32   2  12  7.05 7(e) 309.6 or 309.57... 2 FT 270 + their(a) M1 for 270 + their(a) oe

8 (a) Spinner A Spinner B The diagram shows two fair spinners. Spinner A is numbered 1, 2, 3, 4, 5 and spinner B is numbered 1, 2, 3, 4. The two spinners are spun and the two scores are added. (i) Draw a possibility diagram to show all the possible totals. [2] (ii) Find the probability that the total of the two numbers is (a) 7, ................................................. [1] (b) a square number, ................................................. [1] (c) less than 10. ................................................. [1] (iii) The two spinners are spun 60 times. Calculate the expected number of times the total is 7. ................................................. [1] (b) When a coin is tossed it is equally likely to show heads or tails. When a die is rolled it is equally likely to show a 1, 2, 3, 4, 5 or 6. (i) The die is rolled. Find the probability that the die shows 4. ................................................. [1] (ii) The coin is tossed and the die is rolled. (a) Find the probability that the coin shows tails and the die shows 4. ................................................. [2] (b) Find the probability that the coin shows tails or the die shows 4. ................................................. [2] (c) When the weather is fine, the probability that Jodie goes swimming is 4. 5 1 When the weather is not fine, the probability that Jodie goes swimming is . 10 The probability that the weather will be fine tomorrow is 2. 3 Find the probability that Jodie goes swimming tomorrow. ................................................. [3]

Mark scheme: 8(a)(i) Correct possibility diagram showing all 2 B1 for 20 correct totals twenty totals or 20 correct combinations or possibility diagram with no more than two errors 8(a)(ii)(a) 3 1 oe 20 8(a)(ii)(b) 1 1 oe 5 8(a)(ii)(c) 1 1 8(a)(iii) 9 1 3 FT their × 60 20 8(b)(i) 1 1 oe 6 8(b)(ii)(a) 1 2 1 1 oe M1 for  12 2 6 8(b)(ii)(b) 7 2 1 1 1 1 oe M1 for    12 2 6 2 6 6 If 0 scored SC1 for final answer oe 12 8(c) 17 3 2 4 1 1 M2 for    30 3 5 3 10 or M1 for one of these products

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 work. k = ................................................. [6] (b) NOT TO SCALE 2y cm y cm ( 2y+ 3) cm ( y + 3)cm The area of the larger triangle is 2 cm 2 greater than the area of the smaller triangle. (i) Show that 3y 2 + 3y - 4 = 0 . [4] (ii) Find the area of the smaller triangle. You must show all your work. ......................................... cm2 [4] Question 10 is printed on the next page.

Mark scheme: 9(a) x 2  x  30   0  B3 M1 for (2 x  1)( x  1)  x 2  29 oe B1 for (2 x  1)( x  1)  2 x 2  2 x  x  1 oe soi  x  6  x  5  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 x  6 cao B1 12 or 2 × their x evaluated B1 FT or k  2 x stated 9(b)(i) 1 1 M2 1 1 (2 y  3)2 y  ( y  3) y  2 oe M1 for (2 y  3)2 y or ( y  3) y 2 2 2 2 1 2 1 2 M1 i.e. correct expansions or correct removal of (4 y  6 y )  ( y  3 y )  2 oe fractions which may be seen together 2 2 or 2 y (2 y  3)  y ( y  3)  4 oe Completion to 3 y 2  3 y  4  0 A1 With no errors or omissions 9(b)(ii) 2 B2 2 3 3  4(3)( 4) or B1 for 3  4(3)( 4) 2  3 3 ... 3 ... or for or 2  3 2  3

10 (a) Solve the system of linear equations. 2p + q = 2 1 p - q =- 2 p = ................................................. q = ................................................. [2] (b) Hence, for 0° G u G 360° and 0° G v G 360 ° , solve this system of equations. 2 sin u + cos v = 2 1 sin u - cos v =- 2 u = ................... or u = ................... v = ................... or v = ................... [4]

Mark scheme: 10(a) 1 2 B1 for each [p =] oe [q =] 1 If zero scored, SC1 for 2 values satisfying one 2 of the original equations 10(b) [u =] 30 and 150 4 B1 for each [v =] 0 and 360 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