Cambridge IGCSE Mathematics - International 0607 — 2024 May/June Paper 4 · Variant 1

1 (a) Solve the equations. (i) 3x - 2 = - 14 x = ................................................. [2] (ii) 7x + 11 = 26 - 3 x x = ................................................. [2] (b) Solve the simultaneous equations. You must show all your working. 5x + 3y = - 15 3x + 5y = - 17 x = ................................................. y = ................................................. [4] (c) Solve the inequality. 2x + 1 2 9 ................................................. [4]

Mark scheme: Question Answer Marks Partial Marks 1(a)(i) –4 2 2 14 M1 for 3x = –14 + 2 or x   3 3 1(a)(ii) 15 2 M1 for 7x + 3x = 26 – 11 oe or better 1.5 or oe 10 1(b) Correctly equating coefficients M1 or sketch of one equation with negative slope and y intercept Correct method to eliminate one variable M1 or sketch of other equation with negative slope and y intercept If 0 scored, SC1 for answers that satisfy one x = –1.5 oe A1 equation. SC1 for two correct answers with no working y = –2.5 oe A1 1(c) x > 4 and x < –5 4 M2 for correct sketch showing both answers Mark final answer or M1 for appropriate sketch of y = 2 x  1 OR M2 for 2x + 1 > 9 oe and 2x + 1 < –9 oe or M1 for either correct inequality or for 2x + 1 = 9 and 2x + 1 = –9 oe B1 for x > 4 or x < –5 Mark final answer

More questions on Equations

2 (a) The heights, x cm, of 100 plants are shown in the table. Height (x cm) 0 1 x G 20 20 1 x G 35 35 1 x G 40 40 1 x G 60 60 1 x G 80 Frequency 7 13 20 32 28 (i) Calculate an estimate of the mean height of the plants. ............................................ cm [2] (ii) (a) Complete the cumulative frequency table for the plants. Height (x cm) x G 20 x G 35 x G 40 x G 60 x G 80 Cumulative 7 100 frequency [1] (b) On the grid, draw the cumulative frequency curve. 100 90 80 70 60 Cumulative 50 frequency 40 30 20 10 0 x 0 10 20 30 40 50 60 70 80 Height (cm) [3] (c) Use your cumulative frequency curve to find an estimate for the interquartile range. ............................................ cm [2] (b) The heights, h cm, of 50 different plants are shown in the table, where k is an integer. Height (h cm) Frequency 0 1 h G 20 25 20 1 h G k 15 k 1 h G 80 10 An estimate of the mean height of these plants is 27 cm. Find the value of k. k = ................................................. [3]

Mark scheme: 2(a)(i) 47.4 or 47.375 2 M1 for at least 4 correct mid-points soi 2(a)(ii)(a) [7], 20, 40, 72, [100] 1 In parts (a)(ii)(b) and (a)(ii)(c), marks can only be earned with an increasing curve (a)(ii)(b) Correct curve 3 B1 for horizontal plot correct (20, 7), (35, 20), (40, 40), (60, 72), B1FT for at least 4 vertical plots correct (80, 100) (a)(ii)(c) 24 to 28 2 M1FT for [UQ =] 61 to 63 or [LQ =] 35 to 37 2(b) 44 3 M2 for (20  k ) ( k  80) 10  25   15   10 [  27  50] 2 2 oe (20  k ) ( k  80) or M1 for or 2 2

More questions on Averages and measures of spread

3 y 9 8 7 6 5 4 A 3 2 1 – 9 – 8 – 7 – 6 – 5 – 4 – 3 – 2 – 1 0 1 2 3 4 5 6 7 8 9 x – 1 – 2 – 3 – 4 – 5 – 6 – 7 – 8 – 9 2 (a) Translate triangle A with vector Label the image B. [2] e- 6o. (b) Describe fully the single transformation that maps triangle B onto triangle A. ..................................................................................................................................................... ..................................................................................................................................................... [2] (c) Rotate triangle A through 90° clockwise about (0, 0). Label the image C. [2] (d) Reflect triangle A in the line y = x . Label the image D. [2] (e) Describe fully the single transformation that maps triangle C onto triangle B. ..................................................................................................................................................... ..................................................................................................................................................... [3]

Mark scheme: 3(a) Correct triangle 2  2   k  (–5, 0), (–5, –3), (–4, –3) B1 for translation   or    k   6  3(b) Translation 2 B1 for each  2     6  3(c) Correct triangle 2 B1 for correct rotation, incorrect centre or for (3, 6), (3, 7), (6, 7) correct rotation 90 anticlockwise 3(d) Correct triangle 2 B1 for correct size and orientation, wrong (3, –6), (3, –7), (6, –7) position or reflection in y = –x 3(e) Rotation 3 B1 for each 90 anticlockwise oe [Centre] (4, –2)

More questions on Transformations

4 (a) The price of a coat is $84. The price is reduced by 12%. Find the new price of the coat. $ ................................................. [2] (b) The price of a table is reduced by 25%. The price is now $960. Find the original price of the table. $ ................................................. [2] (c) Samir invests $600 in a bank that pays compound interest at a rate of 5.1% each year. (i) Find the value of Samir’s investment after 4 complete years. $ ................................................. [2] (ii) Find the number of complete years for the value of Samir’s investment to be first worth more than $1000. ................................................. [4] (d) Amir and Bob work together and share their earnings in the ratio 3 : 5. (i) Find the amount Bob receives when their earnings are $120. $ ................................................. [2] (ii) They decide to change the ratio for all further earnings. Amir’s share of the earnings is increased by 20% of his original share. Bob’s share of the earnings is decreased by 20% of his original share. Show that the ratio of their earnings is now 9 : 10. [3]

Mark scheme: 4(a) 73.92 2 (100  12) M1 for 84  soi 100 or B1 for 10.08 4(b) 1280 2 100  25 M1 for x   960 oe 100 4(c)(i) 732.09 2 M1 for 600  1.0514 oe 4(c)(ii) 11 4 B3 for 10.3 or 10.26 to 10.27 1000 or M3 for n log1.051  log 600 or good sketch indicating value between 10 and 11 or correct trials reaching 10 and 11 n 1000 or M2 for 1.051  oe 600 or sketch that could lead to solution e.g. y  1.051x , y  1.67 or at least 3 correct trials with n > 4 or M1 for 600  1.051n  1000 oe or suitable graph e.g. y  1.051x or at least 2 trials with n > 4 4(d)(i) 75 2 5 M1 for [120  ]5  3 oe 4(d)(ii) 3[ k ](1  0.2) : 5[ k ](1  0.2) M2 M1 for 5[ k ](1  0.2) or 3[ k ](1  0.2) or for their (d)(i)(1 – 0.2) Leading to 9:10 A1 No errors seen

More questions on Exponential growth and decay

5 A NOT TO B SCALE 61° 22° D O P C A, B, C and D lie on a circle, centre O. AP is a tangent to the circle at A. OP is perpendicular to AC and AD is parallel to BC. Angle ABC = 61° and angle PAD = 22° . (a) Write down the mathematical name of the cyclic quadrilateral ABCD. ................................................. [1] (b) Complete the statement. Angle OAP = 90° because ......................................................................................................... ..................................................................................................................................................... [1] (c) Find (i) angle ADC Angle ADC = ................................................. [1] (ii) angle ACD Angle ACD = ................................................. [1] (iii) angle ACB Angle ACB = ................................................. [2] (iv) angle OCA. Angle OCA = ................................................. [2]

Mark scheme: 5(a) trapezium 1 5(b) Angle between tangent and radius oe 1 5(c)(i) 119 1 5(c)(ii) 22 1 5(c)(iii) 39 2 M1 for 180 – their (i) – their (ii) 5(c)(iv) 29 2 B1 for 61 or 122 at centre (may be on diagram)

More questions on Circle theorems I

6 y 3 – 3 0 3 x – 3 f( x) = 2 - 1 - 0 .5 x 2 (a) On the diagram, sketch the graph of y = f( x) , for values of x between - 3 and 3. [3] (b) The graph cuts the x-axis at points A and B. Work out the length AB. AB = ................................................. [2] (c) Solve f( x) = 0 .5 . ................................................................................ [2] (d) Write down the coordinates of the minimum point of the graph. (....................... , .......................) [1] (e) The equation f( )x = k has two solutions. Find the range of values of k. ................................................................................ [2]

Mark scheme: 6(a) Correct sketch 3 B2 if peaks below y = 1 (by eye) or rounded at peaks not cusps or outside branches convex from below. or B1 for graph symmetrical about y-axis and in all 4 quadrants 6(b) 4.9[0] or 4.898 to 4.899 2 B1 for 2.45 or –2.45 or 2.449... or –2.449… seen 6(c) –2.24 or –2.236… 2 B1 for each or for both values seen 2.24 or 2.236… 6(d) (0, 1) 1 6(e) k =2 2 B1 for each k < 1

More questions on Graphs of functions

7 (a) Spinner A and spinner B are each fair 5-sided spinners. Spinner A is numbered 1, 2, 2, 3, 4. Spinner B is numbered 1, 2, 3, 4, 4. The two spinners are each spun once and the number on each spinner is recorded. Find the probability that (i) the number on spinner A is 6 ................................................. [1] (ii) the number on spinner B is not 4 ................................................. [1] (iii) the number on spinner A is the same as the number on spinner B ................................................. [3] (iv) the sum of the two numbers is 6. ................................................. [3] (b) (i) On the Venn diagram, shade A , B . U A B [1] (ii) Describe the shaded region using set notation. U A B ................................................. [1] (iii) The Venn diagram below shows the number of elements in each subset. U A B 3 8 4 1 6 9 7 5 C Find n ( A + B) + C l ` j. ................................................. [1]

Mark scheme: 7(a)(i) 0 1 7(a)(ii) 3 1 oe 5 7(a)(iii) 6 3  1 1   2 1   1 1   1 2  oe M2 for                25  5 5   5 5   5 5   5 5  or correct sample space showing all 6 points or M1 for 2 correct products or correct sample space showing at least 3 points or list of all correct pairs 12 If 0 scored, SC1 for 25 7(a)(iv) 6 3  2 2   1 1   1 1  M2 for            25  5 5   5 5   5 5  or correct sample space showing all 6 points or M1 for 2 correct products or correct sample space showing at least 3 or for listing pairs that sum to 6 7(b)(i) 1 7(b)(ii) A  B / 1 7(b)(iii) 8 1

More questions on Introduction to probability

8 8 m A B 73.4° NOT TO 36° SCALE 7 m C D The diagram shows a shape ABDC formed from triangle ABC and a sector of a circle BCD, centre B. (a) Show that BC = 9.0 m , correct to 1 decimal place. [3] (b) Use the sine rule to find angle BCA. Angle BCA = ................................................. [3] (c) Find the area of triangle ABC. ........................................... m2 [2] (d) Find the area of the shaded region. ........................................... m2 [3] (e) Find the perimeter of the shape ABDC. .............................................. m [2]

Mark scheme: 8(a) M2 or M1 for [ BC 2 ]  82  7 2 2 8 7 cos73.4 [ BC  ] 8 2  7 2  2 8 7 cos73.4 M1 for 81.[00…] 9.00[0…] A1 8(b) 8  sin73.4 M2 8 9 [sin...]  M1 for  oe 9 sin BCA sin73.4 58.4 or 58.41… B1 8(c) 26.8 or 26.83… 2 1 M1 for 7 8 sin73.4 oe 2 8(d) 1.64 or 1.641 to 1.642 3 36 M1 for areasector BCD  π 9 9 360 M1 for area triangle BCD  0.5 9 9 sin36 8(e) 29.7 or 29.65 to 29.66 NFWW 2 36 M1 for  2π  9 oe 360

More questions on Non-right-angled triangles

9 NOT TO SCALE 60 cm 40 cm The diagram shows a solid cone with base radius 40 cm and slant height 60 cm. (a) Find the volume of the cone. .......................................... cm3 [3] (b) Show that the total surface area of the cone is 4000 r cm 2 . [2] (c) A mathematically similar cone has a surface area of 1000 r cm 2 . Show that the radius of this cone is 20 cm. [2] (d) A cone with radius 20 cm is removed from the top of the cone with radius 40 cm to leave a solid. Calculate the surface area of the remaining solid. .......................................... cm2 [3]

Mark scheme: 9(a) 74 900 or 74 929 to 74 941.1 3 1 2 2 2 M2 for  π  40  60  40 oe 3 or M1 for 602 – 402 [=2000] oe 9(b) π × 402 + π × 40 × 60 M2 M1 for π × 402 or π × 40 × 60 = 4000π with no errors 9(c) Ratio areas = 4000π : 1000π M1 implies Ratio sides = 2 : 1 oe [r=] 40 × 0.5 = 20 oe A1 ALTERNATIVE (M2) 1000π 4000π M1 for or oe 1000π 4000π 1000π 40  oe 4000π 2  40 4000π  oe = 20 with no errors or    x  1000π 9(d) 11 900 or 11 930 to 11 940 or 3800π 3 M2 for  40 2  20 2  60  40  30  20 or M1 for π  60  40  π  30  20 If 0 scored, SC1 for 3400π or 10 700 or 10 680 to 10 681.4…

More questions on Surface area and volume

10 y varies inversely as the square root of ( x + 1) . y = 18 when x = 3 . (a) (i) Find the value of y when x = 8 . y = ................................................. [3] (ii) Find the value of x when y = 1.5 . x = ................................................. [2] (b) w varies directly as the square root of ( x + 1) . w = 18 when x = 3 . Find the value of wy. wy = ................................................. [3]

Mark scheme: 10(a)(i) 12 3 k M1 for y  oe x  1 A1 for k = 36 OR 8  1 M2 for 18  oe 3  1 y 3  1 or M1 for  oe 18 8  1 10(a)(ii) 575 2 k M1 for x their1 or better 1.5 10(b) 18 NFWW 3 M1 for 18  c  3  1 oe or better their 36 M1 for wy  their 9 x  1  oe x  1

More questions on Proportion

11 f( )x = 3 x - 1 g( )x = 5 - 2 x h( x) = , x ! 1 .5 2x - 3 (a) Find f( 4) . ................................................. [1] (b) Solve f( )x = - 7 . ................................................. [2] (c) Find g -1 ( )x . g -1 ( )x = ................................................. [2] (d) Solve g( x) = 7 h(f( x)) . You must show all your working. x = ................................................. [6]

Mark scheme: 11(a) 11 1 11(b) –2 2 M1 for 3x = – 7 + 1 11(c) 5 x 2 y 5 oe final answer M1 for x = 5 – 2y or   x or 2x = 5 – y 2 2 2 11(d) 1 M1  h(f ( x ))   2(3 x  1)  3 (5  2 x )(6 x  5)  7 A1 All further FTs dep on second stage in correct form (5 – 2x)(ax + b) = k where a, and b are integers or sketch of rectangular hyperbola Correct expansion of brackets M1 30x – 25 – 12x2 + 10x [= 7] or sketch of straight line with negative gradient Correct rearrangement to 3 term quadratic M1 12 x 2  40 x  32  0 oe on one side or graphs intersecting twice in 1st quadrant Correct factorisation M1 (6 x  8)(2 x  4)  0 oe or correct use of formula or correct sketch of the quadratic or solutions indicated at points of intersection 4 B1 Both answers correct 2, oe 3

More questions on Functions