Cambridge IGCSE Mathematics - International 0607 — 2023 Oct/Nov Paper 4 · Variant 2

1 y 6 5 4 V 3 2 1 x – 7 – 6 – 5 – 4 – 3 – 2 – 1 0 1 2 3 4 5 6 – 1 – 2 T – 3 – 4 – 5 – 6 (a) Reflect triangle T in the line x =- 1. Label the image A. [2] 1 (b) Translate triangle T using the vector Label the image B. [2] e4o. (c) Describe fully the single transformation that maps triangle T onto triangle V. ..................................................................................................................................................... ..................................................................................................................................................... [3] 1 (d) Stretch triangle V with factor and invariant line x = 2 . Label the image C. [2] 2

Mark scheme: Question Answer Marks Partial Marks 1(a) triangle (–6, –2) (–4, –2) (–4, –5) 2 B1 for reflection in y = −1 or reflection in x = k 1(b) triangle (3, 2) (5, 2) (3, −1) 2 1 k B1 for translation  or  k 4 1(c) rotation 3 B1 for each 180⁰ [centre] (0, 0) oe OR enlargement [centre] (0, 0) oe [sf] –1 1(d) triangle (−1, 2) (0, 5) (0, 2) 2 B1 for correct shape but translated horizontally from correct position.

More questions on Transformations

2 f ( x) = 2 x + 4 g ( x) = x - 1 h ( )x = x 2 - 3x (a) Find (i) f ( 3) ................................................. [1] (ii) h ( 7) . ................................................. [1] (b) Find the value of x when g ( x) =- 6 . x = ................................................. [1] (c) Find f -1 ( x) . f -1 ( x) = ................................................. [2] (d) Simplify f ( x) # g ( x) + 1. ................................................. [2] (e) Solve h ( g ( x)) = 0 . x = ..................... or ..................... [3]

Mark scheme: 2(a)(i) 10 1 2(a)(ii) 28 1 2(b) −5 1 2(c) x − 4 2 y 4 oe M1 for 2 x = y − 4 or x = 2 y + 4 or = x + 2 2 2 2(d) 2 x 2 + 2 x − 3 2 M1 for 2 x 2 + 4 x − 2 x − 4 [+1] 2 − 3( x − 1) or better 2(e)  x = 1 and  x =  4 3 M1 for ( x − 1) M1 for ( x − 1)( x − 4) = 0 or dep M1 for correct use of formula on their quadratic equation or dep M1 for sketch of their quadratic equation clearly showing 2 intersections with x-axis

More questions on Functions

3 y is inversely proportional to the square of ( x + 1) . (a) When x = 5 , y = 1. Find y in terms of x. y = ................................................. [2] (b) Find y when x = 3 . y = ................................................. [2] (c) Find the value of x when y = ( x + 1) . x = ................................................. [3]

Mark scheme: 3(a) 36 2 k  y =  2 M1 for 2 ( x + 1) ( x + 1) 3(b) 2.25 oe 2 their 36 M1 for or better ( 3 + 1) 2 3(c) 2.3[0…] 3 M2 for x + 1 = 3 their 36 their 36 or M1 for x + 1 = ( x + 1) 2 M1 for sketch of a cubic crossing x axis once with a positive x intercept

More questions on Proportion

4 (a) A group of 10 people were asked the time, correct to the nearest minute, they each spent listening to the news and reading the news on Monday. The results are shown in the table. Minutes listening (x) 1 9 3 3 11 8 2 7 1 4 Minutes reading (y) 5 10 9 7 2 1 3 6 11 5 (i) Complete the scatter diagram. The first six points have been plotted for you. y 12 10 8 Minutes 6 reading 4 2 x 0 2 4 6 8 10 12 Minutes listening [2] (ii) Find the median time spent reading the news on Monday. .......................................... min [1] (iii) Find the equation of the line of regression. Give your answer in the form y = mx + c . y = ................................................. [2] (iv) On Tuesday, each person spends the same time listening to the news as they did on Monday. They each spend 5 minutes longer reading the news than they did on Monday. Write down the equation of the line of regression for Tuesday. y = ................................................. [1] (b) In February Sancho read the news for a total of 8 hours. This was a reduction of 36% from January. Work out how long Sancho read the news in January. ........................................ hours [2] (c) The bar chart shows the number of news articles read one day by each of 23 people. 5 Number of 4 news articles read 3 2 0 1 2 3 4 5 6 7 8 9 10 Number of people Calculate the mean number of articles read. ................................................. [2]

Mark scheme: 4(a)(i) 4 correct points plotted 2 B1 for 3 correct 4(a)(ii) 5.5 1 4(a)(iii)  y =  − 0.323 x + 7.48 2 B1 for [ y =] − 0.323 x + k or [y=] kx + 7.48 or [y=] −0.32 x + 7.5 4(a)(iv)  y =  − 0.323 x + 12.48 1 FT their (k) + 5 4(b) 12.5 2  100 − 36 = M1 for x   8 oe  100  4(c) 3.57 or 3.565... 2 M1 for 5 +5 4 +6 3 +9 2  3 implied by 82

More questions on Scatter diagrams

5 (a) y 8 0 x – 6 7 – 5 1 (i) On the diagram sketch the lines y =- x + 3 , 2y = x + 5 and y = x for - 6 G x G 7 . 2 [4] (ii) Show, by shading, the region that satisfies these inequalities. 1 y 2- x + 3 2y 1 x + 5 y 2 x [2] 2 (b) y 9 0 x – 1 4.7 – 10 f ( x) = ( x - 2) 3 - 5x + 12 for - 1 G x G 4.7 (i) On the diagram, sketch the graph of y = f ( x) . [2] (ii) Write down the coordinates of the local maximum. ( ...................... , ...................... ) [2] (iii) The equation ( x - 2) 3 - 5x + 12 = k has exactly 2 solutions. Find the values of k. k = .................. or k = .................. [2] (iv) g ( x) =- ( x - 1) 2 for - 1 G x G 4.7 On the diagram, sketch the graph of y = g ( x) . [2] (v) Solve f ( x) = g ( x) . x = ................................................. [1]

Mark scheme: 5(a)(i) correct sketch 4 B1 for correct sketch of 2 y = x + 5 1 B1 for correct sketch of y = − x + 3 2 B1 for correct sketch of y = x passing through (0,0) B1 for all intersections in 1st quadrant 5(a)(ii) correct region indicated 2 FT their lines B1 for region satisfying 2 inequalities or for shading shown but region not clearly indicated 5(b)(i) correct sketch 2 M1 for positive cubic curve with a maximum and minimum 5(b)(ii) (0.709, 6.3[0]) 2 B1 for one correct coordinate 5(b)(iii) –2.3[0], their 6.3[0] 2 B1 for each

More questions on Graphs of functions

6 (a) (i) Kayla walks from A to B on a bearing of 105°. She then walks back to A. Calculate the bearing Kayla walks from B to A. ................................................. [2] (ii) The distance from A to B is 1.5 km. (a) It takes Kayla 24 minutes to walk from A to B. Calculate her average speed in km/h. ......................................... km/h [2] (b) Kayla has a map with a scale of 1 : 25 000 showing A and B. Work out the length of AB on the map. Give your answer in centimetres. ............................................ cm [2] (b) A train is 770 m long. The train takes 2 minutes and 36 seconds to travel completely through a tunnel. Its speed through the tunnel is 120 km/h. Work out the length of the tunnel. Give your answer in metres. .............................................. m [4]

Mark scheme: 6(a)(i) 285 2 M1 for 360 – (180 – 105) oe or a sketch with correct indication of 75 or 105 at B 6(a)(ii)(a) 3.75 oe 2 1.5   60  M1 for 24 6(a)(ii)(b) 6 2 1.5  1000  100  M1 for oe 25000 6(b) 4430 4 B3 for 5200 OR  36   2 +  36  60  M1 for 2 + or ( 2  60 ) + 36 or 60 60 36 M1 for their 2  120 [ ] 60 or their ( ( 2  60 ) + 36 )  120 [ ] 36 120  1000 M1 for their 2  60 60 120  1000 or their ( ( 2  60 ) + 36 )  60  60

More questions on Angles

7 A NOT TO SCALE 5 cm 7.63 cm B 7 cm 12° D C In triangle ACD, AB = 5 cm , AD = 7. 63 cm and BD = 7 cm . Angle BDC = 12° . (a) Show that angle ABD = 77.0° correct to 1 decimal place. [3] (b) Calculate the area of triangle ABD. .......................................... cm2 [2] (c) Calculate BC. ............................................ cm [4]

Mark scheme: 7(a) 7 2 + 5 2 − 7.632 M2 M1 for 7.632 = 7 2 + 52 −2 7 5 cos ABD cos ABD = 2 7 5 ABD = 76.96... A1 no errors or omissions 7(b) 17.1 or 17.[0] or 17.04 to 17.05… 2 M1 for 0.5 5 7 sin77 7(c) 1.61 or 1.605 to 1.606… 4 7sin12 M3 for oe sin (77 − 12) BC 7 or M2 for = sin12 sin(77 − 12) or B1 for [ ACD =]65

More questions on Non-right-angled triangles

8 (a) E NOT TO 20 cm SCALE D C M A 24 cm F 10 cm B ABCDE is a pyramid with a rectangular base. AB = 10 cm and BC = 24 cm. The length of each sloping edge is 20 cm. Vertex E is vertically above the centre of the base, M. (i) Calculate the length AC. ............................................ cm [2] (ii) Calculate EM, the height of the pyramid. ............................................ cm [3] (iii) F is the mid-point of BC. Find the angle between EF and the base of the pyramid. ................................................. [3] (b) B A NOT TO 7 cm SCALE Cone A is mathematically similar to cone B. The height of cone A is 7 cm and its volume is 66 cm 3. The volume of cone B is 222.75 cm 3. (i) Find the height of cone B. ............................................ cm [3] (ii) A sphere also has volume 66 cm 3. Calculate the radius of the sphere. ............................................ cm [2]

Mark scheme: 8(a)(i) 26 2 M1 for 10 2 + 24 2 or better 8(a)(ii) 15.2 or 15.19 to 15.20 3 2 2 their 26  M2 for 20 −   2  2 their 26  2 M1 for 20 −   2  8(a)(iii) 71.8 or 71.78 to 71.80… 3 231 M2 for tan EFM = or using their height 5 oe ALT M1 for EF 2 = 20 2 − 12 2 5 M1 for cos EFM = their EF or B1 for identifying correct angle or stating EFM 8(b)(i) 10.5 3 222.75 M2 for 3  7 oe 66 222.75  7 3 66 3 or M1 for oe or =   66  l  222.75 8(b)(ii) 2.51 or 2.506 to 2.507… 2 66  3 M1 for 3 4

More questions on Pythagoras’ theorem

9 (a) A bag contains 3 black discs and 5 white discs. Jani takes a disc from the bag at random. When the disc is black, he does not replace it. When the disc is white, he replaces it in the bag. Jani then takes a second disc at random. (i) Complete the tree diagram for the first and second discs. First disc Second disc Third disc Black ........ Black ........ ........ White Black ........ ........ White ........ White [3] (ii) Jani takes a third disc from the bag at random. Find the probability that he takes 2 black discs and 1 white disc. ................................................. [4] (b) Another bag contains 10 discs. x are red and the rest are green. (i) Write down an expression for the number of green discs. ................................................. [1] (ii) y blue discs are added to the bag. A disc is taken from the bag at random. (a) The probability of taking a red disc from the bag is 1. 3 Show that 3x = 10 + y . [1] (b) The probability of taking a green disc is 2. 9 Write another equation in x and y and find the number of red discs and the number of blue discs. number of red discs, x = ................................................. number of blue discs, y = ................................................. [5]

Mark scheme: 9(a)(i) 3 5 2 5 3 5 3 B1 for each correct pair in correct position , , , , , 8 8 7 7 8 8 9(a)(ii) 365 4 FT their tree diagram probabilities for method or 0.233 marks 1568 3 2 5 3 5 2 5 3 2 M3 for   [ + ]   [ + ]   soi 8 7 6 8 7 7 8 8 7 without extras or M2 for two correct products soi or M1 for one correct product soi or for clear indication on tree diagram of all three combinations. or list of the options 9(b)(i) 10−x 1 9(b)(ii)(a) x 1 1 = oe seen 10 + y 3 leading to 3x = 10 + y 9(b)(ii)(b) [x = or red =] 6 5 B2 for 9 x + 2 y = 70 oe [y = or blue =] 8 10 − x 2 or B1 for = 10 + y 9 M1 for correct method to eliminate one variable A1 for [x = or red =] 6 or [y = or blue =] 8

More questions on Probability of combined events

10 (a) Q NOT TO SCALE O R 135° P S P, Q, R and S are points on the circle centre O. Find angle PQR. Angle PQR = ................................................. [1] (b) C B D NOT TO SCALE O F 62° A E A, B, C, D and E are points on the circle centre O. AC is parallel to ED. Find the obtuse angle BOD. Angle BOD = ................................................. [2] (c) B NOT TO A C SCALE O 7.5 cm E D ABCDE is a regular pentagon. A, B, C, D and E are points on the circle centre O. The length of the perpendicular from O to ED is 7.5 cm. (i) Show that the length of one side of the pentagon is 10.9 cm correct to 3 significant figures. [4] (ii) Calculate the shaded area. .......................................... cm2 [4]

Mark scheme: 10(a) 45 1 10(b) 124 2 B1 for BED = 62 10(c)(i) [EOD=] 72 or [ODE=] 54 B1 seen or implied by 36 2  7.5  tan36 oe M2 halfside 7.5 M1 for tan(36) = or tan54 = 7.5 7.5 halfside or 2  tan54 10.89…. A1 no errors or omissions 10(c)(ii) 13.1 or 13.11 to 13.13… 4 2 2 2  10.9  M1 r = 7.5 + or better or trig method    2  for r 1 1 2 M1  10.9  7.5 or 2  their r  their sin72 oe 2 their 72 2 M1   their r oe 360

More questions on Pythagoras’ theorem and trigonometry

11 (a) Simplify fully ( 64x 6 y 3 ) 3 . ................................................. [3] (b) 3 x # 2 x = 279 936 Find the value of x. x = ................................................. [2] (c) B NOT TO SCALE 15 x + 2 A C 2 x In triangle ABC, AB = BC . The perimeter of triangle ABC is 16 cm. (i) Show that 4x 2 - 1 = 0 . [5] (ii) Find the length of AB. AB = ............................................ cm [2]

Mark scheme: 11(a) 16x 4 y 2 final answer 3 B2 for final answer kx 4 y 2 or 16 kx y 2 or 16 x 4 y k 2 or 4x 2 y ( ) B1 for 16 or x4 or y2 correct in 3 term final answer or M1 for 4 x 2  y or 4096 x12  y 6 seen 11(b) 7 nfww 2 M1 for 128 or 6x or 2187 seen OR log279936 M1 for x = log6 11(c)(i) 15 15 2 M1 + + = 16 x + 2 x + 2 x 30 2 or + = 16 x + 2 x 30 x + 2( x + 2)[ = 16] or better M2 M1 for 30 x + 2( x + 2) x ( x + 2) M1 for common denominator x ( x + 2) oe 30 x + 2 x + 4 = 16 x( x + 2) M1 FT their numerator with correct denominator to fraction removed rearranging to get to 4 x 2 −=1 0 A1 no errors or omissions 11(c)(ii) 6 2 1 M1 for x = or for 6 and 10 as answers 2

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