Algebraic manipulation — Paper 4 · IGCSE Mathematics - International 0607

7 (a) Ali walks for 1 hour at x km/h and then for 2 hours at x + km/h. 4 He walks a total distance of 8 km. Write an equation and solve it to find the value of x. x = … [3] (b) NOT TO SCALE x (x – 2) (x – 2) x 2x x The volume of the cube is equal to the volume of the cuboid. (i) Show that x 3 - 8x 2 + 8x = 0 . [3] (ii) y 40 0 x 7.5 –40 On the diagram, sketch the graph of y = x 3 - 8x 2 + 8x for 0 G x G 7.5 . [2] (iii) Find the volume of the cuboid. … [2]

10 marks

Mark scheme: 7(a)  1  M2  1  x + 2  x +  = 8 oe M1 for 2  x +  oe seen  4   4  2.5 B1 7(b)(i) x 3 = 2 x ( x − 2)( x − 2) oe M1 [( x − 2) 2 = ] x 2 − 2 x − 2 x + 4 B1 Allow – 4x for – 2x – 2x 3 2 2 Allow – 8x2 for – 4x2 – 4x2 or 2 x − 4 x − 4 x + 8 x leading to x 3 − 8 x 2 + 8 x = 0 A1 Final equation reached without any errors or omissions 7(b)(ii) Correct sketch 2 B1 for correct shaped cubic with max 40404040 before min 20202020 0000 0000 2222 4444 6666 -20-20-20-20 -40-40-40-40 7(b)(iii) 318 or 319 or 318.3 to 318.7 2 B1 for 6.83 or 6.828… seen isw use of other values (1.1715…)

This question in 0607/42 Oct/Nov 2017

7 In this question, all lengths are measured in millimetres. 44 A small plastic cup, A, is shown in this diagram. 55 A 28 5 These plastic cups are stacked as shown in the diagram. 5 55 (a) Find the height of a stack of 8 of these cups. … mm [2] (b) Find the number of these cups in a stack that has a total height of 105 mm. … [2] (c) A similar cup, B, has base diameter 42 mm. Find the height of this cup. … mm [2] (d) 2r + 2a h 2r 2 2 r h (3r + 3ar + a ) The formula for the volume of a similar cup is V = . 3 (i) For cup A, show that a = 8 mm. [2] (ii) Find the volume of cup A. … mm3 [2] (iii) Find the volume of cup B. … mm3 [3] 2 2 r h (3r + 3ar + a ) (iv) Rearrange V = to make h the subject. 3 h = … [2]

15 marks

Mark scheme: 7(a) 90 2 M1 for 55 + 5k, k = 7 or 8 7(b) 11 2 M1 for 55 + 5(n – 1) = 105 or better 105 − 55[ + 5] or soi by 10 5 7(c) 82.5 2 42 [ ] M1 for = oe 28 55 7(d)(i) 28 + 2a = 44 oe or 44 – 28 oe seen M1 44 − 28 A1 2a = 16 or oe[= 8] 2 7(d)(ii) 56 900 or 56 900 to 56 920 2 π 2 2 M1 for × (3 × 14 + 3 × 14 × 8 + 8 ) [× 55] 3 7(d)(iii) 192 000 or 192 000 to 192 200 3 3  42  M2 for their (d)(ii) ×   28   42  3  28  3 or M1 for   or    28   42  OR M2 for π 2 2 × their (c) × 3 × 21 + 3 × (8 × 1.5) × 21 + ( 8 × 1.5 ) ( ) 3 or B1 for a =12 7(d)(iv) 3V 2 M1 for 3V = π h 3r 2 + 3ar + a 2 [ h = ] ( ) π(3r 2 + 3ar + a 2 ) V h 3V or = or πh = π 3r 2 + 3ar + a 2 3 3r 2 + 3ar + a 2 ( )

This question in 0607/42 May/June 2018

4 y A NOT TO SCALE D x O B C ABCD is a rectangle. The equation of the line AB is 4x + 3y = 24 . (a) Find the co-ordinates of (i) point A, ( … , … ) [1] (ii) point B, ( … , … ) [1] (iii) the midpoint of AB. ( … , … ) [2] (b) Rearrange the equation 4x + 3y = 24 to make y the subject. y = … [2] (c) Find the equation of the line BC. Give your answer in the form y = mx + c . y = … [3] (d) Find the co-ordinates of (i) point C, ( … , … ) [1] (ii) point D. ( … , … ) [3]

13 marks

Mark scheme: 4(a)(i) (0, 8) 1 4(a)(ii) (6, 0) 1 4(a)(iii) (3, 4) 2 FT their (i) and (ii) B1FT for each co-ordinate 4(b) 4 [ y = ] − x + 8 oe 2 M1 for correct isolating y term or for 3 correct division 4(c) 3 FT their (a)(ii) y = x − 4.5 oe 3 3 4 B2 for y = x + k , k ≠ 0 4 or M1 for gradient = 0.75 oe and M1 for correct subst of their (a)(ii) into y = mx + c 4(d)(i) (0, –4.5) 1 Strict FT their (c) and only if in form y = mx + c 4(d)(ii) (–6, 3.5) 3 FT their (a), (d)(i) B2 for one correct co-ordinate  − 6   6  or M1 for   or   soi  −4.5   4.5 

This question in 0607/41 Oct/Nov 2018

10 (a) Amy buys 3 pencils and 1 ruler and pays 67 cents. Ben buys 2 pencils and 3 rulers and pays 96 cents. Find the cost of 1 pencil and the cost of 1 ruler. You must show all your working. Pencil … cents Ruler … cents [5] (b) In this part, all measurements are in centimetres. NOT TO SCALE x x – 4 x + 1 5x + 3 4 The area of the triangle is the same as the area of the rectangle. (i) Show that 3x 2 - 10x - 48 = 0 . [4] (ii) Factorise 3x 2 - 10x - 48 . … [2] (iii) Find the area of the triangle. … cm2 [2]

13 marks

Mark scheme: 10(a) 3p + r = 67 oe B2 B1 for each 2p + 3r = 96 oe Accept words in equations, using + and =. correctly eliminating one variable M1 [pencil = ] 15 B1 [ruler =] 22 B1 If M0 scored in addition to B0 (for answers) scored then award SC1 for answers satisfying one of their two original equations in 2 variables 10(b)(i) 1  5 x  M2 1  5 x  ( x + )1 x =  + 3  ( x − 4) oe M1 for ( x + )1 x or  + 3  ( x − 4) 2  4  2  4  5 x 2 B1 i.e. correct expansion for rectangle − 5 x + 3 x − 12 oe 4 3 x 2 − 10 x − 48 = 0 reached with no A1 Dependent on B1 errors or omissions including at least one line of working 10(b)(ii) (3 x + 8)( x − 6 ) 2 B1 for 3 x ( x − 6 ) + 8( x − 6 ) or x (3 x + 8) − 6(3 x + 8) or (3 x + a )( x + b ) with ab = – 48 or a + 3b = – 10 10(b)(iii) 21 2 FT 1 2 × (their positive x ) × ( their positive x + 1) if x > 4 M1 12 × (their positive x) × (their positive x + 1) if x > 4

This question in 0607/42 May/June 2019

3 y 4 –3 0 3 x –2 1 f (x) = 3 x ! 1 (1 - x ), (a) On the diagram, sketch the graph of y = f ( x) for values of x between - 3 and 3. [3] (b) Write down the range of f(x) for - 3 G x G 0 . … [2] (c) On the same diagram, sketch the graph of y = x2 for - 2 G x G 2 . [1] 1 2 (d) (i) Solve the equation 3 = x . 1 - x x = … [1] 1 2 u w (ii) The equation 3 = x can be written in the form x - x + 1 = 0 . 1 - x Find the value of u and the value of w. u = … w = … [2]

9 marks

Mark scheme: 3(a) Correct sketch 3 B2 for first branch correct, including gradient 4444 zero at y-intercept or B1 for first branch above x-axis, increasing 3333 and crossing y-axis 2222 1111 B1 for second branch correct 3333 -2-2-2-2 -1-1-1-1 0000 0000 1111 2222 3333 -1-1-1-1 -2-2-2-2 3(b) 0.0357 or 0.03571... - f ( x ) - 1 oe 2 B1 for 0 < f ( x ) or f ( x ) - 1 3(c) Correct sketch 1 Vertex at origin 3(d)(i) –[0].809 or –[0].8087... 1 3(d)(ii) [u = ] 5 2 B1 for each [w = ] 2 or SC1 for answers reversed

This question in 0607/41 Oct/Nov 2019

7 (a) (i) Factorise 2x 2 - 11x - 6 . … [2] (ii) Using your answer to part (i), solve 2x 2 - 11x - 6 1 0. … [2] (b) Solve the equation 3x 2 - x - 5 = 0. Give your answers correct to 2 decimal places. You must show all your working. x = … or x = … [3]

7 marks

Mark scheme: 7(a)(i) (2 x + 1)( x − 6) final answer 2 M1 for (2x + a)(x + b) where ab = –6 or a + 2b = –11 or 2x(x – 6) + x – 6 or x(2x + 1) – 6(2x + 1) or correct answer seen 7(a)(ii) − 0.5 < x < 6 2 FT their (i) only from factors giving positive x2 term B1 for each or –0.5 and 6 seen 7(b) Appropriate sketch indicating answers M1 2 Allow 61 for ( −1) − 4(3)( −5) (one positive and one negative) or correct substitution in formula or correct completion of square 1.47 B2 B1 for each –1.14 or both correct but not rounded to 2dp 1.468… , –1.135…

This question in 0607/42 Oct/Nov 2019

10 All lengths in this question are in metres and all areas are in square metres. 2x + 3 NOT TO SCALE The length of this rectangle is (2x + 3) and the area is 840. (a) Write down an expression, in terms of x, for the width of the rectangle. … [1] (b) The perimeter of the rectangle is 118. Show that 2x 2 - 53x + 336 = 0. [3] (c) Solve the equation 2x 2 - 53x + 336 = 0. Show all your working. x = … or … [3] (d) Find the length and the width of the rectangle. Length = … m Width = … m [2]

9 marks

Mark scheme: 10(a) 840 1 2 x + 3 10(b) 840 M1 2(2x + 3) + 2 their × = 118 oe 2 x + 3 2(2x + 3)2 + 1680 = 118(2x + 3) oe M1 Clearing fractions Correct completion to A1 No errors or omissions 2x2 – 53x + 336 = 0 10(c) (2x – 21)(x – 16) = 0 M1 2 −−( 53) ± ( − 53) − 4(2)(336) or x = 2 × 2 or sketch of parabola (+ve x2, +ve zeros) 10.5, 16 B2 B1 for each 10(d) 35 2 B1 for each 24 If 0 scored, SC1 for a pair of values with a product of 840 or a sum of 59

This question in 0607/43 Oct/Nov 2019

5 (a) Expand the brackets and simplify. (i) 5 ( 2 - p) - 3 ( 3 + 2 p) … [2] (ii) ( 7g - 2h)( 3g + 11h ) … [3] (b) Factorise completely. (i) 2x 2 y 3 - 4x 3 y 2 … [2] (ii) 49t 2 - 9 u 2 … [2] (iii) 6d 2 + d - 2 … [2]

11 marks

This question in 0607/41 May/June 2020

7 (a) (i) Factorise a 2 - b 2 . … [1] (ii) a NOT TO b SCALE a b The diagram shows two squares. The difference between the areas of the squares is 7.41 cm 2. The difference between the lengths of the sides of the squares is 1.3 cm. Find the area of the larger square. … cm2 [5] (b) (i) Factorise x 2 - 5 x - 24 . … [2] (ii) NOT TO (x + 13) cm B SCALE (x + 4) cm A (2x + 1) cm (x + 1) cm The area of rectangle A is 15 cm 2 greater than the area of rectangle B. Find the area of rectangle A. … cm2 [8]

16 marks

Mark scheme: 7(a)(i) ( a − b )( a + b ) 1 7(a)(ii) 12.25 5 B4 for 2a = 7 or 2.6a = 9.1 or better or B3 for 2b = 4.4 or 2.6b = 5.72 or better OR M1 for a 2 − b 2 = 7.41 oe M1 for a – b = 1.3 oe M1 for a + b = 7.41 ÷ 1.3 or for a or b correctly eliminated 7(b)(i) ( x − 8)( x + 3) 2 B1 for ( x + a )( x + b ) with ab = –24 or a + b = –5 or for x ( x + 3) − 8( x + 3) or for x ( x − 8) + 3( x − 8) 7(b)(ii) 204 8 B7 for x = 8 isw x = –3 OR M1 for ( x + 1)( x + 13) or for ( x + 4)(2 x + 1) B1 for 2 x 2 + 8 x + x + 4 B1 for x 2 + x + 13 x + 13 M1 for ( x + 4)(2 x + 1) − ( x + 1)( x + 13) = 15 oe A1 for x 2 − 5 x − 24 = 0 reached without any error or omission M1 for (2 × their x + 1)(their x + 4) evaluated correctly OR M6 for graph(s) which would lead to correct value of x. or M3 for appropriate graph(s) but not leading to value of x. or M1 for setting up input for graphics calculator

This question in 0607/43 Oct/Nov 2020

5 (a) (i) Expand and simplify 2x + 3 . … [2] 2 2 (ii) The equation 4x + 12x + 5 = 0 can be written as 2x + 3 = k . ` j Find the value of k. k = … [1] (iii) Use your answer to part(ii) to solve the equation 4x 2 + 12x + 5 = 0 . x = … or x = … [2] (b) x varies inversely as the square root of (w – 1). When w = 10, x = 2. (i) Find x in terms of w. x = … [2] (ii) Find x when w = 3.25 . x = … [1] (iii) Find w in terms of x. w = … [3]

11 marks

Mark scheme: 5(a)(i) 4 x 2 + 12 x + 9 final answer 2 B1 for three of 4x2, 6x, 6x, 9 or for correct answer seen 5(a)(ii) 4 1 FT their 9 – 5 5(a)(iii) 2 x + 3 = ± their 4 M1 their 4 > 0 1 1 B1 −2 , − oe 2 2 5(b)(i) 6 2 k final answer M1 for [x = ] oe w − 1 w − 1 5(b)(ii) 4 1 FT only incorrect k 5(b)(iii) 2 2 3 M1 for correct multiplication of term in w 36 36 + x  6 + + 1 or or   1 M1 for correct squaring 2 2 x x  x  M1 for correctly isolating w final answer Max M2 if incorrect answer

This question in 0607/42 Feb/March 2022

6 In this question all lengths are in centimetres. NOT TO SCALE 2x – 1 7 5x + 1 13 – x The area of the larger rectangle is 84 cm 2 greater than the area of the smaller rectangle. (a) Show that 5x 2 + 2x - 88 = 0 . [4] (b) Factorise 5x 2 + 2x - 88 . … [2] (c) Find the area of the smaller rectangle. … cm2 [2]

8 marks

Mark scheme: 6(a) (5x + 1)(2x – 1) – 7(13 – x) = 84 oe M1 Correct first statement without brackets expanded 10x2 – 5x + 2x – 1 B1 – 91 + 7x B1 on LHS or 91 – 7x on RHS 10 x 2 + 4 x − 176 = 0 oe A1 leading to 5 x 2 + 2 x − 88 = 0 with no errors or omissions 6(b) ( 5 x + 22 )( x − 4 ) 2 B1 for (5x + a)(x + b) with ab = –88 or a + 5b = 2 or for 5x(x – 4) + 22(x – 4) or for x(5x + 22) – 4(5x + 52) 6(c) 63 2 M1 for solving their factorised quadratic, allowing omission of negative root. FT (13 – their positive root) × 7 1 if < x < 13 2

This question in 0607/42 Feb/March 2022

10 (a) P = 5 Work out the value of P when x =- 18 and y = 28 . P = … [3] (b) Simplify fully. 5 y 4 x # 2 x 3 … [2] (c) Factorise fully. (i) 15ab - 25bc … [2] (ii) 6 x 2 y 5 - 16 x 3 y 3 … [2] (iii) 6cd - 3 - 9d + 2c … [2] (d) Make x the subject of the formula. 2 x 3ax = 1 - a + 2 x = … [4] (e) Solve the inequality. 3 - x 2 1 2 + x … [3]

18 marks

Mark scheme: 10(a) – 84 3 M1 for correct substitution B1 for answer 84 10(b) 10 y 1 2 20 xy 10 xy 20 y 5 y  2 or 3 3 y or 3.3 (or 3.33 or 3.333…)y B1 for or or or 3 6 x 3 x 6 3 final answer or correct answer seen 10(c)(i) 5b(3a – 5c) final answer 2 M1 for b(15a – 25c) or 5(3ab – 5bc) or correct answer seen 10(c)(ii) 2x2y3(3y2 – 8x) final answer 2 M1 for x2y3(6y2 – 16x) or 2y3(3x2y2 – 8x3) or 2x2(3y5 – 8xy3) or better i.e. answers which are correct and have only one common factor left inside brackets e.g. 2x2y(3y4 – 8xy2) or correct answer seen 10(c)(iii) (2c – 3)(3d + 1) final answer 2 M1 for 2c(3d + 1) – 3(3d + 1) or 3d(2c – 3) + 2c – 3 or correct answer seen 10(d) a  2 4 M1 for correctly eliminating fractions [ x  ] oe 2 M1 for correctly expanding brackets 3a  6 a  2 final answer M1 for correctly collecting all terms in x on one side and other terms on other side of equation M1 for correctly isolating x by factorising and dividing Max 3 marks only if final answer is incorrect 10(e) –2 < x < 0.5 final answer 3 M2 for –2 and 0.5 SOI or M1 for correct graph(s) sketched 1  2 x or M1 for  0 oe 2  x or B1 for 0.5 soi

This question in 0607/42 May/June 2022

9 NOT TO SCALE The diagram shows a square of side 2a cm inside a square of side ( 2a + 2x)cm . (a) (i) Find an expression, in terms of a and x, for the area of the shaded region. Give your answer in the form px 2 + qax , where p and q are integers. … [2] (ii) Calculate the area of the shaded region when a = 6 and x = 1. … cm2 [1] (b) Find an expression, in terms of a and x, for the total perimeter of the shaded region. Give your answer in its simplest form. … [2] (c) The numerical value of the shaded area is equal to the numerical value of the perimeter of the shaded region. Find x when a = 10 . You must show all your working. x = … [4]

9 marks

Mark scheme: 9(a)(i) 4 x 2 + 8ax final answer 2 M1 for (2 a + 2 x ) 2 − (2 a ) 2 oe 9(a)(ii) 52 1 FT their (i) if answer positive and if a and x both used 9(b) 8x + 16a or 8(x + 2a) oe final answer 2 M1 for 4(2a + 2 x) + 4(2a ) oe 9(c) 2 cao 4 M1 for their (a)(i) = their (b) M1 for rearranging to 3-term quadratic[=0] with a = 10 substituted, or for appropriate sketch M1 for correct method to solve their 3 term quadratic e.g. ( x + 20)( x − 2) [ = 0] or sketch If 0 scored, SC1 for their (a)(ii) = their (b) 10 For all parts accept decimals or percentages with the usual rules for 3sf Do not penalise incorrect cancelling or converting Do not accept ratios or words

This question in 0607/41 Oct/Nov 2022

6 (a) Simplify. (i) 5 ( 2a + 3) - 3 ( a - 7) … [2] 2x x - 1 (ii) - 3 2 … [2] ab + 3 (b) x = b - 2 Rearrange the formula to make (i) a the subject, a = … [3] (ii) b the subject. b = … [2] (c) Solve. (i) x 12 = 1200 x = … [1] (ii) .12 x = 12 x = … [2] (iii) x + 3 = 7 … [2] (d) Solve by factorising. 6x 2 - 11 x - 10 = 0 x = … or x = … [3]

17 marks

Mark scheme: 6(a)(i) 7a + 36 Final answer 2 B1 for ka + 36 or 7a + k or 10a + 15 – 3a + 21 6(a)(ii) x + 3 2 2  2 x − 3( x − 1) Final answer M1 for orbetter 6 6 6(b)(i) bx − 2 x − 3 3 M1 for x(b – 2) = ab + 3 oe Final answer M1FT for bx – 2x – 3 = ab oe b 6(b)(ii) 2 x + 3 2 M1FT for bx – ab = 2x + 3 oe Final answer OR x − a M1FT for factorising and dividing Max 1 mark if answer incorrect 6(c)(i) 1.81 or 1.805 to 1.806 1 6(c)(ii) 13.6 or 13.62 to 13.63 2 M1 for xlog1.2 = log12 or log1.2 12 or a suitable sketch leading to answer 6(c)(iii) 4, –10 final answer 2 B1 for either seen 6(d) (2x – 5)(3x + 2) [= 0] B2 B1 for (ax + b)(cx + d) where ac = 6 and bd = –10 or ad + bc = –11 or for 3x(2x – 5) + 2(2x – 5) or for 2x(3x + 2) –5(3x + 2) 5 2 B1 oe − oe 2 3

This question in 0607/42 Oct/Nov 2022

12 P = 2n + 1 where n is a positive integer. (a) Show that P2 is always an odd number. [2] (b) P and Q are consecutive odd numbers where Q 2 P . (i) Write down an expression for Q, in terms of n. … [1] (ii) Show that Q 2 - P 2 is always a multiple of 8. [3]

6 marks

Mark scheme: 12(a) 4n² + 4n + 1 M1 OR 2n is even so 2n + 1 is odd 4n² + 4n [has a factor 2 so] is even oe A1 So 4n² + 4n + 1 is odd OR So p2 is odd as odd × odd is odd 12(b)(i) 2n + 3 1 12(b)(ii) 4n² +12n + 9 M1 FT their(b)(i) provided of form an + b or (2n + 3 + (2n + 1)) (2n + 3 – (2n + 1)) 8n + 8 A1 With no errors 8n + 8 has factor of 8 and so is a multiple A1 of 8

This question in 0607/42 Oct/Nov 2022

2 y 1 – 5 0 5 x – 3 1 1 f ( x) = - 2 x x (a) On the diagram, sketch the graph of y = f ( x) for values of x between - 5 and 5. [2] (b) Find f ( - 2) . … [1] (c) Solve the equation f ( x) = 0 . x = … [1] (d) Find the maximum value of f(x). … [1] (e) Write down the equation of each asymptote. … [2] (f) (i) Solve the equation. 1 1 2 - = x - 2 2 x x … [3] 1 1 2 4 2 (ii) The equation - 2 = x - 2 can be rearranged to the form x + ax + bx + c = 0 . x x Find the values of a, b and c. a = … b = … c = … [2]

12 marks

Mark scheme: 2(a) Correct sketch 2 No intersections with y-axis B1 for each branch with no large curl back or feathering. Right hand branch with a maximum or level. 2(b) –[0].75 oe 1 2(c) 1 1 2(d) 0.25 oe 1 Not coordinates 2(e) x = 0 2 B1 for each y = 0 2(f)(i) 0.525 or 0.5248 to 0.5249 3 B2 for one correct or M1 for sketch of y = x2 – 2 added 1.49 or 1.490... to diagram 2(f)(ii) [a =] –2 2 B1 for x −=1 x 4 − 2 x 2 oe [b =] –1 [c =] 1

This question in 0607/43 Oct/Nov 2022

10 (a) Simplify. 3x - 5y + 4x - 6y … [2] (b) Expand. x ( x + 2) … [1] (c) Factorise. 10ab + 8ac - 15b 2 - 12bc … [2] 2 5 (d) - = 3 2x + 1 x - 3 (i) Show that 6x 2 - 7x + 2 = 0 . [4] (ii) Solve 6x 2 - 7x + 2 = 0 . You must show all your working. x = … or x = … [3]

12 marks

Mark scheme: 10(a) 7x – 11y Final answer 2 B1 for 7x – ky or kx – 11y k not zero 10(b) x2 + 2x Final answer 1 10(c) (5b + 4c)(2a – 3b) Final answer 2 M1 for 2a(5b + 4c) – 3b(5b + 4c) or 5b(2a – 3b) + 4c(2a – 3b) or correct answer seen but spoiled 10(d)(i) 2(x – 3) – 5(2x +1) = 3(2x + 1)(x – 3) M1 oe or better 2x – 6 – 10 x – 5 or better B1 [3](2x2 – 6x + x – 3) oe or better B1 completion to 6x2 – 7x + 2 [ = 0] A1 at least one step with no errors or omissions 10(d)(ii) (2x – 1)(3x – 2) [= 0] M2 M1 for pair of brackets giving two terms or sketch of parabola showing two correct positive solutions or sketch of any parabola for +ve x2 2 ( 7) ( 7)  (  7)  4(6)(2) or correct formula with or or 2  6 2  6 ( 7) 2  4(6)(2) seen 1 2 B1 , oe 2 3

This question in 0607/42 May/June 2023

8 (a) v = u + at Find v when u = 60, a =-32 and t = 3 . v = … [2] (b) Solve. (i) 6x + 2 = 9 - 4x x = … [2] (ii) 2x - 3 = 7 … [3] (c) Solve by factorisation. 3x 2 - 11x + 6 = 0 x = … or x = … [2] ax + 3b(d) Rearrange y = to make x the subject. 5x x = … [3] (e) Simplify. ax - 2bx + 3ay - 6by x 2 - 9y 2 … [4]

16 marks

Mark scheme: 8(a) –36 2 M1 for 60 + (– 32) × 3 oe or B1 for 96 seen 8(b)(i) 0.7 oe 2 M1 for 6x + 4x = 9 – 2 or better 8(b)(ii) 5 and -2 nfww 3 B2 for –2 nfww B1 for 5 or M1 for 2x – 3 = –7 or 2x – 3 = ±7 or M1 for a correct diagram 8(c) (3x – 2)(x – 3) M1 2 B1 [ x = ]3 oe , 3 8(d) 3b 3 M1 for 5xy = ax + 3b oe final answer M1FT for 5xy – ax = 3b 5 y − a M1FT for factorising and division Incorrect answers score M2 maximum. 8(e) a − 2b 4 B2 for (x + 3y)(a – 2b) oe final answer or B1 for x(a – 2b) + 3y(a – 2b) oe x − 3 y B1 for (x + 3y)(x – 3y)

This question in 0607/43 Oct/Nov 2023

11 (a) Solve. 3x + 2 2 7x - 8 … [2] (b) Factorise fully. 75x 2 - 3 … [2] (c) Simplify. 2 1 1 (i) + - 3x 6x 5x … [2] 2 x 2 + 3 x - 2bx - 3b (ii) 2 2x - 7x - 15 … [4]

10 marks

Mark scheme: 11(a) x < 2.5 oe final answer 2 M1 for 2 + 8 > 7x – 3x oe or B1 for x * 2.5 where * is =, >, ≤ or ≥ 11(b) 3(5x + 1)(5x – 1) final answer 2 B1 for 3(25x2 – 1) or (15x + 3)(5x – 1) or (15x – 3)(5x + 1) 11(c)(i) 19 2 B1 for any equivalent cao final answer or M1 for correct use of common 30x denominator 20 + 5 − 6 20 x + 5 x − 6 x e.g. , etc. oe 30 x 30 x 2 11(c)(ii) x − b 4 B3 for (x – b)(2x + 3) and (x – 5)(2x + 3) final answer or B2 for (x – b)(2x + 3) x − 5 or for (x – 5)(2x + 3) or B1 for x(2x + 3) – b(2x + 3) or 2x(x – b) + 3(x – b) or x(2x + 3) – 5 (2x + 3) or 2x(x – 5) + 3 (x – 5) or (2x + c)(x + d) where c + 2d = –7 or cd = –15

This question in 0607/42 Feb/March 2024

7 (a) Solve 63 = 8 ( 3 - 2 a) . a = … [3] (b) Solve the simultaneous equations. You must show all your working. p 5 - q = 3 12 q 7 2 p + = 2 8 p = … q = … [3] (c) (i) Factorise c 2 - c - 56 . … [2] (ii) Solve c 2 - c - 56 = 0 . c = … or c = … [1]

9 marks

Mark scheme: 7(a) 39 3 B1 for [63 = ] 24 – 16a  oe or –2.44 16 M1 for their 16a = their 24 – 63 oe or better 7(b) Correctly equating coefficients M1 Allow 1 arithmetic slip Correct method to eliminate M1 Allow 1 further arithmetic slip one variable 1 B1 If 0 scored, SC1 for answers that satisfy one [p = ] oe equation 2 1 [q = ]  oe 4 7(c)(i) ( c  7)( c  8) oe 2 B1 for ( c  a )( c  b ) where ab = –56 or a + b = –1 or c ( c  7)  8( c  7) or c ( c  8)  7( c  8) 7(c)(ii) –7, 8 1 FT their (i)

This question in 0607/42 May/June 2024

11 (a) Solve the equation. 2 log 5 - 5 log 2 = 3 log 4 - 2 logx a Give your answer in the form b, where a, b and c are integers. c x = … [4] (b) Make x the subject of the formula. x y = 2x + 1 x = … [4]

8 marks

Mark scheme: 11(a) 32 2 4 25 64 B3 for log = log oe 5 32 x 2 OR M1 for log p + log q = log pq or p log p − log q = log q M1 for log5 2 or log 2 5 or log 4 3 or log x 2 oe 11(b) y 2 4 M1 for correctly squaring [ x =] oe final answer M1 for correct elimination of fractions 2 1 − 2 y M1 for correct expansion of brackets and collecting terms into form px=q M1 for correct division by a 2-term expression Max 3 marks for incorrect final answer

This question in 0607/41 Oct/Nov 2024

3 (a) Expand. x 2 ( 3x - 2 ) … [1] (b) Expand and simplify. 2 ( 3x - 2 ) - 5 ( 1 - 4x) … [2]

3 marks

Mark scheme: 3(a) 3 x 3 − 2 x 2 final answer 1 3(b) 26x – 9 final answer 2 B1 for 26x + k or kx – 9 or M1 for 6x – 4 – 5 + 20x

This question in 0607/41 May/June 2025

7 Factorise. 3ax + 4 by - 3 ay - 4 bx … [2]

2 marks

Mark scheme: 7 ( x − y )(3a − 4b) final answer 2 M1 for 3a ( x − y ) − 4b( x − y ) or x (3a − 4b) − y (3a − 4b)

This question in 0607/42 Oct/Nov 2025

13 Factorise. 3x ( 2x - y) 2 + 4 ( 2x - y) 3 … [3]

3 marks

Mark scheme: 13 (2 x − y ) 2 (11x − 4 y ) final answer 3 M2 for (2 x − y ) 2 (3 x + 8 x − 4 y ) or M1 for (2 x − y ) 2 (3 x + 4(2 x − y ))

This question in 0607/42 Oct/Nov 2025