Equations — Paper 4 · IGCSE Mathematics (9-1) 0980

6 (a) Expand and simplify. (x + 7)(x - 3) … [2] (b) Factorise completely. (i) 15p 2 q 2 - 25q 3 … [2] (ii) 4fg + 3h - 6gh - 2f … [2] (iii) 81k 2 - m 2 … [2] (c) Solve the equation. x + 2 3 (x - 4) + = 6 5 x = … [4]

12 marks

Mark scheme: 6(a) x 2 + 4 x − 21 final answer 2 B1 for three of x 2, + 7 x, − 3 x, − 21 6(b)(i) 2 2 2 2 2 3 2 2 5 q 3 p − 5 q final answer B1 for 5 3 p q − 5 q or q 15 p − 25 q or ( ) ( ) ( ) q 15 p 2 q − 25 q 2 or 5 q 3 p 2 q − 5 q 2 ( ) ( ) or for correct answer seen 6(b)(ii) ( 2 f − 3h )( 2 g − 1) final answer 2 B1 for 2 g ( 2 f − 3h ) − ( 2 f − 3h ) or or (3h – 2f)(1 – 2g) final answer 2 f ( 2 g − 1) − 3h ( 2 g − 1) or 3h(1 – 2g) – 2f(1 – 2g) or 3h – 2f – 2g(3h – 2f) 6(b)(iii) ( 9 k + m )( 9 k − m ) final answer 2 M1 for (9 + m)(9 – m) or for correct answer seen 6(c) 5.5 4 M1 for 5 × 3 ( x − 4 ) + x + 2 = 5 × 6 M1 for 15 x − 60 + x + 2 = 30 FT their first step x + 2 or 3 x − 12 + = 6 5 If M0M0, SC1 for 3x – 12 + x + 2 = 30 oe M1dep for 16 x = 88 FT their previous steps

This question in 0980/42 May/June 2019

11 Diagram 1 Diagram 2 Diagram 3 Diagram 4 Diagram 5 The sequence of diagrams above is made up of small lines and dots. (a) Complete the table. Diagram 1 Diagram 2 Diagram 3 Diagram 4 Diagram 5 Diagram 6 Number of 4 10 18 28 small lines Number of 4 8 13 19 dots [4] (b) For Diagram n find an expression, in terms of n, for the number of small lines. … [2] (c) Diagram r has 10 300 small lines. Find the value of r. r = … [2] (d) The number of dots in Diagram n is an 2 + bn + 1. Find the value of a and the value of b. a = … b = … [2]

10 marks

Mark scheme: 11(a) 40 54 4 B1 for each 26 34 11(b) n 2 + 3n or n ( n + 3 ) oe 2 B1 for a quadratic expression or for 2nd common difference 2 (at least 2 shown) or for 2 correct equations seen or for subtracting n2 11(c) 100 2 M1 for their (b) = 10300 seen 11(d) 1 2 B1 for each [a = ] oe or M1 for one correct equation 2 or for 2nd difference = 1 soi (at least 2 shown) and 5 [b =] oe 2

This question in 0980/42 May/June 2019

1 (a) The sizes of angles in a quadrilateral are in the ratio 1 | 2 | 3 | 4 . (i) Calculate the size of each angle. … , … , … , … [2] (ii) Write down the mathematical name of a special quadrilateral that can be drawn with these angles. … [1] x + 1 ° (b) The angles of a triangle are x°, and (x + 7) ° . e 2 o Find the value of x. x = … [3] (c) A regular polygon has 72 sides. Find the size of an interior angle. … [3] (d) D NOT TO P y° SCALE v° C x° O 60° u° 20° A w° B Q A, B, C and D lie on the circle, centre O, with diameter AC. PQ is a tangent to the circle at A. Angle PAD = 60° and angle BAC = 20° . Find the values of u, v, w, x and y. u = … , v = … , w = … , x = … , y = … [6] (e) A, B and C lie on the circle, centre O. Angle AOC = (3x + 22) ° and angle ABC = 5x° . Find the value of x. NOT TO SCALE O (3x + 22)° A C 5x° B x = … [4]

19 marks

Mark scheme: Question Answer Marks Partial Marks 1(a)(i) 36, 72, 108, 144 2 M1 for 360 ÷ (1 + 2 + 3 + 4) 1(a)(ii) Trapezium 1 or Cyclic [quadrilateral] 1(b) 69 3 5 x + 15 B2 for = 180 oe 2 x + 1 or M1 for x + + x + 7 = 180 oe 2 1(c) 175 3 360 180 ( 72 − 2 ) M2 for 180 − or for 72 72 360 or M1 for or for 180 (72 – 2) 72 1(d) [u = ] 30 6 B1 for 30 [v = ] 60 B1 for 60 [w = ] 60 B1 for 60 FT their v [x = ] 120 B1 for 120 FT 2 × their w [y = ] 40 B2 for 40 or B1 for angle BDC = 20 or angle ADO = 30 or angle ADB = 70 1(e) 26 4 B3 for 360 – 22 = 10x + 3x oe or better or for 5x + 1.5x = 180 – 11 oe or better or M2 for 360 – (3x + 22) = 2 × 5x oe 1 or for 5x + (3 x + 22) = 180 oe 2 or SC2 for 360 + 22 = 10x + 3x oe or better or M1 for 180 – 5x, 10x or 360 – (3x + 22) correctly placed on the diagram or identified or for angle A + angle C = 5x

This question in 0980/41 Oct/Nov 2019

7 (a) Oranges cost $x per kilogram and apples cost $(x - 0.6) per kilogram. The total cost of 2 kg of oranges and 1.75 kg of apples is $19.20 . Find the value of x. x = … [3] (b) The cost of one ruler is r cents. The cost of one protractor is p cents. The total cost of 5 rulers and 1 protractor is 245 cents. The total cost of 2 rulers and 3 protractors is 215 cents. Write down two equations in terms of r and p and solve these equations to find the cost of one protractor. … cents [5] (c) Carol walks 12 km at x km/h and then a further 6 km at (x - 1 ) km/h. The total time taken is 5 hours. (i) Write an equation, in terms of x, and show that it simplifies to 5x 2 - 23x + 12 = 0 . [3] (ii) Factorise 5x 2 - 23x + 12 . … [2] (iii) Solve the equation 5x 2 - 23x + 12 = 0 . x = … or x = … [1] (iv) Write down Carol’s walking speed during the final 6 km. … km/h [1]

15 marks

Mark scheme: 7(a) 5.4[0] 3 B2 for 3.75x = 19.2[0] + 1.05 or better or M1 for 2x + 1.75(x – 0.6) [ = 19.20] oe or better 7(b) 5r + p = 245 B1 2r + 3p = 215 B1 45 3 Finds p M1 for correctly equating coefficients of r M1 for correct method to eliminate r OR M1 for correctly making r the subject of one of their equations M1 for correctly substituting their correct r to form an equation in p OR Finds r first M1 for correctly eliminating p from their equations M1 for correctly substituting their value of r to find p 7(c)(i) 12 6 M1 + [ = 5] x x − 1 12(x – 1) + 6x = 5x(x – 1) M1 Dependent on previous M1 earned May be over common denominator 5 x 2 − 23 x + 12 = 0 reached, with at A1 least one more line of working and with no errors or omissions 7(c)(ii) (5 x − 3)( x − 4) final answer 2 B1 for (5 x + a )( x + b ) with ab = 12 or a + 5b = – 23 or for 5 x ( x − 4 ) − 3( x − 4 ) or x (5 x − )3 − 4(5 x − 3) 7(c)(iii) 3 1 FT from their two brackets in (c)(ii) oe and 4 5 7(c)(iv) 3 cao 1

This question in 0980/41 Oct/Nov 2019

9 A car hire company has x small cars and y large cars. The company has at least 6 cars in total. The number of large cars is less than or equal to the number of small cars. The largest number of small cars is 8. (a) Write down three inequalities, in terms of x and/or y, to show this information. … , … , … [3] (b) A small car can carry 4 people and a large car can carry 6 people. One day, the largest number of people to be carried is 60. Show that 2x + 3y G 30 . [1] (c) y 10 9 8 7 6 5 4 3 2 1 x 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 By shading the unwanted regions on the grid, show and label the region R that satisfies all four inequalities. [6] (d) (i) Find the number of small cars and the number of large cars needed to carry exactly 60 people. … small cars, … large cars [1] (ii) When the company uses 7 cars, find the largest number of people that can be carried. … [2] Question 10 is printed on the next page.

13 marks

Mark scheme: 9(a) x + y ⩾ 6 oe 3 B1 for each y ⩽ x oe x ⩽ 8 9(b) 4x + 6y ⩽ 60 1 9(c) Correct region indicated cao 6 B1 for x + y = 6 ruled and long enough B1 for x = y ruled and long enough B1 for x = 8 ruled and long enough B2 for 2x + 3y = 30 ruled and long enough or B1 for ruled line through (0, 10) or (15, 0) but not y = 10 or x = 15 9(d)(i) 6, 6 1 9(d)(ii) 34 2 M1 for trying 4x + 6y with (4, 3) or (5, 2) or (6, 1) or (7, 0)

This question in 0980/41 Oct/Nov 2019

9 (a) (i) Write x 2 + 8x - 9 in the form ( x + k) 2 + h . … [2] (ii) Use your answer to part (a)(i) to solve the equation x 2 + 8x - 9 = 0 . x = … or x = … [2] 2 - 7 + 61 - 7 - 61 (b) The solutions of the equation x + bx + c = 0 are and . 2 2 Find the value of b and the value of c. b = … c = … [3] (c) (i) y O x On the diagram, (a) sketch the graph of y = ( x - 1) 2 , [2] 1 (b) sketch the graph of y = x + 1. [2] 2 2 1 (ii) The graphs of y = ( x - 1) and y = x + 1 intersect at A and B. 2 Find the length of AB. AB = … [7] Question 10 is printed on the next page.

18 marks

Mark scheme: 9(a)(i) 2 2 2 2 2 ( x + 4) − 25 B1 for ( x + k ) −−9 (theirk ) or ( x + 4) − h or k = 4 9(a)(ii) x + 4 = [ ± ] 5 M1 FT their (a)(i) –9 and 1 A1 9(b) [b =] 7 3 B1 for [b = ] 7 [c =] –3 M1 for b2 – 4c = 61 9(c)(i)(a) Correct sketch 2 B2 for correct quadratic curve with min touching x-axis 8888 or B1 for parabola vertex downwards 6666 4444 2222 -2-2-2-2 00000000 2222 4444 9(c)(i)(b) Correct sketch 2 B2 for correct straight line intersecting curve on 6666 y-axis 5555 or B1 for straight line with positive gradient and 4444 positive y-intercept 3333 2222 1111 4444 -3-3-3-3 -2-2-2-2 -1-1-1-1 00000000 1111 2222 -1-1-1-1 9(c)(ii) 2.8[0] or 2.795... 7 2 5 B3 for x − x = 0 oe 2 2 1 or M1 for ( x − 1) = x + 1 2 B1 for [(x – 1)2 =] x2 – x – x + 1 AND  5 9  B2 for (0, 1) and  ,  oe  2 4  5 or B1 [x =] 0 and oe 2 AND M1 for (difference in x )2 + (difference in y)2

This question in 0980/42 May/June 2020

5 (a) The diagram shows the graph of y = f ( x) for - 3 G x G 3 . y 20 16 12 8 4 – 3 – 2 – 1 0 1 2 3 x – 4 – 8 – 12 (i) Solve f ( x) = 14 . x = … [1] (ii) By drawing a suitable tangent, find an estimate of the gradient of the graph at the point (-2, 4). … [3] (iii) By drawing a suitable straight line on the grid, solve f ( x) = 2 x - 2 for - 3 G x G 3 . . x = … [3] (b) y A NOT TO B SCALE O x The diagram shows a curve with equation y = 2x 2 - 2x - 7 . The straight line with equation y = 3x + 5 intersects the curve at the points A and B. Find the coordinates of the points A and B. A ( … , … ) B ( … , … ) [5]

12 marks

Mark scheme: 5(a)(i) 2.7 to 2.8 1 5(a)(ii) tangent ruled at x = –2 B1 6 to 10 2 dep on B1 or a close attempt at tangent at x = –2 or M1 for rise/run for their tangent, or close attempt, at any point Must see correct or implied calculation from a drawn tangent After M0, SC1 for gradient of tangent (or close attempt) in range embedded in y = mx + c 5(a)(iii) y = 2x – 2 ruled 3 B2 for correct ruled line and x = –2.9 to –2.8 cao or B1 for short line or for freehand line or broken line or ruled line with gradient 2 or with y-intercept at –2 (but not y = –2) 5(b) A (4, 17) B (–1.5, 0.5) 5 B4 for (–1.5, 0.5) and (4, 17), or for x = 4 and x = –1.5 OR B3 for A(4, 17) or B(–1.5, 0.5) OR M1 for 2x2 –2x – 7 = 3x + 5 oe AND either M2 for (2x + 3)(x – 4) or M1 for 2x(x – 4) + 3(x – 4) or x(2x + 3) – 4(2x + 3) or (2x +c)(x + d) where cd = –12 or c + 2d = –5 [c and d are integers] OR M2 for − their b ± (theirb ) 2 − 4( their a )( their c ) 2( their a ) or M1 for ( their b ) 2 − 4( their a )( their c ) or for p = –their b, r = 2(their a) if in the ௣ ା √௤ ௣ ି √௤ form or ௥ ௥

This question in 0980/41 Oct/Nov 2020

3 (a) Simplify, giving your answer as a single power of 7. (i) 7 5 # 7 6 … [1] (ii) 7 15 ' 7 5 … [1] (iii) 42 + 7 … [1] (b) Simplify. ( 5x 2 # 2xy 4 ) 3 … [3] (c) P = 2 5 # 3 3 # 7 Q = 540 (i) Find the highest common factor (HCF) of P and Q. … [2] (ii) Find the lowest common multiple (LCM) of P and Q. … [2] (iii) P # R is a cube number, where R is an integer. Find the smallest possible value of R. … [2] (d) Factorise the following completely. (i) x 2 - 3x - 28 … [2] (ii) 7 ( a + 2b) 2 + 4a ( a + 2b) … [2] 2 x - 1 1 2 y - x # 3(e) 3 = x 9 Find an expression for y in terms of x. y = … [4]

20 marks

Mark scheme: 3(a)(i) 711 cao 1 3(a)(ii) 710 cao 1 3(a)(iii) 72 cao 1 If answers 11, 10 and 2 in (a) then allow SC1 in this part 3(b) 1000x9y12 final answer 3 B2 for correct answer seen or answer of the form 1000x9yk or 1000xky12 or kx9y12 or B1 for answer with one correct element in product or (10x3y4)[3] seen 3(c)(i) 108 2 M1 for [540 =] 22 [×] 33 [×] 5 or B1 for 108 oe not in prime factor form e.g. 22 × 3 × 9 3(c)(ii) 30 240 2 M1 for (540 × 25 × 33 × 7) ÷ their (c)(i) oe or B1 for answer 30 240 oe not in prime factor form e.g. 25 × 33 × 35 3(c)(iii) 98 2 B1 for 592 704 seen or 26 × 33 × 73 seen or 2 × 72 oe seen 3(d)(i) (x – 7) (x + 4) final answer 2 M1 for x(x – 7) + 4(x – 7) or x(x + 4) – 7 (x + 4) or better or for (x + a)(x + b) where ab = – 28 or a + b = – 3 3(d)(ii) (a + 2b)(11a + 14b) final answer 2 M1 for (a + 2b) (7(a + 2b) + 4a) or (a + pb)(11a + qb) where pq = 28 or 11p + q = 36 If 0 scored, SC1 for a + 2b (11a + 14b) 3(e) 5 x − 1 4 B2 for 2x – 1 = –2x + 2y – x oe [ y = ] oe final answer or B1 for 9x = 32x or better 2 M1dep for correct rearrangement of their 5 term ‘linear’ equation in y and x to make y the subject

This question in 0980/42 May/June 2021

4 (a) Solve. (i) 6 ( 7 - 2)x = 3x - 8 x = … [3] 2x 2 (ii) = x - 5 3 x = … [3] (b) Factorise completely. (i) 2x 2 - 288y 2 … [3] (ii) 5x 2 + 17x - 40 … [2] (c) Solve x 3 + 4x 2 - 17x = x 3 - 9 . You must show all your working and give your answers correct to 2 decimal places. x = … or x = … [5]

16 marks

Mark scheme: 4(a)(i) 10 1 3 M1 for 42 – 12x = 3x – 8 oe or or 3.33[3…] 3 x 8 3 33 or for 7 – 2x = − oe 6 6 M1 for reaching ax = b correctly FT their first step 4(a)(ii) 1 5 3 M1 for 3 × 2x = 2(x – 5) oe –2.5 or −2 or − 2 2 M1 for reaching ax = b correctly FT their first step 4(b)(i) 2(x + 12y)(x – 12y) final answer 3 B2 for (2x + 24y)(x – 12y) or (2x – 24y)(x + 12y) or for 2(x + 12y)(x – 12y) seen OR M2 for k(x + 12y)(x – 12y) or M1 for 2(x2 – 144y2) 4(b)(ii) (5x – 8) (x + 5) final answer 2 M1 for 5x(x + 5) – 8(x + 5) or x (5x – 8)+ 5(5x – 8) or for (5x + a)(x + b) where ab = – 40 or a + 5b = 17 4(c) 4x2 – 17x + 9 [= 0] oe B1 2 B2 FT their 3 term quadratic [ −− ]17 ± ( [ − ]17 ) − 4 ( 4 )( 9 ) 2 B1FT for ( [ − ]17 ) − 4 ( 4) ( 9 ) ) or better 2 × 4 2 − ]17 ) − 4 ( 4 )( 9 )  17  2 ( [ or  x −  oe or  8  4 or better [ −− ]17 + q and B1FT for or 2(4) [ −− ]17 − q or better 2(4) 17 145 17 145 or + oe or − oe or 8 64 8 64 [ −− ]17 [ −− ]17 + q − q 2 2 or 4 4 0.62 and 3.63 cao B2 B1 for each SC1 for 0.6[0] or 0.619 to 0.620 and 3.6[0] or 3.6301 to 3.6302 or 0.62 and 3.63 seen in working or –0.62 and–3.63 as final answers

This question in 0980/41 Oct/Nov 2021

9 (a) NOT TO x cm SCALE ( x + 3)cm This rectangle has perimeter 20 cm. Find the value of x. x = … [3] (b) M y° NOT TO SCALE 20° This rhombus has perimeter 20 cm and angle y is obtuse. M is the midpoint of one of the sides. Find the value of y. y = … [5] (c) r cm NOT TO SCALE z cm 40° This sector of a circle has radius r and perimeter 20 cm. Find the value of z. z = … [6]

14 marks

Mark scheme: 9(a) 3.5 oe 3 M1 for 2(x + x + 3) = 20 oe M1 for correct ax = b for their linear equation 9(b) 116.8 or 116.83 to 116.85 nfww 5 5sin 20 M2 for sin p = 2.5 2.5 5 or M1 for = sin 20 sin p A1 for 43.2 or 43.15 to 43.17 M1dep for 180 – (20 + their 43.2) After 0 scored, SC1 for length of side = 5 9(c) 5.07 or 5.068 to 5.071 6 B3 for 7.41 or 7.412 to 7.413 40 or M2 for r + r + × 2 × π× r = 20 oe 360 40 or M1 for × 2 × π× r oe seen 360 M2 for 2 × 7.41 × sin 20 oe or 7.412 + 7.412 – 2(7.412) cos 40 oe 7.41sin 40 or oe sin70 or M1 for implicit version

This question in 0980/41 Oct/Nov 2021

8 (a) Solve. 10 - 3p = 3 + 11p p = … [2] (b) Make m the subject of the formula. mc 2 - 2k = mg m = … [3] (c) Solve. 1 4 + = 1 x - 3 2x + 3 x = … or x = … [5] (d) Solve the simultaneous equations. You must show all your working. x + 2y = 12 5 x + y 2 = 39 x = …………….. y = ……………… x = …………….. y = ……………… [5] (e) Expand and simplify. ( 2x - 3)( x + 6)( x - 4) … [3]

18 marks

Mark scheme: 8(a) 1 2 M1 for 10  3  11 p  3 p oe or better or 0.5 oe 2 8(b) 2 k 3 M1 for correctly isolating m terms [ m  ] oe final answer 2 M1 for correctly factorising c  g M1 for dividing by a bracket with two terms to the final answer Maximum mark M2 if final answer incorrect 8(c) 0 4.5 oe 5 B4 for 2 x 2  9 x [  0] or 9x – 2x2 [= 0] or better OR M2 for  2 x  3   4  x  3    x  3  2 x  3  or better or M1 for  2 x  3   4  x  3  seen oe or common denominator  x  3  2 x  3  oe B1 for 2 x 2  6 x  3 x  9 or better seen 8(d) y 2  10 y  21[  0] or M2 M1 for y 2  5 12  2 y   39 oe x 2  4 x  12[  0] 12  x  2 or 5 x   39 seen oe 2 2 (y – 3)(y – 7) [= 0] M1 or for correct factors for their 3– term quadratic or (x + 2)(x – 6) [= 0] equation or for correct substitution into quadratic formula or correctly completing the square for their 3– term quadratic equation x = − 2 y = 7 B2 B1 for x = − 2, x = 6 or for y = 7, y = 3 x = 6 y = 3 or for one correct pair of x and y values 8(e) 2 x 3  x 2  54 x  72 final answer 3 B2 correct expansion of three brackets unsimplified or for final answer of correct form with 3 out of 4 terms correct or B1 correct expansion of two brackets with at least three terms out of four correct

This question in 0980/42 May/June 2022

7 f ( x) = 10 - x g ( x) = , x ! 0 h ( x) = 2x j ( x) = 5 - 2 x x 1 (a) (i) Find g b 2 l. … [1] 1 (ii) Find hg b 2 l. … [1] (b) Find x when f ( x) = 7 . x = … [1] (c) Find x when g ( x) = h ( 3) . x = … [2] (d) Find j -1 ( x) . j -1 ( x) = … [2] (e) Write f ( x) + g ( x) + 1 as a single fraction in its simplest form. … [3] 2 2(f) f ( x) - ff ( x) = ax + bx + c ` j Find the values of a, b and c. a = … b = … c = … [4] (g) Find x when h -1 ( x) = 10 . x = … [2]

16 marks

Mark scheme: 7(a)(i) 4 1 7(a)(ii) 16 1 FT 2their 4 7(b) 3 1 7(c) 1 2 2 3 oe M1 for = 2 or better 4 x 7(d) 5 −x 2 M1 for oe final answer x = 5 – 2y or y + 2x = 5 oe 2 y 5 or = − x oe 2 2 7(e) 11x − x 2 + 2 3 x (10 − x ) + 2 + x final answer B2 for oe single fraction x x or B1 for x(10 – x) + 2 + x oe 2 or M1 for 10 − x + + 1 x 7(f) [a =] 1 4 B3 for x 2 − 21x + 100 [b =] –21 OR [c =] 100 2 M1 for (10 − x ) − (10 − (10 − x ) ) oe or better 2 2 B2 for [(10 − x ) ] = 100 − 10 x − 10 x + x or B1 for three out of four terms of [(10 − x ) 2 ] = 100 − 10 x − 10 x + x 2 correct 7(g) 1024 2 M1 for [x =] h(10) oe or better

This question in 0980/41 Oct/Nov 2022

9 (a) NOT TO B 2 cm SCALE x cm A (x – 1) cm (3x + 4) cm The total of the areas of rectangles A and B is 20 cm 2. (i) Show that 3x 2 + 6x - 22 = 0 . [2] (ii) Solve the equation 3x 2 + 6x - 22 = 0 , giving your answers correct to 4 significant figures. You must show all your working. x = … or x = … [4] (iii) Find the perimeter of rectangle B. … cm [1] (b) NOT TO SCALE Area 15 cm2 H cm Area 20 cm2 h cm (y – 2) cm y cm The diagram shows two rectangles where H - h = 1. By forming a quadratic equation and factorising, find the value of y. y = … [7]

14 marks

Mark scheme: 9(a)(i) x ( 3 x + 4 ) + 2 ( x − 1) = 20  M1 Correct expression with brackets unexpanded Leading to 3 x 2 + 6 x − 22 = 0 with no A1 Must see equated to 20 and brackets expanded first to award A1 errors or omissions 9(a)(ii) B2 2 −+6 or − k −6 6 2 − 4(3)( −22) oe B1 for 6 − 4(3)( −22) or or 2.3 2.3 2 22 ( x + 1) = k oe or for = −1 1 + oe 3 –3.887 and 1.887 cao B2 B1 for one correct answer or for answers –3.89 or – 3.88 or -3.886 or –3.8868 to –3.8867 and 1.88 or 1.89 or 1.886 or 1.8867 to 1.8868 or correct answers seen in working or –1.887 and 3.887 answers 9(a)(iii) 5.77 or 5.773 to 5.774 1 FTdep 2(positive x +1) evaluated to 3 sig. fig. or more, dep on x > 1 9(b) y 2 + 3 y − 40  = 0 oe B4 Oe 3 term quadratic M3 for 15 y − 20( y − 2) = y ( y − 2) oe Or 15 20 M2 for − = 1 oe y − 2 y Or M1 for H(y – 2) = 15 or hy = 20 soi ( y + 8)( y − 5) [= 0] oe B2 Strict FT a three term quadratic B1FT for ( y + a )( y + b ) where ab = – 40 or a + b = 3 or y(y – 5) + 8 (y – 5) or y( y + 8 ) − 5 ( y + 8 ) 5 B1

This question in 0980/41 Oct/Nov 2022

9 (a) Simplify. (i) ( 3x 2 y 4 ) 3 … [2] 3 16 - 2 (ii) 16 8 e x y o … [3] (b) (i) Factorise. x 2 - 9 … [1] (ii) Simplify. x 2 - 9 2 xy - 6 y + 5 x - 15 … [3] (c) Solve the simultaneous equations. You must show all your working and give your answers correct to 2 decimal places. 2x + y = 7 y = 5x 2 + 2x - 13 x = … , y = … x = … , y = … [6]

15 marks

Mark scheme: 9(a)(i) 27x6y12 final answer 2 B1 for two terms correct in answer e.g. 27x6yk or 27xky12 or kx6y12 or for correct answer seen then spoilt 9(a)(ii) x 24 y12 3 B2 for final answer with two correct final answer elements 64 64 641 or final answer or or x 24 y12 x 24 y  12 better or for correct answer seen or B1 for 64 or x24 or y12 seen in final answer k or final answer x 24 y 12 or M1 for first correct step seen 3  3  x16 y 8   2  4  eg   or  8 4  or  16   x y   1  2   4096     48 24   x y  9(b)(i) (x + 3)(x – 3) final answer 1 9(b)(ii) x  3 3 M2 for (x – 3)(2y + 5) final answer or M1 for 2y(x – 3) + 5(x – 3) 2 y  5 or x (2y + 5) – 3( 2y + 5) 9(c) 5x2 + 4x – 20 [= 0] oe M2 M1 for 7 – 2x = 5x2 + 2x – 13 oe seen or 2  7  y   7  y  5y2 – 78y + 221 [= 0] oe or y  5    2    13 oe seen  2   2  2 M2 FT their 3-term quadratic 4 4  4(5)( 20) oe 2(5) 2 or M1 for (4)  4(5)( 20) or better or 2 4 q 4 q 4  4  or for or   4   oe 2  5 2  5 10  10  2  4  or for x  oe    10  x = 1.64 y = 3.72 B2 B1 for one correct pair or both x-values and correct or both y – values correct x = – 2.44 y = 11.88

This question in 0980/42 May/June 2023

2 (a) s = at 2 2 Find the value of s when a = 9.8 and t = 20 . s = … [2] (b) Solve. 5 ( 4y - 3) = 15 y = … [3] (c) Expand and simplify. 3 ( 5x - 8) - 2 ( 3x - 7) … [2] (d) Rearrange A = 2 b 2 - 3c 3 to make c the subject. c = … [3] (e) Factorise completely. 6pq - 4q - 3p + 2 … [2]

12 marks

Mark scheme: 2(a) 1960 2 1 M1 for  9.8  202 oe 2 2(b) 3 3 M1 for a first correct step, e.g. 1.5 or 1½ or 20y – 15 = 15 or 4y – 3 = 3 2 M1FTdep for a second correct step, e.g. 20y = 30 or 4y = 6 15 15 or y – = oe 20 20 2(c) 9x – 10 final answer 2 B1 for kx – 10 or 9x + c or M1 for 15x – 24 or –6x + 14 or B1 for correct answer seen and then spoiled 2(d) 2 3 2b − A 3 oe final answer 3 M1 for isolating 3c3, 3c3 = 2b2 – A oe or for A 2b 2 3 A 2b 2 3 = − c or = + c 3 3 −3 −3 M1FT for isolating c3, follow through their first step dep on a 3-term expression with a kc3 term M1FT taking the cube root to the final answer, follow through their previous step Maximum of two marks if answer incorrect 2(e) (2q – 1)(3p – 2) or (1 – 2q)(2 – 2 M1 for 2q(3p – 2) – [1](3p – 2) 3p) final answer or 3p(2q – 1) – 2(2q – 1) or for correct answer seen then spoiled

This question in 0980/41 Oct/Nov 2023

7 (a) Solve 3x - 8 = 6 - 4x . x = … [2] (b) Factorise fully 10a 2 + 5a . … [2] (c) Factorise fully ( 2x - 3) 2 - 9 . … [2] 1 1 x (d) f ( )x = , x ! g ( )x = 3 4x - 1 4 (i) Find f ( 4) . … [1] (ii) Find gg ( 2) . … [2] (iii) Find k when g ( k) = f ( 7) . … [2]

11 marks

Mark scheme: 7(a) 2 2 M1 for 3 x  4 x  6  8 or better 7(b) 5a  2 a  1 final answer 2 B1 for a 10 a  5  or 5(2a2 +a) or 5a  2 a  1 then spoilt 7(c) 4 x  x  3  final answer 2 M1 for  (2 x  3)  3  (2 x  3)  3  or better or for 4 x 2  6 x  6 x  9 [ 9] oe or better 7(d)(i) 1 1 oe 15 7(d)(ii) 19 683 2 3 x B1 for g(9), 39 or 3 seen 7(d)(iii) −3 2 k 1 k 3 M1 for 3  or 3  3 27 or answer g(–3)

This question in 0980/42 May/June 2024

1 (a) (i) Write 70 as a product of its prime factors. … [2] (ii) Find the highest common factor (HCF) of 70 and 112. … [2] (iii) Find the lowest common multiple (LCM) of 70x 4 y 2 and 112x 3 y 5. … [2] (b) Simplify. (i) a 12 ' a 4 … [1] 5 bc (ii) # 2b 20 … [2] (c) Solve. 4 + 2x = 15 x = … [2] (d) Solve. 34 + 2x = 4 - x 5 x = … [3] 3(e) P = d + m2 (i) Find P when d = 7 and m = -8. P = … [2] (ii) Rearrange the formula to make m the subject. m = … [3]

19 marks

Mark scheme: Question Answer Marks Partial Marks 1(a)(i) 2 × 5 × 7 [=70] 2 B1 for 2, 5, 7 1(a)(ii) 14 2 M1 for [112 = ] 24 × 7 oe or for answer 2 × 7 1(a)(iii) 560x4y5 2 B1 for answer kx4y5 or for answer 560xayb or for correct answer seen then spoiled 1(b)(i) a8 1 1(b)(ii) c 2 5 bc final answer M1 for or better 8 40b 1(c) 11 2 15 5.5 or or 5½ M1 for 2x = 15 – 4 oe or 2 + x = oe 2 2 1(d) –2 3 M1 for 34 + 2x = 5(4 – x) oe or better M1 dep for reaching ax = b FT their first step 1(e)(i) 11 2 3 2 M1 for 7 + ( −8) oe 1(e)(ii)  ( P − d )3 oe final answer 3 B1 for P – d = 3 m2 oe M1 for cube both sides M1 for square root leading to final answer

This question in 0980/41 Oct/Nov 2024

10 (a) NOT TO SCALE ( x + 1)cm ( 2x + 3)cm This rectangle has area 190 cm2. (i) By forming and solving an equation, show that x = 8.5 . [4] (ii) Work out the perimeter of the rectangle. … cm [2] (b) A r cm NOT TO SCALE 50° O B The diagram shows a sector OAB of a circle, with centre O, and a chord AB. The shaded segment has area 30 cm2. (i) Show that r = 23.7 cm, correct to 1 decimal place. [4] (ii) Calculate the perimeter of the shaded segment. … cm [4]

14 marks

Mark scheme: 10(a)(i) 2x2 + 5x –187 [= 0] M2 M1 for (2x + 3)(x + 1) = 190 (2x –17)(x + 11) [= 0] oe M1 Leading to x = 8.5 with no errors A1 10(a)(ii) 59 2 M1 for 6 × 8.5 + 8 oe or 6x + 8 oe or B1 for 9.5 and 20 10(b)(i) 50 1 M3 π  r2 – r2  sin50 = 30 oe 360 2 50 M1 for π  r2 360 1 M1 for r2  sin50 oe 2 23.70[9] to 23.72… A1 must see at least 4 sig figs 10(b)(ii) 40.7 or 40.8 or 40.71 to 40.75… 4 M2 for 2 × 23.7 × sin 25 oe or 23.72 + 23.72 −2 23.7  23.7cos50 oe 23.7 sin 50 or oe  180 − 50  sin    2  x or M1 for = sin25 oe 23.7 or for 23.7 2 + 23.7 2 −2 23.7  23.7cos50 oe AB 23.7 or = oe sin 50  180 − 50  sin    2  AND 50 M1 for × 2 × π  23.7 oe 360

This question in 0980/41 Oct/Nov 2024

9 Solve the simultaneous equations. You must show all your working. 2w - 3y = 11 3w + y = 11 w = … y = … [3]

3 marks

Mark scheme: 9 Correct elimination of one M1 variable [w =] 4 A2 A1 for one correct [y =] –1 If A0 scored, SC1 for answers satisfying one of the original equations

This question in 0980/42 May/June 2025

10 A group of 12 adults and 9 children travel on a bus. The cost of an adult ticket is $n. The cost of a child ticket is $( n - 10 ) . The total cost of the tickets is $277.50 . Find the cost of one adult ticket. $ … [3]

3 marks

Mark scheme: 10 17.5[0] cao 3 M1 for 12n + 9(n – 10) = 277.50 oe M1 dep on their equation using n and n – 10 for simplifying their equation correctly to an = b

This question in 0980/42 May/June 2025

24 Ahmed walks 2 km at a speed of x km/h. He then walks a further 3 km at a speed of ( x + 1 ) km/h. 1 The total time he takes to walk the 5 km is 1 hours. 4 (a) Show that 5x 2 - 15 x - 8 = 0 . [5] (b) Find the value of x. Show all your working and give your answer correct to 2 decimal places. x = … [3]

8 marks

Mark scheme: 24(a) 2 3 5 M2 2 3 + = oe M1 for seen or seen x x + 1 4 x x + 1 2 × 4(x + 1) + 3 × 4x = 5x(x + 1) M1 Correctly removing algebraic fractions or use of or common denominator from their three-term 2 ( x + 1) 3 x 5 equation with two fractions with different + = algebraic denominators x ( x + 1) x ( x + 1) 4 8x + 8 + 12x = 5x2 + 5x oe M1 Correctly multiplying their brackets and clearing algebraic fractions from their three-term equation with two fractions with different algebraic denominators Leading to 5x2 – 15x – 8 = 0 A1 With no errors or omissions 24(b) 2 B2 2 [ −−]15 + ([ − ]15) − 4(5)( −8) B1 for ([ −]15) −−oe4 5 8 oe 2(5) 15 + p 15 − p or for oe or oe or 2(5) 2(5) 2 3 8  3  2 + + oe  3    or for  x −  oe 2 5  2   2  3.46 B1

This question in 0980/42 May/June 2025

24 Martha walks a distance of 10 km at a speed of x km/h. She then runs a distance of 5 km at a speed of ( x + 4 ) km/h. The total time taken for the whole journey is 3.5 hours. (a) Write down an expression in terms of x for the time Martha is walking. … h [1] (b) Show that 7x 2 - 2 x - 80 = 0 . [4] (c) Solve 7x 2 - 2 x - 80 = 0 , giving your answers correct to 2 decimal places. You must show all your working. x = … or x = … [3] (d) Calculate the difference between the time Martha is walking and the time she is running. Give your answer in hours and minutes correct to the nearest minute. … h … min [3]

11 marks

Mark scheme: 24(a) 10 1 x 24(b) their10 + 5 = 7 oe M1 x x + 4 2 20 x + 80 + 10 x = 7 x 2 + 28 x oe M2 Strict FT for correctly clearing fractions from their three-term equation with two algebraic denominators in x and x + 4 and expanding all brackets Strict M1FT for correctly expressing their two algebraic fractions with two denominators in x and x + 4 as a single fraction or with a common denominator within a correct equation or for correctly clearing fractions from their three-term equation with two algebraic denominators in x and x + 4 but not all brackets expanded Leading to 7 x 2 − 2 x − 80 = 0 A1 No errors or omissions 24(c) 2 B2 2 −−( 2 )  ([ − ]2) − 4 ( 7 )( −80 ) or B1 for ([ −]2) − 4 ( 7 )( −80 ) oe or for 2 ( 7 ) −−( 2) − p −−( 2) + p oe or for oe oe 2(7) 2(7)  2  2 or x −    14  –3.24 and 3.53 B1 24(d) 2h 10min 3 B2 for 2.168 to 2.18 [h] or for 130.08 to 130.8 [min] or for 2hours 10.08 min to 2 hours 10.8 min OR 10 5 M2 for − their positive x their positive x + 4 or 10 5 M1 for or their positive x their positive x + 4

This question in 0980/41 Oct/Nov 2025

26 Solve the simultaneous equations. You must show all your working. y = 2 x 2 - 3x - 7 y = 2 x - 7 x = … , y = … x = … , y = … [4]

4 marks

Mark scheme: 26 M1 for 2 x 2 − 3 x − 7 = 2 x − 7 oe or for 2 x 2 − 5 x  = 0 or better or M2 2 2 y 2 + 18 y + 28 = 0 or better y = 2  y + 7  − 3  y + 7  − 7      2   2  x = 0, y = –7 B2 B1 for x = 0, y = –7, or for x = 0 and x = 2.5 or x = 2.5, y = –2 for x = 2.5, y = –2 or for y = –2 and y = –7 If M1B0 or M2B0 scored then SC1 for correct substitution seen of both of their x-values or their y-values into y = 2 x 2 − 3 x − 7 or y = 2 x − 7

This question in 0980/41 Oct/Nov 2025