Equations — Paper 3 · IGCSE Mathematics 0580

7 (a) Rajeesh thought of a number. For He multiplied this number by 2. Examiner's He then added 10. Use The answer was 42. (i) What was the number Rajeesh first thought of? Answer(a)(i) [1] (ii) Simon thought of a number x. He multiplied this number by 3 and then added 8. Write down an expression in x for his answer. Answer(a)(ii) [2] (b) Simplify − 8a + 7b − a − 2b. Answer(b) [2] (c) Factorise fully 6a − 9a2 . Answer(c) [2] (d) Make t the subject of the formula v = u + at. Answer(d) t= [2] (e) Solve the simultaneous equations 8x + 2y = 13, 3x + y = 4. Answer(e) x = , y = [4]

13 marks

Mark scheme: 7 a) i) 16 1 ii) 3x + 8 o.e. 2 M1 for 3x. allow n instead of x. deduct 1 for ‘= x’ or ‘= 0’ or = any number, but allow a different letter b) -9a 1 +5b 1 c) 3a(2 – 3a) 2 M1 for any correct partial factorisation d) v - u 2 M1 for v – u seen o.e. a e) (x=) 2.5 2 M1 for correct multiplication of LHS of one or both equations to equalise coefficients or for a recognisable attempt to eliminate one variable (y=) -3.5 2 M1 for correct substitution of their other value or M2 correct matrix method 13

This question in 0580/31 Oct/Nov 2004

2 (a) Complete the table of values for y = 1 + 2x – x2. For Examiner's Use x − 3 − 2 − 1 0 1 2 3 4 5 y − 14 − 7 1 − 2 − 14 [3] (b) Draw the graph of y = 1 + 2x – x2 on the grid below. y 4 2 x –3 –2 –1 0 1 2 3 4 5 –2 –4 –6 –8 –10 –12 –14 [4] (c) Use your graph to find the solutions to the equation 1 + 2x – x2 = 0. Answer (c) x = or x = [2] (d) (i) On the grid, draw the line of symmetry of the graph. [1] (ii) Write down the equation of this line of symmetry. Answer(d)(ii) [1]

11 marks

Mark scheme: 2 (a) –2 1 2 –7 3 B2 for 3 correct, B1 for 1 or 2 correct (b) 9 correct points P3 f.t. P2 f.t. for 7 or 8 correct, P1 f.t. for 5 or 6 plotted correct limit for acurracy is ½ small square smooth curve drawn C1 must go through the 9 correct points not dependent on P3 (c) –0.4 ( ± /0.1) 1 please note no f.t. on this part 2.4 ( ± 0.1) 1 (d) (i) correct line drawn 1 accept dotted/dashed line must be full length from (1, –14) to (1,2) (ii) x = 1 1 f.t. f.t. from (d)(i) if x = k any reference to y is X 11

This question in 0580/31 May/June 2005

6 (a) For Examiner's NOT TO Use x cm SCALE 2x cm The perimeter of the rectangle in the diagram above is 36 centimetres. (i) Find the value of x. Answer(a)(i) x = [2] (ii) Using this value of x, calculate the area of the rectangle. Answer(a)(ii) cm2 [2] (b) 4z + 2 NOT TO 3y y + 3 SCALE 10z – 1 The diagram above shows another rectangle. (i) In this rectangle 3y = y + 3. Solve the equation to find y. Answer(b)(i) y = [2] (ii) Write down an equation in z. Answer(b)(ii) [1] (iii) Solve the equation in part (b)(ii) to find z. Answer(b)(iii) z = [3] (c) For 4a+b Examiner's Use NOT TO a–b 3 SCALE 17 The diagram above shows another rectangle. (i) Write down two equations in a and b. Answer(c)(i) [2] (ii) Solve these two equations simultaneously to find a and b. Answer(c)(ii) a = b = [3]

15 marks

Mark scheme: 6 (a) (i) 6 2 M1 for 6x = 36 or 3x = 18 o.e. (ii) 72 2 f.t. f.t. is 2 x (a)(i) x (a)(i) M1 (f.t.) for 6 x 12, 2 x 36, 2 x 6 x 6 (b) (i) 1.5 or 1 ½ or 3/2 2 M1 for 3y – y = 3 o.e. [unknown on one side] (ii) 4z + 2 = 10z – 1 1 accept any equivalent equation in z if (b)(ii) is left blank may recover mark if 4z + 2 = 10z – 1 seen in (b)(iii) (iii) 0.5 or ½ or 3/6 3 B1 for correct single z term B1 for correct single constant term (c) (i) a – b = 3 o.e. 4a + b = 17 o.e. } 1,1 if (c)(i) is left blank may recover mark(s) 5a = 20 } with a – b = 3, 4a + b = 17, 5a = 20 seen in (c)(ii) 4a + b + 3 = a – b + 17 (ii) (a=) 4 and (b=) 1 3 2 for either (a=) 4 or (b=) 1 or M1 (f.t.) for correctly eliminating one of the variables 15

This question in 0580/31 May/June 2005

2 (a) Complete the table for the equation y = − x2 + x + 2. For Examiner's Use x −3 −2 −1 0 1 2 3 4 y −10 0 2 2 0 [3] (b) On the grid below draw the graph of y = − x2 + x + 2. y 3 2 1 x 3 2 1 0 1 2 3 4 1 2 3 4 5 6 7 8 9 10 [4] (c) On the grid, draw the line of symmetry of your graph. [1] (d) Use your graph to find the maximum value of y. Answer(d) y = [1] (e) Draw the line y = 1 on the grid. [1] (f) Write down the two values of x for which − x2 + x + 2 = 1. Answer(f) x = or x = [2]

12 marks

Mark scheme: 2 (a) –4 –4 –10 3 1 for each correct entry (b) 1 P3ft P2 for 6 or 7 correct. ft 8 correctly plotted points, within square. P1 for 4 or 5 correct. ft 2 Allow small errors in the points Smooth curve through 8 points C1 provided shape is maintained. (c) x = 0.5 drawn. 1 must be from (0.5, –9) to curve at least (d) 2.2 to 2.4 1ft (e) y = 1 drawn. 1 must touch curve as min. length (f) (x =) –0.7 to –0.5 1 (x =) 1.5 to 1.7 1 12

This question in 0580/31 Oct/Nov 2006

36 , (x ≠ 0). For3 (a) Complete the table for the function y = x Examiner's Use x −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 y −7.2 −9 −18 18 9 7.2 [3] 36 (b) On the grid below, draw the graph of y = for −6 x −1 and 1 x 6. x y 40 30 20 10 x –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 –10 –20 –30 –40 [4] (c) Use your graph to find x when y = 21. Answer(c) x = [1] (d) Complete the table for the function y = x2. For Examiner's Use x −6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 y 25 16 4 1 1 4 16 25 [2] (e) On the same grid, draw the graph of y = x2 for −6 x 6. [4] 36(f) Write down the co-ordinates of the point of intersection of the graphs of y = and y = x2. x Answer(f)( , ) [1]

15 marks

Mark scheme: 3 (a) –6, –12, –36, 36, 12, 6 B3 B1 for ± 36, B1 for ± 12, B1 for ± 6 SC1 for any 3 correct (b) 12 points plotted P3 correct points ft within 1 mm P2 for 10 or 11, P1 for 8 or 9, P1 for 1 correct branch 2 curves drawn C1 must be smooth branches of rectangular hyperbola (c) 1.6 to 1.8 B1 ft (d) 36, 9, 0, 9, 36 B2 B1 for 4 correct (e) 13 points plotted P3 correct points ft within 1 mm P2 for 11 or 12 P1 for 9 or 10 curve drawn C1 must be smooth parabola (f) 3.3, 10.9 B1ft x from 3.2 to 3.4, y from 10.0 to 12.0 [15]

This question in 0580/31 Oct/Nov 2007

10 For Examiner's Use Diagram 1 Diagram 2 Diagram 3 Diagram 4 Diagram 5 Look at the sequence of five diagrams above. Diagram 1 has 2 dots and 1 line. Diagram 2 has 6 dots and 7 lines. The numbers of dots and lines in each of the diagrams are shown in the table below. Diagram number 1 2 3 4 5 6 7 Number of dots 2 6 12 20 30 Number of lines 1 7 17 31 49 (a) Fill in the empty spaces in the table for Diagrams 6 and 7. [4] (b) How many dots are there in Diagram n? Answer(b) [2] (c) The number of lines in Diagram n is 2n2 – 1. Which diagram has 287 lines? Answer(c) [2]

8 marks

Mark scheme: 10 (a) 42, 56 B1B1 cao 71, 97 B1B1 cao (b) n (n + 1) oe B2 M1 for attempt at length x width involving n or n'th (n'th + 1) or k (k + 1) where k is any variable (c) 12 B2 M1 for 2 n² – 1 = 287 [8]

This question in 0580/31 Oct/Nov 2007

4 (a) Solve the equations For Examiner's (i) 3x − 4 = 14, Use Answer(a)(i) x = [2] y +1 (ii) = 2 , 5 Answer(a)(ii) y = [2] (iii) 3(2z − 7) − 2(z − 3) = −9. Answer(a)(iii) z = [3] (b) Donna sent p postcards and q letters to her friends. (i) The total number of postcards and letters she sent was 12. Write down an equation in p and q. Answer(b)(i) [1] (ii) A stamp for a postcard costs 25 cents and a stamp for a letter costs 40 cents. She spent 375 cents on stamps altogether. Write down another equation in p and q. Answer(b)(ii) [1] (iii) Solve these equations to find the values of p and q. Answer(b)(iii) p = and q = [3]

12 marks

Mark scheme: 4 (a) (i) 3x = 14 + 4 oe M1 (x =) 6 A1cao SC2 for 6 www (ii) y + 1 = 2 x 5 oe M1 (y =) 9 A1cao SC2 for 9 www (iii) 6z - 21 - 2z + 6 (= -9) B1 4z = 6 B1ft ft their expansion but must be 4 terms z = 1.5 B1cao (b) (i) p + q = 12 B1 (ii) 25p + 40q = 375 B1 (iii) correct method M1 multiply and subtract, substitution p = 7 A1 q = 5 A1 SC3 for p=7 and q=5 www [12]

This question in 0580/31 May/June 2008

8 (a) The width of a rectangle is x centimetres. For Examiner's The length of the rectangle is 3 centimetres more than the width. Use Write down an expression, in terms of x, for (i) the length of the rectangle, Answer(a)(i) cm [1] (ii) the area of the rectangle. Answer(a)(ii) cm2 [1] (iii) The area of the rectangle is 7 square centimetres. Show that x2 + 3x − 7 = 0. Answer (a)(iii) [1] (b) (i) Complete the tables of values for the equation y = x2 + 3x − 7. x −5 −4 −3 −2 −1 0 1 2 y 3 −7 −9 −7 3 [3] (ii) On the grid below, draw the graph of y = x2 + 3x − 7 for −5 Y x Y 2. For y Examiner's Use 4 2 x –5 –4 –3 –2 –1 0 1 2 A –2 –4 –6 –8 –10 [4] (c) (i) Use your graph to find the solutions to the equation x2 + 3x − 7 = 0. Answer(c)(i) x = or x = [2] (ii) Find the length of the rectangle in part (a). Answer(c)(ii) cm [1] (d) The point A(1, −1) is marked on the grid. (i) Draw a straight line through A with a gradient of 2. [1] (ii) Write down the equation of this line in the form y = mx + c. Answer(d)(ii) y = [2]

16 marks

Mark scheme: 8 (a) (i) x + 3 B1 (ii) x (x + 3) or x² +3x B1 ft from their (a)(i) (iii) x² +3x = 7 x² +3x - 7 = 0 E1 both lines seen (b) (i) -3, -9, -3 B3 B1, B1, B1 (ii) 8 points correctly plotted P3 ft P2ft or 6 or 7, P1ft for 4 or 5 (+/- 1/2 small square) smooth curve C1 (must go below y = -9) IGCSE – May/June 2008 0580/0581 03 (c) (i) 1.5 to 1.6 B1 ft -4.5 to -4.6 B1 ft ft is their intersections with the x-axis (ii) 4.5 to 4.6 B1 ft ft is their positive (c)(i) + 3 (d) (i) correct line L1 long enough to cross y axis (+/- 1/2 small square) (ii) (y =) 2x - 3 B1,B1ft B1 for 2 (as coefficient of x) B1 ft for their intersection with the y-axis [16]

This question in 0580/31 May/June 2008

6 (a) 2y = 75 − 7x (i) Find y when x = 7. Answer(a)(i) y = [2] (ii) Find x when y = 6. Answer(a)(ii) x = [2] (b) Make x the subject of the equation 2y = 75 − 7x. Answer(b) x = [2] (c) Solve these simultaneous equations. 4x − y = 45 7x + 2y = 75 Answer(c) x = y = [3]

9 marks

Mark scheme: 6 (a) (i) ( y =)13 W2 M1 for (2y =) 75 − 7 × 7 (ii) ( x =) 9 W2 M1 for 7x = 75 − 12 or −7x = 12 − 75 (b) 75 − 2 y or 2y−75−7 W2 M1 for 7x + 2y = 75. 7 7x = 75 − 2y or −7x = 2y − 75 or −7x − 2y = −75 IGCSE – October/November 2008 0580 and 0581 03

This question in 0580/31 Oct/Nov 2008

3 For 6 (a) Complete the table of values for the function y = , x ≠ 0. Examiner's x Use x −3 −2.5 −2 −1.5 −1 −0.5 −0.3 0.3 0.5 1 1.5 2 2.5 3 y −1 −1.2 −2 −3 −6 3 2 1.5 1 [3] 3 (b) On the grid below, draw the graph of y = for −3 Y x Y −0.3 and 0.3 Y x Y 3. x y 10 9 8 7 6 5 4 3 2 1 x –3 –2 –1 0 1 2 3 –1 –2 –3 –4 –5 –6 –7 –8 –9 [5] –10 3 For (c) Use your graph to solve the equation = 7. Examiner's x Use Answer(c) x = [1] 2 x (d) Complete the table of values for y = − 1 . 3 x −3 0 3 y [2] 2 x (e) On the grid, draw the straight line y = − 1 for −3 Y x Y 3. [2] 3 2 x (f) Write down the co-ordinates of the points where the line y = − 1 intersects 3 3 the graph of y = . x Answer(f) ( , ) and ( , ) [2]

15 marks

Mark scheme: 6 (a) –1.5 –10 10 6 1.2 3 B2 for 3 or 4 correct, B1 for 2 correct (b) 14 points plotted accurately P3ft P2ft for 11, 12 or 13 points, P1ft for 8, 9 or 10 2 smooth correct curves C1 No part across y-axis B1 Indep (c) 0.4 to 0.5 1 (d) −3 −1 1 2 B1 for 2 correct (e) Ruled line from (−3, −3) to (3, 1) 2 SC1 for freehand or short ruled line – must meet curve twice or P1 for their 3 points plotted (f) (−1.5, −2) and (3, 1) 1, 1

This question in 0580/31 May/June 2010

7 S = a + 4d (a) Find S when a = 17 and d = − 5. Answer(a) S = [2] (b) Find d when S = 37 and a = 5. Answer(b) d = [2] (c) Make d the subject of the formula S = a + 4d . Answer(c) d = [2]

6 marks

Mark scheme: 7 (a) −3 2 1 for correct substitution seen (b) 8 2 M1 for 37−5 =4d oe S − a (c) 2 M1 for one correct step seen 4 275 × 4 × 3 6

This question in 0580/31 May/June 2010

7 (a) Solve the equation. For 4x + 3 = 2 + 6x Examiner's Use Answer(a) x = [2] (b) Simplify. 7(3x – 4y) – 3(5x + 2y) Answer(b) [2] (c) Factorise completely. 6g2 – 3g3 Answer(c) [2]

6 marks

Mark scheme: 7 (a) 0.5 or 1/2 2 M1 for collecting terms correctly (b) 6x – 34y or 2(3x – 17y) 2 B1 for 21x – 28y or B1 for –15x – 6y or B1 for 6x or B1 for –34y (c) 3g²(2 – g) cao 2 B1 for correct partial factorising IGCSE – October/November 2010 0580 31

This question in 0580/31 Oct/Nov 2010

11 Roberto earns a total of $p per week. For He works for t hours each week and is paid a fixed amount per hour. Examiner's He also receives a bonus of $k every week. Use The formula for p is p = 8t + k. (a) Write down how much Roberto is paid per hour. Answer(a) $ [1] (b) (i) Find how much Roberto earns in a week when he works for 40 hours and his bonus is $35. Answer(b)(i) $ [2] (ii) Find how many hours Roberto works in a week when he earns $288 and his bonus is $24. Answer(b)(ii) h [3] (c) Make t the subject of the formula. Answer(c) t = [2]

8 marks

Mark scheme: 11 (a) 8 1 (b) (i) 355 2 M1 for 8 × 40 + 35 seen or better ( 288 − 24) (ii) 33 3 M2 for 8 or B1 for 264 seen p − k (c) t = 2 B1 mark for a correct step 8

This question in 0580/31 Oct/Nov 2010

7 Alex has d dollars to spend. For He buys a book which costs $9 less than 2 times d. Examiner's Use (a) Write down an algebraic expression, in terms of d, for the cost of the book. Answer(a) $ [2] (b) The actual cost of the book is $7.80. Find the value of d. Answer(b) d = [2] (c) How much does Alex have left after buying the book? Answer(c) $ [1]

5 marks

Mark scheme: 7 (a) 2d – 9 2 SC1 for 9 – 2d (b) 8.4(0) 2 M1 for their (a) = 7.8(0) (c) 0.6(0) 1ft ft their (b) – 7.80, only if positive

This question in 0580/33 Oct/Nov 2010

8 The area, A, of a sector of a circle of radius r is given by the formula below. For Examiner's Use π r 2 A = 5 (a) Calculate the area when the radius is 7.5 cm. Answer(a) cm2 [2] (b) Make r the subject of the formula. Answer(b) r = [3] (c) Calculate r when A = 4.8 cm2. Answer(c) r = cm [2]

7 marks

Mark scheme: 8 (a) 35.3 art 2 M1 for substituting r = 7.5 in formula 5 A (b) 3 M1 for correctly multiplying by 5 π M1 for correctly dividing by π M1 for correctly taking a square root (c) 2.76 art cao 2 M1 for substituting 4.8 in their (b) or if working backwards from original formula, substituting and reaching r2 = 5 × 4.8 ÷ π IGCSE – October/November 2010 0580 33

This question in 0580/33 Oct/Nov 2010

7 (a) Solve the equations. For Examiner's (i) 2x + 3 = 15 O x Use Answer(a)(i) x = [2] 2 y − 1 (ii) = 7 3 Answer(a)(ii) y = [2] 1 (iii) 2 = u − 1 Answer(a)(iii) u = [3] (b) Write down equations to show the following. For Examiner's (i) p is equal to r plus two times q. Use Answer(b)(i) [1] (ii) k is equal to the square of the sum of l and m. Answer(b)(ii) [2] (c) Pierre walks for 2 hours at w km/h and then for another 3 hours at (w –1) km/h. The total distance of Pierre’s journey is 11.5 km. Find the value of w. Answer(c) w = [4]

14 marks

Mark scheme: 7 (a) (i) 4 2 M1 for 2 x + x = 15 − 3 or better (ii) 11 2 M1 for 2 y − 1 = 7 × 3 or 23y = 7 + 13 or better (iii) 1.5 oe 3 M1 for 2(u − )1 = 1 A1 for 2u – 2 = 1 (b) (i) p = 2q + r or p = r + 2q oe 1 (ii) k = (l + m)2 2 SC1 for (l + m 2) or for k = l + m (c) 2.9 cao www 4 4 M1 for 2w or 3(w – 1) M1 for 2w + 3(w – 1) = 11.5 A1 for 2w + 3w = 11.5 + 3 or better

This question in 0580/33 May/June 2011

5 For y Examiner's Use 6 5 4 3 2 1 x –4 –3 –2 –1 0 1 2 3 4 5 6 –1 –2 l –3 (a) Find the gradient of the line l. Answer(a) [2] (b) (i) Complete the table below for x + 2y = 6 . x 0 2 y 0 [3] (ii) On the grid, draw the line x + 2y = 6 for −4 Y x Y 6 . [2] (c) The equation of the line l is 4x + 3y = 4. Use your diagram to solve the simultaneous equations 4x + 3y = 4 and x + 2y = 6 . Answer(c) x = y = [2]

9 marks

Mark scheme: − 4 rise 5 (a) oe, –1.2 to –1.4 2 B1 for attempt at 3 run (b) (i) 3, 2, 6 3 B1 for each value (ii) Correct continuous line 2ft Minimum length (0,3) to (6,0) B1 for plotting their 3 points (c) x = −2, y = 4 2ft B1 for their x, B1 for their y from their intersections

This question in 0580/31 Oct/Nov 2011

7 (a) Solve the equation 2(x + 4) = 3(x + 2) + 8 . For Examiner's Use Answer(a) x = [3] (b) Make z the subject of za + b = 3 . Answer(b) z = [2] (c) Find x when 2x3 = 54 . Answer(c) x = [2] (d) A rectangular field has a length of x metres. For The width of the field is (2x – 5) metres. Examiner's Use (i) Show that the perimeter of the field is (6x – 10) metres. Answer (d)(i) [2] (ii) The perimeter of the field is 50 metres. Find the length of the field. Answer(d)(ii) length = m [2]

11 marks

Mark scheme: 7 (a) −6 www 3 M2 for 8 = x + 6 + 8 or better or –x + 8 = 6 + 8 or better M1 for 2x + 8 or 3x + 6 or 3x + 14 3 − b 3 b b 3 (b) or − 2 B1 for 3 – b seen or z + = a a a a a 54 (c) 3 2 B1 for or better 2 SC1 for embedded answer ie 2 × 33 = 54 or 2 × 3 × 3 × 3 = 54 (d) (i) x + x + 2x − 5 + 2x − 5 = 6x – 10 2 M1 accept 2x + 2(2x – 5) or 2(x + 2x – 5) E1 dep (ii) 10 2 M1 for 6x – 10 = 50  0 

This question in 0580/31 Oct/Nov 2011

4 (a) Expand and simplify 3(2x + y) + 5(x – y). For Examiner's Use Answer(a) [2] (b) Expand x2(3x – 2y). Answer(b) [2] (c) Factorise completely 4y2 – 10xy. Answer(c) [2] 4 x 2 (d) y = 3 (i) Find the value of y when x = O3 . Answer(d)(i) y = [2] (ii) Make x the subject of the formula. Answer(d)(ii) x = [3]

11 marks

Mark scheme: 4 (a) 11x − 2y final answer 2 B1 for 6x + 3y or 5x − 5y or 11x or −2y in working (b) 3x3 − 2x2 y final answer 2 B1 for 3x3 ± jx2y or kx3 − 2x2y (c) 2y(2y − 5x) final answer 2 B1 for y(4y − 10x) or 2(2y2 − 5xy) or SC1 for 2y(2y + 5x) or SC1 for 2y(2y − 5x) in working but then spoilt 4 × ( −3) 2 (d) (i) 12 2 M1 for or better in working. 3 3y (ii) (x) = final answer oe 3 Maximum of M2 from 4 M1 for × by 3 M1 for ÷ by 4 M1 for square root h

This question in 0580/32 Oct/Nov 2011

6 (a) Complete the table for y = 4 + 2x O x2. For Examiner's Use x –2 –1 0 1 2 3 4 y 1 5 1 [2] (b) On the grid, draw the graph of y = 4 + 2x O x2 for –2 Y x Y 4 . y 6 5 4 3 2 1 x –2 –1 0 1 2 3 4 –1 –2 –3 –4 [4] (c) (i) Draw the line of symmetry of the graph. [1] (ii) Write down the equation of this line of symmetry. Answer(c)(ii) [1] (d) Use your graph to solve the equation 4 + 2x O x2 = 0. Answer(d) x = or x = [2]

10 marks

Mark scheme: 6 (a) –4, …, 4, …, 4, …, –4 2 B1 for both –4s B1 for both 4s (b) 7 points plotted ft 3ft P2 for 5 or 6 points plotted ft P1 for 3 or 4 Reasonable curve through at least 6 1ft Only ft if shape parabola points (c) (i) The line x = 1 drawn 1ft (ii) x = 1 1ft (d) –1.4 to –1.1, 3.1 to 3.4 2ft B1 B1ft if not in these ranges

This question in 0580/33 Oct/Nov 2011

4 In this question all the measurements are in centimetres. For Examiner's Use 11 – x NOT TO 2x + 3 SCALE 3x The diagram shows a triangle with sides of length 2x + 3, 11 – x and 3x. (a) Explain why x must be less than 11. Answer(a) [1] (b) Write down an expression, in terms of x, for the perimeter of the triangle. Give your answer in its simplest possible form. Answer(b) [2] (c) The perimeter of the triangle is 32 cm. (i) Write down an equation in terms of x and solve it. Answer(c)(i) x = [3] (ii) Work out the length of the shortest side of the triangle. Answer(c)(ii) cm [2]

8 marks

Mark scheme: 4 (a) If x is more than 11 then 11 – x 1 would be negative oe (b) 14 + 4x cao 2 M1 for 2x + 3 + 11 – x + 3x accept 2(2x + 7) (c) (i) 4.5 cao 3 B1ft for “their (b)” = 32 M1ft for collecting their like terms correctly to give simplified expression of form ax = b b OR M1ft x = a (ii) 6.5 2ft M1ft for clear attempt at substituting their (c)(i) into 2 or more sides of triangle IGCSE – May/June 2012 0580 31 5 (a) Correct diagram: 4 rows & 6 1 columns

This question in 0580/31 May/June 2012

8 (a) Complete the table of values for y = x2 – 2x + 5 . For Examiner's Use x –3 –2 –1 0 1 2 3 4 5 y 20 8 8 20 [3] (b) On the grid, draw the graph of y = x2 – 2x + 5 for −3 Y x Y 5 . y 22 20 18 16 14 12 10 8 6 4 2 x –4 –3 –2 –1 0 1 2 3 4 5 6 [4] (c) (i) On the grid, draw the line of symmetry of the graph. [1] (ii) Write down the equation of the line of symmetry. Answer(c)(ii) [1] (d) (i) On the grid, draw the line y = 12 . [1] For Examiner's (ii) Use your graph to solve the equation x2 – 2x + 5 = 12 . Use Answer(d)(ii) x = or x = [2] (e) The equation of a straight line is y = 6 – 3x . (i) Write down the gradient of this line. Answer(e)(i) [1] (ii) Write down the co-ordinates of the point where this line crosses the y-axis. Answer(e)(ii) ( , ) [1] (iii) Write down the equation of a line parallel to y = 6 – 3x . Answer(e)(iii) [1] (f) Simplify 3(2x + 1) O=2(6 – 3x) . Answer(f) [2]

17 marks

Mark scheme: 8 (a) (20) 13 (8) 5 4 5 (8) 13 (20) 3 B2 for 4 correct B1 for 2 or 3 correct or a correct substitution seen (b) correctly plotting 9 points and 4 P3 for correctly plotting 9 points, P2 for correctly connecting with a smooth curved line plotting 7 or 8 points and P1 for 5 or 6 points C1 for a smooth curve (c) (i) correct line of symmetry cao 1 (ii) x = 1 1ft ft their line (d) (i) correct line 1 (ii) –1.9 to –1.7 and 3.7 to 3.9 1ft,1ft SC1 for correct co-ordinates (e) (i) –3 cao 1 (ii) (0,6) cao 1 (iii) y = c – 3x 1 c can be any number except 6 (f) 12x – 9 or 3(4x – 3) 2 B1 for 6x + 3, –12 + 6x, 12x or –9 IGCSE – May/June 2012 0580 33

This question in 0580/33 May/June 2012

9 (a) Each day from Monday to Saturday Caroline buys a newspaper, costing d cents. For On Sunday she buys a newspaper costing 160 cents. Examiner's Use The total amount she spends on newspapers in a week is 430 cents. (i) Write down an equation in d, to show this information. Answer(a)(i) [1] (ii) Solve your equation to find d. Answer(a)(ii) d = [2] (iii) The price of the Sunday newspaper is increased by 15%. Calculate the price of the Sunday newspaper after this increase. Answer(a)(iii) cents [2] (b) Potatoes cost p cents per kilogram and carrots cost c cents per kilogram. (i) Bernard buys 3 kilograms of potatoes and 2 kilograms of carrots. An expression for the amount he spends is 3p + 2c. He spends 92 cents on these items. Write down an equation, in p and c, to show this. Answer(b)(i) [1] (ii) Eleanor buys 2 kilograms of potatoes and 5 kilograms of carrots. She spends 153 cents on these items. Write down an equation, in p and c, to show this. Answer(b)(ii) [2] (iii) Solve your equations to find p and c. Answer(b)(iii) p = c = [4]

12 marks

Mark scheme: 9 (a) (i) 6d + 160 = 430 oe 1 (ii) 45 2ft Ft for pd + q = r p, q and r ≠ 0 and p ≠ 1 M1ft for1st step correct SC1 for 270 (iii) 184 or $1.84 2 M1 for 1.15 × 160 oe SC1 for answer 1.84 (b) (i) 3p + 2c = 92 oe 1 Final answer (ii) 2p + 5c = 153 oe 2 B1 for 2p + 5c seen IGCSE – October/November 2012 0580 32 (iii) (p =) 14 (c =) 25 cao 4 M2ft for correct method to eliminate 1 variable A1 for a correct answer If not M2 M1 for 2 equations with common coefficients of p or c seen or M1 for correct rearrangement to p= or c= seen

This question in 0580/32 Oct/Nov 2012

4 (a) Solve the following equations. For Examiner's (i) 6x O 2 = 2x + 8 Use Answer(a)(i) x = [2] (ii) 4(2y O 3) = 24 Answer(a)(ii) y = [3] (b) Solve the simultaneous equations. 5x + 9y = O21 12x O 2y = 44 Answer(b) x = y = [4]

9 marks

Mark scheme: 4 (a) (i) 2.5 or 5/2 or 2 ½ 2 M1 6x − 2x = 8 + 2 or better (ii) 4.5 or 9/2 or 4 ½ 3 M1 8y − 12 or 2y − 3 = 6 M1 8y = 36 ft or 2y=9 ft their first step (b) (x =) 3, (y =) −4 4 M1 coefficient of x or y the same dep M1 for addition or subtraction A1 for 1 correct answer (their first answer)

This question in 0580/33 Oct/Nov 2012

10 (a) In 2001 Arnold was x years old. For Examiner′s Ken is 34 years younger than Arnold. Use (i) Complete the table, in terms of x, for Arnold’s and Ken’s ages. 2001 2013 Arnold’s age x Ken’s age [3] (ii) In 2013 Arnold is three times as old as Ken. Write down an equation in x and solve it. Answer(a)(ii) x = … [4] (b) Solve the simultaneous equations. For Examiner′s Use 3x + 2y = 18 2x – y = 19 Answer(b) x = … y = … [3] _____________________________________________________________________________________ Question 11 is printed on the next page.

10 marks

Mark scheme: 10 (a) (i) x +12 in each part allow correct unsimplified x – 34 x − 22 1,1,1 terms (ii) x +12 = 3(x – 22) 1ft accept x +12 = 3x – 66 or (x+12) / 3 = x – 22 39 cao 3 M1 for their 3x – 66 seen M1 for correctly collecting terms from ax + b = cx + d a,b,c,d ≠ 0 (e) 8 3 M1 for correct method to eliminate one − 3 variable. A1 for x or y correct. 2

This question in 0580/31 May/June 2013

2 Three children have some marbles. For Examiner′s Shireen has m marbles. Use Nazaneen has three times as many marbles as Shireen. Karly has 4 more marbles than Shireen. (a) Write down an expression, in terms of m, for (i) the number of marbles Nazaneen has, Answer(a)(i) … [1] (ii) the number of marbles Karly has. Answer(a)(ii) … [1] (b) The three children have a total of 84 marbles between them. (i) Write down an equation in m. Answer(b)(i) … [1] (ii) Solve your equation. Answer(b)(ii) m = … [2] (c) Shireen weighs the 84 identical marbles. Their total weight is 4.2 kg. Calculate, in grams, the weight of one marble. Answer(c) … g [2] (d) The children now decide to share the 84 marbles in the ratio Shireen : Nazaneen : Karly = 2 : 7 : 3 . Calculate the number of marbles each receives. Answer(d) Shireen … Nazaneen … Karly … [3] _____________________________________________________________________________________

10 marks

Mark scheme: 2 (a) (i) 3m 1 (ii) m + 4 1 (b) (i) m + 3m + m + 4 = 84 oe isw 1ft ft m + (a)(i) + (a)(ii) = 84 if and only if (a)(i) and (a)(ii) are both in terms of m (ii) 16 2 M1ft for “5”m = “80” i.e. pm = q (could be seen in bi) May be implied by a correct answer (c) 50 2 M1 for 4.2/84 × 1000 or better SC1 for figs ‘5’ or 4200 seen (d) [Shireen =] 14 1 if M0 then M1 for 84/(2 + 7 + 3) or [Nazaneen =] 49 1 better [Karly =] 21 1 and / or SC1 3 correct answers in wrong order. IGCSE – May/June 2013 0580 32

This question in 0580/32 May/June 2013

8 (a) Simplify the following expressions. For Examiner′s Use (i) 3m – 5m + 6m Answer(a)(i) … [1] (ii) 5e – 4f – 3e – 6f Answer(a)(ii) … [2] (b) s = u + at (i) Calculate the value of s when u = 27, a = –2 and t = 15. Answer(b)(i) s = … [2] (ii) Make t the subject of the formula s = u + at. Answer(b)(ii) t = … [2] (c) Solve the simultaneous equations. 5x + 2y = 4 4x – y = 11 Answer(c) x = … y = … [3] _____________________________________________________________________________________

10 marks

Mark scheme: 8 (a) (i) 4m 1 (ii) 2e – 10f 2 B1 for ae – 10f or 2e ± bf (a,b ≠ 0) (b) (i) –3 2 M1 for 27 + (–2) × 15 or better (ii) s − u s u 2 M1 first step correct [t=] or − a a a SC1 for s – u ÷ a www (c) [x =] 2, [y =] –3 3 M1 for correct method to eliminate one variable. A1 for x or y correct IGCSE – May/June 2013 0580 33

This question in 0580/33 May/June 2013

5 (a) The cost, $C, of a party for n people is calculated using the following formula. For Examiner′s Use C = 130 + 4n (i) Calculate C when n = 25. Answer(a)(i) … [2] (ii) Eurdley has a party which costs $1138. How many people is this party for? Answer(a)(ii) … [2] (b) Solve the following equations. (i) 3x = 27 Answer(b)(i) x = … [1] (ii) 8y – 4 = 24 Answer(b)(ii) y = … [2] (iii) 4(5q – 2) = 72 Answer(b)(iii) q = … [3] (c) Solve the simultaneous equations. 6x + 8y = –31 14x – 5y = 46 Answer(c) x = … y = … [4] _____________________________________________________________________________________

14 marks

Mark scheme: 5 (a) (i) 230 2 M1 for 130 + 4 × 25 or better (ii) 252 2 M1 for 4n = 1138 – 130 or better Or (1138 – 130) / 4 or better (b) (i) 9 1 (ii) 3.5 2 M1 for 8y = 24 + 4 or better Or y – 4/8 = 24/8 or better (iii) 4 3 M1 for first correct step M1FT for second correct step (c) x = 1.5 or 3/2 4 M1 for correctly equating one set of y = –5 coefficients. M1 for correct method to eliminate one variable. A1 for x = 1.5 A1 for y = –5

This question in 0580/31 Oct/Nov 2013

7 (a) Complete the table of values for y = x2 – x + 2 . For Examiner′s Use x –3 –2 –1 0 1 2 3 4 y 8 2 4 [3] (b) On the grid, draw the graph of y = x2 – x + 2 for −3 Y x Ğ 4 . y 16 14 12 10 8 6 4 2 x –3 –2 –1 0 1 2 3 4 [4] (c) Write down the equation of the line of symmetry of the graph. For Examiner′s Use Answer(c) … [1] (d) (i) On the grid, draw the line y = 9 . [1] (ii) Solve the equation x2 – x + 2 = 9 . Answer(d)(ii) x = … or x = … [2] _____________________________________________________________________________________

11 marks

Mark scheme: 7 (a) 14, 4, 2, 8, 14 3 B2 for 4 correct B1 for 2 or 3 correct (b) 8 points correctly plotted P3FT P2FT for 6 or 7 points correctly plotted P1FT for 4 or 5 points correctly plotted Smooth and correct curve through all C1 correct points 1 (c) x = 0.5 or x = 1 2 (d) (i) y = 9 ruled 1 (ii) –2.15 to –2.25 1FT 3.15 to 3.25 1FT

This question in 0580/31 Oct/Nov 2013

3 (a) Sweets are sold in packets. For Examiner′s There are n sweets in each packet. Use (i) Maya has 4 packets of sweets and 21 extra sweets. Write an expression, in terms of n, for the number of sweets Maya has. Answer(a)(i) … [1] (ii) Tassos has 5n + 3 sweets. Roma has 3n + 27 sweets. Tassos and Roma each have the same number of sweets. Write down an equation, in terms of n, and solve it. Answer(a)(ii) n = … [3] (iii) Work out the number of sweets Tassos and Roma have altogether. Answer(a)(iii) … [1] (b) A different packet of sweets contains 6 red sweets, 10 yellow sweets and 4 green sweets. Simon takes one sweet from the packet at random. (i) Write down the colour of sweet Simon is most likely to take. Answer(b)(i) … [1] (ii) On the probability scale, draw an arrow to show the probability that Simon’s sweet is yellow. 0 1 [1] (iii) Write down the probability that Simon’s sweet is green. Answer(b)(iii) … [1] (iv) Write down the probability that Simon’s sweet is red or yellow. Answer (b)(iv) … [1] _____________________________________________________________________________________

9 marks

Mark scheme: 3 (a) (i) 4n + 21, final answer 1 (ii) 5n + 3 = 3n + 27 1 [n =] 12 2 M1 for 5n – 3n = 27 – 3 or better (iii) 126 1FT (b) (i) yellow 1 (ii) arrow pointing at 0.5 1 4 (iii) o.e. or 0.2 or 20% 1 20 16 (iv) o.e. or 0.8 or 80% 1FT SC1 for 4 out of 20 and 16 out of 20 20 IGCSE – October/November 2013 0580 32

This question in 0580/32 Oct/Nov 2013

10 (a) Solve the equation. For Examiner′s 6(x – 2) = 9 Use Answer(a) x = … [2] (b) Expand and simplify. 8(n – 1) – 2(3n + 5) Answer(b) … [2] (c) Factorise completely. 10p2 + 5p3 Answer(c) … [2]

6 marks

Mark scheme: 10 (a) 3.5 2 M1 for 6x – 12 = 9 or better 9 or x – 2 = or better 6 (b) 2n – 18 or 2 ( n – 9 ) final answer 2 B1 for 8n – 8 or –6n –10 or 2n or –18 (c) 5p2(2 + p) final answer 2 M1 for any correct incomplete factorisation or 5p2(2 + p) seen in working

This question in 0580/32 Oct/Nov 2013

86 (a) (i) Complete the table of values for y = , x ≠ 0 . x x –8 –4 –2 –1 1 2 4 8 y –2 2 [3] 8 (ii) On the grid, draw the graph of y = for –8 Ğ x Ğ –1 and 1 Ğ x Ğ 8 . x y 8 6 4 2 x –8 –6 –4 –2 0 2 4 6 8 –2 –4 –6 –8 [4] (iii) Write down the order of rotational symmetry of your graph. Answer(a)(iii) … [1] (b) (i) Complete this table of values for y = 1.5x + 3 . x –6 –4 –2 0 2 y –6 3 [2] (ii) On the grid, draw the graph of y = 1.5x + 3 . [1] 8 (c) Use your graphs to solve the equation = 1.5x + 3 . x Answer(c) x = … or x = … [2] (d) Write down the gradient of the graph of y = 1.5x + 3 . Answer(d) … [1] __________________________________________________________________________________________

14 marks

Mark scheme: 6 (a) (i) −1, −4, −8, 8, 4, 1. 3 1 for each symmetrical pair (ii) 8 points correctly plotted, within ½ square. 3FT B2FT for 6 or 7 correct Or B1 FT for 4 or 5 correct 2 smooth correct curves, not joined 1 (iii) 2 1 IGCSE – May/June 2014 0580 31 (b) (i) −3 0 6 2 B1 for two correct (ii) Correct ruled line 1 (c) 1.4 to 1.6 and −3.6 to −3.4 1FT,1FT FT from their graph ±0.1 (d) 1.5 1

This question in 0580/31 May/June 2014

8 (a) Write down an expression for the total mass of c cricket balls, each weighing 160 grams, and f footballs, each weighing 400 grams. Answer(a) … grams [2] (b) Expand and simplify. 3(2x – 5y) – 4(x – 2y) Answer(b) … [2] (c) Factorise completely. 5x2y – 20x Answer(c) … [2] (d) Solve the simultaneous equations. 3x + 4y = 7 4x – 3y = 26 Answer(d) x = … y = … [4] __________________________________________________________________________________________

10 marks

Mark scheme: 8 (a) 160c + 400f final answer 2 B1 for 160c or 400f seen (b) 2x – 7y final answer www 2 B1 for 2 x or –7y or 6x – 15y or –4x + 8y www (c) 5x(xy – 4) final answer 2 B1 for 5( x 2 y − 4 x ) or x ( 5 xy − 20 ) IGCSE – May/June 2014 0580 31 (d) [x=] 5 [y=] – 2 4 M1 for correctly equating one set of coefficients M1 for correct method to eliminate one variable A1 for correct x or y If zero scored SC1 for 2 values satisfying one of the original equations Alternative method M1 for correct rearrangement of one equation x = (7 – 4y) ÷ 3 or y = (7 – 3x) ÷ 4 or x = (26 + 3y) ÷ 4 or y = (4x – 26) ÷ 3 M1 for correct substitution in other equation 4(7 – 4y) ÷ 3 – 3y = 26 4x – 3(7 – 3x) ÷ 4 = 26 3(26 + 3y) ÷ 4 + 4y = 7 3x + 4(4x – 26) ÷ 3 = 7 (7 – 4y) ÷ 3 = (26 + 3y) ÷ 4 (7 – 3x) ÷ 4 = (4x – 26) ÷ 3 A1 for correct x or y If zero scored SC1 for 2 values satisfying one of the original equations

This question in 0580/31 May/June 2014

6 (a) Complete the table of values for y = x2 + 2x – 3 . x –4 –3 –2 –1 0 1 2 3 4 y 0 –3 –4 –3 0 5 21 [2] (b) On the grid, draw the graph of y = x2 + 2x – 3 for –4 Ğ x Ğ 4 . y 25 20 15 10 5 x –4 –3 –2 –1 0 1 2 3 4 –5 [4] (c) On the grid, draw the line y = 10 . [1] (d) Use your graphs to solve the equation x2 + 2x – 3 = 10 for –4 Y x Y 4 . Answer(d) x = … [1] __________________________________________________________________________________________

8 marks

Mark scheme: 6 (a) 5 12 2 B1, B1 (b) 9 points plotted correctly 3FT B2FT for 7 or 8 points correctly plotted B1FT for 5 or 6 points correctly plotted correct smooth curve through all 1 9 correct points (c) correct ruled line 1 minimum length must touch y axis and curve (d) 2.7 to 2.8 1FT FT their curve and ruled line IGCSE – May/June 2014 0580 32

This question in 0580/32 May/June 2014

7 (a) NOT TO 5p + 3r 7p – 6r SCALE p + 2r Write an expression for the perimeter of this triangle. Give your answer in its simplest form. Answer(a) … [2] (b) Another triangle has a perimeter 12w – 2z . Calculate this perimeter when w = 16 and z = –3. Answer(b) … [2] (c) Solve. (i) 5a = 32 Answer(c)(i) a = … [1] (ii) 5b + 23 = 8 Answer(c)(ii) b = … [2] (iii) 5c + 7 = 2(c – 10) Answer(c)(iii) c = … [3] (d) (i) Multiply out the brackets. 8(2x + 3) Answer(d)(i) … [1] (ii) Factorise completely. 6x2 – 12x Answer(d)(ii) … [2] (e) Write each expression in its simplest form. (i) 3q4 × 5q2 Answer(e)(i) … [2] (ii) t 8 ÷ t 2 Answer(e)(ii) … [1] __________________________________________________________________________________________

16 marks

Mark scheme: 7 (a) 13p – r Final Answer 2 B1 for either 13p or – r in the answer or 13p – r spoilt (b) 198 2 M1 for 12 × 16 – 2 × –3 or B1 for 192 or + 6 or – (–6) seen (c) (i) 6.4 or 6 2 1 5 (ii) 2 M1 for first correct step, i.e. 5b = 8 – 23 or better, –3 23 8 or b + = or better 5 5 (iii) 3 B1 for 2c – 20 –9 M1FT for correctly collecting cs on one side and numbers on the other, e.g. 5c – 2c = –7 – 20 or better (d) (i) 16x + 24 1 (ii) 6x (x – 2) 2 B1 for x(6x – 12), 6(x2 – 2x), 2(3x² – 6x), 3( 2x² – 4x), 2x (3x – 6) or 3x(2x – 4) (e) (i) 15q6 2 B1 for 15qn (n not 0) or kq6 ( k not 0) (ii) t6 1

This question in 0580/32 May/June 2014

7 Today it is Simon’s birthday. (a) Simon is x years old. Katy is twice as old as Simon. Bob is 8 years younger than Simon. (i) Write expressions, in terms of x, for the ages of Katy and Bob. Answer(a)(i) Katy … Bob … [2] (ii) The sum of their three ages is 40 years. Write an equation in terms of x. Answer(a)(ii) … [1] (iii) Solve your equation for x. Answer(a)(iii) x = … [2] (b) Simon’s birthday cake weighs 600 grams. 1 He eats of the cake. 8 Katy eats 25% of the cake. Bob eats 0.3 of the cake. Find the weight of the cake that is left. Answer(b) … g [4] (c) Aunty Millie gives Simon $150 for his birthday. He invests the money in a bank at a rate of 6% per year compound interest. Calculate the total amount Simon will have after 3 years. Answer(c) $ … [3] (d) One of Simon’s presents is a bag of sweets. He decides to eat the sweets in a sequence. On day 1 he eats 1 sweet, on day 2 he eats 5 sweets, on day 3 he eats 9 sweets and so on. (i) Describe in words the rule for continuing the sequence 1, 5, 9, 13, 17 … . Answer(d)(i) … [1] (ii) Write down an expression for the number of sweets he eats on day n. Answer(d)(ii) … [2] __________________________________________________________________________________________

15 marks

Mark scheme: 7 (a) (i) 2x 1, 1 x – 8 (ii) x + 2x + x – 8 = 40 or better 1FT FT if algebraic (iii) 12 cao 2 M1 FT for ax = b and a and b not zero (b) 195 cao 4 B1 for 75 B1 for 150 B1 for 180 (c) 178.65 3 M2 for 150 × 1.063 oe or 178.7 or or 179 M1 for 150 × 1.06 × 1.06 (d) (i) Add 4 oe 1 (ii) 4n – 3 oe, final answer 2 M1 for 4n + k (k not -3), qn–3 (q not 0 or 4) seen

This question in 0580/33 May/June 2014

20 .6 (a) (i) Complete the table of values for y = x x –8 –5 –4 –2.5 2.5 4 5 8 y –2.5 –4 8 4 [2] 20 (ii) On the grid, draw the graph of y = for –8 Y x Y –2.5 and 2.5 Y x Y 8. x y 9 8 7 6 5 4 3 2 1 x –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 –1 –2 –3 –4 –5 –6 –7 –8 –9 [4] 20 (iii) By drawing a suitable line on your graph solve the equation = 6. x Answer(a)(iii) x = … [2] (b) x –8 0 8 y 1 (i) Complete the table for y = x – 1. [2] 2 1 (ii) On the grid, draw the graph of y = x – 1 for –8 Y x Y 8. [1] 2 1 (iii) Write down the gradient of y = x – 1. 2 Answer(b)(iii) … [1] 20 1 (c) Write down the values of x at the points of intersection of the graphs of y = and y = x – 1. x 2 Answer(c) x = … and x = … [2] __________________________________________________________________________________________

14 marks

Mark scheme: 6 (a) (i) –5 –8 5 2.5 2 B1 for 3 correct (ii) 8 points correctly plotted B3FT B2FT for 6 or 7 correct points Correct curve 1 B1FT for 4 or 5 correct points (iii) Ruled line y = 6 drawn 1 Independent marks 3.1 to 3.6 1 (b) (i) –5 –1 3 2 B1 for 2 correct (ii) Ruled correct line 1 (iii) 1 1 oe 2 (c) 7.2 to 7.6 1FT –5.2 to –5.6 1FT

This question in 0580/32 Oct/Nov 2014

2 (a) Simplify. 7e – 5f + 4e – f Answer(a) … [2] (b) Find the value of 8g – 9h when g = 5 and h = –3. Answer(b) … [2] (c) Solve the equation. 4x – 7 = 29 Answer(c) x = … [2] (d) Simplify. k 4 ÷ k 11 Answer(d) … [1] (e) Pens cost p cents and pencils cost w cents. (i) Aisha buys 3 pens and 5 pencils for $2.20 . Complete the equation representing this cost in cents. Answer(e)(i) 3p + 5w = … [1] (ii) Bishen buys 4 pens and 10 pencils for $3.50 . Write down an equation representing this cost in cents. Answer(e)(ii) … [1] (iii) Solve your equations to find the value of p and the value of w. Answer(e)(iii) p = … w = … [3] __________________________________________________________________________________________

12 marks

Mark scheme: 2 (a) 11e – 6f as final answer 2 B1 for either 11e or –6f in their final answer (b) 67 2 B1 for 8 × 5 – 9 × –3 or 40 or +27 (c) 9 2 M1 for algebraic first step correct 4x = 29 + 7 7 29 or x – = or better 4 4 (d) k–7 oe 1 (e) (i) 220 1 (ii) 4p + 10w = 350 1 (iii) [p =] 45, [w =] 17 3 M1FT for correct elimination of one variable from their equations A1 for p = 45 A1 for w = 17 If zero scored, SC1FT for 2 values satisfying one of their original equations

This question in 0580/32 May/June 2015

4 Three friends are going on holiday. They travel by plane. (a) Ahmed’s suitcase has mass m kilograms. (i) The mass of Sonia’s suitcase is 5 kg more than the mass of Ahmed’s suitcase. Write down an expression, in terms of m, for the mass of Sonia’s suitcase. Answer(a)(i) … kg [1] (ii) The mass of Hala’s suitcase is twice the mass of Ahmed’s suitcase. Write down an expression, in terms of m, for the mass of Hala’s suitcase. Answer(a)(ii) … kg [1] (iii) The total mass of the three suitcases is 47 kg. Write down an equation in terms of m. Answer(a)(iii) … [1] (iv) Solve your equation and find the mass of each suitcase. Answer(a)(iv) Ahmed’s suitcase … kg Sonia’s suitcase … kg Hala’s suitcase … kg [3] (b) Each friend carries one bag of hand luggage onto the plane. (i) The rule for the maximum size of each bag of hand luggage is length + width + height  115 cm. The measurements of Ahmed’s bag are shown in the table. Length Width Height 550 mm 395 mm 200 mm Can Ahmed carry this bag onto the plane? Explain your answer. Answer(b)(i) … because … … [2] (ii) The mass of Ahmed’s bag is 5 kg, correct to the nearest kilogram. Write down the upper bound of the mass of his bag. Answer(b)(ii) … kg [1] (c) The friends change money from dollars into euros (€) to spend on holiday. The exchange rate is $1 = €0.68 . (i) Sonia changes $150 into euros. Work out how many euros she receives. Answer(c)(i) € … [1] (ii) Hala pays €25.50 for a meal. Work out how much this is in dollars. Answer(c)(ii) $ … [2] __________________________________________________________________________________________

12 marks

Mark scheme: 4 (a) (i) m + 5 1 (ii) 2m 1 (iii) m + m + 5 + 2m = 47 isw 1FT FT m + their (a)(i) + their (a)(ii) = 47 isw or 4m + 5 = 47 isw (iv) 10.5 3 M1FT for correct first step to solve their (a)(iii) 15.5 A1FT for m = 10.5 21 (b) (i) Yes, [total = ] 114.5 [cm] 2 M1 for 55 + 39.5 + 20 oe or for 1145 mm (ii) 5.5 1 (c) (i) 102 1 (ii) 37.5[0] 2 M1 for 25.5[0] ÷ 0.68

This question in 0580/31 Oct/Nov 2015

9 (a) Expand and simplify. 2(3x + 2) – 4(x + 1) Answer(a) … [2] (b) Factorise completely. 3y2 – 6y Answer(b) … [2] (c) Make b the subject of the formula. b a = – 5 4 Answer(c) b = … [2] (d) Solve the simultaneous equations. You must show all your working. 3x + 2y = 11 6x – y = 32 Answer(d) x = … y = … [3] __________________________________________________________________________________________

9 marks

Mark scheme: 9 (a) 2x final answer 2 M1 for 6x + 4 or –4x – 4 (b) 3y(y – 2) final answer 2 B1 for 3(y2 – 2y) or y(3y – 6) (c) 4a + 20 or 4(a + 5) 2 M1 for a + 5 = b or 4a = b – 20 4 (d) Correct working and 3 M1 for correctly eliminating one variable [x =] 5, [y =] –2 A1 for x = 5 A1 for y = –2 If zero scored, SC1 for 2 values satisfying one of the original equations SC1 if no working shown, but 2 correct answers given

This question in 0580/31 Oct/Nov 2015

4 (a) Solve. (i) 29 – x = 18 Answer(a)(i) x = … [1] (ii) 4(2y + 7) = 164 Answer(a)(ii) y = … [3] (b) Simplify. 6x4 × 8x Answer(b) … [2] (c) Find (i) 81, Answer(c)(i) … [1] (ii) 73, Answer(c)(ii) … [1] (iii) 80. Answer(c)(iii) … [1] (d) (i) Write 6751 correct to the nearest hundred. Answer(d)(i) … [1] (ii) Write 0.25 as a fraction. Answer(d)(ii) … [1] (iii) Write 0.06 as a percentage. Answer(d)(iii) … % [1] (iv) Write 687 000 000 in standard form. Answer(d)(iv) … [1] __________________________________________________________________________________________

13 marks

Mark scheme: 4 (a) (i) 11 1 (ii) 17 3 M1 for 8y + 28 = 164 or 2y + 7 = 41 M1 FT for a correct further step (b) 48x5 2 M1 for 48xk or jx5 (c) (i) 9 1 Accept ± 9 (ii) 343 1 (iii) 1 1 (d) (i) 6800 1 1 (ii) 1 Accept equivalent fraction 4 (iii) 6 1 (iv) 6.87 ×108 1

This question in 0580/33 Oct/Nov 2015

6 (a) Solve these equations. (i) x + 7 = 15 x = … [1] (ii) 5(3x + 8) = 10 x = … [3] (b) A club is arranging transport for its members. Speedy Coaches charge $625 plus $15 per member. The total cost, in dollars, for x members is given by the expression 15x + 625. (i) Sporty Coaches charge $117 plus $19 per member. Write an expression for the total cost, in dollars, for x members. … [2] (ii) The total cost is the same for both Speedy Coaches and Sporty Coaches. Write down an equation and solve it to find x. x = … [3]

9 marks

Mark scheme: 6 (a) (i) 8 1 (ii) –2 3 M1 for first step correctly completed M1FT for second step correctly completed (b) (i) 19x + 117 2 B1 for 19x + c or mx + 117 (ii) 15x + 625 = their (b)(i) 1 127 2 M1FT for the first correct step of their linear equation

This question in 0580/31 May/June 2016

9 (a) Complete the table of values for y = 8 + 7x − x2. x 0 1 2 3 4 5 6 7 8 y 8 18 18 8 [3] (b) On the grid, draw the graph of y = 8 + 7x − x2 for 0 G x G 8. y 22 20 18 16 14 12 10 8 6 4 2 x 0 1 2 3 4 5 6 7 8 [4] (c) Write down the co-ordinates of the highest point of the curve. ( … , … ) [1] (d) (i) On the grid, draw the line y = 16. [1] (ii) Use your line to solve the equation 8 + 7x − x2 = 16. x = … or x = … [2]

11 marks

Mark scheme: 9 (a) … 14 … 20 20 … 14 … 0 3 B2 for 3 or 4 correct B1 for 2 correct (b) Completely correct curve 4 B3FT for 8 or 9 points correctly plotted or B2FT for 6 or 7 points correctly plotted or B1FT for 4 or 5 points correctly plotted (c) (3.5, h) 1 20 < h ⩽ 20.4 (d) (i) Correct ruled line 1 (ii) 1.4 5.6 1, 1FT FT their graph and line

This question in 0580/31 May/June 2016

6 y 6 4 2 x 0 –5 –4 –3 –2 –1 1 2 –2 –4 –6 –8 –10 (a) On the grid, (i) draw the line y = 3, [1] (ii) draw the line that is perpendicular to the line y = 3 that passes through the point (1, −4). [2] (b) Complete the table of values for y = 2 − 3x − x2. x −5 −4 −3 −2 −1 0 1 2 y −2 2 2 −2 [2] (c) On the grid, draw the graph of y = 2 − 3x − x2 for - 5 G x G 2 . [4] (d) Write down the co-ordinates of the highest point of the graph of y = 2 − 3x − x2. ( … , … ) [1] (e) Use your graphs to solve the equation 2 − 3x − x2 = 3. x = … or x = … [2]

12 marks

Mark scheme: 6 (a) (i) Ruled continuous line y = 3 1 (ii) Ruled continuous line x = 1 2 B1 for (1, –4) plotted or B1 for any line perpendicular to their y = 3 drawn (b) –8, 4, 4, –8 2 B1 for 3 correct (c) Completely correct curve 4 B3FT for 7 or 8 points correctly plotted B2FT for 5 or 6 points correctly plotted B1FT for 3 or 4 points correctly plotted (d) (–1.5, 4.1 to 4.4) 1 (e) –2.5 to –2.7 and –0.3 to –0.5 2FT FT intersection of their (a)(i) with their curve B1FT for one correct

This question in 0580/32 May/June 2016

9 (a) p = 4r − 3t (i) Calculate the value of p when r = 5 and t = −6. p = … [2] (ii) Make r the subject of the formula p = 4r − 3t. r = … [2] (b) Expand the brackets and simplify. 4(3x − 2) − 3(x − 5) … [2] (c) Factorise completely. 12ab − 20a2 … [2]

8 marks

Mark scheme: 9 (a) (i) 38 2 M1 for 4 × 5 – 3 × −6 or better or B1 for 20 or 18 or –18 seen p + 3t p 3t (ii) oe 2 M1 for 4r = p + 3t or = r − 4 4 4 (b) 9x + 7 final answer 2 B1 for 12x – 8 or –3x + 15 or 9x or + 7 seen in working (c) 4a(3b – 5a) final answer 2 M1 for a(12b – 20a) or 4(3ab – 5a2) or 2a(6b – 10a) or 2(6ab – 10a2)

This question in 0580/32 May/June 2016

4 (a) Complete the table of values for y = x 2 - 5x + 3 . x –1 0 1 2 3 4 5 y 3 –1 –1 3 [2] (b) On the grid, draw the graph of y = x 2 - 5x + 3 for -1 G x G 5 . y 10 8 6 4 2 x –1 0 1 2 3 4 5 –2 –4 [4] (c) Write down the equation of the line of symmetry of the graph of y = x 2 - 5x + 3 . … [1] (d) Write down the co-ordinates of the point where the line y = 4 - x (i) crosses the x-axis, ( … , … ) [1] (ii) crosses the y-axis. ( … , … ) [1] (e) On the grid, draw the line y = 4 - x . [1] (f) Write down the co-ordinates of the points of intersection of the graph of y = x 2 - 5x + 3 and the line y = 4 - x . ( … , … ) ( … , … ) [2]

12 marks

Mark scheme: 4 (a) 9, –3, –3 2 B1 for 9 or –3 and –3 (b) Correct curve 4 B3FT for 6 or 7 correctly plotted points or B2FT for 4 or 5 correctly plotted points or B1FT for 2 or 3 correctly plotted points (c) x = 2.5 1 (d) (i) (4, 0) 1 (ii) (0, 4) 1

This question in 0580/32 Oct/Nov 2016

8 (a) Complete the table of values for y = x 2 - 2x . x - 3 - 2 - 1 0 1 2 3 4 y 3 - 1 3 [3] (b) On the grid, draw the graph of y = x 2 - 2x for - 3 G x G 4 . y 16 14 12 10 8 6 4 2 x 0 –3 –2 –1 1 2 3 4 –2 [4] (c) On the grid, draw the line y = 6 . [1] (d) Use your graph to solve the equation x 2 - 2x = 6 . Give your answers correct to 1 decimal place. x = … or x = … [2] Question 9 is printed on the next page.

10 marks

Mark scheme: 8 (a) 15 8 … 0 … 0 … 8 3 B1 for 8 and 8 in the correct place B1 for 0 and 0 in the correct place B1 for 15 in the correct place (b) Correct curve 4 B3FT for 7 or 8 correctly plotted points or B2FT for 5 or 6 correctly plotted points or B1FT for 3 or 4 correctly plotted points (c) Correct ruled line 1 (d) –1.8 or –1.7 or –1.6 2FT B1FT for one correct 3.6 or 3.7 or 3.8 or B1FT for both correct answers as co-ordinates or B1FT for both answers correct to more than 1dp

This question in 0580/33 Oct/Nov 2016

2 (a) Simplify. 5a + 6a - a … [1] (b) 3f – 4g NOT TO SCALE 5f + 2g Write an expression for the perimeter of the rectangle. Give your answer in its simplest form. … [3] (c) (i) Work out the value of 5x + 10 y when x = 7 and y = 9 . … [2] (ii) Work out the value of 4r 2 - pr when p = 3 and r = 5 . … [2] (d) Solve. 5 3x - 6 = 75 ^ h x = … [3] (e) Mr and Mrs Barker have three children, Molly, Dean and Raul. Age, in terms of x Molly’s age is x years x Dean is 5 years younger than Molly x - 5 Raul is 4 years older than Molly Mr Barker is 4 times older than Molly Mrs Barker is 6 years younger than Mr Barker (i) Complete the table with expressions in terms of x. [2] (ii) The total of the five ages is 125 years. Write down an equation in terms of x and show that it simplifies to 11x - 7 = 125 . [1] (iii) Solve the equation 11x - 7 = 125 to find Molly’s age. Molly’s age = … years [2]

16 marks

Mark scheme: 2(a) 10a final answer 1 2(b) 16f – 4g final answer 3 M2 for 2 × (5f + 2g) + 2×(3f − 4g) oe or or 4(4f – g) final answer B1 for 10f +4g or 6f −8g or 8f −2g or 16f + kg or kf – 4g 2(c)(i) 125 2 M1 for 5 × 7 + 9 × 10 or better 2(c)(ii) 85 2 M1 for 4 × 52 – 3 × 5 or better 2(d) 7 3 M1 for 15x – 30 [= 75] or 3x – 6 = 15 M1FT for correct second step 2(e)(i) x + 4 2 B1 for any two correct 4x 4x – 6 2(e)(ii) x + x–5 + x+4 + 4x + 4x–6 = 125 1 2(e)(iii) 12 2 7 125 M1 for 11x = 125 + 7 or x – = 11 11 or better

This question in 0580/32 May/June 2017

8 (a) Simplify. (i) 8p + 2r + 4p − 9r … [2] (ii) 4x3 × 6x2 … [1] (b) Write down an expression, in terms of x and y, for the total cost of x cakes at 90 cents each and y drinks at 75 cents each. … cents [2] (c) Factorise completely. 12p2 − 8p … [2] (d) Solve. 4(7r − 3) = 128 r = … [3] (e) Solve the simultaneous equations. You must show all your working. 4x + 3y = 43 6x + 7y = 92 x = … y = … [4] Question 9 is printed on the next page.

14 marks

Mark scheme: 8(a)(i) 12p – 7r final answer 2 B1 for 12p + jr or kp –7r j, k can be 0 or 12p + –7r 8(a)(ii) 24x5 final answer 1 8(b) 90x + 75y final answer 2 B1 for 90x + jy or kx +75y j, k can be 0 or 0.9x + 0.75y 8(c) 4p(3p – 2) final answer 2 B1 for 4(3p2 – 2p) or p (12p – 8) or 2(6p2 – 4p) or 2p(6p – 4) 8(d) 5 3 M1 for first correct step M1FT for second correct step 8(e) Correctly equating one set of M1 coefficients Correct method to eliminate one M1 Dependent on the coefficients being the same for variable one of the variables. Correct consistent use of addition or subtraction using their equations. [x = ] 2.5 A1 [y = ] 11 A1 If zero scored, SC1 if no working shown, but 2 correct answers given or SC1 for 2 values satisfying one of the original equations

This question in 0580/32 Oct/Nov 2017

5 (a) A small box contains n biscuits. (i) A medium box contains 10 more biscuits than the small box. Write an expression, in terms of n, for the number of biscuits in the medium box. … [1] (ii) A large box contains twice as many biscuits as the medium box. Write an expression, in terms of n, for the number of biscuits in the large box. … [1] (iii) There are 52 biscuits in the large box. Write down an equation, in terms of n, and solve it. n = … [3] (iv) Olga buys a small box and a medium box of biscuits. How many biscuits does she have altogether? … [1] (b) In the large box, 13 of the 52 biscuits are chocolate. Leo takes a biscuit from the box at random. (i) Find the probability that Leo’s biscuit is chocolate. Give your answer as a fraction in its lowest terms. … [2] (ii) On the probability scale, draw an arrow to show the probability that Leo’s biscuit is not chocolate. 0 1 [1] (c) The mass of the large box of biscuits is 450 g. Work out the total mass of 6 large boxes of biscuits. Give your answer in kilograms. … kg [2] (d) The mass, m grams, of the small box of biscuits is 120 g, correct to the nearest 10 g. Complete the statement about the value of m. … G m 1 … [2]

13 marks

Mark scheme: 5(a)(i) n + 10 1 5(a)(ii) 2(n + 10) oe isw 1FT 5(a)(iii) their (ii) = 52 M1 16 final answer B2 M1 for 2n = 52 – 20 or n = 26 – 10 or better 5(a)(iv) 42 1FT FT 2 × their (iii) + 10 5(b)(i) 1 2 13 cao B1 for oe soi 4 52 5(b)(ii) 3 1 Correct arrow at 4 5(c) 2.7[00] 2 B1 for answer figs 27 or for 0.45 seen 5(d) 115 2 B1 for one correct or both values 125 correct but reversed

This question in 0580/33 Oct/Nov 2017

9 (a) Factorise. y2 + 8y … [1] (b) Expand the brackets and simplify. 3(2x – 1) – 4(x – 5) … [2] (c) Make p the subject of the formula k = 5m + 7p. p = … [2] (d) Solve the simultaneous equations. You must show all your working. 3x + 2y = 6 2x – 3y = 17 x = … y = … [4]

9 marks

Mark scheme: 9(a) y(y + 8) final answer 1 9(b) 2x + 17 final answer 2 B1 for 6x – 3 or –4x + 20 or 2x + j or kx + 17 as final answer 9(c) k − 5 m 2 k 5m oe final answer M1 for 7p = k – 5m or = + p 7 7 7 9(d) Correctly equating one set of coefficients M1 Correct method to eliminate one variable M1 Dependent on the coefficients being the same for one of the variables. Correct consistent use of addition or subtraction using their equations. x = 4 A1 y = –3 A1 If zero scored, SC1 if no working shown, but 2 correct answers given or SC1 for 2 values satisfying one of the original equations.

This question in 0580/33 Oct/Nov 2017

9 (a) Solve the equation 3(2x − 4) = 4(x + 7). x = … [3] (b) Beindu goes to the market to buy apples and bananas. She can buy • 7 apples and 4 bananas for 85 cents or • 3 apples and 8 bananas for 93 cents. Apples cost a cents each and bananas cost b cents each. (i) This information can be used to write down two equations. One of these is 7a + 4b = 85. Write down the other equation. … = … [2] (ii) Solve these two simultaneous equations. You must show all your working. a = … b = … [3]

8 marks

Mark scheme: 9(a) 20 nfww 3 B2 for 6x – 4x = 28 + 12 or better or B1 for 6x – 12 or 4x + 28 or B1FT for correct ax = b after incorrect expansions first step 9(b)(i) 3a + 8b = 93 2 B1 for 3a + 8b 9(b)(ii) For correctly eliminating one variable M1 For correct method to equate coefficients and eliminate one variable [a =] 7 A1 [b =] 9 A1 If 0 scored SC1 for 2 values satisfying one of the original equations SC1 if no working shown, but 2 correct answers given

This question in 0580/32 May/June 2018

4 (a) Solve these equations. (i) 3x = 18 x = … [1] (ii) 8x - 15 = 6x + 2 x = … [2] (b) Factorise. 5x - 15 … [1] (c) Simplify. 2x - 6y + 3x + 2y … [2] (d) Find the value of 5u - 2v when u = 11 and v =- 3 . … [2] (e) Make p the subject of this formula. H = 7p - 3 p = … [2] (f) (i) Find the value of k when x 10 ' x k = x 3 . k = … [1] (ii) Find the value of n when y 10 # y n = 1. n = … [1]

12 marks

Mark scheme: 4(a)(i) 6 1 4(a)(ii) 8.5 2 M1 for 8x – 6x = 2 + 15 or better 4(b) 5(x – 3) final answer 1 4(c) 5x – 4y final answer 2 B1 for 5x + ky or kx – 4y (k could be 0) 4(d) 61 2 B1 for 55 or 6 or M1 for 5 × 11 – 2 × –3 4(e) H + 3 2 M1 for correct first step p = oe final answer 7 4(f)(i) 7 1 4(f)(ii) –10 1

This question in 0580/33 May/June 2018

66 (a) Complete the table of values for y = , x =Y 0 . x x - 6 - 4 - 3 - 2 - 1 1 2 3 4 6 y - .15 - 3 3 1.5 [3] 6 (b) On the grid, draw the graph of y = for - 6 G x G - 1 and 1 G x G 6 . x y 6 5 4 3 2 1 x – 6 – 5 – 4 – 3 – 2 – 1 0 1 2 3 4 5 6 – 1 – 2 – 3 – 4 – 5 – 6 [4] (c) On the grid, draw the line y =- 5 . [1] 6 (d) Use your graph to solve the equation =- 5 . x x = … [1]

9 marks

Mark scheme: 6(a) –1 … –2 … –6 … 6 … 2 … 1 3 B2 for 4 or 5 correct or B1 for 2 or 3 correct 6(b) correct smooth curves 4 B3FT for 9 or 10 points plotted correctly B2FT for 7 or 8 points plotted correctly B1FT for 5 or 6 points plotted correctly FT their table 6(c) correct continuous ruled line 1 6(d) –1.2 oe 1 or FT their line and their graph

This question in 0580/33 May/June 2018

8 (a) Simplify. 4c + 2d - c + 6 d … [2] (b) h = 5m - 2n Calculate h when m = 4 and n = -6. … [2] (c) Solve. 7(x - 3) = 56 x = … [2] (d) Make t the subject of the formula r = 6t + 7. t = … [2] (e) The diagram shows a triangle. x° NOT TO SCALE (x + 15)° 3x° Use the diagram to write down an equation and solve it to find the value of x. x = … [4] Question 9 is printed on the next page.

12 marks

Mark scheme: 8(a) 3c + 8d 2 B1 for 3c or 8d 8(b) 32 2 M1 for 5 × 4 – 2 × –6 or better 8(c) 11 2 M1 for x – 3 = 8 or 7x – 21 = 56 or better 8(d) r − 7 2 r 7 oe M1 for 6t = r – 7 or = t + 6 6 6 8(e) 3x + x + x + 15 = 180 or better 4 M1 for 3x + x + x + 15 or better leading to M1 for their expression = 180 [x = ] 33 M1 for rearranging their equation to ax = b If 0 scored, SC2 for 33 nfww

This question in 0580/31 Oct/Nov 2018

3 (a) Simplify. (i) 16c - 5d - 4c + 4d … [2] (ii) 4x3 # 2x7 … [2] (b) Solve. 3x - 2 = 5x + 1 x = … [2] (c) Factorise completely. 3x2y - 5xy … [2] (d) Make r the subject of the formula. T = 3(r + 5) r = … [2]

10 marks

Mark scheme: 3(a)(i) 12c – d final answer 2 B1 for 12c or − d 3(a)(ii) 8x10 final answer 2 B1 for 8xn or kx10 (n and k ≠ 0) 3(b) 1 2 M1 for −2 −1 = 5x − 3x or better −1.5 or −1 oe nfww or 3x − 5x = 1 + 2 or better 2 3(c) xy(3x – 5) final answer 2 B1 for y(3x2– 5x) or x(3xy – 5y) or correct answer spoilt 3(d) T 2 M1 for first correct step e.g. T = 3r + 15 or [r =] − 5 oe nfww T 3 = r + 5 final answer 3

This question in 0580/32 Oct/Nov 2018

8 (a) Complete the table of values for y = 8x - x 2 . x 0 1 2 3 4 5 6 7 8 y 0 12 15 15 12 [3] (b) On the grid, draw the graph of y = 8x - x 2 for 0 G x G 8 . y 18 16 14 12 10 8 6 4 2 0 x 1 2 3 4 5 6 7 8 [4] (c) Write down the equation of the line of symmetry of this graph. … [1] (d) Use the graph to solve 8x - x 2 = 10 . x = … or x = … [2]

10 marks

Mark scheme: 8(a) 7 16 7 0 3 B2 for 2 or 3 correct B1 for 1 correct 8(b) Correct curve 4 B3FT for 8 or 9 points plotted correctly or B2FT for 6 or 7 points plotted correctly or B1FT for 4 or 5 points plotted correctly 8(c) x = 4 1 8(d) 1.45 to 1.65 and 6.35 to 6.55 2 B1 for each or both correct as co-ordinates

This question in 0580/32 Oct/Nov 2018

8 (a) y 4 L 3 2 1 –2 –1 0 1 2 3 4 5 6 7 8 x –1 Line L is drawn on the grid. Find the equation of line L. Give your answer in the form y = mx + c. y = … [3] - 7. (b) The points (9, a) and (b, 3) lie on the line y = 23 x Work out the value of (i) a, a = … [2] (ii) b. b = … [2] (c) (i) Complete the table of values for y = x (3 - x). x -4 -2 -1 0 1 2 4 y -10 0 2 -4 [3] (ii) On the grid, draw the graph of y = x (3 - x) for - 4 G x G 4. y 5 -4 -2 0 2 4 x -5 -10 -15 -20 -25 -30 [4] (iii) Write down the co-ordinates of the highest point of the graph for - 4 G x G 4. ( … , … ) [1]

15 marks

Mark scheme: 8(a) 1 3 1 [ y = ] − x + 3 B2 for [ y = ] − x + c 2 2 or rise 1 M1 for or m = ± oe run 2 and B1 for [ y = ] kx + 3 , k ≠ 0 or c = 3 8(b)(i) –1 2 2 M1 for [ a = ] × 9 − 7 or better 3 8(b)(ii) 15 2 2 M1 for 3 = b − 7 or better 3 8(c)(i) –28, –4, 2 3 B1 for each 8(c)(ii) correct smooth curve 4 B3FT for 6 or 7 correct plots or B2FT for 4 or 5 correct plots or B1FT for 2 or 3 correct plots 8(c)(iii) (1.5 , 2.25) 1 accept (x, y) where 1 < x < 2 and 2 < y < 4

This question in 0580/32 Feb/March 2019

9 (a) Simplify 8a + 3b - 2a + b. … [2] (b) Calculate the value of 4x2 + xy when x = 3 and y = -2. … [2] (c) Solve these equations. x (i) = 20 4 x = … [1] (ii) 3x - 5 = 16 x = … [2] (iii) 5(2x + 1) = 27 x = … [3] (d) Make r the subject of this formula. p = 3r - 5 r = … [2] Question 10 is printed on the next page.

12 marks

Mark scheme: 9(a) 6a + 4b final answer 2 B1 for 6a + kb or ka + 4b 9(b) 30 2 M1 for 4 × 32 + 3 × −2 or better 9(c)(i) 80 1 9(c)(ii) 7 2 M1 for 3x = 16 + 5 or x – 53 = 163 or better 9(c)(iii) 2.2 oe 3 M1 for 10x + 5 [= 27] or 2 x + 1 = 275 M1 for second correct step 9(d) p + 5 p 5 2 p 5 or + final answer M1 for p + 5 = 3r oe or = r − 3 3 3 3 3

This question in 0580/31 May/June 2019

4 (a) Complete the table of values for y = 5 + 2 x - x 2 . x -2 -1 0 1 2 3 4 y 2 5 6 -3 [2] (b) On the grid, draw the graph of y = 5 + 2x - x 2 for - 2 G x G 4 . y 7 6 5 4 3 2 1 0 –2 –1 1 2 3 4 x –1 –2 –3 [4] (c) (i) On the grid, draw the line of symmetry. [1] (ii) Write down the equation of the line of symmetry. … [1] (d) Use your graph to find the solutions of the equation 5 + 2x - x 2 = 4 . x = … or x = … [2] (e) (i) On the grid, draw a line from (- 1, 2) to (1, 6) . [1] (ii) Find the equation of this line in the form y = mx + c . y = … [3]

14 marks

Mark scheme: 4(a) −3 5 2 2 B1 for 2 correct 4(b) Correct curve 4 B3FT for 6 or 7 points correct B2FT for 4 or 5 points correct B1FT for 2 or 3 points correct 4(c)(i) Ruled line x = 1 drawn 1 4(c)(ii) x = 1 1 4(d) −0.5 to −0.3 and 2.3 to 2.5 2 B1 for each If 0 scored, B1 for y = 4 drawn 4(e)(i) Correct ruled continuous line 1 4(e)(ii) [y =] 2x + 4 3 B2 for [y =] 2x + k rise or M1 for run B1 for kx + 4 , k ≠ 0, or c = 4

This question in 0580/33 May/June 2019

6 (a) Complete the table of values for y = x 2 - 5x + 3 . x −1 0 1 2 3 4 5 6 y −1 −3 −3 −1 3 [2] (b) On the grid, draw the graph of y = x 2 - 5x + 3 for - 1 G x G 6 . y 10 9 8 7 6 5 4 3 2 1 x – 1 0 1 2 3 4 5 6 – 1 – 2 – 3 – 4 [4] (c) Use your graph to solve the equation x 2 - 5x + 3 = 0 . x = … or x = … [2]

8 marks

Mark scheme: 6(a) 9, 3, 9 2 B1 for two correct 6(b) Correct curve 4 B3FT for 7 or 8 correctly plotted points or B2FT for 5 or 6 correctly plotted points or B1FT for 3 or 4 correctly plotted points 6(c) 0.6 to 0.8, 4.2 to 4.4 2 FT their curve B1 for each

This question in 0580/32 Oct/Nov 2019

5 (a) Simplify. 9x - 2y - 5x - y … [2] (b) P = 4ab + 3b 2 Work out the value of a when P = 35 and b = 5. a = … [3] (c) Solve. (i) 10x = 5 x = … [1] (ii) 7x - 3 = 2x + 11 x = … [2] (iii) 3 (2 x - 1) = 27 x = … [3] (d) Rearrange T = 5 (p + 2 ) to make p the subject. p = … [2] (e) Solve the simultaneous equations. You must show all your working. 3x - y = 22 x + 2 y = 5 x = … y = … [3]

16 marks

Mark scheme: 5(a) 4x – 3y final answer 2 B1 for 4x or – 3y or 4x – 3y not as final answer 5(b) −2 3 M1 for 35 = 4 × a × 5 + 3 × 52 or better M1 for 35 – their 3 × 52 = their 4 × 5 × a or better or M1 for P − 3b 2 = 4 ab or better M1 for 35 – 3 × 52 = 4 × 5 × a or better 5(c)(i) 1 1 [0].5 or 2 5(c)(ii) 2.8 oe 2 M1 for 7x − 2x = 11 + 3 or better 5(c)(iii) 5 3 M1 for correct first step i.e. 6x – 3 [= 27] or 2x – 1 = 9 M1 for correct second step leading to ax = b 5(d) T T − 10 2 T [p =] − 2 or final answer M1 for = p + 2 oe or T = 5p + 10 5 5 5 5(e) Correct method to eliminate one variable M1 [x =] 7 A1 [y =] −1 A1 If 0 scored, SC1 for two values that satisfy one of the original equations or SC1 if no working shown, but 2 correct answers given

This question in 0580/33 Oct/Nov 2019

4 (a) Complete the table of values for y = 7 + 2 x - x 2 . x -2 -1 0 1 2 3 4 y -1 8 7 -1 [2] (b) On the grid, draw the graph of y = 7 + 2x - x 2 for - 2 G x G 4 . y 9 8 7 6 5 4 3 2 1 – 2 – 1 0 1 2 3 4 x – 1 – 2 [4] (c) Write down the equation of the line of symmetry of the graph. … [1] (d) Use your graph to solve the equation 7 + 2x - x 2 = 0 . x = … or x = … [2]

9 marks

Mark scheme: 4(a) 4 7 4 2 B1 for one correct 4(b) Correct curve 4 B3FT for 6 or 7 points correct or B2FT for 4 or 5 points correct or B1FT for 2 or 3 points correct 4(c) x = 1 oe 1 4(d) −1.9 to −1.7 and 3.7 to 3.9 2 B1 for each

This question in 0580/31 May/June 2020

8 (a) Simplify 3c - 5d - c + 2d . … [2] (b) Solve the equation 12x - 7 = 23 . x = … [2] (c) Multiply out. 9(3 - x) … [1] ( a + b) h (d) A = 2 Work out the value of h when A = 38.64 , a = 5.5 and b = 3.7 . h = … [3] (e) Alphonse is x years old and Beatrice is y years old. Three times Alphonse’s age is equal to 5 times Beatrice’s age. Twice Beatrice’s age is 4 years more than Alphonse’s age. (i) Use this information to write down two equations in x and y. … … [2] (ii) Find the age of Alphonse and the age of Beatrice. Alphonse … years old Beatrice … years old [3]

13 marks

Mark scheme: 8(a) 2c – 3d final answer 2 B1 for 2c or −3d 8(b) [x =] 2.5 2 7 23 M1 for 12x = 23 + 7 or x − = 12 12 8(c) 27 – 9x 1 8(d) [h =] 8.4 3 B2 for 2 × 38.64 38.64 = 4.6h or 77.28 = 9.2h or 5.5 + 3.7 (5.5 + 3.7) h or B1 for 38.64 = 2 2A or M1 for [h=] a + b 8(e)(i) 3x = 5y oe 2 B1 for each 2y = x + 4 oe 8(e)(ii) [x =] 20 3 M1 for correctly eliminating one variable [y =] 12 B1 for one correct

This question in 0580/31 May/June 2020

9 (a) Complete the table of values for y = x 2 - 3x - 6 . x -3 -2 -1 0 1 2 3 4 5 6 y 12 -2 -2 12 [3] (b) On the grid, draw the graph of y = x 2 - 3x - 6 for - 3 G x G 6 . y 14 12 10 8 6 4 2 -3 -2 -1 0 1 2 3 4 5 6 x -2 -4 -6 -8 -10 [4] (c) Write down the equation of the line of symmetry of the graph. … [1] (d) Use your graph to solve the equation x 2 - 3x - 6 = 0 . x = … or x = … [2] Question 10 is printed on the next page.

10 marks

Mark scheme: 9(a) 4 –6 –8 –8 –6 4 3 B2 for 4 or 5 correct or B1 for 2 or 3 correct 9(b) Correct curve 4 B3FT for 9 or 10 points correctly plotted or B2FT for 7 or 8 points correctly plotted or B1FT for 5 or 6 points correctly plotted 9(c) x = 1.5 1 9(d) −1.4 4.4 2 B1 for each

This question in 0580/32 May/June 2020

10 (a) Solve these equations. (i) 5x =- 30 x = … [1] (ii) 4x - 2 = 28 x = … [2] (iii) 3( 2x + 7) = 12 x = … [3] (b) Solve the simultaneous equations. You must show all your working. 5x - 2y = 44 2x + 3y = 10 x = … y = … [4]

10 marks

Mark scheme: 10(a)(i) −6 1 10(a)(ii) 7.5 or 7 12 2 M1 for 4x = 28 + 2 oe 10(a)(iii) −1.5 or −1 12 3 M1 for 6x + 21 [= 12] or 2x + 7 = 12 ÷ 3 or better M1 for their 6x = 12 – their 21 or 2x = their 4 – 7 10(b) Equating coefficients of one variable M1 Accept any correct method such as by multiplying the equations by substitution e.g. scalars e.g. M1 for rearranging one equation to make 15x – 6y = 132 either x or y the subject 4x + 6y = 20 M1 for substituting their rearranged equation into the other equation Adding or subtracting equations to M1 A1 A1 as in other method eliminate one variable e.g. 19x = 152 If 0 scored, SC1 for two values that satisfy one equation or SC1 for two correct answers [x =] 8 A1 without any correct working [y =] −2 A1

This question in 0580/32 May/June 2020

4 (a) Simplify. 6a - 3b + 2a - 4b … [2] (b) Expand. 5 ( x - 3) … [1] (c) Solve these equations. x (i) = 18 3 x = … [1] (ii) 5x + 18 = 8 x = … [2] (iii) 12x - 3 = 4x + 21 x = … [2] (d) 6 10 # 6 x = 6 2 Find the value of x. x = … [1] (e) The Fraser family and the Singh family go to the cinema. The Fraser family buys 6 adult tickets and 2 child tickets for $124. The Singh family buys 3 adult tickets and 5 child tickets for $100. Find the price of an adult ticket and the price of a child ticket. Adult ticket $ … Child ticket $ … [5]

14 marks

Mark scheme: 4(a) 8a – 7b 2 B1 for 8a or –7b in final answer or for 8a – 7b seen then spoilt 4(b) 5x – 15 1 4(c)(i) 54 1 4(c)(ii) −2 2 M1 for 18 8 5x = 8 – 18 or x + = or better oe 5 5 4(c)(iii) 3 2 M1 for 12x – 4x = 21 + 3 or better oe 4(d) −8 cao 1 4(e) 17.5[0] 5 B1 for 6a + 2c = 124 B1 for 3a + 5c = 100 9.5[0] oe M1FT for a correct method to eliminate one variable A1 for 17.5[0] A1 for 9.5[0] If 0 scored after B0, B1 or B2, SC1 for two values that satisfy one of the/their original equations or family

This question in 0580/31 Oct/Nov 2020

159 (a) Complete the table of values for y = . x x - 5 - 3 - 2 - 1 1 2 3 5 y - 15 15 [3] 15 (b) On the grid, draw the graph of y = for - 5 G x G - 1 and 1 G x G 5 . x y 16 14 12 10 8 6 4 2 0 x – 5 – 4 – 3 – 2 – 1 1 2 3 4 5 – 2 – 4 – 6 – 8 – 10 – 12 – 14 – 16 [4] (c) On the grid, draw the line y = 6 . [1] 15 (d) Use your graph to solve = 6 . x x = … [1]

9 marks

Mark scheme: 9(a) −3 −5 −7.5 7.5 5 3 3 B2 for 4 or 5 correct or B1 for 2 or 3 correct 9(b) Correct curve 4 B3FT for 7 or 8 points plotted correctly or B2FT for 5 or 6 points plotted correctly or B1FT for 3 or 4 points plotted correctly 9(c) Correct ruled line 1 9(d) 2.5 or 2.4 to 2.6 1 FT their line (y = k) and their curve

This question in 0580/31 Oct/Nov 2020

9 (a) Simplify. 4x + 3y + 2x - 8y … [2] (b) A pen costs 60 cents and a ruler costs 29 cents. Write down an expression for the total cost, in cents, of x pens and y rulers. … cents [2] (c) Solve. 5 ( 2x + 4) = 85 x = … [3] (d) (i) 2 8 # 2 m = 2 6 Work out the value of m. m = … [1] (ii) 5 n ' 5 4 = 5 6 Work out the value of n. n = … [1] (e) A plant costs p dollars and a bush costs b dollars. Ana buys 2 plants and 4 bushes for $42. Paola buys 7 plants and 9 bushes for $107. Write down a pair of simultaneous equations and solve them to find the value of p and the value of b. You must show all your working. p = … b = … [6] Question 10 is printed on the next page.

15 marks

Mark scheme: 9(a) 6x – 5y final answer 2 B1 for 6x or – 5y in final answer or for 6x – 5y seen then spoilt 9(b) 60x + 29y final answer 2 B1 for 60x or 29y in final answer or for 60x + 29y seen then spoilt or for 60p + 29r or [$] 0.60x + 0.29y 9(c) 6.5 oe 3 M1 for a first correct step e.g. 10x + 20 = 85 or 2x + 4 = 17 M1FT for a second correct step e.g. 10x = 65 or 2x = 13 9(d)(i) –2 cao 1 9(d)(ii) 10 cao 1 9(e) 2p + 4b = 42 and 7p + 9b = 107 B2 B1 for each Correctly equating one set of M1 FT coefficients Correct method to eliminate one M1 FT variable Dependent on the coefficients being the same for one of the variables. Correct consistent use of addition or subtraction using their equations [p =] 5 A1 [b =] 8 A1 If M0 scored, SC1 for 2 values satisfying one of the/their original equations or SC1 if no working, but 2 correct answers

This question in 0580/32 Oct/Nov 2020

10 (a) Complete the table of values for y = x 2 - 4x - 3 . x - 2 - 1 0 1 2 3 4 5 y 2 - 3 - 6 - 6 - 3 2 [2] (b) On the grid, draw the graph of y = x 2 - 4x - 3 for - 2 G x G 5 . y 10 8 6 4 2 – 2 – 1 0 1 2 3 4 5 x – 2 – 4 – 6 – 8 [4] (c) Use your graph to solve the equation x 2 - 4 x - 3 = 0 . x = … or x = … [2]

8 marks

Mark scheme: 10(a) 9 –7 2 B1 for each 10(b) Correct curve 4 B3FT for 7 or 8 points correctly plotted B2FT for 5 or 6 points correctly plotted B1FT for 3 or 4 points correctly plotted 10(c) – 0.8 to –0.5 and 4.5 to 4.8 2 B1 for each

This question in 0580/32 Oct/Nov 2020

7 Prakash buys 45 flowers from a shop. (a) Special offer Buy 3 bunches of flowers for the price of 2 bunches. Each bunch has 5 flowers. The price of one bunch of flowers is $2.68 . Using the special offer, work out how much Prakash pays for the 45 flowers. $ … [3] (b) 15 of the flowers are red. 18 of the flowers are orange. The rest of the flowers are yellow. Write down the ratio of the number of red flowers : orange flowers : yellow flowers. Give your answer in its simplest form. … : … : … [2] (c) Prakash gives the 45 flowers to his family. He gives his grandmother x flowers. He gives his mother 8 more flowers than his grandmother. He gives his cousin 6 fewer flowers than his grandmother. He gives his sister twice as many flowers as he gives his cousin. (i) Use this information to show that 5x - 10 = 45 . [3] (ii) Solve the equation 5x - 10 = 45 . x = … [2] (iii) Find the number of flowers that Prakash gives to his cousin. … [1]

11 marks

Mark scheme: 7(a) 16.08 3 45 2 M2 for × × 2.68 oe 5 3 45 2 45 or M1 for × or × 2.68 oe 5 3 5 7(b) 5 : 6 : 4 2 B1 for 15 : 18 : (45−18−15) or better 7(c)(i) x + x + 8 + x − 6 + 2 ( x − 6 ) = 45 M3 M2 for x + x + 8 + x − 6 + 2 ( x − 6 ) = 45 or M1 for two of x + 8, x − 6, 2 ( x − 6 ) leading to 5 x − 10 = 45 with no errors seen 7(c)(ii) 11 2 M1 for x − 2 = 9 or 5 x = 55 7(c)(iii) 5 1 FT their(c)(ii) −6

This question in 0580/32 Feb/March 2021

7 (a) Martin, Suki and Pierre make clocks. In one week • Martin makes x clocks. • Suki makes 3 fewer clocks than Martin. • Pierre makes twice as many clocks as Suki. (i) Write an expression for the total number of clocks they make in one week. Give your expression in its simplest form. … [3] (ii) The total number of clocks they make in one week is 35. (a) Work out the value of x. x = … [3] (b) Work out how many more clocks Pierre makes than Martin. … [2] (b) 12 11 1 10 2 9 3 8 4 7 5 6 (i) Complete the clock diagram to show the time 2.30 pm. [1] (ii) Calculate the obtuse angle between the hands of the clock at 2.30 pm. … [2] (c) Work out the number of seconds in 10 days. Give your answer in standard form. … seconds [2] (d) A clock is started at 15 00. The clock is not working correctly and is slow. The clock loses 8 minutes every hour so after one hour the clock shows 15 52. What time will the clock show 3 12 hours after it is started? … [2] (e) The times on two clocks are checked regularly. One clock is checked every 6 days. The other clock is checked every 8 days. Both clocks are checked on 1st January 2021. Find the number of days during 2021 when both clocks will be checked on the same day. [There are 365 days in 2021.] … [4]

19 marks

Mark scheme: 7(a)(i) 4x− 9 cao 3 B2 for x + ( x − 3) + 2( x − 3) oe or B1 for k ( x− )3 seen k = 1,2 or 3 oe 7(a)(ii)(a) 11 nfww 3 M1 for their (a)(i) = 35 M1 for rearranging their( ax + b) = 35 b 35 to ax = 35 − b or x + = or better a a 7(a)(ii)(b) 5 2 FT their x M1 for (their x − 3) × 2 soi or B1 for [Pierre makes] 16 7(b)(i) Half-past two shown correctly on clock 1 face 7(b)(ii) 105 2 3.5 M1 for [× 360 ] oe 12 7(c) 8.64 × 105 2 M1 for 60 × 60 × 24 ×10 or B1 for figs 864 If 0 scored, SC1 for correctly changing their answer into standard form provided their answer >10 000 7(d) 18 02 2 1 M1 for 3 2× 8 or B1 for 1736 seen 7(e) 16 cao 4 B1 for LCM=24 soi 365 364 365 364 M1 for or or or 24 24 48 48 A1 for 15 or 15.2 or 15.16 to 15.17 or 15.20 to 15.21 If A0 scored, SC1 for 7.60[4..] or 7.58[3..] and 8 final answer

This question in 0580/31 May/June 2021

74 (a) Put a ring around the fraction that is equivalent to . 12 35 20 49 82 64 62 36 84 144 110 [1] (b) Write these numbers in order, starting with the smallest. 7 8 2 0.6 58% 12 13 3 … 1 … 1 … 1 … 1 … [2] smallest (c) Write 0.724 as a fraction in its simplest form. … [1] (d) The mass, m grams, of a ball is 415 g, correct to the nearest 5 grams. Complete the statement about the value of m. … G m 1 … [2] (e) Ruth uses three-quarters of a bag of flour to make one cake. Work out the number of bags of flour she needs to buy to make 7 cakes. … [3] (f) A tin of soup costs $t and a packet of biscuits costs $p. (i) 3 tins of soup and 2 packets of biscuits cost $15.50 . Complete the equation. 3t + 2p = … [1] (ii) 5 tins of soup and 4 packets of biscuits cost $28.50 . Write down another equation in terms of t and p. … [1] (iii) Solve the two simultaneous equations. You must show all your working. t = … p = … [3]

14 marks

Mark scheme: 4(a) 49 1 84 4(b) 7 8 2 2 B1 for 4 in correct order 58% 0.6 or M1 for 0.583[...], [0.6], 0.58, 0.615[…] 12 13 3 or 0.61 or 0.62, 0.66[...] or 0.67 or 0.667 or 0.7 oe 4(c) 181 1 cao 250 4(d) 412.5 417.5 2 B1 for each If 0 scored, SC1 for both correct but reversed 4(e) 6 3 3 M1 for 7 × oe 4 1 A1 for 5.25 or 5 4 4(f)(i) 15.5[0] 1 4(f)(ii) 5t + 4p = 28.5[0] 1 4(f)(iii) For correct method to eliminate one M1 FT their two linear equations variable [t =] 2.5 A1 [p =] 4 A1 If 0 scored, SC1 for 2 values satisfying one of the correct equations or their (f)(i) or (f)(ii) or SC1 if no working shown, but 2 correct answers

This question in 0580/32 May/June 2021

4 (a) Simplify. 3a - 5b + 2a + b … [2] (b) P = 3x 2 - xy Find the value of y when P = 90 and x = 5. y = … [3] (c) Factorise completely. (i) 6x - 18 … [1] (ii) 25x 2 + 10 x … [2] (d) T = 8d - 3 Make d the subject of this formula. d = … [2] (e) Solve these equations. x (i) = 12 6 x = … [1] (ii) 7x - 4 = 3x + 2 x = … [2]

13 marks

Mark scheme: 4(a) 5a – 4b final answer 2 B1 for 5a or – 4b in final answer or for 5a – 4b seen then spoilt 4(b) −3 3 M1 for 90 = 3 × 52 – 5y oe or better M1FT for 90 – 3 × 52 = −5y oe or better 4(c)(i) 6(x – 3) final answer 1 4(c)(ii) 5x(5x + 2) final answer 2 B1 for 5(5x2 + 2x) or x(25x + 10) or 5x(5x + 2) seen then spoilt 4(d) T + 3 2 M1 for a correct first step d = oe final answer T 3 8 either T + 3 = 8d oe or = d − oe 8 8 4(e)(i) 72 1 4(e)(ii) 1 3 2 M1 for 7x – 3x = 2 + 4 oe or better 1.5 or 12 or 2

This question in 0580/33 May/June 2021

9 The table shows some values for y = x 2 + x - 5 . x -4 -3 −2 −1 0 1 2 3 y 7 -3 -5 -5 7 (a) Complete the table. [2] (b) Draw the graph of y = x 2 + x - 5 for - 4 G x G 3 . y 8 7 6 5 4 3 2 1 – 4 – 3 – 2 – 1 0 1 2 3 x – 1 – 2 – 3 – 4 – 5 – 6 – 7 [4] (c) Write down the equation of the line of symmetry of this graph. … [1] (d) Use the graph to solve the equation x 2 + x - 5 = 0 . x = … or x = … [2]

9 marks

Mark scheme: 9(a) 1 −3 1 2 B1 for 2 correct 9(b) Correct curve 4 B3FT for 7 or 8 points correctly plotted or B2FT for 5 or 6 points correctly plotted or B1FT for 3 or 4 points correctly plotted 9(c) x = −[0].5 oe 1 9(d) −2.9 to −2.7 1.7 to 1.9 2 B1FT for each

This question in 0580/33 May/June 2021

10 (a) Complete the table of values for y = 4 + 3 x - x 2 . x -2 -1 0 1 2 3 4 y 0 4 6 0 [2] (b) On the grid, draw the graph of y = 4 + 3x - x 2 for - 2 G x G 4 . y 8 6 4 2 - 2 - 1 0 1 2 3 4 x - 2 - 4 - 6 - 8 [4] (c) The line y = 2 x - 1 is drawn on the grid. Use your graph to solve the equation 4 + 3x - x 2 = 2x - 1. x = … or x = … [2]

8 marks

Mark scheme: 10(a) –6, 6, 4 2 B1 for 2 correct 10(b) Correct curve 4 B3FT for 6 or 7 points correctly plotted or B2FT for 4 or 5 points correctly plotted or B1FT for 2 or 3 points correctly plotted 10(c) 2.7 to 2.9 and –1.9 to –1.7 2 FT their curve B1 for one correct If 0 scored, SC1 for both correct or FT answers as coordinates

This question in 0580/31 Oct/Nov 2021

5 (a) Simplify. 5a - 3b + 7a + 2b … [2] (b) Find the value of 8x - 3y when x = 5 and y =- 2 . … [2] (c) Solve. 6x - 3 = 2x + 8 x = … [2] (d) P = t6 - 11 Make t the subject of this formula. t = … [2] (e) Solve the simultaneous equations. You must show all your working. 3x - 4y = 30 2x + 5y =- 3 x = … y = … [4]

12 marks

Mark scheme: 5(a) 12a – b final answer 2 B1 for 12a or – b in final answer or for correct answer spoilt 5(b) 46 2 M1 for 8 × 5 – 3 × −2 or B1 for 40 or [+] 6 5(c) 2.75 or 2 34 2 M1 for 6x – 2x = 3 + 8 or better 5(d) P + 11 2 P 11 [t =] oe final answer M1 for P + 11 = 6t or = t − 6 6 6 5(e) Correctly equating one set of coefficients M1 Correct method to eliminate one variable M1 Dependent on the coefficients being the same for one of the variables Correct consistent use of addition or subtraction using their equations [x =] 6 A1 [y =] −3 A1 If 0 scored, SC1 for two values that satisfy one of the original equations SC1 if no working shown, but 2 correct answers given

This question in 0580/32 Oct/Nov 2021

10 (a) Complete the table of values for y = x 2 - 5x - 2 . x - 2 - 1 0 1 2 3 4 5 6 y 4 - 2 - 8 - 8 - 2 4 [2] (b) On the grid, draw the graph of y = x 2 - 5x - 2 for - 2 G x G 6 . y 14 12 10 8 6 4 2 – 2 – 1 0 1 2 3 4 5 6 x – 2 – 4 – 6 – 8 – 10 [4] (c) On the grid, draw the line y = 2 . [1] (d) Use your graph to solve the equation x 2 - 5 x - 2 = 2 . x = … or x = … [2]

9 marks

Mark scheme: 10(a) 12 −6 −6 2 B1 for 1 or 2 correct 10(b) Correct curve 4 B3FT for 8 or 9 points correctly plotted or B2FT for 6 or 7 points correctly plotted or B1FT for 4 or 5 points correctly plotted 10(c) Ruled line y = 2 drawn 1 10(d) −0.9 to −0.5 2 FT y = 2 and their curve B1 for each 5.5 to 5.9

This question in 0580/32 Oct/Nov 2021

9 (a) Simplify. 3g + 7g - 4g … [1] (b) Solve. 4x + 5 = 27 x = … [2] (c) 6 p # 6 3 = 6 17 Work out the value of p. p = … [1] (d) Mia buys 4 calculators and 2 pens for $20.60 . Heidi buys 5 calculators and 3 pens for $26.90 . Write down a pair of simultaneous equations and solve them to find the cost of a calculator and the cost of a pen. Calculator $ … Pen $ … [6]

10 marks

Mark scheme: 9(a) 6g 1 9(b) 5.5 2 4 x 5 27 M1 for 4x = 27– 5 or + = 4 4 4 or better 9(c) 14 1 9(d) 4c + 2p = 20.60 B1 5c + 3p = 26.90 B1 correctly equating one set of coefficients M1 correct method to eliminating one variable M1 Dependent on the coefficients being the same for one of the variables Correct consistent use of addition or subtraction using their equations [c =] 4 A1 [p =] 2.30 A1 If M0 scored, SC1 for two values that satisfy one of the original or FT equations SC1 if no working shown, but 2 correct answers given If A0A0 working in cents SC1 for final answers of 400 and 230

This question in 0580/33 Oct/Nov 2021

6 (a) A football team has w wins and d draws. The team scores 3 points for each win and 1 point for each draw. Write an expression, in terms of w and d, for the total number of points scored by the team. … [2] (b) Athletic, Rovers and United are three football teams. Athletic have a point score of x. Rovers have 12 points more than Athletic’s point score. United have 3 points fewer than twice Athletic’s point score. The total point score of all three teams is 121. Use this information to write down an equation in terms of x. Solve your equation to work out the point score for each team. Athletic … points Rovers … points United … points [5] (c) Simplify. (i) 4a - 3b + 5a + 6b … [2] (ii) 6 ( 2x + 1) - 5 ( x - 2) … [2] (d) Solve the simultaneous equations. You must show all your working. 3x + 5y = 11 2x - 3y = 20 x = … y = … [4]

15 marks

Mark scheme: 6(a) 3w + [1]d final answer 2 B1 for 3w or [1]d in final answer 6(b) 28 5 M2 for x + x + 12 + 2x – 3 = 121 or better 40 or B1 for x + 12 or 2x – 3 or 4x + 9 53 M1 for 4x + 9 = 121 or better or for simplifying their equation to ax + b = 121or better M1 for solving their linear equation if 0 scored then SC1 for 3 numbers adding to 121 6(c)(i) 9a + 3b final answer 2 B1 for 9a or 3b in final answer or 9a + 3b seen and spoilt 6(c)(ii) 7x + 16 final answer 2 B1 for 12x + 6 or −5x + 10 or 5x – 10 or for 7x or 16 in the final answer 6(d) Correctly equating one set of coefficients M1 correct method to eliminate one variable M1 [x =] 7 A1 [y =] −2 A1 If M0 scored, SC1 for 2 values satisfying one of the original equations or no working shown but 2 correct answers given

This question in 0580/31 May/June 2022

8 The grid shows a line L. y 6 5 L 4 3 2 1 – 6 – 5 – 4 – 3 – 2 – 1 0 1 2 3 4 5 6 x – 1 – 2 – 3 – 4 – 5 – 6 (a) Find the equation of line L. Give your answer in the form y = mx + c . y = … [2] (b) (i) Complete the table of values for y = 2x + 5 . x −5 −3 0 y −5 5 [1] (ii) On the grid, draw the graph of y = 2x + 5 . [1] (c) Write down the coordinates of the point which lies on both line L and the graph of y = 2x + 5 . ( … , … ) [1] (d) Write down the equation of the line that is parallel to y = 2x + 5 and passes through the point (0, 18). … [1]

6 marks

Mark scheme: 8(a) 1 2 M1 for rise ÷ run [y =] – + 2 or for [y =] kx + 2 (k ≠ 0) 2x 1 or [y =] – 2x + j oe 8(b)(i) −1 1 8(b)(ii) Correct ruled line on grid 1 8(c) −1.2, 2.6 1 FT Line L and their (b)(ii) 8(d) y = 2x + 18 1

This question in 0580/31 May/June 2022

129 (a) Complete the table of values for y = , x ! 0 . x x -6 -4 -3 -2 -1 1 2 3 4 6 y -3 -6 6 3 [3] 12 (b) On the grid, draw the graph of y = for - 6 G x G - 1 and 1 G x G 6 . x y 12 10 8 6 4 2 – 6 – 5 – 4 – 3 – 2 – 1 0 1 2 3 4 5 6 x – 2 – 4 – 6 – 8 – 10 – 12 [4] (c) On the grid, draw the line y = 5 . [1] 12 (d) Use your graph to solve the equation = 5 . x x = … [1]

9 marks

Mark scheme: 9(a) −2 … −4 … −12 12 … 4 … 2 3 B2 for 4 or 5 correct B1 for 2 or 3 correct 9(b) Correct curve 4 B3FT for 9 or 10 points plotted correctly B2FT for 7 or 8 points plotted correctly B1FT for 5 or 6 points plotted correctly 9(c) Correct ruled line drawn 1 9(d) 2.4 1 FT their graph and y = 5

This question in 0580/31 May/June 2022

5 (a) Work out the number of days in seven weeks. … days [1] (b) The summit of Mount Everest is 8848 metres above sea level. Ayding Lake is 154 metres below sea level. Work out the difference in height between these places. … m [1] (c) Find two integers that have a sum of - 12 and a product of 32. … and … [1] 3 (d) Write as 8 (i) a decimal, … [1] (ii) a percentage. … % [1] (e) Write down the reciprocal of 1. 9 … [1] (f) Find the value of (i) 45, … [1] (ii) 3 512 . … [1] (g) (i) Write 587 000 in standard form. … [1] (ii) Calculate 4.9 # 10 - 3 + 8.1 # 10 - 4 . Give your answer in standard form. … [1] (h) The height, h metres, of a fence post is 2.43 m, correct to the nearest centimetre. Complete the statement about the value of h. … G h 1 … [2]

12 marks

Mark scheme: 5(a) 49 1 5(b) 9002 1 5(c) −8 and −4 1 5(d)(i) [0].375 1 5(d)(ii) 37.5 1 5(e) 9 1 5(f)(i) 1024 1 5(f)(ii) 8 1 5(g)(i) 5.87 × 105 cao 1 5(g)(ii) 5.71 × 10−3 cao 1 5(h) 2.425 2.435 2 B1 for each If zero scored, SC1 for 242.5 ⩽ h < 243.5 or for both correct but reversed

This question in 0580/33 May/June 2022

7 (a) Simplify. 5g - 3h - 7g + 6h … [2] (b) j = 4k + 7m Find the value of j when k =- 5 and m = 6 . j = … [2] (c) Factorise completely. 14x 3 + 49x … [2] (d) Solve. 8 ( 3t - 9) = 108 t = … [3] (e) (i) 9 24 ' 9 w = 9 5 Find the value of w. w = … [1] (ii) 4x 2 = 256 Find the value of x. x = … [1] (f) Ranjit’s age is x years. Suzi’s age is 3 times Ranjit’s age. Juan’s age is 4 years more than Suzi’s age. The total of their ages is 46 years. Use this information to write down an equation and solve it to find the value of x. x = … [4]

15 marks

Mark scheme: 7(a) –2g +3h final answer 2 B1 for –2g or 3h in final answer or –2g+ 3h seen then spoilt 7(b) 22 2 M1 for 4  –5 + 7  6 or B1 for –20 or [+]42 7(c) 7x(2x2 + 7) final answer 2 B1 for 7(2x3 +7x) or x(14x2 + 49) or correct answer seen then spoilt 7(d) 7.5 3 M1 for a first correct step 24t – 72 = 108 or 3t –9 =13.5 M1FT for a second correct step e.g. 24t =180 or 3t =22.5 7(e)(i) 19 1 7(e)(ii) 8 1 7(f) x + 3x + 3x + 4 = 46 4 M2 for a correct equation which would or 7x + 4 = 46 lead to 7x + 4 = 46 leading to x = 6 or B1 for 3x or 3x + 4 seen M1 for 7x = 42 or for rearranging their equation to ax = b B1 for [x =] 6

This question in 0580/31 Oct/Nov 2022

5 (a) Apples cost $a per kilogram and bananas cost $b per kilogram. Lee buys 6 kg of apples and 8 kg of bananas. Write down an expression, in terms of a and b, for the total cost, in dollars, of the apples and the bananas. $ … [2] (b) Cara is one year older than twice Asher’s age. Brice is 22 years older than Asher. The total of their three ages is 167. Find the age of each person. Asher … years Brice … years Cara … years [4] (c) Solve the simultaneous equations. You must show all your working. 3x + 2y = 21 2x - 5y = 33 x = … y = … [4]

10 marks

Mark scheme: 5(a) 6a + 8b or 2(3a +4b) 2 B1 for 6a or 8b in final answer final answer or correct answer seen then spoilt 5(b) 36 4 B3 for [Asher]36 or [Brice]58 or [Cara]73 58 73 OR M2 for a correct equation which would lead to 4a + 23 = 167 or for (167 – 23) ÷ 4 oe or B1 for 2a + 1 or a + 22 seen or (167 – 23) oe M1 for 4a = 144 or for rearranging their equation to ka = j 5(c) Correctly equating one set of M1 coefficients Correct method to eliminate one M1 Dependent on the coefficients being the variable same for one of the variables Correct consistent use of addition or subtraction using their equations [x =] 9 A1 [y =] −3 A1 If M0 scored, SC1 for 2 values satisfying one of the original equations

This question in 0580/32 Oct/Nov 2022

9 (a) NOT TO 2y SCALE 3y Write down an expression for the area of this rectangle. Give your answer in its simplest form. … [2] (b) In this part, all measurements are in centimetres. NOT TO SCALE x + 70 3x – 10 4x – 50 The perimeter of the triangle is 526 cm. Find the value of x. x = … [3]

5 marks

Mark scheme: 9(a) 6y 2 cao 2 M1 for 2 y  3 y or for final answer ky 2 9(b) 64.5 3 M2 for 4 x + 3x + x = 526 − 70 + 10 + 50 or better OR M1 for 3 x − 10 + x + 70 + 4 x − 50  = 526  or better or for their ax + b = k leading to k − b x = a

This question in 0580/32 Feb/March 2023

8 (a) T = 5P + 3Q Find the value of T when P = 6 and Q = 8 . T = … [2] (b) Simplify. 3a - 7b + 2a + 4b … [2] (c) Multiply out. 5 ( 2x - 3y) … [1] (d) Solve. 5x - 1 = 3x + 19 x = … [2] (e) Make t the subject of the formula p = t5 - 3 . t = … [2] (f) Entry to a castle costs $x for an adult and $y for a child. Entry for 2 adults and 3 children costs $15.00 . Entry for 3 adults and 5 children costs $23.50 . Write down a pair of simultaneous equations to show this information and solve them to find the value of x and the value of y. You must show all your working. x = … y = … [6]

15 marks

Mark scheme: 8(a) 54 2 M1 for 5×6 + 3×8 or 30 or 24 8(b) 5a – 3b final answer 2 B1 for 5a or – 3b in final answer or for correct answer seen and spoilt 8(c) 10x – 15y final answer 1 8(d) 10 2 M1 for 5x – 3x = 19 + 1 or better 8(e) p  3 2 M1 for p + 3 = 5t or 5p t 53 oe [t=] oe final answer 5 8(f) 2x + 3y = 15 and 3x + 5y = 23.5 B2 B1 for each correctly equating one set of M1 FT coefficients correct method to eliminate one M1 FT variable Dependent on the coefficients being the same for one of the variables Correct consistent use of addition or subtraction using their equations [x =] 4.5 A1 [y =] 2 A1 If M0 scored, SC1 for 2 values satisfying one of correct equations or their equations

This question in 0580/32 May/June 2023

1 (a) Write the number six and a half million in figures. … [1] (b) Write 37 508 correct to the nearest thousand. … [1] (c) 6 9 100 28 31 1000 32 36 From this list of numbers, write down (i) a factor of 18 … [1] (ii) a multiple of 12 … [1] (iii) a square number … [1] (iv) a prime number … [1] (v) an irrational number. … [1] (d) Put one pair of brackets in each statement to make it correct. (i) 24 - 4 # 3 + 2 = 62 [1] (ii) 24 - 4 # 3 + 2 = 4 [1] 3 (e) Write as a decimal. 4 … [1] 3(f) Work out of 126. 7 … [1] (g) Write down the value of the reciprocal of 0.5 . … [1] 2 1(h) Without using a calculator, work out 5 - 2 . 3 5 You must show all your working and give your answer as a mixed number in its simplest form. … [3]

15 marks

Mark scheme: Question Answer Marks Partial Marks 1(a) 6 500 000 1 1(b) 38 000 1 1(c)(i) 6 or 9 1 1(c)(ii) 36 1 1(c)(iii) 9 or 36 1 1(c)(iv) 31 1 1(c)(v) 1000 1 1(d)(i) (24 − 4)  3 + 2 = 62 1 1(d)(ii) 24 − 4  (3 + 2) = 4 1 1(e) 0.75 1 1(f) 54 1 1(g) 2 1 1(h) 17 11 B1 Correct step for dealing with mixed or 17 k 11k 3 5 numbers, allow e.g. or 3k 5 k 85 33 M1 FT Correct method to find a common and denominator 15 15 3 157 cao A1 Alternative methods 2 1 10 3 7 3 − B1, and M1, 3 15 cao 3 5 15 15 A1 510 and 3 M1, 3 157 cao B1 A1 15 215 10 3 7 and M1, 3 15 cao B1A1 15 15

This question in 0580/31 Oct/Nov 2023

7 (a) Simplify. 5a + 3b + 2a - 4b … [2] (b) P = 8x + 3y Find the value of x when P = 21 and y =-5 . x = … [2] (c) Make v the subject of the formula S = kv2 . v = … [2] (d) Multiply out and simplify. ( x - 3)( x + 5) … [2] (e) Nasser has x marbles. Selina has 15 more marbles than Nasser. Hanif has 3 times as many marbles as Selina. In total they have 150 marbles. Find the value of x. x = … [5]

13 marks

Mark scheme: 7(a) 7a – b final answer 2 B1 for 7a or –b in final answer or 7a – b seen then spoilt 7(b) 4.5 2 M1 for 21 = 8x + 3  −5 oe or better 7(c) S 2 2 S final answer M1 for v = or S = k  v k k 7(d) x2 + 2x – 15 final answer 2 B1 for three correct terms from x2 – 3x + 5x – 15 7(e) 18 5 B1 for x + 15 or 3  their (x + 15) oe M1 for x + their (x + 15) + their (3( x + 15)) = 150 or better M1 for 5x + 60 = 150 or better or their linear equation simplified to ax + b = 150 M1 for their [ax + b = c] 150 − b solved to x = a

This question in 0580/31 Oct/Nov 2023

3 (a) Write the number fourteen thousand and ninety-seven in figures. … [1] (b) Write down a common multiple of 17 and 5. … [1] (c) Write 0.25 as a percentage. … % [1] (d) Find the value of (i) 75 … [1] (ii) 80. … [1] 5 (e) Ranjit buys some plants and sells of them. 11 He sells 190 plants. Work out how many plants he buys. … [2] (f) Factorise completely. 15x 3 y - 3x … [2] (g) Make n the subject of the formula V = 3n + t . n = … [2] (h) 7 15 ' 7 x = 7 9 Find the value of x. x = … [1]

12 marks

Mark scheme: 3(a) 14 097 1 3(b) Any correct multiple i.e. 85k 1 3(c) 25 1 3(d)(i) 16 807 1 3(d)(ii) 1 1 3(e) 418 2 M1 for 190 ÷ 5 soi by 38 3(f) 3x(5x2y – 1) final answer 2 B1 for 3(5x3y – x) or for x(15x2y – 3) or correct answer spoilt 3(g) V − t 2 V t oe final answer M1 for V – t =3n or = n + 3 3 3 3(h) 6 1

This question in 0580/32 Oct/Nov 2023

4 (a) Complete the table of values for y = x 2 - 4x - 2 . x -2 -1 0 1 2 3 4 5 y 3 -2 -5 -5 -2 3 [2] (b) On the grid, draw the graph of y = x 2 - 4x - 2 for - 2 G x G 5 . y 10 9 8 7 6 5 4 3 2 1 x -2 -1 0 1 2 3 4 5 -1 -2 -3 -4 -5 -6 [4] (c) Use your graph to solve the equation x 2 - 4x - 2 = 0 . x = … or x = … [2]

8 marks

Mark scheme: 4(a) 10 –6 2 B1 for each 4(b) Correct curve 4 B3FT for 7 or 8 points correctly plotted or B2FT for 5 or 6 points correctly plotted or B1FT for 3 or 4 points correctly plotted 4(c) –0.6 to –0.3, 4.3 to 4.6 2 FT their curve B1 for each

This question in 0580/32 Oct/Nov 2023

6 (a) Simplify. a + 4 a - 3a … [1] (b) Simplify. 8b - 4 # 7b … [1] (c) 4x + 3 x + 7 3x - 9 NOT TO SCALE 9x + 8 7x + 3 The perimeter of this shape is equal to the perimeter of a square. Find an expression for the length of one side of the square. Give your answer in its simplest form. … [4] (d) Victoria buys 5 cups of tea and 4 cakes for $15.69 . Isabella buys 3 cups of tea and 7 cakes for $17.97 . Write down a pair of simultaneous equations and solve them to find the cost of one cup of tea and the cost of one cake. You must show all your working. Tea $ … Cake $ … [6]

12 marks

Mark scheme: 6(a) 2a final answer 1 6(b) – 20b final answer 1 6(c) 6x + 3 final answer 4 24 x+12 B3 for seen or 6x + 3 seen then 4 spoilt 24 x + k kx+12 or B2 for 24x + 12 or or seen 4 4 k≠0 or M2 for 3 x – 9 + 4 x + 3 + x + 7 + 9 x + 8 + 7 x + 3 oe 4 or M1 for 3x – 9 + 4x + 3 + x + 7 + 9x + 8 + 7x + 3 oe or B1 for 24x + k or kx + 12 seen k≠0 6(d) 5t + 4c = 15.69 oe B1 3t + 7c = 17.97 oe B1 correctly equating one set of M1 coefficients correct method to eliminate one M1 variable t = 1.65 A1 c = 1.86 A1 If A0 scored, SC1 for 2 values satisfying one of the original equations or if no working shown, but 2 correct answers given

This question in 0580/32 Oct/Nov 2023

7 These are the first four diagrams in a sequence. The diagrams are made using dots and lines. Diagram 1 Diagram 2 Diagram 3 Diagram 4 (a) Complete the table. Diagram 1 2 3 4 Number of small squares 2 4 6 Number of dots 6 9 12 Number of lines 7 12 17 [2] (b) Complete this statement. A diagram in this sequence cannot have 51 small squares because … … [1] (c) An expression for the number of dots in Diagram n is 3n + 3 . Which diagram has 249 dots? … [2] (d) (i) Find an expression, in terms of n, for the number of lines in Diagram n. … [2] (ii) Find the number of lines in Diagram 41. … [1]

8 marks

Mark scheme: 7(a) 8 2 B1 for 2 correct in correct position in 15 table 22 7(b) 51 is an odd number oe 1 7(c) 82 2 M1 for 3n + 3 = 249 or 3(n + 1) = 249 or better 7(d)(i) 5n + 2 oe final answer 2 B1 for 5n + j or kn + 2 k ≠ 0 as final answer or for 5n + 2 oe seen then spoilt 7(d)(ii) 207 1 FT dep on their(d)(i) in the form an + b with a ≠ 0

This question in 0580/33 Oct/Nov 2023

8 (a) Expand and simplify. (i) 4 ( x + 3) + 2 ( x - 1) … [2] (ii) ( m - 6)( m - 4) … [2] (b) Make t the subject of the formula p = t4 + 3 . t = … [2] (c) In this part, all measurements are in centimetres. 4x - 15 NOT TO 2x + 1 SCALE x The perimeter of this triangle is 49 cm. Work out the value of x. x = … [3]

9 marks

Mark scheme: 8(a)(i) 6x + 10 final answer 2 B1 for 4x + 12 or 2x – 2 or for 6x or 10 in the final answer or for 6x + 10 seen then spoilt 8(a)(ii) m2 – 10m + 24 final answer 2 B1 for m2 – 6m – 4m + 24 with at least three terms correct 8(b) p − 3 2 p 3 oe final answer M1 for p – 3 = 4t or = t + 4 4 4 8(c) 9 3 M2 for 4x + 2x + x = 49 + 15 – 1 or better or M1 for 4x – 15 + 2x + 1 + x [=49] oe

This question in 0580/33 Oct/Nov 2023

10 y NOT TO SCALE x (–1, 0) (3, 0) The sketch shows the graph of y = x 2 - 2x - 3 . The graph crosses the x-axis at ( - 1, 0) and (3, 0). (a) Find the equation of the line of symmetry of the graph. … [1] (b) (i) The point A with coordinates (6, k) lies on the graph. Show that the value of k is 21. [1] (ii) The point B with coordinates (p, 21) also lies on the graph. Find the value of p. p = … [1] (c) Write down the y-coordinate of the point where the graph crosses the y-axis. … [1]

4 marks

Mark scheme: 10(a) x = 1 oe 1 10(b)(i) 62 – 2 × 6 – 3 [leading to 21] 1 10(b)(ii) –4 1 10(c) –3 1

This question in 0580/33 Oct/Nov 2023

5 (a) (i) Complete the table of values for y =- x 2 + 5x + 7 . x -1 0 1 2 3 4 5 6 y 11 11 1 [3] (ii) On the grid, draw the graph of y =- x 2 + 5x + 7 for - 1 G x G 6 . y 14 13 12 11 10 9 8 7 6 5 4 3 2 1 – 1 0 1 2 3 4 5 6 x [4] (iii) (a) Write down the equation of the line of symmetry of the graph. … [1] (b) The points ( -8, -97) and ( t, -97) also lie on the graph of y =- x 2 + 5 x + 7 . Use symmetry to find the value of t. t = … [1] (b) Write down the gradient of the line y = 9x - 4 . … [1] (c) Write down the equation of a line parallel to y =-5x + 19 . y = … [1] (d) y 7 6 L 5 4 3 2 1 – 4 – 3 – 2 – 1 0 1 2 3 4 x – 1 – 2 – 3 Find the equation of line L in the form y = mx + c . y = … [2] (e) Make x the subject of the formula y = mx + c . x = … [2]

15 marks

Mark scheme: 5(a)(i) 1, 7, 13, 13, 7 3 B2 for 3 or 4 correct or B1 for 1 or 2 correct 5(a)(ii) Completely correct curve 4 B3FT for 7 or 8 correctly plotted points or B2FT for 5 or 6 correctly plotted points or B1FT for 3 or 4 correctly plotted points 5(a)(iii)(a) x = 2.5 oe 1 5(a)(iii)(b) 13 1 5(b) 9 1 5(c) y = −5 x + k where k  19 1 5(d)  y =  − x + 2 final answer 2 B1 for  y =  − x + c or  y =  mx + 2 where m is their gradient 5(e) − c 2 M1 for a correct first step  x = y oe final answer m y c y −=c mx or = x + m m

This question in 0580/32 Feb/March 2024

7 (a) P = 3a + 5 Find the value of P when a = 2 . P = … [1] (b) Solve these equations. (i) 7x =-42 x = … [1] (ii) 9 ( 8x - 7) = 72 x = … [3] (c) 5 8 # 5 k = 5 -24 Find the value of k. k = … [1] (d) Solve the simultaneous equations. -6x - y = 13 8x + y =- 51 x = … y = … [2] (e) n is an integer where n 2- 3 and n G 1. Write down all the possible values of n. … [2] (f) A boy walks for 35 minutes at x metres per minute. He then runs for t minutes at 160 metres per minute. Write down an expression, in terms of x and t, for the total distance, in metres, the boy travels. … m [2] (g) NOT TO SCALE x – 2 x + 5 Find an expression for the area of this rectangle. Give your answer in the form x 2 + ax + b . … [3] Question 8 is printed on the next page.

15 marks

Mark scheme: 7(a) 11 1 7(b)(i) −6 1 7(b)(ii) 1.875 oe 3 M1 for a correct first step e.g. 8 x − 7 = 8 or 72 x− 63 = 72 M1FT for a correct second step e.g. 8 x = 15 or 72 x = 135 7(c) −32 1 7(d)  x =  − 19  y = 101 2 B1 for  x =  − 19 B1 for  y = 101 7(e) −2, −1, 0, 1 2 B1 for 3 correct and no extras or 4 correct and one extra 7(f) 35x + 160t final answer 2 B1 for 35x or 160t seen in final answer or 35 x + 160t seen and spoilt 7(g) x 2 + 3 x − 10 final answer 3 B2 for x 2 + 5 x − 2 x − 10 with at least 3 terms correct or B1 for ( x + 5 )( x − 2 ) oe

This question in 0580/32 Feb/March 2024

8 The table shows some values for y = x 2 - x - 3 . x - 3 - 2 - 1 0 1 2 3 4 y - 1 - 1 9 (a) Complete the table. [3] (b) On the grid, draw the graph of y = x 2 - x - 3 for - 3 G x G 4 . y 9 8 7 6 5 4 3 2 1 x – 3 – 2 – 1 0 1 2 3 4 – 1 – 2 – 3 – 4 [4] (c) Write down the coordinates of the lowest point on the graph. ( … , … ) [1] (d) Use your graph to solve the equation x 2 - x - 3 = 7 . x = … or x = … [2]

10 marks

Mark scheme: 8(a) 9, 3, 3, 3, 3 3 B2 for 3 or 4 correct or B1 for 1 or 2 correct 8(b) Completely correct curve 4 B3FT for 7 or 8 correctly plotted points or B2FT for 5 or 6 correctly plotted points or B1FT for 3 or 4 correctly plotted points 8(c)  0.5,k  3.6  k 3 1 8(d) 3.7  2.7 2 B1FT for each

This question in 0580/31 May/June 2024

2 (a) Simplify. (i) 5a - 6a + 3a … [1] (ii) 6x 2 - 6x - 4x 2 - x … [2] (b) Find the value of c 2 + d 2 when c = 7 and d =- 5 . … [2] (c) The time, T minutes, to cook a chicken with a mass of m kg is T = 35m + 20 . (i) Make m the subject of the formula. m = … [2] (ii) Find the mass of a chicken that takes 83 minutes to cook. … kg [2] (d) Solve these simultaneous equations. You must show all your working. 5x - 6y = 24 15x + 8y = 33 x = … y = … [3]

12 marks

Mark scheme: 2(a)(i) 2a final answer 1 2(a)(ii) 2x2 – 7x final answer 2 B1 for 2x2 or −7x in final answer or for 2x2 – 7x seen then spoilt. 2(b) 74 2 B1 for 49 or 25 2(c)(i) T  20 2 T 20 [m =] final answer M1 for T – 20 = 35m or  m  35 35 35 2(c)(ii) 1.8 2 FT their (c)(i) for 2 marks or 1 mark M1 for (83 – 20) ÷ 35 2(d) Correctly eliminates one variable M1 Making the coefficients the same for one of the variables and correct consistent use of addition or subtraction using their equations alternative substitution method. M1 for correct rearrangement of one equation to make either x or y the subject and correct substitution of their rearrangement into 2nd equation. [x =] 3 A1 If A0 scored SC1 for 2 values satisfying one of the original equations. [y =] −1.5 A1

This question in 0580/32 May/June 2024

6 (a) In a sport, teams are given points using the formula number of points = number of wins # 4 + number of draws # 2 + bonus points. One team has 15 wins, 7 draws and 6 bonus points. Calculate the total number of points for this team. … [2] (b) Solve. x = 18 2 x = … [1] (c) Solve. 4x + 12 = 18 x = … [2] (d) Expand and simplify. 6 ( 3x - 4) + 5 ( x - 2) … [2] (e) T = 5r - 6 Make r the subject of this formula. r = … [2] (f) Bo has a green bag and a blue bag. Each bag contains some marbles. The green bag has x marbles. There are 5 times as many marbles in the blue bag than in the green bag. Bo now adds 6 marbles to each bag. There are now 4 times as many marbles in the blue bag than in the green bag. Use this information to write down an equation and solve it to find the value of x. x = … [5]

14 marks

Mark scheme: 6(a) 80 2 M1 for 15×4 + 7×2 + 6 oe 6(b) 36 1 6(c) 1 12 or 1.5 2 M1 for 4x = 18 – 12 or x  12  18 oe 4 4 6(d) 23x – 34 final answer 2 M1 for 23x or – 34 in the final answer or for 18x – 24 or 5x – 10 or 23x – 34 seen then spoilt 6(e) T  6 2 T 6 [r = ] oe final answer M1 for 5r = T + 6 or  r  5 5 5 6(f) 5x + 6 = 4(x + 6) oe B2 B1 for 5x or 5x + 6 or x + 6

This question in 0580/33 May/June 2024

9 (a) Line L has a gradient of 4 and passes through the point (0, 3). Write down the equation of line L in the form y = mx + c . y = … [1] (b) Line G has the equation y = 2 - 6x . Line G passes through the point (a, 5). Find the value of a. a = … [3] (c) (i) Complete the table of values for y = x 2 - 6 . x -4 -3 -2 -1 0 1 2 3 4 y 10 -2 -5 -5 -2 10 [2] (ii) On the grid, draw the graph of y = x 2 - 6 for - 4 G x G 4 . y 10 9 8 7 6 5 4 3 2 1 – 4 – 3 – 2 – 1 0 1 2 3 4 x – 1 – 2 – 3 – 4 – 5 – 6 – 7 [4] (iii) Write down the equation of the line of symmetry of the graph. … [1] (iv) Use your graph to solve the equation x 2 - 6 = 0 for x 2 0 . x = … [1]

12 marks

Mark scheme: 9(a) [y =] 4x + 3 1 9(b) – 12 oe 3 B1 for 5 = 2 – 6a M1 for rearranging their linear equation to ra = s 9(c)(i) 3 –6 3 2 B1 for 2 correct 9(c)(ii) correct curve 4 B3FT for 8 or 9 points plotted accurately or B2FT for 6 or 7 points plotted accurately or B1FT for 4 or 5 points plotted accurately 9(c)(iii) x = 0 1 9(c)(iv) 2.3 to 2.6 1 FT their point of intersection

This question in 0580/33 May/June 2024

1 (a) Write the number six million and thirty in figures. … [1] (b) Write 7.896 correct to 2 decimal places. … [1] (c) 8 24 25 36 39 41 48 From this list of numbers, write down (i) a multiple of 16 … [1] (ii) a factor of 24 … [1] (iii) a cube number … [1] (iv) a prime number. … [1] (d) Put one pair of brackets into this calculation to make it correct. 10 - 12 ' 4 + 2 = 8 [1] (e) By writing each number in the calculation correct to 1 significant figure, find an estimate for the value of 596 # 0.047 . 8.65 You must show all your working. … [2] (f) Calculate ( 8 # 10 6 ) # ( 3 # 10 -2) . Give your answer in standard form. … [2] (g) 216 = 2 3 # 3 3 Write 2160 as a product of its prime factors. … [1]

12 marks

Mark scheme: Question Answer Marks Partial Marks 1(a) 6 000 030 1 1(b) 7.90 cao 1 1(c)(i) 48 1 1(c)(ii) 8 or 24 1 1(c)(iii) 8 1 1(c)(iv) 41 1 1(d) 10 – 12 ÷ (4 + 2) 1 1(e) 600  0.05 M1 9 10 A1 If 0 scored SC1 for 2 correct roundings or for all correct but with any trailing zeros 1(f) 2.4 × 105 2 B1 for correct value but not in standard form 1(g) 24 × 33 × 5 1

This question in 0580/31 Oct/Nov 2024

10 (a) Solve. 4x - 7 = 3 x = … [2] (b) Simplify. 2 (i) `x6j … [1] (ii) `5x 3 y 4j # `2 x 2 y 2j … [2] (c) Expand and simplify. (i) 4a + 5 - 2 ( a - 1) … [2] (ii) ( d + 7)( d - 3) … [2]

9 marks

Mark scheme: 10(a) 2.5 oe 2 7 3 M1 for 4x = 3 + 7 or for x − = oe 4 4 10(b)(i) x12 final answer 1 10(b)(ii) 10x5y6 final answer 2 B1 for two of 10, x5, y6 correct in final answer or for 10x5y6 seen then spoilt 10(c)(i) 2a + 7 final answer 2 B1 for 2a or + 7 seen in final answer or for 2a + 7 seen then spoilt 10(c)(ii) d 2 + 4d – 21 final answer 2 B1 for d 2 + 7d – 3d – 21 with at least 3 terms correct

This question in 0580/31 Oct/Nov 2024

2 (a) Write the number 845 024 in words. … [1] (b) Write these numbers in order, starting with the smallest. 7 15 39% 0.388 18 40 … 1 … 1 … 1 … [2] smallest (c) Find the value of (i) 45 … [1] (ii) 60 … [1] (iii) 16 # 49 . … [1] (d) Solve. 6x + 5 = 29 x = … [2] (e) By writing each number in the calculation correct to 1 significant figure, find an estimate for the value of 2.7 # 42.4 . 8.6 - 4.3 You must show all your working. … [2] (f) 3 5 # 3 x = 3 20 Find the value of x. x = … [1] (g) Simplify ( x4 ) 3. … [1] (h) A boat to town A leaves a port every 16 minutes. A boat to town B leaves the same port every 34 minutes. Both boats leave the port at 07 30. Work out the next time both boats leave the port together. … [3]

15 marks

Mark scheme: 2(a) Eight hundred and forty five thousand 1 and twenty four 2(b) 15 7 2 0.388 39% 40 18 B1 for 3 in the correct order Or M1 for 0.389 or 0.3888 or 0.3889 and 0.39 and 0.375 2(c)(i) 1024 1 2(c)(ii) 1 1 2(c)(iii) 28 1 2(d) 4 2 5 29 M1 for 6x = 29 – 5 or x + = oe 6 6 2(e) 3  40 M1 9 – 4 24 A1 If 0 scored SC1 for 3 correct from 3, 40, 9 and 4 or for all correct but with trailing zeros 2(f) 15 1 2(g) x12 1 2(h) 12 02 3 B2 for 272 or 4 h 32 mins or M1 for 272k or 2  2  2  2  17 oe or [16 = ] 2  2  2  2 and [34 = ] 2  17 or 2 correct factor trees/tables of both 16 and 34 OR M2 for listing times/multiples of both 16 and 34 to at least 1202 or 272 or M1 for listing at least the next 2 of each or 1 full list

This question in 0580/33 Oct/Nov 2024

4 (a) n = 15 r + 20c Find the value of r when n = 180 and c = 3 . r = … [2] (b) Factorise completely. 20p 2 q - 5p … [2] (c) Apples cost 35 cents each and bananas cost 14 cents each. Write down an expression for the total cost, in cents, of x apples and y bananas. … cents [2] (d) Solve the simultaneous equations. You must show all your working. 8x + 3y = 59 5x + 7y = 83 x = … y = … [4]

10 marks

Mark scheme: 4(a) 8 2 M1 for 180 = 15r + 20 × 3 or better 4(b) 5p (4pq – 1) 2 B1 for 5 (4p2q – p) or p (20pq – 5) or 5p (4pq – 1) seen then spoilt 4(c) 35x + 14y 2 B1 for 35x or 14y in final answer or for 35x + 14y seen then spoilt 4(d) correctly equating one set of coefficients M1 correct method to eliminate one variable M1 x = 4 A1 y = 9 A1 If A0 scored SC1 for 2 values satisfying one of the original equations

This question in 0580/33 Oct/Nov 2024

6 Raj thinks of a negative number, n. He adds 10 to n and then multiplies by 5. The answer is 30. Work out the value of n. n = … [2]

2 marks

Mark scheme: 6 –4 2 B1 for 6 seen or M1 for ( n + 10 )  5 = 30 or better

This question in 0580/32 Feb/March 2025

17 (a) Solve. x = 18 3 x = … [1] 6 9 11 (b) y = 6 6 Find the value of y. y = … [1] (c) Simplify. a 6 b -2 a 4 b 3 … [2]

4 marks

Mark scheme: 17(a) 54 1 17(b) −2 1 17(c) a 2 2 −5 2 B1 for a 2 b k or a k b − 5 as final answer 5 or a b final answer or for correct answer spoilt b

This question in 0580/32 Feb/March 2025

7 Solve. 3y = 18 y = … [1]

1 marks

Mark scheme: 7 6 1

This question in 0580/31 May/June 2025

8 (a) T = 3 ( 5P - 8) + 4 Find the value of T when P = 12 . T = … [2] (b) W = t4 + 8 Find the value of t when W = 369 . t = … [2]

4 marks

Mark scheme: 8(a) 160 2 M1 for 3(5×12 – 8) + 4 oe or better 8(b) 90.25 2 M1 for 369 – 8 = 4t oe or better 369 8 or = t + oe or better 4 4 W − 8 or [= t] oe or better 4

This question in 0580/31 May/June 2025

14 (a) P = 6 a + 5b Find the value of b when P = 25 and a = 3 . b = … [2] (b) Make T the subject of the formula W = kT + y . T = … [2]

4 marks

Mark scheme: 14(a) 1.4 2 M1 for 25 = 6 +3 5b or better 14(b) W − y W y 2 W y T =  or T =  − M1 for W − y = kT or = T + k k k k k final answer

This question in 0580/32 May/June 2025

20 Solve. (a) x + 3 = 3 x = … [1] 5y (b) = 8 3 y = … [2]

3 marks

Mark scheme: 20(a) 0 1 20(b) 4 2 y 8 4.8 or 4 M1 for 5 y = 3  8 or = or better 5 3 5

This question in 0580/32 May/June 2025

10 Solve. 4p + 11 = 25 p = … [2]

2 marks

Mark scheme: 10 1 7 2 11 25 3 or 3.5 or M1 for 4p = 25 – 11 oe or p + = 2 2 4 4 oe

This question in 0580/33 May/June 2025

8 Solve. (a) 8x = 32 x = … [1] (b) 6x - 3 = 12 x = … [2] (c) Represent the inequality -4 G x 1 2 on the number line. x – 5 – 4 – 3 – 2 – 1 0 1 2 3 4 [2]

5 marks

Mark scheme: 8(a) 4 1 8(b) 2.5 oe 2 3 12 M1 for 6x = 12 + 3 oe or x – = oe 6 6 8(c) 2 B1 for correct line with circles at both ends or for correct circles but line missing –5 –4 –3 –2 –1 0 1 2 3 4

This question in 0580/31 Oct/Nov 2025

5 Solve. (a) 8x = 5 x = … [1] (b) x + 8 = 5 x = … [1]

2 marks

Mark scheme: 5(a) 5 1 or 0.625 8 5(b) −3 1

This question in 0580/32 Oct/Nov 2025

12 27 (a) Complete the table of values for y = . x x -4 -3 -2 -1 1 2 3 4 y -4 -6 6 4 [2] 12 (b) On the grid, draw the graph of y = for - 4 G x G -1 and 1 G x G 4 . x y 12 10 8 6 4 2 – 4 – 3 – 2 – 1 0 1 2 3 4 x – 2 – 4 – 6 – 8 – 10 – 12 [4] 12 (c) Use your graph to write down the solution of the equation = 10 . x x = … [1] Question 28 is printed on the next page.

7 marks

Mark scheme: 27(a) −3 −12 12 3 2 B1 for 2 correct 27(b) Correct curve 4 B3FT for 7 points correctly plotted B2FT for 5 points correctly plotted B1FT for 3 points correctly plotted 27(c) 1.1 to 1.3 1 FT their curve

This question in 0580/32 Oct/Nov 2025

12 (a) Simplify. 5b - 8c + 2 b - 3 c … [2] (b) q = 3r + 5t Find the value of t when q = 37 and r = 4 . t = … [2]

4 marks

Mark scheme: 12(a) 7b – 11c final answer 2 B1 for 7b or – 11c in final answer or for 7b – 11c seen then spoilt 12 (b) 5 2 M1 for 37 = 3 × 4 + 5t oe

This question in 0580/33 Oct/Nov 2025

24 Make p the subject of the formula m = 9p + 32 . p = … [2]

2 marks

Mark scheme: 24 m − 32 2 m 32 p = final answer oe M1 for m – 32 = 9p or = p + oe 9 9 9

This question in 0580/33 Oct/Nov 2025