Algebraic manipulation — 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

3 (a) Kinetic energy, E, is related to mass, m, and velocity, v, by the formula For Examiner's 1 Use E = mv2. 2 (i) Calculate E when m = 5 and v = 12. Answer(a)(i) E= [2] (ii) Calculate v when m = 8 and E = 225. Answer(a)(ii) v = [2] (iii) Make m the subject of the formula. Answer(a)(iii) m = [2] (b) Factorise completely xy2 – x2y. Answer(b) [2] (c) Solve the equation 3(x – 5) + 2(14 – 3x) = 7. Answer(c) x = [3] (d) Solve the simultaneous equations 4x + y = 13, 2x + 3y = 9. Answer(d) x = y = [3]

14 marks

Mark scheme: 1 3 (a) (i) 360 B2 M1 for × 5 × 122oe 2 (ii) 7.5oe B2 M1 for 225 /4 oe (implied by 56.25) 2 E 1 2 1 (iii) or E v B2 B1 for 2E or E or division by v2 2 v 2 2 (b) xy( y – x) final answer B2 B1 for x(y2 – xy) or y(xy – x2) SC1 for xy(y + x) (c) 3x – 15 + 28 – 6x (= 7) MA1 13 – 3x (= 7) M1ft Independent ax + b (=7) from their expansion x= 2 A1cao www 3 (d) Equating coefficients of x or y, or M1 equivalent method. or a correctly substituted substitution. E.g. 5y = 5 oe or 10x = 30 oe A1 y = 13 – 4x ⇒ 2x + 3(13 – 4x) = 9 x = 3, y = 1 A1 www 3 [14]

This question in 0580/31 May/June 2007

4 r 2r 5πr 2 The area of the shape is given by the formula A = . 2 (a) Calculate the area when r = 3 cm. Answer(a) A = cm2 [2] (b) Calculate the value of r when A = 200 cm2. Answer(b) r = cm [3] (c) Make r the subject of the formula. Answer(c) [3]

8 marks

Mark scheme: 4 (a) 70.7 art B2 M1 for 5 x π x 3² / 2 or better (b) 5.05 art B3 M1 for 200 = 5 x π x r² / 2 oe M1 for (r² =) 400 / 5π oe (c) (r =) √2A/5π B3 M1 for any correct x or ÷ of 1 term 2A = 5πr² MA1 for r² = 2A / 5π M1 for square root at end [8]

This question in 0580/31 Oct/Nov 2007

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

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

6 (a) The formula for finding the interior angle of a regular polygon with n sides is given below. For Examiner's 180( n − 2) Use Interior angle = n (i) Find the size of the interior angle of a regular polygon with 9 sides. Answer(a)(i) [2] (ii) Multiply out the brackets. 180(n – 2) Answer(a)(ii) [1] (iii) A regular polygon has an interior angle of 156°. How many sides does this polygon have? Answer(a)(iii) [3] (b) Solve the simultaneous equations. 3x + 5y = 9 x + 2y = 4 Answer(b) x = y = [3]

9 marks

Mark scheme: 6 (a) (i) 140 2 M1 for 180 × (9 – 2) ÷ 9 or better (ii) 180n – 360 1 (iii) 15 3 M2 for 360 ÷ (180 – 156) or M1 for 156n = their (a)(ii) and M1dep for pn = q from their linear expression (b) (x =) –2, (y =) 3 3 M1 for equating coefficients of x or y and adding or subtracting, allow 1 error A1 for 1 correct

This question in 0580/32 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 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

9 (a) Factorise completely 3x2 + 12x. For Examiner's Use Answer(a) [2] (b) Find the value of a3 + 3b2 when a = 2 and b = −2 . Answer(b) [2] (c) Simplify 3x4 × 2x3. Answer(c) [2]

6 marks

Mark scheme: 9 (a) 3x(x + 4) 2 B1 for 3(x2 + 4x) or B1 for x(3x + 12) or B1 for 3x(x + 4) seen (if not final answer) (b) 20 2 B1 for 8 or 12 seen (c) 6x7 2 B1 for kx7 or for 6xk , k ≠ 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

5 (a) A = 1 (a + b)h Examiner'sFor 2 Use Work out the value of A when a = 9.6, b = 12.4 and h = 7.5 . Answer(a) [2] (b) (i) Expand x(x2 – 3y). Answer(b)(i) [2] (ii) Expand and simplify 4(2w – 3) + 5(w – 2). Answer(b)(ii) [2] (c) A quadrilateral has sides x, 2x, y and 3y. (i) Write down and simplify a formula for the perimeter, p, of the quadrilateral. Answer(c)(i) p = [2] (ii) Make y the subject of the formula in part (c)(i). For Examiner's Use Answer(c)(ii) y = [2] (d) Joseph is 3 times as old as Amy. In 5 years time Joseph will be 2 times as old as Amy. (i) Amy is now n years old. Write down an equation in n connecting the ages of Joseph and Amy in 5 years time. Answer(d)(i) [2] (ii) Solve the equation to find n. Answer(d)(ii) n = [3]

15 marks

Mark scheme: 15 (a) 82.5 2 M1 for 2 (9.6 + 12.4) × 7.5 or better (b) (i) x3 − 3xy final ans 2 B1 for x 3 or −3xy seen (ii) 13w − 22 final ans 2 B1 for 13w or −22 or 8w − 12 or 5w − 10 seen (c) (i) (p =) 3x + 4y final ans 2 B1 for 3x or 4y seen or x + 2x + y + 3y seen p−3x 2ft p 3 x (ii) (y =) 4 oe B1ft for 4y = p − 3x or = + y 4 4 (d) (i) 2(n + 5) = 3n + 5 oe 2 B1 for 2(n + 5) or 2n + 10 or 3n + 5 seen or B1 for any different letter to n in 2(n + 5) = 3n + 5 oe (ii) (n =) 5 cao 3 M1 for clearing bracket M1 for an = b

This question in 0580/32 May/June 2012

7 (a) The cost, $C, of hiring a meeting room for n people is calculated using the formula For Examiner's Use C = 80 + 5n. (i) Calculate C when n = 12. Answer(a)(i) [2] (ii) Maria pays $230 to hire the meeting room. Work out the number of people at the meeting. Answer(a)(ii) [2] (iii) Make n the subject of the formula C = 80 + 5n. Answer(a)(iii) n = [2] (b) Expand and simplify 2(3x + 4) – 3(2 – x) . Answer(b) [2] (c) Solve the simultaneous equations. 3x + y = 13 2x + 3y = 18 Answer(c) x = y = [3]

11 marks

Mark scheme: 7 (a) (i) 140 2 M1 for 80 + 5 × 12 or better (ii) 30 2 M1 for (230 – 80) ÷ 5 or 150 seen C − 80 C 80 −C (iii) or − 16 or 2 M1 for C – 80 = 5n 5 5 − 5 C 80 5n final answer Or M1 for = + or better 5 5 5 (b) 9x + 2 final answer 2 M1 for 9x + k or mx + 2 or 6x + 8 or – 6 + 3x or 9x + 2 spoilt (c) x = 3, y = 4 3 M1 for correct method to eliminate one variable A1 x = 3 A1 y = 4

This question in 0580/31 Oct/Nov 2012

10 (a) (i) Find the highest common factor (HCF) of 24 and 36. For Examiner′s Use Answer(a)(i) … [2] (ii) Factorise. 24x + 36y Answer(a)(ii) … [1] (b) Simplify. (i) w + 8k – 5w + 2k Answer(b)(i) … [2] (ii) (x4)5 Answer(b)(ii) … [1] (c) Here are the fi rst four terms of a sequence. 7 11 15 19 Find the nth term of this sequence. Answer(c) … [2] (d) Solve the simultaneous equations. 3x + y = 8 x + 5y = 5 Answer(d) x = … y = … [3]

11 marks

Mark scheme: 10 (a) (i) 12 2 B1 for any other common factor other than 1 (ii) 12(2x + 3y) cao 1 (b) (i) 10k – 4w 2 B1 for either 10k ± nw or qk – 4w p,q ≠ 0 (ii) x20 1 (c) 4n + 3 oe final answer 2 B1 for 4n + c or kn + 3 , k ≠ 0 (d) [x] = 2.5, [y] = 0.5 3 M1 for correct method to eliminate one variable. A1 for x or y correct.

This question in 0580/32 May/June 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

8 Here is a sequence of patterns made using identical polygons. For Examiner′s Use Pattern 1 Pattern 2 Pattern 3 (a) Write down the mathematical name of the polygon in Pattern 1. Answer(a) … [1] (b) Complete the table for the number of vertices (corners) and the number of lines in Pattern 3, Pattern 4 and Pattern 7. Pattern 1 2 3 4 7 Number of vertices 8 14 Number of lines 8 15 [5] (c) (i) Find an expression for the number of vertices in Pattern n. Answer(c)(i) … [2] (ii) Work out the number of vertices in Pattern 23. Answer(c)(ii) … [1] (d) Find an expression for the number of lines in Pattern n. For Examiner′s Use Answer(d) … [2] (e) Work out an expression, in its simplest form, for (number of lines in Pattern n) – (number of vertices in Pattern n). Answer(e) … [2] _____________________________________________________________________________________ Question 9 is printed on the next page.

13 marks

Mark scheme: 8 (a) Octagon 1 (b) [Pattern 3] 20 and 22 1 [Pattern 4] 26, 29 1, 1 [Pattern 7] 44, 50 1, 1 (c) (i) 6n + 2 oe final answer 2 B1 for 6n + a or bn + 2 b ≠ 0 (ii) 140 oe 1FT ft linear expression in (c)(i) (d) 7n + 1 oe final answer 2 B1 for 7n + c or dn + 1 d ≠ 0 (e) n – 1 final answer 2FT B1FT for n + j or kn 1 k ≠ 0 3V 2 3V 3V

This question in 0580/33 Oct/Nov 2013

1 For 9 (a) The formula for the volume, V, of a cone with radius r, and height h, is V = 3 πr2h . Examiner′s Use (i) To make r the subject of this formula, the fi rst step is 3V = πr2h. Show the remaining steps to make r the subject of this formula. Answer(a)(i) r = … [2] (ii) An ice-cream cone has a volume of 141 cm3 and height 15 cm. Show that the radius of the cone is 3 cm, correct to the nearest whole number. Answer(a)(ii) [2] (b) The open end of an ice-cream cone is a circle of radius 3 cm. Calculate the circumference of this circle. Answer(b) … cm [2] (c) The volume of a ball of ice-cream is 113 cm3. The ball of ice-cream costs $2.15 . Calculate the cost of 1 cm3 of the ice-cream. Give your answer in cents, correct to 1 decimal place. Answer(c) … cents [3]

9 marks

Mark scheme: 3V 3V 3V 9 (a) (i) [r =] 2 B1 for [r2 =] or seen or better πh π h 3 x141 (ii) [r =] M1FT their formula πx15 [r =] 2.99… A1 (b) 18.9 or 18.8 or 18.849 to 18.852 2 M1 for 2 × π × 3 oe (c) 1.9 [cents] cao 3 M1 for 2,15 (or 215) ÷ 113 A1 for 0.019 (0…) or 1.9 (0…) soi

This question in 0580/33 Oct/Nov 2013

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

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

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

8 (a) (i) A = 4πr2 Work out the value of A when r = 5.6 . Give your answer correct to 1 decimal place. Answer(a)(i) A = … [2] (ii) Simplify the expression. 2a – b + 5a – 3b Answer(a)(ii) … [2] (iii) Solve the equation. x = 6 3 Answer(a)(iii) x = … [1] (iv) Solve the equation. x – 2 = 9 Answer(a)(iv) x = … [1] (b) Solve the simultaneous equations. You must show all your working. 2x + 3y = 4 3x – 4y = 23 Answer(b) x = … y = … [4]

10 marks

Mark scheme: 8 (a) (i) 394.1 cao 2 M1 for 394[. …] or 4 × π × 5.62 (ii) 7a − 4b final answer 2 B1 for either 7a or −4b in their final answer (iii) 18 1 (iv) 11 1 (b) [x =] 5 4 M1 for correctly equating one set of [y =] –2 coefficients M1 for correct method to eliminate one Working must be shown variable 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

This question in 0580/33 May/June 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

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

3 The diagram shows a cylindrical flower vase with radius, r, and height, h. The volume, V, of the vase is V = r r 2 h . NOT TO SCALE h The surface area, A, of the vase is A = 2 r rh + r r 2 . (a) The vase has radius 4 cm and height 15 cm. r (i) Calculate the volume of the vase. Write down the units of your answer. … … [3] (ii) Calculate the surface area of the vase. … cm2 [2] (b) Make h the subject of the formula A = 2 r rh + r r 2 . h = … [2] (c) Factorise completely. 2r rh + r r 2 … [2] (d) Another cylindrical flower vase has radius 6 cm and height 22.5 cm. (i) For this vase and the vase in part (a) the ratio of the radii is 4 : 6 and the ratio of the heights is 15 : 22.5 . Write these ratios in their simplest form. 4 : 6 = … : … 15 : 22.5 = … : … [2] (ii) Write down a mathematical word to complete the statement. The ratios show that the two vases are … [1]

12 marks

Mark scheme: 3 (a) (i) 754 or 753.9 to 754.1 2 M1 for π × 42 × 15 or better cm3 or cubic centimetres 1 Independent mark (ii) 427 or 427.2 to 427.312 2 M1 for 2 × π × 4 × 15 + π × 42 or better A − πr 2 (b) oe final answer 2 B1 for A – πr2 = 2πrh or better 2πr or A π r 2 = h + or better 2π r 2π r (c) πr(2h + r) final answer 2 B1 for π(2rh + r2) or r(2πh + πr) (d) (i) 2 : 3 1 2 2 : 3 1 Accept 1 : 1.5 or : 1 3 (ii) Similar 1

This question in 0580/33 May/June 2016

6 (a) Here are the first four terms of a sequence. 18 25 32 39 (i) Write down the next term. … [1] (ii) Explain how you worked out your answer. … [1] (b) The nth term of another sequence is n2 + 3. Write down the first three terms of this sequence. … , … , … [2] (c) Simplify. (i) 6a + 5h − 4a − 8h … [2] (ii) 5(x + 3) + 4(2x − 6) … [2] (d) Factorise. 6g + 15 … [1] (e) A rectangle has length (x + 6) cm and width 5 cm. The area of this rectangle is 85 cm2. Find the value of x. x = … [3]

12 marks

Mark scheme: 6 (a) (i) 46 1 (ii) Add 7 oe 1 (b) 4, 7, 12 2 M1 for 2 correct or 3 , 4 , 7 (c) (i) 2a – 3h final answer 2 B1 for 2a or −3h (ii) 13x – 9 final answer 2 M1 for 5x + 15 or 8x – 24 or 13x or −9 (d) 3( 2g + 5) final answer 1 (e) 11 nfww 3 M2 for 5x = 55 or x + 6 = 17 or M1 for 5x + 30 [ = 85] or 5 (x + 6 ) [ = 85] or M1 for correct first step of incorrect linear equation if of the form ax + b = 85, a ≠ 1

This question in 0580/31 Oct/Nov 2016

6 (a) A regular hexagon has side length h. Write down an expression, in terms of h, for the perimeter of the hexagon. … [1] (b) A square has side length x. Write down an expression, in terms of x, for (i) the perimeter of the square, … [1] (ii) the area of the square. … [1] (c) In this part, all measurements are in centimetres. (2x + 1) NOT TO (x + 3) SCALE A rectangle has length (2x + 1) and width (x + 3 ) . The perimeter of the rectangle is 53. Work out the value of x. x = … [5] (d) Simplify. 5a + 4b - 2a - b + 3a - 2b … [2] (e) Multiply out the brackets. (i) 5 (x - 4) … [1] (ii) x (x 2 + 3) … [2] (f) Factorise completely. 8x 2 - 4x … [2]

15 marks

Mark scheme: 6 (a) 6h oe 1 (b) (i) 4x oe 1 (ii) x2 oe 1 (c) 7.5 5 M1 for 2x + 1 + x + 3 + 2x + 1 + x + 3 oe M1 for 6x + 8 or their expression simplified correctly M1 for their 6x + 8 = 53 M1 for a correct first step in solving their linear equation (d) 6a + b final answer 2 B1 for 6a or [+] b (e) (i) 5x – 20 final answer 1 (ii) x3 + 3x final answer 2 B1 for x3 or [+] 3x (f) 4x(2x – 1) final answer 2 B1 for x(8x – 4) or 4(2x2 – x) or 2(4x2 – 2x) or 2x(4x – 2)

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) Multiply out the brackets and simplify. 5 2x + 3 - 2 x + 4 ^ h ^ h … [2] (b) (i) An equilateral triangle has side length 2x. Write down an expression, in terms of x, for the perimeter of the triangle. Give your answer in its simplest form. … [1] (ii) A square has a perimeter of 20a. Write down an expression, in terms of a, for the length of one side of the square. Give your answer in its simplest form. … [1] (c) The diagram shows a rectangle. 3y + 1 NOT TO SCALE 2y + 5 Find an expression, in terms of y, for the perimeter of the rectangle. Give your answer in its simplest form. … [3] (d) One mint costs m cents. One toffee costs 6 cents more than one mint. The cost of 3 mints and 7 toffees is 182 cents. Write an equation, in terms of m, and solve it to find the cost of one mint. Cost of one mint = … cents [5]

12 marks

Mark scheme: 8(a) 8x + 7 final answer 2 B1 for 10x + 15 or –2x – 8 or 8x + j or kx + 7 as final answer 8(b)(i) 6x final answer 1 8(b)(ii) 5a final answer 1 8(c) 10y + 12 or 2(5y + 6) 3 M1 for 2(3y + 1) + 2(2y + 5) oe final answer B1 for 10y + j or ky + 12 (k≠0) 8(d) 7(m + 6) + 3m = 182 or 2 B1 for m + 6 7m + 42 + 3m = 182 or 7t + 3m = 182 14 3 M1 for 7m + 42 [+ 3m = 182] M1 for 7m + 3m = 182 − 42 or better OR M2 for [m=] (182 – (6 × 7)) / (7 + 3) or better or M1 for 182 – (6 × 7) or better

This question in 0580/31 Oct/Nov 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

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

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

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

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

9 (a) c = 5 a - 2b (i) Find the value of c when a = 8 and b = −3. … [2] (ii) Make a the subject of the formula c = 5 a - 2 b . a = … [2] (b) Factorise 3x + 12 . … [1] (c) Expand x (2 y + x) . … [2] (d) Cara has n pencils. Alice has twice as many pencils as Cara. Leon has three more pencils than Alice. The three children have a total of 58 pencils. Use this information to write down an equation and solve it to find the value of n. n = … [4]

11 marks

Mark scheme: 9(a)(i) 46 2 M1 for 5 × 8 – 2 × –3 or better 9(a)(ii) c + 2b c 2b 2 c 2b oe or + oe M1 for c + 2b = 5a oe or = a − oe 5 5 5 5 5 final answer 9(b) 3(x + 4) final answer 1 9(c) 2xy + x2 final answer 2 B1 for 2xy or x2 or for 2xy + x2 not as final answer 9(d) n + 2n + 2n + 3 = 58 4 M2 for any correct equation which would or 5n + 3 = 58 lead to 5n + 3 = 58 leading to [n = ] 11 or B1 for 2n or 2n + 3 seen M1 for 5n = 55 or for rearranging their linear equation to an = b B1 for [n =]11

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

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

5 (a) T = 3a 2 b Find the value of T when a = 4 and b = 5. T = … [2] (b) (i) Multiply out the brackets. x ( 3 - 5 x) … [2] (ii) Factorise fully. 5x - 20x 2 … [2] (c) NOT TO SCALE 3a + b 4a - 5b a + 2b Find an expression for the perimeter of this triangle. Give your answer in its simplest form. … [3]

9 marks

Mark scheme: 5(a) 240 2 M1 for 3 × 42 × 5 oe 5(b)(i) 3x – 5x2 final answer 2 B1 for 3x or – 5x2 5(b)(ii) 5x(1 – 4x) final answer 2 B1 for 5(x – 4x2) or x(5 – 20x) 5(c) 8a – 2b 3 M1 for 3a + b + 4a – 5b + a + 2b B1 for 8a or – 2b

This question in 0580/32 May/June 2020

3 (a) NOT TO SCALE 6 m 8 m The diagram shows a rectangular patio with sides 6 m and 8 m. (i) Work out the perimeter of the patio. … m [1] (ii) Henri covers the patio floor with square tiles. The tiles are 0.5 m by 0.5 m. Work out the number of tiles he needs. … [2] (b) The diagram shows the net of a solid on a 1 cm2 grid. (i) Write down the mathematical name for the solid. … [1] (ii) Work out the volume of the solid. … cm3 [2] (c) A square has perimeter 12x. Find an expression, in terms of x, for the area of the square. Give your answer in its simplest form. … [3] (d) B NOT TO SCALE A C 10 cm The diagram shows a semicircle with diameter AC. B is a point on the circumference and AB = BC. Work out the area of triangle ABC. … cm2 [3]

12 marks

Mark scheme: 3(a)(i) 28 1 3(a)(ii) 192 2 8 6 M1 for × oe 0.5 0.5 or B1 for 16 and 12 or 4 tiles = 1 m2 soi 3(b)(i) Cuboid 1 3(b)(ii) 10 2 M1 for 5 × 2 [× 1] 3(c) 9x 2 3  12 x  2 M2 for   oe  4  12 x or M1 for oe 4 If 0 scored, SC1 for final answer kx 2 3(d) 25 3 B1 for height is 5 [cm] 1 M1 for × 10 × 5 oe 2

This question in 0580/33 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

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

7 (a) W = 3a + 5c Find the value of W when a = 6 and c = 2 . W = … [2] (b) Factorise completely. 12b + 8b 2 … [2] (c) Make m the subject of the formula y = 4m - p . m = … [2] (d) Find the value of x when 5 x # 5 3 = 5 12 . x = … [1] (e) Find the value of (i) 30, … [1] (ii) 5 - 2 . … [1] (f) In this part, all measurements are in centimetres. (3x – 1) NOT TO SCALE (2x + 5) The diagram shows a kite with sides ( 2x + 5) and ( 3x - 1) . The perimeter of the kite is 33 cm. Work out the length of a shorter side. … cm [5]

14 marks

Mark scheme: 7(a) 28 2 M1 for 3 × 6 + 5 × 2 or better or B1 for 18 or 10 seen 7(b) 4b(3 + 2b) final answer 2 B1 for final answer 2(6b + 4b2) or 4(3b + 2b2) or 2b(6 + 4b) or b(12 + 8b) or 4b(3 + 2b) seen then spoilt 7(c) y + p y p 2 y p or + oe final answer M1 for 4m = y + p or = m − 4 4 4 4 4 7(d) 9 cao 1 7(e)(i) 1 1 7(e)(ii) 1 1 or 0.04 25 7(f) 6.5 5 M2 for 10x + 8 = 33 or 5x + 4 = 16.5 or M1 for [2 ×] (3x – 1 + 2x + 5) oe M1 for a correct first step in solving their linear equation A1 for [x =] 2.5 If A0 scored, SC1 for 3x – 1 correctly evaluated FT their positive x

This question in 0580/33 Oct/Nov 2020

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

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

1 (a) 14 17 25 27 30 36 48 From the list, write down (i) the square root of 289, … [1] (ii) a factor of 81, … [1] (iii) a common multiple of 3 and 5. … [1] (b) A, B and C are three consecutive whole numbers. • A is a prime number. • B is a cube number. • C is a square number. • A + B + C is less than 40. Find A, B and C. A = … B = … C = … [2] (c) Put one pair of brackets into each of these calculations to make them correct. (i) 4 # 3 + 7 ' 2 = 20 [1] (ii) 51 - 12 ' 3 + 6 = 19 [1] (d) Write down (i) the reciprocal of 8, … [1] (ii) the value of 140. … [1] (e) Calculate. (i) 54 … [1] (ii) 3 6859 … [1] 1 (iii) 16 - 2 … [1]

12 marks

Mark scheme: Question Answer Marks Partial Marks 1(a)(i) 17 1 1(a)(ii) 27 1 1(a)(iii) 30 1 1(b) 7, 8, 9 2 M1 for any 2 conditions in final answer from: A prime or B cube or C square or consecutive A + B + C < 40 1(c)(i) 4 × (3 + 7) ÷ 2 = 20 1 1(c)(ii) (51 – 12) ÷ 3 + 6 = 19 1 1(d)(i) 1 1 or 0.125 8 1(d)(ii) 1 1 1(e)(i) 625 1 1(e)(ii) 19 1 1(e)(iii) 1 1 or 0.25 4

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

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

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

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

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

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

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

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

14 (a) Factorise. 8x 2 - 2x … [2] (b) Expand the brackets and simplify. 5 ( 2m - 1) + 3 ( m + 7) … [2]

4 marks

Mark scheme: 4 x 2 − x or x ( 8 x − 2 )14(a) 2 x ( 4 x − 1) final answer 2 B1 for 2 ( ) or 2 x ( 4 x − 1) seen then spoilt 14(b) 13m + 16 final answer 2 M1 for 10m − 5 or 3 m + 21 or for 13m or +16 in the final answer or for 13m + 16 seen then spoilt

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

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

11 (a) Expand and simplify. 5 ( x - 2) + 3 ( x - 7) … [2] (b) Factorise. 4a 2 + 16a … [2]

4 marks

Mark scheme: 11(a) 8x – 31 final answer 2 B1 for 5x – 10 or 3x – 21 or for 8x – 31 seen then spoilt or for 8x or –31 in the final answer 11(b) 4a(a + 4) final answer 2 B1 for 4(a2 + 4a) or a(4a + 16) or 2a(2a + 8) as answers or for 4a(a + 4) seen and spoilt

This question in 0580/33 May/June 2025

17 Factorise. 21x - 7xy … [2]

2 marks

Mark scheme: 17 7x(3 – y) 2 B1 for 7(3x – xy) or x(21 – 7y) or for 7x(3 – y) seen then spoilt

This question in 0580/31 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

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

2 marks

Mark scheme: 19 11x + 2 final answer 2 B1 for 15x + 6 or – 4x – 4 or for 11x + 2 seen then spoilt or for 11x or 2 in the final answer

This question in 0580/33 Oct/Nov 2025