Algebraic manipulation — Paper 1 · IGCSE Mathematics 0580

12 Make s the subject of the formula p = st – q. Answer s = [2]

2 marks

Mark scheme: 12 (s =) (p + q)/t or p + q oe 2 B1 for p + q seen or correct ÷ by t t or p/t = s – q/t or (p –q)/t SC1 for p + q/t or p/t + q

This question in 0580/11 May/June 2005

17 (a) Expand the bracket and simplify the expression For Examiner's 7x + 5 – 3(x – 4). Use Answer (a) [2] (b) Factorise 5x2 – 7x. Answer (b) [1]

3 marks

Mark scheme: 17 (a) 4x + 17 final answer 2 B1 for –3x + 12 or 4x or +17 seen (+17 strictly www) (b) x (5x – 7) 1 condone missing final bracket

This question in 0580/11 May/June 2005

2 Factorise 3xy – 2x. Answer [1]

1 marks

Mark scheme: 2 x(3y – 2) 1

This question in 0580/11 Oct/Nov 2005

8 The formula for the perimeter, P, of a rectangle with length a and width b is P = 2a + 2b. Make a the subject of the formula. Answer a = [2]

2 marks

Mark scheme: 8 P – 2b = 2a M1 P − 2b A1 oe 2

This question in 0580/11 Oct/Nov 2005

16 Simplify the following expressions. (a) 9r – 4s – 6r + s Answer(a) [1] (b) q4 ÷ q3 Answer(b) [1] (c) p6 × p−2 Answer(c) [1]

3 marks

Mark scheme: 16 (a) 3r – 3s or 3(r – s) 1 (b) q or q1 1 (c) p4 1

This question in 0580/11 Oct/Nov 2005

8 Factorise completely 2x2 – 6xy. Answer [2]

2 marks

Mark scheme: 8 2x(x – 3y) 2 M1 for 2(x2 – 3xy) or x(2x – 6y) or 2x(……)

This question in 0580/11 May/June 2006

6 Factorise completely 2a2b − 6a. For Examiner's Use Answer [2]

2 marks

Mark scheme: 6 2a( ab − 3 ) final answer. 2 SC1 for 2( a 2 b − 3a ) or a( 2ab − 6 ) or 2a( ab + 3 ) or 2a( ab – 6 ) final answer.

This question in 0580/11 Oct/Nov 2006

8 Simplify 3x2y × x4y2. Answer [2]

2 marks

Mark scheme: 8 3x 6y 3 or 3(x 2y) 3 2 SC1 for x 6or y 3 seen in final answer

This question in 0580/11 Oct/Nov 2006

17 The surface area of a sphere with radius r is A = 4πr2. (a) Calculate the surface area of a sphere with a radius of 5 centimetres. Answer(a) cm2 [1] (b) Make r the subject of the formula A = 4πr2. Answer(b) r = [2]

3 marks

Mark scheme: 17 (a) art 314 1 A A (b) oe 2 M1 for seen 4π 4π

This question in 0580/11 Oct/Nov 2006

5 Factorise completely 4xy – 2x. For Examiner's Use Answer [2]

2 marks

Mark scheme: 5 2x(2y − 1) final answer 2 SC1 for x(4y − 2) or 2(2xy − x) or 2x(2y +1) Or SC1 for 2x(2y – 1) not as final answer.

This question in 0580/11 Oct/Nov 2007

5 Expand the brackets and simplify 5x − 6(3x − 2). Answer [2]

2 marks

Mark scheme: 5 12 − 13x cao final answer 2 W1 for (+)12 or −13x seen anywhere

This question in 0580/11 Oct/Nov 2008

10 (a) Expand and simplify 5(3c – 4d) – 8c. Answer(a) [2] (b) Factorise pq – q2. Answer(b) [1]

3 marks

Mark scheme: 10 (a) 7c − 20d www final answer 2 M1 for 15c − 20d − 8c or better or W1 for 7c or − 20d seen as terms in final answer (b) q(p − q) www 1 IGCSE – May/June 2009 0580, 0581 11

This question in 0580/11 May/June 2009

12 z = 2x – y For Examiner's (a) Find z when x = –3 and y = 7. Use Answer(a) z = [1] (b) Make x the subject of the formula. Answer(b) x = [2]

3 marks

Mark scheme: 12 (a) (z =) − 13 1cao z + y z y (b) (x =) oe final answer 2 M1 for z + y = 2x or = x − or 2 2 2 –2x = –z – y ± z ± y SC1 for answer of form ± 2

This question in 0580/11 May/June 2009

15 Simplify (a) 3p × 5p3, Answer(a) [2] (b) 24q2 ÷ 8q –3. Answer(b) [2]

4 marks

Mark scheme: 15 (a) 15p4 final answer 2 W1 for 15pn (n ≠ 0) or kp4 (k ≠ 0) (b) 3q5 final answer 2 W1 for 3qn (n ≠ 0) or kq5 (k ≠ 0)

This question in 0580/11 May/June 2009

md 15 J = 3 (a) Find the value of d when J = 35 and m = 7. Answer(a) d = [2] (b) Make d the subject of the formula. Answer(b) d = [2]

4 marks

Mark scheme: 15 7d (a) 15 2 M1 for 35 = or better. 3 3 J J d (b) (d =) 2 M1 for 3J = md or = m m 3 3

This question in 0580/11 Oct/Nov 2009

17 (a) Factorise 5x2 + 4xy. Answer (a) [1] (b) Simplify completely 7(2x + y) − 3(3x − 2y). Answer (b) [3]

4 marks

Mark scheme: 17 (a) x(5x + 4y) final answer 1 Ignore check by expansion. (b) 5x + 13y www 3 W1 for 14x + 7y and W1 for –9x + 6y If zero ww SC1 for 5x or (+)13y in answer

This question in 0580/11 Oct/Nov 2009

md 15 J = 3 (a) Find the value of d when J = 32 and m = 8. Answer(a) d = [2] (b) Make d the subject of the formula. Answer(b) d = [2]

4 marks

Mark scheme: 8d 15 (a) 12 2 M1 for 32 = or better. 3 3 J J d (b) (d =) 2 M1 for 3J = md or = m m 3 3

This question in 0580/12 Oct/Nov 2009

17 (a) Factorise 3mp + 7p2. Answer (a) [1] (b) Simplify completely 8(3m + p) − 5(2m − 3p). Answer (b) [3]

4 marks

Mark scheme: 17 (a) p(3m + 7p) final answer 1 Ignore check by expansion. (b) 14m + 23p www 3 W1 for 24m + 8p and W1 for –10m + 15p If zero ww SC1 for 14m or (+)23p in answer

This question in 0580/12 Oct/Nov 2009

10 Simplify (a) p 3 × p 4 , Answer(a) [1] (b) 12q 8 ÷ 3q 2 . Answer(b) [2]

3 marks

Mark scheme: 10 (a) p7 1 (b) 4q6 2 W1 for 4qn or kq6 k ≠ 0

This question in 0580/11 May/June 2010

18 (a) Factorise 3 y − 7 xy . Examiner's Use Answer(a) [1] (b) Expand the brackets and simplify completely. p ( 4 p + 5 r ) + 2 r ( 6 p + r ) Answer(b) [3]

4 marks

Mark scheme: 18 (a) y(3y – 7x) final answer 1 (b) 4p2 + 17pr + 2r2 final answer 3 W2 for 2 correct terms in answer. W1 for 1 correct term in answer. OR M1 for 4p2 + 5pr and M1 ind for 12pr + 2r2

This question in 0580/11 May/June 2010

3 Factorise completely 6mp − 9pq. Answer [2]

2 marks

Mark scheme: 3 3p(2m − 3q) final answer 2 W1 for 3(2mp − 3pq) or p(6m − 9q) or 3p(am ± bq) where a and b are integers. 7 2 × 4 5 × 1 8 5

This question in 0580/12 May/June 2010

13 Expand the following expressions. (a) 5(3 – 4h) Answer(a) [1] (b) 4 d ( 6 d 2 + e 2 ) Answer(b) [2]

3 marks

Mark scheme: 13 (a) 15 – 20h final answer 1 (b) 24d3 + 4de2 final answer 2 W1 for 24d3 or (+)4de2 seen

This question in 0580/13 May/June 2010

15 Simplify the following expressions. (a) 6r + 2s + s – 4r Answer(a) [1] 2 2 (b) 4f – 3g + 4g – 9f Answer(b) [2]

3 marks

Mark scheme: 15 (a) 2r + 3s final answer 1 (b) g – 5f 2 final answer 2 W1 for g or for – 5f 2 seen

This question in 0580/13 May/June 2010

8 (a) Factorise completely. For 8pq + 12pr Examiner's Use Answer(a) [2] (b) Use your answer to part (a) to make p the subject of the formula below. s = 8pq + 12pr Answer(b) p = [1]

3 marks

Mark scheme: 8 (a) 4p(2q + 3r) 2 B1 for p(8q + 12r) or 2p(4q + 6r) or 4p(aq + br) a, b integers or 4(2pq + 3pr) s (b) (p =) oe 1ft ft if p is a common factor in (a) or in working in (b) 4( 2 q + 3r )

This question in 0580/12 Oct/Nov 2010

5 Multiply out the brackets. For x(2x + y) Examiner's Use Answer [2]

2 marks

Mark scheme: 5 2x2 + xy final answer 2 B1 for 2x2 or xy seen in working

This question in 0580/13 Oct/Nov 2010

11 Factorise completely. 3ac – 6ad Answer [2]

2 marks

Mark scheme: 11 3a(c − 2d) 2 B1 for a(3c − 6d) or 3(ac − 2ad) or 3a(jc − kd) where j and k are non-zero. 8 1 3

This question in 0580/13 Oct/Nov 2010

15 (a) Expand the brackets and simplify. For 3(2x O 5y) O 4(x O y) Examiner'sUse Answer(a) [2] (b) Factorise completely. 6x2 O 9xy Answer(b) [2]

4 marks

Mark scheme: 15 (a) 2x − 11y final answer 2 M1 for 6x − 15y or − 4x + 4y or better seen or B1 for 2x ± jy or kx − 11y. (b) 3x(2x − 3y) final answer 2 B1 for 3(2x2 − 3xy) or x(6x − 9y) or 3x(2x − by) or 3x(ax − 3y) (a, b ≠ 0) x

This question in 0580/12 May/June 2011

4 Factorise completely. 2xy – 4yz Answer [2]

2 marks

Mark scheme: 4 2y(x − 2z) 2 B1 for y(2x − 4z) or 2(xy − 2yz)

This question in 0580/13 May/June 2011

x 6 Make x the subject of the formula. y = + 5 3 Answer x = [2]

2 marks

Mark scheme: 6 (x =) 3(y − 5) oe final answer 2 M1 for correct first move x y − 5 = or 3y = x + 15 3 M1 for their correct second move

This question in 0580/13 May/June 2011

14 (a) Expand and simplify. 2(3x – 2) + 3(x – 2) Answer(a) [2] (b) Expand. x(2x2 – 3) Answer(b) [2]

4 marks

Mark scheme: 14 (a) 9x − 10 final answer 2 B1 for 6x − 4 or 3x − 6 or for answer of 9x + j, or kx − 10 (b) 2x3 – 3x final answer 2 B1 for answer in form 2x3 + m or n – 3x

This question in 0580/11 Oct/Nov 2011

3 Simplify the expression. 7x + 11y + x – 6y Answer [2]

2 marks

Mark scheme: 3 8x + 5y cao 2 B1 8x or 5y in final answer

This question in 0580/13 Oct/Nov 2011

14 Factorise completely. For Examiner's 5g2h + 10hj Use Answer [2]

2 marks

Mark scheme: 14 5h(g2 + 2j) 2 B1 for 5(g2h + 2hj) or for h(5g2 + 10j)

This question in 0580/13 Oct/Nov 2011

16 Simplify 4x4 × 5x5. Answer [2]

2 marks

Mark scheme: 16 20x9 cao 2 B1 for kx9or 20xk

This question in 0580/13 Oct/Nov 2011

5 w = 3a – 5b Calculate w when a = 2 and b = –3. Answer w = [2]

2 marks

Mark scheme: 5 21 2 M1 for 2 × 3 – 5 × (–3) or better or B1 for 6 and –15 i.e. both terms evaluated

This question in 0580/11 May/June 2012

11 (a) Write down all the factors of 15. Answer(a) [1] (b) Factorise completely. 15p2 + 24pt Answer(b) [2]

3 marks

Mark scheme: 11 (a) 1, 3, 5, 15 1 (b) 3p(5p + 8t) final answer 2 B1 for answer of 3(5p2 + 8pt) or p(15p + 24t) or SC1 for correct answer seen in working IGCSE – May/June 2012 0580 11

This question in 0580/11 May/June 2012

16 (a) Solve the equation 5(x − 3) = 21 . Answer(a) x = [2] (b) Make x the subject of the equation y = 3x − 2 . Answer(b) x = [2]

4 marks

Mark scheme: 16 (a) 7.2 oe 2 M1 for 5x – 15 = 21 or x – 3 = 215 y + 2 (b) [x =] 3 2 M1 for 3x = y + 2 or – 3x = –2 – y

This question in 0580/13 May/June 2012

18 (a) Find the value of 7p – 3q when p = 8 and q = O5 . Answer(a) [2] (b) Factorise completely. 3uv + 9vw Answer(b) [2]

4 marks

Mark scheme: 18 (a) 71 2 M1 for 7×8 – 3×–5 or B1 56 and –15 (b) 3v (u + 3w) final answer 2 B1 for 3(uv + 3vw) or v (3u + 9w) As final answer

This question in 0580/13 May/June 2012

7 Simplify the following expression. For 3j – 4k – 2 + 5j + k – 6 Examiner's Use Answer [2]

2 marks

Mark scheme: 7 8j – 3k – 8 2 B1 for two correct terms in final answer final answer or for correct answer seen then spoilt

This question in 0580/11 Oct/Nov 2012

13 (a) Factorise 9y + 12 . For Examiner's Use Answer(a) [1] (b) Expand a(a2 – 7) . Answer(b) [2]

3 marks

Mark scheme: 13 (a) 3(3y + 4) final answer 1 (b) a3 – 7a final answer 2 B1 for a3 or – 7a in final answer or for correct answer seen then spoilt 24

This question in 0580/11 Oct/Nov 2012

6 The formula for changing a temperature in Celsius to a temperature in Fahrenheit is F = 1.8C + 32 . Make C the subject of the formula. Answer C = [2]

2 marks

Mark scheme: 6 F −32 [C=] oe 1.8 2 M1 for first or second step correct e.g. F – 32 = 1.8 C final ans.  

This question in 0580/12 Oct/Nov 2012

17 (a) Factorise completely. For Examiner's 6x2 – 8xy Use Answer(a) [2] (b) Simplify the following expression. 28a5 Q 4aO2 Answer(b) [2]

4 marks

Mark scheme: 17 (a) 2x (3x – 4y) final ans. 2 M1 for x (6x − 8y) or 2 (3x 2 − 4xy) (b) 7a7 final ans. 2 M1 for 7a k or ka 7 k ≠ 0 for both cases

This question in 0580/12 Oct/Nov 2012

19 (a) Multiply out the brackets. 5 ( x + 3 ) Answer(a) … [1] (b) Factorise completely. 12xy – 3x2 Answer(b) … [2] (c) Solve. 5x – 24 = 51 Answer(c) x = … [2] _____________________________________________________________________________________ Question 20 is printed on the next page.

5 marks

Mark scheme: 19 (a) 5x + 15 final answer 1 (b) 3x( 4y – x ) final answer 2 B1 for 3( 4xy – x2 ) or x( 12y – 3x ) (c) 15 2 M1 for a correct first step

This question in 0580/11 May/June 2013

9 Factorise completely. 6xy2 – 8y Answer … [2] _____________________________________________________________________________________

2 marks

Mark scheme: 9 2y (3xy − 4) 2 B1 for 2 (3xy 2 − 4y) or y (6xy − 8)

This question in 0580/13 May/June 2013

15 (a) 5x × 53 = 510 Find the value of x. Answer(a) x = … [1] (b) Simplify. 12h3 ÷ 4h–2 Answer(b) … [2] _____________________________________________________________________________________

3 marks

Mark scheme: 15 (a) [x =] 7 1 (b) 3h 5 2 B1 for 3h n (n ≠ 0) or kh 5

This question in 0580/13 May/June 2013

7 Simplify the following expression. 5a – 3b – 2a – b Answer … [2] _____________________________________________________________________________________

2 marks

Mark scheme: 7 3a – 4b 2 B1 for answer 3a ± jb or ka – 4b Final Answer or SC1 for answer reached in working then spoilt

This question in 0580/11 Oct/Nov 2013

22 (a) Factorise completely. 6ab – 24bc Answer(a) … [2] (b) Rearrange the following formula to make m the subject. m j = n – k Answer(b) m = … [2] _____________________________________________________________________________________

4 marks

Mark scheme: 22 (a) 6b( a – 4c) 2 B1 for answer 6( ab – 4bc ) or 3b( 2a – 8c ) Final answer or 2b( 3a – 12c ) or b( 6a – 24c ) (b) n (j + k) or nj + nk oe 2 M1 for one correct step of a two-step method Final answer or SC1 for [m] = k + jn or [m] = j+ kn

This question in 0580/11 Oct/Nov 2013

15 c = 10d + 3 For Examiner′s Use (a) Find the value of c when d = 2.3 . Answer(a) c = … [1] (b) Make d the subject of the formula. Answer(b) d = … [2] _____________________________________________________________________________________

3 marks

Mark scheme: 15 (a) 26 1 c − 3 3 −c (b) or oe final answer 2 M1 for one correct step of a two step method. 10 − 10

This question in 0580/12 Oct/Nov 2013

21 (a) Simplify. 3x – 5y + 8x – 2y Answer(a) … [2] (b) Expand and simplify. 4(2a – 3b) – 5(a – 2b) Answer(b) … [2] _____________________________________________________________________________________

4 marks

Mark scheme: 21 (a) 11x − 7y final answer 2 B1 for 11x ± my or nx − 7y (b) 3a − 2b final answer 2 B1 for 8a −12b or −5a + 10b or 3a ± pb or qa −2b

This question in 0580/13 Oct/Nov 2013

1 14 V = Ah 3 (a) Find V when A = 15 and h = 7 . Answer(a) V = … [1] (b) Make h the subject of the formula. Answer(b) h = … [2] __________________________________________________________________________________________

3 marks

Mark scheme: 14 (a) 35 1 1 3V (b) or 3VA–1 2 M1 for multiplying by 3 or for dividing by A 3 or M1 for dividing by A 63− 61 61 M2 for × 100 oe or 100 ×100 oe

This question in 0580/12 May/June 2014

26 (a) Factorise completely. 15a3 – 5ab Answer(a) … [2] (b) Simplify. 3x2y3 × x4y Answer(b) … [2] (c) Multiply out the brackets and simplify. 3(x – 2) – 4(2x – 3) Answer(c) … [2] (d) Solve the equation. 8x + 9 = 3(x + 8) Answer(d) x = … [3] __________________________________________________________________________________________

9 marks

Mark scheme: 26 (a) 5a(3a2−b) 2 B1 for a(15a2 − 5b) or 5(3a3− ab) (b) 3x6 y4 2 B1 for x6 or y4 in a product on answer line (c) 6 – 5x as final answer nfww 2 B1 for 3x – 6 or – 8x + 12 seen or SC1 for 6 or – 5x seen in final answer nfww (d) 3 nfww 3 M2 for 5x = 15 or B1 for 3x +24 seen or M1 for 8x – 3x = 3 × 8 − 9 or better. If zero, SC1 for answer [x =] − 1 5

This question in 0580/13 May/June 2014

8 Factorise completely. 8w2x – 12wy Answer … [2] __________________________________________________________________________________________

2 marks

Mark scheme: 8 4 w(2 wx − 3 y ) 2 B1 for 4 (2 w 2 x − 3 wy ) Final answer or w(8 wx − 12 y ) or 2 w(4 wx − 6 y )

This question in 0580/11 Oct/Nov 2014

11 Simplify. 10x – 15 – 6x + 8 Answer … [2] __________________________________________________________________________________________

2 marks

Mark scheme: 11 4x – 7 2 B1 for answer 4x + k Final answer or answer jx – 7 where j ≠ 0 or correct answer seen then spoilt

This question in 0580/11 Oct/Nov 2014

9 Make y the subject of the formula. 8 + 5y – 3x = 0 Answer y = … [2] __________________________________________________________________________________________

2 marks

Mark scheme: 9 3 x − 8 2 B1 for 5y = 3x – 8 or −5y = 8 – 3x oe 5 3 x + 8 −x3 − 8 If B0 SC1 for or 5 5  

This question in 0580/12 Oct/Nov 2014

12 (a) Simplify 5k – 7k + 4k. Answer(a) … [1] (b) Find the value of 8x – 3y when x = –2 and y = –5. Answer(b) … [2] __________________________________________________________________________________________

3 marks

Mark scheme: 12 (a) 2k 1 (b) −1 2 B1 for −16 or −15 or 15 seen in the working.

This question in 0580/12 Oct/Nov 2014

16 (a) Simplify. 5j + 2k + j – 3k Answer(a) … [2] (b) Factorise. 5p + 10 Answer(b) … [1] __________________________________________________________________________________________

3 marks

Mark scheme: 16 (a) 6 j − k 2 B1 for 6 j ± ak or bj − k (a and b ≠ 0) (b) 1 5( p + 2 )

This question in 0580/13 Oct/Nov 2014

18 (a) Solve the simultaneous equations. You must show all your working. 4x + 2y = 31 6x – 2y = 34 Answer(a) x = … y = … [2] (b) Factorise 14p2 + 21pq. Answer(b) … [2] __________________________________________________________________________________________

4 marks

Mark scheme: 18 (a) [x =] 6.5 [y =] 2.5 2 B1 for x = 6.5 B1 for y = 2.5 If zero scored, SC1 for correct substitution and evaluation to find other variable or SC1 no working, 2 correct answers given. (b) 7p(2p + 3q) 2 B1 for 7(2p2 + 3pq) or p(14p + 21q)

This question in 0580/12 Feb/March 2015

19 Idris has c toy cars. Fadl has twice as many cars as Idris. Baasim has three more cars than Fadl. (a) Write down an expression, in terms of c, to complete each statement. Fadl has … cars. Baasim has … cars. [2] (b) Write down an expression, in terms of c, for the total number of cars the three children have. Give your answer in its simplest form. Answer(b) … [2] __________________________________________________________________________________________

4 marks

Mark scheme: 19 (a) 2c 1 2c + 3 1FT FT is their 2c + 3 provided linear (b) 5c + 3 2FT M1 for c + their 2c + their(2c+3) provided linear

This question in 0580/12 Feb/March 2015

10 Simplify. 6uw–3 × 4uw6 Answer … [2] __________________________________________________________________________________________

2 marks

Mark scheme: 10 24u2w3 final answer 2 B1 for 2 correct elements in final answer AB

This question in 0580/11 May/June 2015

13 (a) Factorise. 3w2 – 2w Answer(a) … [1] (b) Expand and simplify. x (2x + 3) + 5(x – 7) Answer(b) … [2] __________________________________________________________________________________________

3 marks

Mark scheme: 13 (a) w (3w – 2) 1 (b) 2x2 + 8x – 35 final answer 2 B1 for 2 terms correct in final answer or M1 for 2x2 + 3x or 5x – 35

This question in 0580/11 May/June 2015

5 Factorise completely. 9x2 – 6x Answer … [2] __________________________________________________________________________________________

2 marks

Mark scheme: 5 3x(3x – 2) final answer 2 B1 for x(9x – 6) or 3(3x2 – 2x)

This question in 0580/13 May/June 2015

13 Rearrange the formula to make y the subject. ty R = 4 Answer y = … [2]

2 marks

Mark scheme: 4 R R 1 13 [ y = ] 2 M1 for a correct first step: 4R = ty or = y t t 4

This question in 0580/12 Oct/Nov 2015

23 (a) Expand. - 4 (2w - 5 ) Answer(a) … [1] (b) Factorise. 6x 2 - x Answer(b) … [1] (c) A = 2pq + 3pr Find A when p = 7, q = 5 and r = − 2 . Answer(c) A = … [2]

4 marks

Mark scheme: 23 (a) −8w + 20 final answer 1 (b) x(6x – 1) 1 (c) 28 2 M1 for 2 × 7 × 5 + 3 × 7 × (–2) or for 70 or −42 seen

This question in 0580/12 Oct/Nov 2015

9 Simplify. 1 – 2u + u + 4 Answer … [2] __________________________________________________________________________________________

2 marks

Mark scheme: 9 5 – u final answer 2 B1 for 5 + ku or j – u , k ≠ 0 as final answer

This question in 0580/13 Oct/Nov 2015

17 (a) Multiply out. 3(x + 7) Answer(a) … [1] (b) Factorise completely. 2x – 4x2 Answer(b) … [2] __________________________________________________________________________________________

3 marks

Mark scheme: 17 (a) 3x + 21 final answer 1 (b) 2x (1 − 2x) final answer 2 B1 for 2(x − 2x2) or x(2 – 4x) as final answer

This question in 0580/13 Oct/Nov 2015

7 Rearrange the formula to make w the subject. 5w - 3y + 7 = 0 w = … [2]

2 marks

Mark scheme: 3 y 7 3 y 7 7 w = oe 2 M1 for 5w + 7 = 3y or 5w −3 y = −7 or w − + = 0 5 5 5

This question in 0580/12 Feb/March 2016

21 (a) Factorise completely. 18x 2 - 24x … [2] (b) Expand the brackets. x ^3 x - 4h … [2]

4 marks

Mark scheme: x (18 x − 24 )21 (a) 6 x ( 3 x − 4 ) final answer 2 M1 for 6( 3 x 2 − 4 x ) or or 2 x ( 9 x − 12 ) or 3 x ( 6 x − 8 ) or 2( 9 x 2 − 12 x ) or 3( 6 x 2 − 8 x ) (b) 3 x 2 − 4 x final answer 2 M1 for 3x 2 − kx or kx 2 − 4 x or correct answer seen and then spoilt Qu Answer Mark Part marks 5 2 5 2

This question in 0580/12 Feb/March 2016

qx 13 y = p Write x in terms of p, q and y. x = … [2]

2 marks

Mark scheme: py13 2 M1 for one correct step final answer q

This question in 0580/11 May/June 2016

7 Simplify. (a) 3f + 4f – 2f … [1] (b) g3 × g5 … [1]

2 marks

Mark scheme: 7 (a) 5f final answer 1 (b) g8 final answer 1

This question in 0580/12 May/June 2016

21 (a) Simplify. 3p + 5p + p … [1] (b) Multiply out the brackets. 4(q – 3) … [1] (c) Factorise completely. 10t + 15t2 … [2] (d) Solve the simultaneous equations. You must show all your working. 3x + y = 7 5x – y = 17 x = … y = … [2]

6 marks

Mark scheme: 21 (a) 9p final answer 1 (b) 4q – 12 final answer 1 (c) 5t(2 + 3t) final answer 2 M1 for t(10 + 15t) or 5(2t + 3t2) (d) [x = ] 3, [y = ] –2 2 B1 for one correct with working with supporting working If zero scored, SC1 for 2 values satisfying one of the original equations or SC1 if no working shown, but 2 correct answers given

This question in 0580/12 Oct/Nov 2016

9 Factorise completely. 15c 2 - 5c … [2]

2 marks

Mark scheme: 9 5c(3c – 1) final answer 2 B1 for 5(3c2 – c) or c(15c – 5)

This question in 0580/12 Feb/March 2017

18 (a) Expand the brackets and simplify. 4 (5w + 3 ) - 2 (w - 1 ) … [2] (b) Simplify. (w5)2 … [1]

3 marks

Mark scheme: 18 (a) 18w + 14 final answer 2 M1 for 20 w + 12 or −2 w + 2 or answer 18 w + k or kw + 14 (b) w10 1 3

This question in 0580/12 Feb/March 2017

8 Factorise completely. 12n2 - 4mn … [2]

2 marks

Mark scheme: 8 4n(3n – m) final answer 2 B1 for 4(3n2 – mn) or n(12n – 4m) or 2n(6n − 2m) or 2(6n2 – 2mn)

This question in 0580/11 May/June 2017

11 Expand the brackets and simplify. 7 (2x + 3y ) - x (14 - y) … [2]

2 marks

Mark scheme: 11 21y + xy or y(21 + x) 2 B1 for 14x + 21y or −14x + xy final answer or answer of ky + xy

This question in 0580/11 May/June 2017

17 A = 4 rr 2 Make r the subject of this formula. r = … [2]

2 marks

Mark scheme: 17 A 1 A 2 A or oe M1 for r2 = or 2r π = A 4π 2 π 4π A A or 4r2 = or πr2 = π 4

This question in 0580/11 May/June 2017

6 Factorise. 14x - 21y … [1]

1 marks

Mark scheme: 6 7(2x – 3y) final answer 1

This question in 0580/12 May/June 2017

11 Find the value of 5a - 3b when a = 7 and b =- 2 . … [2]

2 marks

Mark scheme: 11 41 2 M1 for 5(7) – 3(– 2)

This question in 0580/12 May/June 2017

9 Factorise completely. 4x2 - 8xy … [2]

2 marks

Mark scheme: 9 4 x ( x − 2 y ) final answer 2 2 M1 for 4 x − 2 xy or x ( 4 x − 8 y ) ( ) or 2 2 x 2 − 4 xy or 2x (2x – 4y) ( )

This question in 0580/13 May/June 2017

21 Simplify. (a) 6w0 … [1] (b) 5x3 − 3x3 … [1] (c) 3y6 # 5y-2 … [2]

4 marks

Mark scheme: 21(a) 6 1 21(b) 2x 3 final answer 1 21(c) 15y 4 final answer 2 B1 for 15 ky or ky 4 as final answer (k ≠ 0)

This question in 0580/13 May/June 2017

7 Factorise completely. 12x 2 + 15xy - 9x … [2]

2 marks

Mark scheme: 7 3x ( 4x + 5y – 3) final answer 2 B1 for 3(4x2 + 5xy – 3x) or x(12x + 15y – 9) allow in working or correct answer spoiled If zero scored, SC1 for 3x(4x + 5y – 3) with only 2 correct elements in the brackets, allow in working

This question in 0580/11 Oct/Nov 2017

17 Simplify. (a) m5 2 ^ h … [1] (b) 4x 3 y # 5 x 2 y … [2]

3 marks

Mark scheme: 17(a) m10 final answer 1 17(b) 20x5y2 final answer 2 B1 for 2 out of 3 elements correct in final answer or correct answer spoiled

This question in 0580/11 Oct/Nov 2017

11 Expand and simplify. 6(2y - 3) - 5(y + 1) … [2]

2 marks

Mark scheme: 11 7y − 23 final answer 2 M1 for 12 y − 18 or −5 y − 5 or B1 for answer 7y − k or cy − 23 c ≠ 0

This question in 0580/11 May/June 2018

22 Factorise completely. (a) 10 + 16w … [1] (b) 12tx - 8t2 … [2]

3 marks

Mark scheme: 22(a) 2 ( 5 + 8w ) 1 22(b) 4t ( 3 x − 2t ) 2 B1 for answer 4(3tx − 2t 2 ) or t (12 x − 8t ) 2(6tx − 4t 2 ) or 2t (6 x − 4t )

This question in 0580/11 May/June 2018

2 Simplify. 7g – g + 2g … [1]

1 marks

Mark scheme: 2 8g 1

This question in 0580/12 May/June 2018

3 Expand. 7(x – 8) … [1]

1 marks

Mark scheme: 3 7x – 56 final answer 1

This question in 0580/12 May/June 2018

10 Factorise completely. 4xy 2 - 6y 3 … [2]

2 marks

Mark scheme: 10 2y2( 2x – 3y) final answer 2 B1 for 2y( 2xy – 3y2) or 2(2xy2 – 3y3) or y(4xy – 6y2) or y2(4x – 6y)

This question in 0580/12 May/June 2018

13 Make y the subject of the equation 5x - 2y + 7 = 0 . y = … [2]

2 marks

Mark scheme: 13 5 x + 7 2 M1 for 2y = 5x + 7 or −2y = −5x – 7 [y =] oe 5 7 2 or x – y + = 0 2 2

This question in 0580/12 May/June 2018

2 Factorise. w + w3 … [1]

1 marks

Mark scheme: 2 w(1 + w2) final answer 1

This question in 0580/13 May/June 2018

23 (a) Expand the brackets and simplify fully. 5 (x - 3) + 2 (3x + 1) … [2] (b) Solve the simultaneous equations. You must show all your working. 4x - y = 14 3x + 2y = 5 x = … y = … [3]

5 marks

Mark scheme: 23(a) 11x – 13 final answer 2 B1 for 5x – 15 or 6x + 2 or for answer of 11x + j or kx – 13 23(b) Correctly eliminating one variable M1 [x = ] 3 A1 [y = ] –2 A1 If zero scored, SC1 for 2 values satisfying one of the original equations or for 2 correct answers with no working

This question in 0580/11 Oct/Nov 2018

5 Expand. 2x 3 - x 2 ^ h … [2]

2 marks

Mark scheme: 5 6x – 2x3 final answer 2 B1 for 6x or –2x3

This question in 0580/12 Oct/Nov 2018

5 Factorise. y - 2y 2 … [1]

1 marks

Mark scheme: 5 y(1 – 2y) final answer 1

This question in 0580/13 Oct/Nov 2018

10 Simplify. 2p - q - 3q - 5p … [2]

2 marks

Mark scheme: 10 −3p – 4q final answer 2 B1 for −3p or −4q

This question in 0580/13 Oct/Nov 2018

19 Factorise completely. 8g 2 - 4g … [2]

2 marks

Mark scheme: 19 4 g ( 2 g − 1) 2 M1 for 2 g ( 4 g − 2 ) , g ( 8 g − 4 ) or 4 2g 2 − g or 2 4 g 2 − 2 g or ( ) ( ) 22g(2g – 1) as final answers

This question in 0580/12 Feb/March 2019

3 Factorise 5y - 6py. … [1]

1 marks

Mark scheme: 3 y (5 – 6p) final answer 1

This question in 0580/11 May/June 2019

16 (a) Expand. x 2 ( x - 7) … [2] (b) Factorise. y 2 + y … [1]

3 marks

Mark scheme: 16(a) x3 – 7x2 final answer 2 B1 for x 3 − jx k j ≠ 0 or nx p − 7 x 2 n ≠ 0 as answers If 0 scored, SC1 for correct answer seen then spoilt 16(b) y(y + 1) final answer 1

This question in 0580/12 May/June 2019

3 Factorise 2x 2 - x . … [1]

1 marks

Mark scheme: 3 x(2x – 1) 1

This question in 0580/13 May/June 2019

9 Simplify. 4x - 12y + 10x + 25y … [2]

2 marks

Mark scheme: 9 14x + 13y final answer 2 B1 for 14x or 13y as answers If 0 scored, SC1 for correct answer seen then spoilt

This question in 0580/13 May/June 2019

14 Rearrange 2 (w + h ) = P to make w the subject. w = … [2]

2 marks

Mark scheme: 14 P P − 2 h 2 P [w =] – h or final M1 for w + h = or 2w + 2h = P 2 2 2 answer

This question in 0580/13 May/June 2019

4 Factorise 5p + pt. … [1]

1 marks

Mark scheme: 4 p(5 + t) final answer 1

This question in 0580/11 Oct/Nov 2019

12 Simplify 5c - d - 3d - 2 c . … [2]

2 marks

Mark scheme: 12 3c – 4d final answer 2 B1 for 3c + kd or kc – 4d

This question in 0580/11 Oct/Nov 2019

1 Simplify 5t + 4t - 2t . … [1]

1 marks

Mark scheme: Question Answer Marks Partial Marks 1 7t 1

This question in 0580/12 Oct/Nov 2019

4 Factorise 12x + 15. … [1]

1 marks

Mark scheme: 4 3(4x + 5) final answer 1

This question in 0580/13 Oct/Nov 2019

12 Expand and simplify (x + 3)(x + 5). … [2]

2 marks

Mark scheme: 12 x2 + 8x + 15 final answer 2 M1 for three terms correct from x2 + 5x + 3x + 15

This question in 0580/13 Oct/Nov 2019

18 (a) Factorise completely. 3x 2 - 12xy … [2] (b) Expand and simplify. ( m - 3)( m + 2) … [2]

4 marks

Mark scheme: 18(a) 3x(x – 4y) final answer 2 B1 for 3(x2 – 4xy) or x(3x – 12y) 18(b) m2 – m – 6 final answer 2 M1 for 3 terms from m2, –3m, +2m, –6

This question in 0580/12 Feb/March 2020

13 Factorise completely. 21a 2 + 28ab … [2]

2 marks

Mark scheme: 13 7a(3a + 4b) final answer 2 B1 for partial factorisation 7(3a2 + 4ab) or a(21a + 28b)

This question in 0580/11 May/June 2020

17 Expand and simplify. ( x - 5)( x - 7) … [2]

2 marks

Mark scheme: 17 x2 – 12x + 35 2 B1 for any three of x2, –5x, –7x, +35

This question in 0580/11 May/June 2020

6 Simplify. 5w + 3h - 7w + 8h … [2]

2 marks

Mark scheme: 6 11h − 2 w final answer 2 M1 for 11h + kw or kh − 2 w

This question in 0580/13 May/June 2020

17 Expand the brackets and simplify. 4 ( 2m + 3) - 5 ( m - 2) … [2]

2 marks

Mark scheme: 17 3m + 22 final answer 2 B1 for 8m + 12 or –5m + 10 or jm + 22 or 3m + k as final answer

This question in 0580/13 Oct/Nov 2020

12 Factorise completely. 9t 2 w - 3t … [2]

2 marks

Mark scheme: 12 3 t ( 3 tw − 1 ) final answer 2 B1 for 3 ( 3t 2 w − t ) or for t ( 9 tw − 3 )

This question in 0580/12 Feb/March 2021

24 Expand and simplify. 5 ( 2x - 7) - 3 ( x - 5) … [2]

2 marks

Mark scheme: 24 7x – 20 final answer 2 B1 for 10x – 35 or 3x – 15 or −3x + 15 or for 7x or −20 in the final answer

This question in 0580/12 May/June 2021

15 Expand and simplify. 6 ( t - q) - 2 ( t - 3q ) … [2]

2 marks

Mark scheme: 15 4t final answer 2 B1 for 6t – 6q or – 2t + 6q or 2t – 6q or for 4t or 0q in the final answer

This question in 0580/13 May/June 2021

14 (a) Factorise completely. 18x 2 - 12x … [2] (b) Expand and simplify. ( x + 5)( x - 3) … [2]

4 marks

Mark scheme: 14(a) 6x (3x – 2) final answer 2 B1 for correct answer spoilt or for correct partial factorisation as answer e.g. 2 (9x2 – 6x) or 3 (6x2 – 4x) or 6 (3x2 – 2x) or 2x (9x – 6) or 3x (6x – 4) or x (18x – 12) 14(b) x2 + 2x – 15 final answer 2 M1 for 3 correct terms from x2 – 3x + 5x – 15 soi

This question in 0580/11 Oct/Nov 2021

23 (a) Simplify. 32g 32 ' 4g 4 … [2] (b) Factorise completely. 10j - 15 j 2 … [2] (c) Expand the brackets and simplify. ( x + 7)( x + 3) … [2]

6 marks

Mark scheme: 23(a) 8g28 final answer 2 B1 for kg28 or 8gk as final answer or correct answer seen and spoilt 23(b) 5j (2 – 3j) final answer 2 B1 for 5(2j– 3j2) or j(10 – 15j) as final answer or correct answer spoilt 23(c) x2 + 10x + 21 final answer 2 B1 for x2 + 7x + 3x + 21 with at least three terms correct

This question in 0580/13 Oct/Nov 2021

19 Simplify. (a) 5f + 7g - 8f + 2g … [2] (b) h 2 # h 5 … [1] (c) 16x 2 # 5y 0 … [2]

5 marks

Mark scheme: 19(a) −3 f + 9 g final answer 2 B1 for −f3 or 9g or correct answer spoilt 19(b) h7 final answer 1 19(c) 20x 2 2 0 B1 for 16 x = 4 x soi or y = 1 soi

This question in 0580/12 Feb/March 2022

23 (a) Expand and simplify. ( x + 3)( x - 5) … [2] (b) Renuka’s teacher asks her to factorise completely 8x 2 - 12x . Renuka writes 2x ( 4x - 6) as her answer. Explain why she does not score full marks and give the correct answer. Reason … Correct answer … [2]

4 marks

Mark scheme: 23(a) x 2 − 2 x − 15 final answer 2 B1 for x 2 − 5 x + 3 x − 15 with at least 3 terms correct or for correct answer seen and spoilt 23(b) Correct reason 2 B1 for each 4 x ( 2 x − 3 )

This question in 0580/12 Feb/March 2022

16 v = 3 - 5t (a) Work out the value of v when t = 4. v = … [1] (b) Make t the subject of the formula. t = … [2]

3 marks

Mark scheme: 16(a) −17 1 16(b) 3 v v  3 2 M1 for or final answer 5 5 v 3 5t = 3 – v or v – 3 = −5t or   t 5 5

This question in 0580/11 May/June 2022

20 (a) Simplify. 3 ( 2a - b) - b … [2] (b) Factorise. x 2 - 8xy … [1]

3 marks

Mark scheme: 20(a) 6a – 4b final answer 2 B1 for 6a or −4b in final answer or for 6 a  4b spoilt 20(b) x(x – 8y) final answer 1

This question in 0580/12 May/June 2022

3 Simplify. 3x - 4x + 7x … [1]

1 marks

Mark scheme: 3 6x 1

This question in 0580/13 May/June 2022

15 Factorise completely. 18px - 27p … [2]

2 marks

Mark scheme: 15 9p(2x – 3) final answer 2 B1 for 9(2px – 3p) or p(18x – 27) or 3p(6x – 9) or 9p(2x – 3) seen and spoilt

This question in 0580/13 May/June 2022

8 (a) Simplify. 6a + 3b - 2a - 5b … [2] 1 2 (b) s = 5t + at 2 Find the value of s when t = 6 and a = 3 . s = … [2]

4 marks

Mark scheme: 8(a) 4a – 2b or 2(2a – b) final answer 2 B1 for 4a or – 2b in final answer or 4a + – 2b as final answer or for correct answer seen and spoilt 8(b) 84 2 B1 for 30 or 54 1 or M1 for 5 × 6 + × 3 × 62 oe 2

This question in 0580/12 Oct/Nov 2022

16 (a) Expand. x ( x + 8) … [2] (b) Factorise completely. 6a - 3ab … [2] (c) Solve. 5x - 6 = x + 3 x = … [2]

6 marks

Mark scheme: 16(a) x2 + 8x final answer 2 B1 for x2 or +8x in final answer or correct answer spoilt 16(b) 3a(2 − b) final answer 2 B1 for 3(2a −ab) or a(6 − 3b) final answer or correct answer spoilt 16(c) 1 2 M1 for 5x – x = 3 + 6 or better 2.25 or 2 4

This question in 0580/12 Oct/Nov 2022

11 (a) Expand the brackets and simplify. (i) 3 ( 2d - 3) + 4 ( d + 1) … [2] (ii) ( x - 4)( x - 3) … [2] (b) Factorise 8p - 3pq . … [1]

5 marks

Mark scheme: 11(a)(i) 10d – 5 or 5(2d – 1) final answer 2 B1 for 6d – 9 or 4d + 4 or 10d + – 5 or for correct answer seen and spoilt 11(a)(ii) x2 – 7x + 12 final answer 2 B1 for x2 –4x –3x + 12 with at least three terms correct 11(b) p(8 – 3q) final answer 1

This question in 0580/13 Oct/Nov 2022

m23 Make m the subject of the formula q = + r . 7 m = … [2] Question 24 is printed on the next page.

2 marks

Mark scheme: 23 7(q – r) or 7q – 7r final answer 2 m M1 for q – r = or 7q = m + 7r 7

This question in 0580/13 Oct/Nov 2022

14 Factorise completely. 8g - 2g 2 … [2]

2 marks

Mark scheme: 14 2g(4 – g) final answer 2 B1 for 2(4g – g2) or for g(8 – 2g) or for 2g(4 – g) seen then spoiled

This question in 0580/12 Feb/March 2023

9 Simplify. 3a - 5b - a - 6b … [2]

2 marks

Mark scheme: 9 2a – 11b final answer 2 B1 for 2a or −11b in final answer or for 2a – 11b seen then spoilt

This question in 0580/11 May/June 2023

16 Factorise completely. 8x 2 - 20x … [2]

2 marks

Mark scheme: 16 4x(2x – 5) final answer 2 B1 for x(8x – 20) or 4(2x2 – 5x) or 2x(4x – 10) or 4x(2x – 5) seen and then spoilt

This question in 0580/11 May/June 2023

19 Simplify. 18x 12 ' 3x 3 … [2]

2 marks

Mark scheme: 19 6x9 final answer 2 B1 for kx9 or 6xk, k ≠ 0 as final answer or for correct answer seen and spoilt

This question in 0580/12 May/June 2023

19 Expand and simplify. ( x - 5)( x + 8) … [2]

2 marks

Mark scheme: 19 x2 + 3x – 40 2 B1 for x2 – 5x + 8x – 40 with at least 3 terms correct

This question in 0580/13 May/June 2023

9 v = u + at Find the value of v when u = 30 , a =- 2 and t = 7 . v = … [2]

2 marks

Mark scheme: 9 16 2 M1 for 30 – 2  7 or B1 for –14

This question in 0580/12 Oct/Nov 2023

18 Expand and simplify. 2 ( t + w) + 3 ( w - t) … [2]

2 marks

Mark scheme: 18 5w – t final answer 2 B1 for 2t + 2w or 3w – 3t or for 5w – t seen then spoiled or for 5w or – t in the final answer

This question in 0580/12 Oct/Nov 2023

11 Factorise completely. 15v 2 - 3v … [2]

2 marks

Mark scheme: 11 3v(5v – 1) final answer 2 B1 for 3(5v2 – v) or v(15v – 3) as final answer or 3v(5v – 1) seen and spoilt

This question in 0580/13 Oct/Nov 2023

9 Simplify 4m + 7k - m + 3k . … [2]

2 marks

Mark scheme: 9 3m + 10k final answer 2 B1 for 3m or 10k in final answer or for 3m + 10k seen and spoilt

This question in 0580/12 Feb/March 2024

14 Factorise. 3x 3 - 7xy … [1]

1 marks

Mark scheme: 14 2 1 x 3 x − 7 y ( )

This question in 0580/12 Feb/March 2024

15 Factorise completely. 36x 2 + 40x … [2]

2 marks

Mark scheme: 9 x 2  10 x15 4 x  9 x  10  2 B1 for x  36 x  40  or 4   final answer or 2 x 18 x  20  or for 4 x  9 x  10  seen then spoilt

This question in 0580/11 May/June 2024

14 Factorise completely. 20x - 90 x 2 … [2]

2 marks

Mark scheme: 14 10x(2 – 9x) final answer 2 B1 for answer of 10(2x – 9x2) or x(20 – 90x) or 2x(10 –45x) or 5x(4 – 18x) or 10x(2 – 9x) seen and spoilt.

This question in 0580/12 May/June 2024

20 Expand and simplify. ( x - 4)( x - 7) … [2]

2 marks

Mark scheme: 20 x2 – 11x + 28 final answer 2 B1 for x2 – 4x – 7x + 28 with at least 3 terms correct

This question in 0580/12 May/June 2024

3 Simplify. 7x - 8y - x - y … [2]

2 marks

Mark scheme: 3 6x – 9y or 3(2x – 3y) final answer 2 B1 for 6x or – 9y in final answer or 6x – 9y seen then spoilt

This question in 0580/13 May/June 2024

13 Factorise completely. 4x 2 y - 5xy 2 … [2]

2 marks

Mark scheme: 13 xy(4x – 5y) final answer 2 B1 for y(4x2 – 5xy) or x(4xy – 5y2) or xy(4x – 5y) seen then spoilt

This question in 0580/13 May/June 2024

13 (a) Find the value of 6c + 7d when c = 3 and d =- 4 . … [2] (b) Solve. 6x + 8 = 11x + 4 x = … [2]

4 marks

Mark scheme: 13(a) –10 2 B1 for 18 or –28 or M1 for 6 × 3 + 7 × −4 13(b) 4 2 M1 for 8 – 4 = 11x – 6x or better or 0.8 5

This question in 0580/11 Oct/Nov 2024

21 (a) Factorise. 28x - 35 … [1] r (b) Make r the subject of the formula T = - p . 4 r = … [2]

3 marks

Mark scheme: 21(a) 7(4x – 5) final answer 1 21(b) 4T + 4p or 4(T + p) 2 r M1 for T + p = or 4T = r – 4p final answer 4

This question in 0580/11 Oct/Nov 2024

8 Simplify. 8c - d - 3c + 3d … [2]

2 marks

Mark scheme: 8 5c + 2d final answer 2 B1 for 5c or 2d in the final answer or for 5c + 2d seen then spoilt

This question in 0580/12 Oct/Nov 2024

7 Simplify. 3p - t – p - 4t … [2]

2 marks

Mark scheme: 7 2p – 5t final answer 2 B1 for 2p or −5t in final answer or for 2 p – 5t seen then spoilt

This question in 0580/13 Oct/Nov 2024

14 Expand 4 ( x - 3) . … [1]

1 marks

Mark scheme: 14 4x – 12 final answer 1

This question in 0580/13 Oct/Nov 2024

12 (a) Kat has a method for finding the difference between two square numbers, a 2 - b 2 . Her method is (the sum of a and b) # (the difference between a and b) . She shows her method for 17 2 - 13 2 . 17 2 - 13 2 = ( 17 + 13) # ( 17 - 13) = 30 # 4 = 120 Work out 29 2 - 21 2 using Kat’s method. … [2] (b) In this part, all lengths are in centimetres. 17 NOT TO SCALE 17 13 13 100 The diagram shows a prism. Work out the volume of the prism. … cm3 [2]

4 marks

Mark scheme: 12(a) ( 29 + 21)  ( 29 − 21) M1 = 50  8 400 A1 12(b) 12000 2 2 2 M1 for 17 − 13 [ ×100] oe ( )

This question in 0580/12 Feb/March 2025

21 X = 1 w 2 p 3 (a) Find the value of X when w = 5 and p = 6. X = ĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭ [2] (b) Make p the subject of the formula. p = ĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭĭ [2]

4 marks

Mark scheme: 21(a) 50 2 1 2 M1 for  5  6 or better 3 21(b) 3 X −2 2 M1 for 3 X = w 2 p p = or p = 3 Xw w 2 X 1 2 1 or = p or Xw−= p 2 final answer w 3 3

This question in 0580/12 Feb/March 2025

23 (a) Factorise. 9x - 6 xy … [2] (b) Expand and simplify. ( 2x + 3)( x - 4) … [2]

4 marks

Mark scheme: 23(a) 3 x ( 3 − 2 y ) final answer 2 B1 for answer of 3 ( 3 x − 2 xy ) or x ( 9 − 6 y ) or for 3 x ( 3 − 2 y ) seen then spoilt 23(b) 2 x 2 − 5 x − 12 final answer 2 B1 for 2 x 2 + 3 x − 8 x − 12 with at least 3 terms correct or for 2 x 2 − 5 x − 12 seen then spoilt

This question in 0580/12 May/June 2025

h 22 g = - 8 3 Rearrange the formula to make h the subject. h = … [2]

2 marks

Mark scheme: 22 3(g + 8) or 3g + 24 2 h M1 for g + 8 = or 3g = h – 24 3

This question in 0580/13 May/June 2025

25 Expand and simplify. ( x + 3)( x - 2) … [2]

2 marks

Mark scheme: 25 x2 + x – 6 final answer 2 B1 for x2 + 3x – 2x – 6 oe with at least three terms correct

This question in 0580/11 Oct/Nov 2025

25 (a) Expand and simplify. ( x - 5)( x + 2) … [2] (b) Factorise. 5x 3 + 10 x … [2]

4 marks

Mark scheme: 25(a) x2 – 3x – 10 final answer 2 B1 for x2 – 5x + 2x – 10 with at least 3 terms correct 25(b) 5x(x2 + 2) final answer 2 B1 for 5(x3 + 2x) or x(5x2 + 10) or 5x(x2 + 2) seen and spoilt

This question in 0580/12 Oct/Nov 2025

12 (a) Simplify. 3y - 4y + 2y … [1] (b) Solve. (i) x + 5 = 19 x = … [1] (ii) 6x - 5 = 7 x = … [2]

4 marks

Mark scheme: 12(a) y 1 12(b)(i) 14 1 12(b)(ii) 2 2 5 7 M1 for 6x = 7 + 5 or x − = or better 6 6

This question in 0580/13 Oct/Nov 2025

17 (a) k x # k 5 = k 20 Find the value of x. x = … [1] (b) Expand and simplify. ( x + 5)( x - 4) … [2] (c) Factorise. 21r 3 - 7r … [2]

5 marks

Mark scheme: 17(a) 15 1 17(b) x2 + x – 20 final answer 2 B1 for x2 – 4x + 5x – 20 with at least 3 terms correct 17(c) 7r(3r2 – 1) final answer 2 B1 for 7(3r3 – r) or r(21r2 – 7) or for 7r(3r2 – 1) seen then spoilt

This question in 0580/13 Oct/Nov 2025