Indices I — Paper 1 · IGCSE Mathematics 0580

10 Simplify the following expressions. For Examiner's (a) a2 × a5 Use Answer(a) [1] (b) b4 ÷ b3 Answer(b) [1]

2 marks

Mark scheme: 10 (a) a7 1 (b) b 1 Allow b1

This question in 0580/11 Oct/Nov 2004

10 Simplify (a) p2 × p3, Answer(a) [1] (b) q3 ÷ q−4, Answer(b) [1] (c) (r2)3. Answer(c) [1]

3 marks

Mark scheme: 10 (a) p5 1 (b) q7 1 (c) r6 1 19 IGCSE – May/June 2006 0580 and 0581 01

This question in 0580/11 May/June 2006

16 Write down the value of x when For Examiner's (a) 2 x = 8, Use Answer(a) x= [1] x 1 (b) 3 = . 81 Answer(b) x= [1]

2 marks

Mark scheme: 16 (a) 3 or 2 3 = 8 1 SC1 for 2 3 and 3 −4 in the answer spaces −4 1 (b) −4 or 3 = 81

This question in 0580/11 Oct/Nov 2006

15 Simplify (a) a0, Answer(a) [1] 2 (b) ( 3x ) Answer(b) [1] -2  3  (c)   .   x  Answer(c) [2]

4 marks

Mark scheme: 15 (a) 1 1 (b) x 6 1 1 3 or better. E.g. ( x3) 2 )2 (c) x92 2 M1 for ( x B1 if answer contains x 2 as numerator or 3 2(or 9) as denominator. 15

This question in 0580/11 Oct/Nov 2007

16 Simplify For Examiner's Use 0  1  (a)   ,  p  Answer(a) [1] (b) q4 × q7, Answer(b) [1] _3 (c) ( r 2 ) . Answer(c) [1]

3 marks

Mark scheme: 16 (a) 1 1 (b) q 11 1 (c) −6 1 r or 6 1 r

This question in 0580/11 Oct/Nov 2008

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

1 Examiner's 12 (a) = 3x . Use 27 Write down the value of x. Answer(a) x = [1] (b) Simplify (i) p7 × p-2 , Answer(b)(i) [1] (ii) m3 ÷ m7. Answer(b)(ii) [1]

3 marks

Mark scheme: 12 (a) −3 1 (b) (i) p5 1 1 (ii) m–4 or 4 1 m IGCSE – May/June 2010 0580 12

This question in 0580/12 May/June 2010

11 Simplify (a) m 3 × m −5 , Answer(a) [1] (b) 15 k 8 ÷ 3k 2 . Answer(b) [2]

3 marks

Mark scheme: 1 11 (a) m–2, o.e. 1 m 2 (b) 5k6 2 W1 for 5kn (n ≠ 0) or mk6 (m ≠ 0).

This question in 0580/13 May/June 2010

14 Simplify the following. For Examiner's (a) 80 Use Answer(a) [1] (b) (x5)2 Answer(b) [1] (c) p–3 ÷ p4 Answer(c) [1]

3 marks

Mark scheme: 14 (a) 1 1 (b) x10 1 −7 1 (c) p or 7 1 p

This question in 0580/11 Oct/Nov 2010

12 (a) Write down the value of x when (i) 5x ÷ 52 = 54, Answer(a)(i) x = [1] 1 (ii) = 7 x. 49 Answer(a)(ii) x = [1] (b) Write down the value of 3p0. Answer(b) [1]

3 marks

Mark scheme: 12 (a) (i) (x=) 6 1 (ii) (x=) −2 1 (b) 3 1 1 × 7 + 2 × 5 1 × 7 2 × 5

This question in 0580/12 Oct/Nov 2010

8 (a) Find m when 4m × 42 = 412. Answer(a) m = [1] (b) Find p when 6 p ÷ 67 = 62. Answer(b) p = [1]

2 marks

Mark scheme: 8 (a) 10 1 (b) 9 1

This question in 0580/11 May/June 2011

9 Use your calculator to find the value of (a) 30 × 2.52, Answer(a) [1] (b) 2.5 – 2. Answer(b) [1]

2 marks

Mark scheme: 9 (a) 6.25 cao 1 (b) 0.16 cao 1

This question in 0580/12 Oct/Nov 2011

9 Simplify (a) a0 , Answer(a) [1] (b) b3 × b-5 . Answer(b) [1]

2 marks

Mark scheme: 9 (a) 1 1 (b) b–2 1 accept 21 b

This question in 0580/13 May/June 2012

23 (a) Simplify y 0. Answer(a) … [1] 1 (b) Make v the subject of E = 2 mv 2. Answer(b) v = … [3] _____________________________________________________________________________________

4 marks

Mark scheme: 23 (a) 1 1 2 E E E 2 E or or (b) [v =] 1 3 M2 for v2 = m 5.0 m m m 2 E or M1 for mv2 = 2E or ½ v2 = m IGCSE – May/June 2013 0580 12

This question in 0580/12 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

18 Simplify the following. (a) x5 × x2 Answer(a) … [1] (b) 20y4 ÷ 4y–2 Answer(b) … [2] _____________________________________________________________________________________

3 marks

Mark scheme: 18 (a) x7 1 (b) 5y6 2 B1 for 5ym or ky6 in answer m ≠ 0, k ≠ 0

This question in 0580/12 Oct/Nov 2013

16 (a) Simplify the expressions. (i) p 3 × p 7 Answer(a)(i) … [1] (ii) t 5 ÷ t 8 Answer(a)(ii) … [1] k (b) (h3) = h12 Find the value of k. Answer(b) k = … [1] __________________________________________________________________________________________

3 marks

Mark scheme: 16 (a) (i) p10 1 1 (ii) t−3 or 3 1 t (b) 4 1

This question in 0580/11 May/June 2014

5 Write down the value of 70. Answer … [1] __________________________________________________________________________________________

1 marks

Mark scheme: 5 1 1

This question in 0580/13 May/June 2014

5 Work out the value of 34 ÷ 3–2. Give your answer as an ordinary number. Answer … [2] __________________________________________________________________________________________

2 marks

Mark scheme: 1 5 729 2 B1 for 81 or seen in the working or 0.111 … 9 or B1 for 36 in the working or on the answer line.

This question in 0580/12 Oct/Nov 2014

13 (a) Simplify (i) x0, Answer(a)(i) … [1] (ii) m4 × m3. Answer(a)(ii) … [1] (b) Solve 5x3 = 40 . Answer(b) x = … [1] __________________________________________________________________________________________

3 marks

Mark scheme: 13 (a) (i) 1 1 (ii) m7 1 (b) 2 1 1000

This question in 0580/12 Feb/March 2015

r 6 .3 Simplify 2 r Answer … [1] __________________________________________________________________________________________

1 marks

Mark scheme: 3 r 4 1

This question in 0580/11 Oct/Nov 2015

11 (a) Solve. 3x 2 = 108 Answer(a) x = … [1] (b) w 6 # w k = w 18 Find the value of k. Answer(b) k = … [1]

2 marks

Mark scheme: 11 (a) 6 cao 1 (b) 12 final answer 1 6 

This question in 0580/12 Oct/Nov 2015

12 (a) Write down the value of 170. … [1] (b) Explain why 17 is irrational. … [1]

2 marks

Mark scheme: 12 (a) 1 1 (b) cannot be written as a fraction oe 1 Qu Answer Mark Part marks 5 2 12 4

This question in 0580/12 Feb/March 2016

4 Simplify. n2 × n5 … [1]

1 marks

Mark scheme: 4 n7 final answer 1

This question in 0580/13 Oct/Nov 2016

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

4 Simplify. (x2) 5 … [1]

1 marks

Mark scheme: 4 x10 1

This question in 0580/11 May/June 2017

16 (a) 5x ÷ 25 = 56 Find the value of x. x = … [1] (b) Solve the simultaneous equations. You must show all your working. 2x + 3y = 16 - 2x + 5y = 24 x = … y = … [2]

3 marks

Mark scheme: 16(a) 8 1 16(b) [x = ] 0.5 1 [y = ] 5 1 If zero scored, SC1 for correct substitution and evaluation to find the other variable

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

3 Work out. 2 -4 # 2 5 … [1]

1 marks

Mark scheme: 3 2 1

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

4 Simplify. y 4 # y 5 … [1]

1 marks

Mark scheme: 4 y9 1

This question in 0580/13 Oct/Nov 2017

18 (a) Simplify. 4 (x3) … [1] w 1 (b) 4 = 16 Find the value of w. w = … [1]

2 marks

Mark scheme: 18(a) x12 1 18(b) −2 1

This question in 0580/11 May/June 2018

4 Find the value of p when 5 p ' 5 8 = 5 13 . p = … [1]

1 marks

Mark scheme: 4 21 1

This question in 0580/12 May/June 2018

12 Simplify. (a) ( y 5 ) 3 … [1] (b) w 7 ' w - 2 … [1]

2 marks

Mark scheme: 12(a) y15 cao 1 12(b) w9 cao 1

This question in 0580/12 Feb/March 2019

11 Simplify. (a) t21 ÷ t7 … [1] (b) (u5)5 … [1]

2 marks

Mark scheme: 11(a) t14 final answer 1 11(b) u25 final answer 1

This question in 0580/11 May/June 2019

19 Simplify. (a) 5m 2 # 2m 3 … [2] 3 (b) x 8 ` j … [1]

3 marks

Mark scheme: 19(a) 10m 5 final answer 2 B1 for 10 m k or km 5 as final answer 19(b) x 24 final answer 1

This question in 0580/12 May/June 2019

13 (a) x 3 # x 6 = xm Find the value of m. m = … [1] n (b) x 2 = x 12 ` j Find the value of n. n = … [1]

2 marks

Mark scheme: 13(a) 9 cao 1 13(b) 6 cao 1

This question in 0580/13 Oct/Nov 2019

16 (a) Simplify. 5 # x 0 … [1] (b) 9 12 ' 9 w = 9 4 Find the value of w. w = … [1]

2 marks

Mark scheme: 16(a) 5 1 16(b) 8 1

This question in 0580/12 Feb/March 2020

15 Simplify. 4p 5 q 3 # p 2 q -4 … [2]

2 marks

Mark scheme: 15 4p7q−1 2 4 pb B1 for 4p7qa or 4pbq−1 or q

This question in 0580/11 May/June 2020

20 Simplify 8t 8 ' 4t 4 . … [2]

2 marks

Mark scheme: 20 2t4 2 B1 for 2tn or kt4 (n,k ≠ 0)

This question in 0580/12 May/June 2020

14 Simplify. (a) p 2 # p 4 … [1] (b) m 15 ' m 5 … [1] (c) ( k3 ) 5 … [1]

3 marks

Mark scheme: 14(a) p 6 1 14(b) m10 1 14(c) k 15 1

This question in 0580/13 May/June 2020

18 (a) Simplify. (i) x 12 ' x 3 … [1] 5 (ii) ( y2 ) … [1] p 1 (b) 3 = 81 Find the value of p. p = … [1]

3 marks

Mark scheme: 18(a)(i) 𝑥ଽ cao 1 18(a)(ii) 𝑦ଵ଴ cao 1 18(b) −4 1

This question in 0580/12 Feb/March 2021

117 (a) Write as a power of 2. 2 # 2 # 2 # 2 # 2 … [1] (b) (i) 3 18 ' 3 t = 3 6 Find the value of t. t = … [1] (ii) Simplify. 8w 10 # 6w 5 … [2]

4 marks

Mark scheme: 17(a) 2–5 1 17(b)(i) 12 1 17(b)(ii) 48w15 final answer 2 B1 for answer 48wk or kw15 (k ≠ 0)

This question in 0580/11 May/June 2021

14 Find the value of x when 7 x ' 7 4 = 7 9 . x = … [1]

1 marks

Mark scheme: 14 13 cao 1

This question in 0580/12 May/June 2021

16 (a) Solve. 7x + 18 = 4 x = … [2] y 6 18 (b) 7 # 7 = 7 Find the value of y. y = … [1]

3 marks

Mark scheme: 16(a) −2 2 18 4 M1 for 7x = 4 − 18 or x + = 7 7 16(b) 12 cao 1

This question in 0580/12 Oct/Nov 2021

15 Write 0.0001 as a power of 10. … [1]

1 marks

Mark scheme: 15 10–4 1

This question in 0580/12 Feb/March 2022

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

17 9 x # 9 2 = 9 12 Find the value of x. x = … [1]

1 marks

Mark scheme: 17 10 cao 1

This question in 0580/12 May/June 2022

12 Simplify. (a) y 3 ' y 5 … [1] (b) 7x 0 … [1]

2 marks

Mark scheme: 12(a) 1 1 y−2 or final answer y 2 12(b) 7 1

This question in 0580/13 May/June 2022

20 (a) 5 8 ' 5 x = 5 2 Find the value of x. x = … [1] 3 (b) Simplify ( x5 ) . … [1]

2 marks

Mark scheme: 20(a) 6 1 20(b) x15 1

This question in 0580/12 Oct/Nov 2022

15 Simplify d 8 ' d 2 . … [1]

1 marks

Mark scheme: 15 d 6 1

This question in 0580/13 May/June 2023

21 5 7 ' 5 x = 5 3 Find the value of x. x = … [1]

1 marks

Mark scheme: 21 4 cao 1

This question in 0580/12 May/June 2024

4 Find the value of (a) 103 … [1] (b) 3 27 … [1] (c) 70. … [1]

3 marks

Mark scheme: 4(a) 1000 1 4(b) 3 1 4(c) 1 1

This question in 0580/12 Feb/March 2025

3 Write down the value of (a) 36 … [1] (b) 103. … [1]

2 marks

Mark scheme: 3(a) 6 1 3(b) 1000 cao 1

This question in 0580/12 May/June 2025

3 (a) Write down 169 . … [1] (b) Work out the value of 24. … [1] (c) Work out the value of 10 -2 . … [1] (d) Work out -18 ' -4 . … [1] (e) Work out 1.6 # 0. 02 . … [1]

5 marks

Mark scheme: 3(a) 13 1 3(b) 16 1 3(c) 1 1 or 0.01 100 3(d) 9 1 1 or 4.5 or 4 2 2 3(e) 0.032 1

This question in 0580/13 May/June 2025

17 Simplify. (a) y 4 # y 6 … [1] p 5 (b) p 8 … [1] 3 (c) `w4j … [1]

3 marks

Mark scheme: 17(a) y10 1 17(b) 1 1 p–3 or p 3 17(c) w12 1

This question in 0580/13 May/June 2025

23 (a) a 8 # a x = a 15 Find the value of x. x = … [1] (b) ( b 7 ) y = b 21 Find the value of y. y = … [1]

2 marks

Mark scheme: 23(a) 7 cao 1 23(b) 3 cao 1

This question in 0580/11 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