Indices II — Paper 1 · IGCSE Mathematics 0580

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

14 (a) Find the value of p when p3 = −27. For Examiner's Use Answer(a) p = [1] −1 1 (b) Find the value of q when q = . 6 Answer(b) q = [1] (c) Simplify 8s 2 ÷ 2s −1. Answer(c) [2]

4 marks

Mark scheme: 14 (a) –3 final answer 1 (b) 6 final answer 1 4 (c) 4s3 or − 3 final answer 2 W1 for 4sn (n ≠ 0) or ks3 (k ≠ 0) seen s

This question in 0580/11 Oct/Nov 2009

14 (a) Find the value of p when p3 = −27. For Examiner's Use Answer(a) p = [1] −1 1 (b) Find the value of q when q = . 6 Answer(b) q = [1] (c) Simplify 8s 2 ÷ 2s −1. Answer(c) [2]

4 marks

Mark scheme: 14 (a) –3 final answer 1 (b) 6 final answer 1 4 (c) 4s3 or − 3 final answer 2 W1 for 4sn (n ≠ 0) or ks3 (k ≠ 0) seen s 8d

This question in 0580/12 Oct/Nov 2009

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

18 (a) Which two of these have the same value? For Examiner's Use 2 1 2 2 2 5–2 0.22 5 c 2 m c 5 m Answer(a) … and … [2] (b) Simplify. (i) a6 × a3 Answer(b)(i) … [1] (ii) 24b16 ÷ 6b4 Answer(b)(ii) … [2] _____________________________________________________________________________________

5 marks

Mark scheme: 18 (a) 5–2 and 0.22 2 M1 for any two correct decimal values seen with the correct expression e.g. 0.04, 0.4, 0.25, 0.16, 0.04 (b) (i) a9 1 (ii) 4b12 2 B1 for 4bk or B1 for kb12 where k is an integer ( k ≠ 0 )

This question in 0580/11 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

11 Simplify. x 3 y 4 # x 5 y 3 … [2]

2 marks

Mark scheme: 11 x8y7 final answer 2 B1 for answer x 8 y k or kx y 7 (k ≠ 0)

This question in 0580/12 Feb/March 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

22 Solve. (a) 4x = 10 x = … [1] (b) 5(x + 8) = 75 x = … [2] (c) 37 ÷ 3x = 9 x = … [1]

4 marks

Mark scheme: 1 22 (a) 2.5 or 2 1 2 (b) 7 2 M1 for 5 x + 40 = [ 75] or x + 8 = 75 ÷ 5 or better (c) 5 1

This question in 0580/12 Feb/March 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

a 5 .3 Simplify 2 a … [1]

1 marks

Mark scheme: 3 a 3 cao 1

This question in 0580/11 Oct/Nov 2018

17 Simplify 2x 3 # 3x 2 . … [2]

2 marks

Mark scheme: 17 6x5 final answer 2 B1 for kx5 or 6xk

This question in 0580/11 Oct/Nov 2019

23 Simplify 3x 3 # 4x 4 . … [2] Question 24 is printed on the next page.

2 marks

Mark scheme: 23 12x 7 final answer 2 B1 for 12xj or kx7 (j,k ≠ 0) as final answer

This question in 0580/13 May/June 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

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