Indices II — Paper 4 · IGCSE Mathematics 0580

2 (a) Solve. x = 49 7 x = … [1] (b) Simplify. (i) x0 … [1] (ii) x 7 # x 3 … [1] 6 2 3x (iii) ^ -4h x … [2] (c) (i) Factorise completely. 2x 2 - 18 … [2] (ii) Simplify. 2x 2 - 18 x 2 + 7x - 30 … [3]

10 marks

Mark scheme: 2(a) 343 1 2(b)(i) 1 1 2(b)(ii) x10 final answer 1 2(b)(iii) 9x16 final answer 2 B1 for x12 or x16 or (3x8)2 seen 2(c)(i) 2(x – 3)(x + 3) final answer 2 M1 for (2x + 6)(x – 3) or (2x – 6)(x + 3) or (x – 3)(x + 3) 2(c)(ii) 2( x + 3) 2 x + 6 3 M2 for (x + 10)(x – 3) or or x + 10 x + 10 M1 for (x + a)(x + b) where ab = –30 final answer nfww or a + b = 7

This question in 0580/43 Oct/Nov 2017

4 (a) Simplify. (i) (3p2)5 … [2] (ii) 18x2y6 ' 2xy2 … [2] 5 -2 (iii) c m m … [1] (b) In this part, all measurements are in metres. 5x – 9 NOT TO w SCALE 3x + 7 The diagram shows a rectangle. The area of the rectangle is 310 m2. Work out the value of w. w = … [4]

9 marks

Mark scheme: 4(a)(i) 243p10 final answer 2 B1 for answer 243pk or kp10 (k ≠ 0) 4(a)(ii) 9xy4 final answer 2 B1 for answer with two correct elements in correct form of expression 4(a)(iii) m 2 1 final answer 25 4(b) 10 4 B2 for x = 8 or for [length of rectangle =] 31 or M1 for 5x – 9 = 3x + 7 oe or better 310 M1 for (3 × theirx + 7) 310 or (5 × theirx − 9) Alt method using simultaneous eqns M1 for 5xw – 9w = 310 and 3xw + 7w = 310 M1 for equating coefficients of xw M1 for subtraction to eliminate term in xw

This question in 0580/42 May/June 2018

2 (a) Solve. 5x - 17 = 7x + 3 x = … [2] (b) Find the integer values of n that satisfy this inequality. - 7 1 4n G 8 … [3] (c) Simplify. (i) a 3 # a 6 … [1] (ii) (5xy 2 ) 3 … [2] 1 12 - 3 27x (iii) 3 f 64y p … [3]

11 marks

Mark scheme: 2(a) –10 2 M1 for –17 – 3 = 7x – 5x oe or better 2(b) −1, 0, 1, 2 final answer 3 B2 for 3 correct values and no incorrect values or 4 correct values and one incorrect value 7 or M2 for − < n - 2 oe 4 7 or M1 for − < n - k or k < n- 2 oe 4 2(c)(i) a9 1 2(c)(ii) 125x3y6 final answer 2 B1 for 2 correct elements if in form kxnym 2(c)(iii) []1 3 4 [ −1] 4 y  3 x  4 final answer B2 for  [1]  oe seen 3 x  4 y  OR B1 for 3x4 or 4y[1] and 13 3  64 y  M1 for  12  oe  27 x  64 y [1] 0.333 x − 4 If 0 scored, SC1 for or seen 27 x 4 0.25 y − 1

This question in 0580/43 May/June 2019

5 (a) Write as a single fraction in its simplest form. x + 3 x - 2 - x - 3 x + 2 … [4] 12 2k (b) 2 ' 2 = 32 Find the value of k. k = … [2] (c) Expand and simplify. ( y + 3 )( y - 4 )( 2y - 1) … [3] (d) Make x the subject of the formula. 3 + x x = y x = … [3]

12 marks

Mark scheme: 5(a) 10 x 10 x 4 M1 for common denominator( x − 3 )( x + 2 ) or ( x − 3 )( x + 2 ) x 2 − x − 6 isw final answer M1 for ( x + 3 )( x + 2 ) − ( x − 2 )( x − 3 ) isw B1 for correct numerator in terms of x only 5(b) 14 2 k 12 k 2 2 M1 for 12 = 5 or 2 = oe 5 −2 2 4096 142 or or 12 – 5 or 212 ÷ 2 [= 32] seen 32 5(c) 2 y 3 − 3 y 2 − 23 y + 12 final answer 3 B2 for correct unsimplified expanded expression or for simplified four-term expression of correct form with 3 terms correct or B1 for correct expansion of 2 of the brackets with at least 3 terms correct 5(d) 3 3 M1 for xy = 3 + x [ x = ] final answer y − 1 x 3 M1 for xy – x = 3 or x – = y y M1 for factorising and dividing

This question in 0580/42 Feb/March 2020

3 (a) s = ut + 12 at 2 Find the value of s when u = 5.2 , t = 7 and a = 1.6 . s = … [2] (b) Simplify. (i) 3a - 5b - a + 2b … [2] 5 9x (ii) # 3x 20 … [2] (c) Solve. 15 (i) =- 3 x x = … [1] (ii) 4 ( 5 - 3)x = 23 x = … [3] (d) Simplify. 2 ( 27x 9) 3 … [2] (e) Expand and simplify. (3x - 5y)(2x + y) … [2]

14 marks

Mark scheme: 3(a) 75.6 2 1 M1 for 5.2 × 7 + × 1.6 × 72 2 3(b)(i) 2a – 3b final answer 2 B1 for answer 2a + kb or ka – 3b or for 2a – 3b seen in working 3(b)(ii) 3 2 45 x B1 for oe single fraction 4 60 x 3(c)(i) −5 1 3(c)(ii) 1 3 23 −0.25 or – M1 for 20 – 12x = 23 or for 5 – 3x = 4 4 M1 for correct completion to ax = b FT their first step 3(d) 9x6 2 B1 for 9xk or kx6 3(e) 6x2 – 7xy – 5y2 2 M1 for 3 terms out of 4 from 6x2 – 10xy + 3xy – 5y2

This question in 0580/41 May/June 2020

8 (a) Factorise completely. 3a 2 b - ab 2 … [2] (b) Solve the inequality. 3x + 12 1 5x - 3 … [2] (c) Simplify. 3 3x 2 y 4 ` j … [2] (d) Solve. 2 6 = x 2 - x x = … [3] (e) Expand and simplify. ( x - 2)( x + 5)( 2x - 1) … [3] (f) Alan invests $200 at a rate of r% per year compound interest. After 2 years the value of his investment is $206.46 . (i) Show that r 2 + 200r - 323 = 0 . [3] (ii) Solve the equation r 2 + 200r - 323 = 0 to find the rate of interest. Show all your working and give your answer correct to 2 decimal places. r = … [3]

18 marks

Mark scheme: 8(a) ab(3a – b) final answer 2 B1 for a(3ab – b2) or b(3a2 – ab) or ab(3a – b) seen 8(b) x > 7.5 final answer 2 B1 for 12+3 < 5x – 3x oe 8(c) 27x6y12 2 B1 for two of 27, x6 and y12 correct 8(d) 1 3 M2 for 4 = 6x + 2x or better 0.5 or 2 or M1 for 2(2 – x) = 6x oe 8(e) 2x3 + 5x2 – 23x + 10 final answer 3 B2 for correct expansion of three brackets unsimplified B1 for correct expansion of two brackets with at least 3 terms correct 8(f)(i) 2 M1  r  200  1 +  = 206.46 oe  100  2r r 2 M1 1 + + oe 100 100 2 r2 + 200r – 323 = 0 A1 Correct solution reached with no errors or omissions seen If 0 scored, SC1 for 200( n ) 2 = 206.46 8(f)(ii) 2 B2 2 −200 + 200 − 4(1)( −323) B1 for 200 − 4(1)( − 323) or (r + 100)2 2 × 1 −200 + q 2 B1 for or r = 323 + 100 – 100 2 × 1 OR  206.46  B2 for 100  − 1     200  206.46 or B1 for 200 1.60 cao final answer B1

This question in 0580/41 Oct/Nov 2020

11 f ( x) = 2 x - 1 g ( x) = x 2 + 2x h ( x) = 4x j ( x) = 2x (a) Find the value of (i) h(3), … [1] (ii) fh(3). … [1] (b) Solve the equation gf ( x) = 0 . x = … or x = … [4] (c) p -1 ( x) = f ( x) Find p(x). … [2] 1(d) h ( x) j ( x) = 2 Find the value of x. x = … [3]

11 marks

Mark scheme: 11(a)(i) 64 1 11(a)(ii) 127 1 FT 2 × their (a)(i) – 1 4 11(b) 1 M1 for ( 2 x − 1) 2 + 2(2 x − 1) ± oe nfww 2 2 B1 for 4 x − 2 x − 2 x + 1 or ( 2 x − 1)( 2 x −+1 2 ) B1 for 4 x 2 − 1 [= 0] or ( 2 x − 1)( 2 x + 1) [= 0] OR M1 for x(x + 2) = 0 (solving g(x) = 0) A1 for x = 0 or –2 B1 for 2x – 1 = 0 or 2x – 1 = –2 11(c) x + 1 2 M1 for oe final answer y 1 2 y + 1 = 2 x or = x − or x = 2 y − 1 2 2 11(d) 1 3 1 − oe nfww B2 for 3 x = − oe 6 2 OR 1 x M1 for 2 2 x × 2 x oe or 4 2 × 4 x oe or 8x oe 1 1 1 − − − M1 for 2 2 or 4 4 or 8 6 soi

This question in 0580/43 Oct/Nov 2021

6 (a) Solve. 4x + 15 = 9 x = … [2] (b) Factorise. a 2 - 9 … [1] (c) Write as a single fraction in its simplest form. 4a 3ad ' 5 10c … [3] (d) 5 n + 5 n + 5 n + 5 n + 5 n = 5 m Find an expression for m in terms of n. m = … [2] (e) Solve by factorisation. 4x 2 + 8x - 5 = 0 x = … or x = … [3] (f) (i) y is directly proportional to ( x + 3) 3 . When x = 2 , y = 13.5 . Find x when y = 108 . x = … [3] (ii) g is inversely proportional to the square of d. When d is halved, the value of g is multiplied by a factor n. Find n. n = … [2] (g) Expand and simplify. ( 2x + 3)( x - 1)( x + 3) … [3] dy 2(h) Find the derivative, , of y = 3x + 4x - 1. dx … [2]

21 marks

Mark scheme: 6(a) 1 3 2 15 9 –1.5 or –1 or – M1 for 4x = 9 – 15 or x + = 2 2 4 4 6(b) (a – 3)(a + 3) final answer 1 6(c) 8c 3 8 ac 40 c final answer B2 for or 3d 3ad 15 d 4 2 or c seen 1 3d or for correct answer seen then spoiled 4 a 10c 8 ac 3ad or M1 for  or  oe 5 3ad 10c 10c 6(d) n + 1 final answer 2 M1 for 5  5 n or 5n+1 seen 6(e) (2x – 1)(2x + 5) [= 0] oe B2 M1 for 2x(2x + 5) – [1](2x + 5) [ = 0] or 2x(2x – 1) + 5(2x – 1) [ = 0] or for (2x + m)(2x + n) [ = 0] with and mn = –5 or n + m = 4 1 1 5 B1 or 0.5 and –2.5 or –2 or – 2 2 2 6(f)(i) 7 3 M1 for y = k(x + 3)3 or better M1 for 108 = their k(x + 3)3 6(f)(ii) 4 2 2  1  M1 for   oe  2  k or oe seen or better 1 2 d 4 6(g) 2x3 + 7x2 – 9 final answer 3 B2 for correct expansion unsimplified or for simplified 4 term expression of correct form with 3 terms correct or B1 for one pair of brackets expanded with at least 3 terms out of 4 correct 6(h) 6x + 4 2 B1 for 6x or 4 or 6x + 4 with one extra term seen

This question in 0580/42 Oct/Nov 2022

2 (a) Simplify fully. (i) p 3 # p 11 … [1] 18 m 6 (ii) 2 3m … [2] 1 27x 9 y 27 - 3 (iii) e 64 o … [3] (b) A sequence has nth term 3n 2. Write down the first 3 terms of this sequence. … , … , … [2] (c) Find the nth term for each of these sequences. (i) 13, 16, 19, 22, 25, … … [2] (ii) 3, 17, 55, 129, 251, … … [2] (d) Solve. 3x - 22 = 23 4 x = … [3] (e) Use the quadratic formula to solve 3x 2 + 8x - 20 = 0 . Show all your working and give your answers correct to 2 decimal places. x = … , x = … [4]

19 marks

Mark scheme: 2(a)(i) p14 final answer 1 2(a)(ii) 6m4 final answer 2 B1 for 6mk or km4 in final answer or correct answer seen and spoilt 2(a)(iii) 4 4 x −3 y −9 3 B2 for correct answer seen and spoilt or final answer or 2 correct elements in final answer 3 9 3x y 3 4 3 or B1 for one of or oe or x3 or y9 seen 3 4 2(b) 3, 12, 27 2 B1 for 12 or 27 2(c)(i) 3n + 10 oe final answer 2 B1 for 3n + k oe or jn + 10 oe (j ≠ 0) or for correct expression shown in working and then spoilt 2(c)(ii) 2n3 + 1 oe final answer 2 B1 for 3rd diff = 12 (both needed) or for cubic answer or for correct expression shown in working and then spoilt 2(d) 38 3 M2 for 3x = 4 × 23 + 22 or M1 for 3x – 22 = 4 × 23 3 x 22 or for = 23 + oe 4 4 2(e) 2 B2 2 −8 8 − 4(3)( −20) B1 for 8 − 4(3)( −20) oe 2  3 −+8 q −−8 q 2 or oe or oe or both −8 8 ( −20) 2  3 2  3 or  − 2  3 4  32 3 or better – 4.24, 1.57 final answers B2 B1 for each If B0, SC1 for answers – 4.2 or –4.23 or –4.240 to – 4.239 and 1.6 or 1.572 to 1.573 or – 4.24 and 1.57 seen in working or for –1.57 and 4.24 as final answer

This question in 0580/43 Oct/Nov 2022

9 (a) Simplify. (i) ( 3x 2 y 4 ) 3 … [2] 3 16 - 2 (ii) 16 8 e x y o … [3] (b) (i) Factorise. x 2 - 9 … [1] (ii) Simplify. x 2 - 9 2 xy - 6 y + 5 x - 15 … [3] (c) Solve the simultaneous equations. You must show all your working and give your answers correct to 2 decimal places. 2x + y = 7 y = 5x 2 + 2x - 13 x = … , y = … x = … , y = … [6]

15 marks

Mark scheme: 9(a)(i) 27x6y12 final answer 2 B1 for two terms correct in answer e.g. 27x6yk or 27xky12 or kx6y12 or for correct answer seen then spoilt 9(a)(ii) x 24 y12 3 B2 for final answer with two correct final answer elements 64 64 641 or final answer or or x 24 y12 x 24 y  12 better or for correct answer seen or B1 for 64 or x24 or y12 seen in final answer k or final answer x 24 y 12 or M1 for first correct step seen 3  3  x16 y 8   2  4  eg   or  8 4  or  16   x y   1  2   4096     48 24   x y  9(b)(i) (x + 3)(x – 3) final answer 1 9(b)(ii) x  3 3 M2 for (x – 3)(2y + 5) final answer or M1 for 2y(x – 3) + 5(x – 3) 2 y  5 or x (2y + 5) – 3( 2y + 5) 9(c) 5x2 + 4x – 20 [= 0] oe M2 M1 for 7 – 2x = 5x2 + 2x – 13 oe seen or 2  7  y   7  y  5y2 – 78y + 221 [= 0] oe or y  5    2    13 oe seen  2   2  2 M2 FT their 3-term quadratic 4 4  4(5)( 20) oe 2(5) 2 or M1 for (4)  4(5)( 20) or better or 2 4 q 4 q 4  4  or for or   4   oe 2  5 2  5 10  10  2  4  or for x  oe    10  x = 1.64 y = 3.72 B2 B1 for one correct pair or both x-values and correct or both y – values correct x = – 2.44 y = 11.88

This question in 0580/42 May/June 2023

5 (a) Simplify 25x 6 2. … [2] (b) These are the first five terms of a sequence. 1 1 6 36 216 6 Find the nth term of the sequence. … [2] (c) Expand and simplify. ( x + 4)( x - 3)( 3x - 1) … [3] 7 2(d) (i) Show that ( 3x + 5) + = x simplifies to 2x + x - 3 = 0 . x - 2 [4] (ii) Solve by factorisation 2x 2 + x - 3 = 0 . x = … or x = … [3] (e) A solid cylinder has base radius x and height 3x. The total surface area of the cylinder is the same as the total surface area of a solid hemisphere of radius 5y. 2 75y 2 Show that x = . 8 [The surface area, A, of a sphere with radius r is A = 4rr 2 .] [4]

18 marks

Mark scheme: 5(a) 125x 9 final answer 2 B1 for answer 125 kx or m x 9 or for correct answer seen then spoilt 5(b) 6 n  2 oe final answer 2 B1 for answer of form 6k oe  k  1  or answer of the form   oe  6  or for correct answer seen 5(c) 3x3 + 2x2 –37x + 12 final answer 3 B2 for correct expansion of three brackets unsimplified or for simplified four-term expression of correct form with 3 terms correct or B1 for correct expansion of two brackets with at least 3 terms out of 4 correct 5(d)(i) eliminates the fraction correctly M1 eg (3x + 5) (x – 2) + 7 = x (x – 2) 3x2 + 5x –6x – 10 + 7 = x2 – 2x oe B2 B1 for 3x2 + 5x –6x –10 [ + 7] oe seen with at least 3 terms correct leading to 2x2 + x – 3 = 0 A1 dep on M1 B2 with no errors or omissions 5(d)(ii)  2 x  3  x  1 M2 or M1 for (2x + a)(x + b) where ab = −3 or 2b + a = [+]1 or for partial factors 2x(x – 1) + 3(x – 1) or x(2x + 3) –[1](2x + 3) −1.5 oe and +1 B1 5(e) 2 M1 [TSA cylinder =] 2x  2x  3 x 2 4(5 y ) 2 M1 [TSA hemisphere=] (5 y )  2 Leading to M1 dep M1M1 2x 2  6x 2  50y 2  25y 2 oe 2 75 y 2 A1 dep on M1M1M1 x  8

This question in 0580/42 May/June 2024

21 Simplify. (a) 3t 5 # 5t 3 … [2] 5 (b) `64u 36j6 … [2]

4 marks

Mark scheme: 21(a) 15t8 final answer 2 B1 for answer kt8 or 15tk (k > 0) or correct answer seen 21(b) 32u30 final answer 2 B1 for answer ku30 or 32uk (k > 0) or correct answer seen

This question in 0580/42 May/June 2025