Proportion — Paper 4 · IGCSE Mathematics - International 0607

6 y varies inversely as the square of x. y = 32 when x = 2. (a) Find the value of y when x = 4. y = … [3] (b) Find the value of x when y = 512. x = … [2] (c) Find x in terms of y. x = … [3]

8 marks

Mark scheme: 6(a) 8 3 k M1 for y = oe x 2 A1 for k = 128 OR  4  2 M2 for 32 ÷  oe  2  1 y 4 2 or M1 for = oe 32 1 2 2 6(b) 1 2 their k 32 [±] oe M1 for x2 = oe or 2 × oe 2 512 512 6(c) 128 3 M1 for multiplication by x2 [x = ±] oe final answer M1 for division by y or for square root y

This question in 0607/41 May/June 2017

6 (a) (i) x is proportional to v. Write down an expression for x in terms of v and a constant c. x = … [1] (ii) y is proportional to v2. Write down an expression for y in terms of v and a constant k. y = … [1] (iii) d = x + y Write down an expression for d in terms of v, c and k. d = … [1] (b) The table shows two values of v and the corresponding values of d. v d 12 750 20 2050 Using your answer to part (a)(iii), (i) show that 125 = 2c + 24k, [1] (ii) write down a second equation connecting c and k. … [1] (c) Solve the simultaneous equations in part (b) to find the value of c and the value of k. c = … k = … [3] (d) Find the value of d when v = 40. d = … [2]

10 marks

Mark scheme: 6(a)(i) [x =] cv oe 1 6(a)(ii) [y =] kv2 oe 1 6(a)(iii) [d =] cv + kv2 or v(c + kv) oe 1 FT 6(b)(i) 750 = 12c + 122 k oe M1 isw any cancelling 6(b)(ii) 2050 = 20c + 202 k oe 1 isw any cancelling 6(c) [c =] 2.5 oe cao 3 M1 for correctly eliminating one variable from [k =] 5 cao their equations in this part. or sketches of lines A1 for either solution If zero scored SC1 for their values satisfying one equation. 6(d) 8100 2 M1 for correct substitution of 40 into their (a)(iii) containing their values of c and k.

This question in 0607/42 May/June 2017

3 (a) y varies directly as the square root of x. y = 32 when x = 16. (i) Find y in terms of x. y = … [2] (ii) Find the value of y when x = 4. y = … [1] (iii) Find x in terms of y. x = … [2] (b) p varies inversely as q + 2 . p = 3 when q = 2. Find the value of p when q = 4. p = … [3]

8 marks

Mark scheme: 3(a)(i) 8 x oe 2 M1 for y = k x 3(a)(ii) 16 1 3(a)(iii) 2 2 2 2 FT only if wrong k, k numeric and k ≠ 1 y  y  y 2 or   or 2 y 2 64  8  8 M1 for = x or y = their k x k ≠ 1 or ( ) their k better y 2  y  2 SC1 for answer or   2 k  k  3(b) 2 3 12 12 M2 for p = oe or p = oe q + 2 4 + 2 k or M1 for p = q + 2 OR 3 ( 2 + 2 ) M2 for p = 4 + 2 or M1 for p ( 4 + 2 ) = 3 ( 2 + 2 )

This question in 0607/42 May/June 2018

4 (a) y varies directly as the square of (x + 2). When x = 3, y = 100. (i) Find an equation connecting x and y. … [2] (ii) Find the value of y when x = 18. … [1] (iii) Find the values of x when y = 25. … [2] (b) z varies inversely as w. When w = A, z = 18. A Find the value of z when w = . 9 … [2]

7 marks

Mark scheme: 4(a)(i) y = 4(x + 2)2 2 B1 for y = k(x + 2)2 4(a)(ii) 1600 1 FT (their k) × 202 dep on k(x + 2)2 4(a)(iii) 1 9 2 9 oe , − oe B1 for 0.5 or − oe 2 2 2 or M1 for 25 = (their k)(x + 2)2 4(b) 54 2 B1 for 3 soi by answer 6

This question in 0607/42 Oct/Nov 2018

6 y is inversely proportional to x. When x = 9 , y = 6 . (a) (i) Find an equation connecting x and y. … [2] (ii) Calculate y when x = 30 . … [1] (iii) Calculate x when y = 15 . … [2] (b) For the three variables x, y and z, z is also proportional to (y + 5) . When x = 9 , z = 33 . Find an equation connecting x and z. … [2]

7 marks

Mark scheme: 6(a)(i) 18 2 k y = oe M1 for y = oe x x 6(a)(ii) 3.29 or 3.286... 1 FT wrong k only 6(a)(iii) 1.44 oe 2 their18 (their18) 2 M1 for x = or 225 = 15 x 6(b)  18  2 M1 for z = K ( their ( a ( i ) ) + 5 ) K≠1 z = 3 + 5 oe    x  or for z = 3(y + 5)

This question in 0607/41 May/June 2019

12 (a) y varies directly as the square root of (x + 1). y = 8 when x = 24. (i) Find the value of y when x = 15. y = … [3] (ii) Find the value of x when y = 16. x = … [2] (b) Find the next term in each of the following sequences. (i) 18, 13, 8, 3, –2, … … [1] (ii) 3, 6, 11, 18, 27, … … [1] (iii) –1000, 100, –10, 1, … … [1] (iv) 0, 0, 0, 6, 24, 60, … … [2]

10 marks

Mark scheme: 12(a)(i) 6.4 3 M2 for y =1.6 x +1 or M1 for y = k x + 1 OR 8 16 M2 for y = 25 8 y or M1 for = 25 16 12(a)(ii) 99 2 16 FT M1 for x + 1 = oe their 1.6 only FT x + 1 12(b)(i) –7 1 12(b)(ii) 38 1 12(b)(iii) –0.1 oe 1 12(b)(iv) 120 2 B1 for row of 0 6 12 18 reached or M1 for (n − 2)3 − (n − 2) or ( n − 1)( n − 2)( n − 3) oe

This question in 0607/43 May/June 2019

6 (a) y is inversely proportional to the square of x. (i) When x = 2, y = 8. Find y in terms of x. y = … [2] (ii) Find the value of y when x = 4. y = … [1] (iii) Find the value of x when y = 128. x = … [2] (b) r is directly proportional to the cube of (p + 1) . When p = 1, r = 16. Find the value of r when p = 4. r = … [3]

8 marks

Mark scheme: 6(a)(i) 32 2 k M1 for y = x 2 x 2 6(a)(ii) 2 1 k FT their k dependent on x 2 6(a)(iii) 1 2 2 their 32 1 [±] M1 for x = soi by oe 2 128 4 6(b) 250 3 B2 for r = 2(p + 1)3 or M1 for r = k(p + 1)3 oe OR r 16 M2 for = oe (4 + 1) 3 (1 + 1) 3

This question in 0607/42 Feb/March 2021

8 (a) y is inversely proportional to the square root of x. When x = 25, y = 0.05 . 1 (i) Show that y = . 4 x [2] (ii) Find y when x = 9. … [1] (iii) Find x in terms of y. x = … [2] 1 (iv) Find x when y = . 2 … [1] (b) b is inversely proportional to a3. When a = P, b = 24. Find b when a = 2P. … [2]

8 marks

Mark scheme: 8(a)(i) k M1 0.05 = oe 25 1 A1 k = 0.25 and y = 4 x 8(a)(ii) 1 1 [ ± ] oe 12 8(a)(iii) 1 1 2 M1 for 4 y x = 1 or better or oe 16y 2 ( 4y ) 2 2 1 or for y = 16 x 8(a)(iv) 1 1 oe cao 4 8(b) 3 2 B1 for 23 soi

This question in 0607/42 May/June 2021

7 y varies inversely as the square of x. y = 5 when x = 3 . (a) (i) Find y in terms of x. y = … [2] (ii) Find the value of x when y = 20 . x = … [2] (b) z varies directly as the square root of y. z = 12 when y = 9 . Use your answer to part (a)(i) to find z in terms of x. z = … [3]

7 marks

Mark scheme: 7(a)(i) 45 2 k [ y = ] M1 for y = oe x 2 x 2 7(a)(ii) [ ± ]1.5 oe 2 2 their 45 M1 for x = or better 20 7(b) 45 3 B2 for z = 4 y oe [ z = ]4 oe x 2 or M1 for z = k y

This question in 0607/43 Oct/Nov 2021

5 (a) (i) Expand and simplify 2x + 3 . … [2] 2 2 (ii) The equation 4x + 12x + 5 = 0 can be written as 2x + 3 = k . ` j Find the value of k. k = … [1] (iii) Use your answer to part(ii) to solve the equation 4x 2 + 12x + 5 = 0 . x = … or x = … [2] (b) x varies inversely as the square root of (w – 1). When w = 10, x = 2. (i) Find x in terms of w. x = … [2] (ii) Find x when w = 3.25 . x = … [1] (iii) Find w in terms of x. w = … [3]

11 marks

Mark scheme: 5(a)(i) 4 x 2 + 12 x + 9 final answer 2 B1 for three of 4x2, 6x, 6x, 9 or for correct answer seen 5(a)(ii) 4 1 FT their 9 – 5 5(a)(iii) 2 x + 3 = ± their 4 M1 their 4 > 0 1 1 B1 −2 , − oe 2 2 5(b)(i) 6 2 k final answer M1 for [x = ] oe w − 1 w − 1 5(b)(ii) 4 1 FT only incorrect k 5(b)(iii) 2 2 3 M1 for correct multiplication of term in w 36 36 + x  6 + + 1 or or   1 M1 for correct squaring 2 2 x x  x  M1 for correctly isolating w final answer Max M2 if incorrect answer

This question in 0607/42 Feb/March 2022

6 y varies inversely as ( 2x - 1) 2 . y = 4 when x = 3 . (a) Find the value of y when x = 2.5 . y = … [3] (b) Find the values of x when y = 16 . x = … or x = … [4]

7 marks

Mark scheme: 6(a) 6.25 oe 3 k M1 for y = (2 x − 1) 2 A1 for k = 100 OR 4  (2 −3 1) 2 M2 for y = (2  2.5 − 1) 2 or M1 for 4  (2 −3 1) 2 = y  (2  2.5 − 1) 2 6(b) 1.75, –0.75 oe 4 B3 for 2 x −=1 2.5 or (4x + 3)(4x – 7) [=0] oe or correct formula their ( k ) or M2 for (2 x − 1) = 16 or 16x2 – 16x – 21 [=0] oe 2 their k or M1 for (2 x − 1) = 16 2 or 16 ( 2 x − 1) = theirk Graphical method. M3 for graph(s) indicating both answers or M2 for graph(s) which could lead to both answers 100 e.g y = and y = 16. (2 x − 1) 2 100 or M1 for appropriate graph e.g y = (2 x − 1) 2

This question in 0607/41 Oct/Nov 2022

10 (a) A machine lays a pipe of length 2.5 km in 18 hours. The machine always works at the same rate. Calculate the time it takes to lay a pipe of length 4 km. … hours [2] (b) t varies inversely as the square root of x. x varies directly as the square of y. When x = 4, t = 3 . When y = 4, x = 81. ty = h Find the value of h. h = … [5]

7 marks

Mark scheme: 10(a) 28.8 2 4 M1 for  18 oe 2.5 10(b) 8 5 6 16 x oe B4 for ty =  oe 3 x 81 2 2 36 16 x or t y =  oe x 81 6 81 2 B3 for t = oe and x = y oe x 16 6 81 2 or B2 for t = oe or x = y x 16 oe k or M1 for t = oe or x = ky2 oe x

This question in 0607/43 Oct/Nov 2022

7 y varies inversely as the cube root of x. y = 10 when x = 8. (a) Find y in terms of x. y = … [3] (b) Find the value of x when y = 8. x = … [2] (c) w varies as the square of y. w = 18 when y = 3. Find w in terms of x. Give your answer in the form w = pxq , where p and q are constants. w = … [4]

9 marks

Mark scheme: 7(a) 20 3 k y = M1 for y = 3 x 3 x B1 for k = 10  3 8 7(b) 15.625 oe 2 3 their 20 M1 for x = 8 or M1 for 10  3 8 = 8  3 x oe 7(c) − 2 4 B2 for w = 2 y 2 w = 800 x 3 cao or B1 for w = ky 2 and   20   2 M1 for w = their k  their  3     x  

This question in 0607/42 Feb/March 2023

3 y is inversely proportional to the square of ( x + 1) . (a) When x = 5 , y = 1. Find y in terms of x. y = … [2] (b) Find y when x = 3 . y = … [2] (c) Find the value of x when y = ( x + 1) . x = … [3]

7 marks

Mark scheme: 3(a) 36 2 k  y =  2 M1 for 2 ( x + 1) ( x + 1) 3(b) 2.25 oe 2 their 36 M1 for or better ( 3 + 1) 2 3(c) 2.3[0…] 3 M2 for x + 1 = 3 their 36 their 36 or M1 for x + 1 = ( x + 1) 2 M1 for sketch of a cubic crossing x axis once with a positive x intercept

This question in 0607/42 Oct/Nov 2023

10 y varies inversely as the square root of ( x + 1) . y = 18 when x = 3 . (a) (i) Find the value of y when x = 8 . y = … [3] (ii) Find the value of x when y = 1.5 . x = … [2] (b) w varies directly as the square root of ( x + 1) . w = 18 when x = 3 . Find the value of wy. wy = … [3]

8 marks

Mark scheme: 10(a)(i) 12 3 k M1 for y  oe x  1 A1 for k = 36 OR 8  1 M2 for 18  oe 3  1 y 3  1 or M1 for  oe 18 8  1 10(a)(ii) 575 2 k M1 for x their1 or better 1.5 10(b) 18 NFWW 3 M1 for 18  c  3  1 oe or better their 36 M1 for wy  their 9 x  1  oe x  1

This question in 0607/41 May/June 2024

8 (a) The amount charged for electricity in one month is $E. $E is the sum of a fixed charge $f and a cost of $d for each unit of electricity used. Find a formula for the amount charged in one month when u units of electricity are used. … [2] (b) Write as a single fraction in its simplest form. x 2 x 5x - + 2 3 18 … [2] (c) Solve 7n - 9 2 21 + 2 n . … [2] (d) Solve the simultaneous equations. You must show all your working. 2x + 15y = –57 20x + 3y = 18 x = … y = … [3] (e) y is proportional to the square of ( x - 3) . y = 5 when x = 7 . Find the value of y when x = 27 . y = … [3]

12 marks

Mark scheme: 8(a) E = du + f final answer 2 M1 for du + f 8(b) x 2 M1 for correct use of common denominator eg final answer 9 9 x 12 x 5 x − + 18 18 18 8(c) n  6 final answer 2 M1 for 7n − 2n *21 + 9 or better * can be = or any inequality 8(d) correctly equating one set of M1 coefficients Or correctly making x or y the subject of an equation and correct substitution x = 1.5 A2 A1 for each y = −4 If M0 scored SC1 for correct substitution and evaluation to find the other variable. or SC1 if no working shown, but 2 correct answers given. 8(e) 180 3 5 2 M2 for y = their ( x − 3) oe 16 OR M1 for y = k ( x − 3) 2 5 A1 for k = 16

This question in 0607/43 Oct/Nov 2024

14 y varies inversely as the square root of ( x - 1 ) . y = 1 when x = 5. (a) Find y in terms of x. y = … [2] (b) w varies directly as y2. w = 45 when y = 3. Use your answer to part (a) to find w in terms of x. Give your answer in the form w = p ( x - 1 ) q , where p and q are constants. w = … [4]

6 marks

Mark scheme: 14(a) 2 4 2 k y = or oe B1 for y = ( x − 1) x − 1 ( x − 1) 14(b) w = 20( x − 1)−1 4  2  2 B3 for w = 5   or better   ( x − 1)   OR B1 for w = 5 y 2  2  2 M1 for w = their 5  their     ( x − 1)  

This question in 0607/42 May/June 2025

12 w ∝ x + 1 When x = 3, w = 8. Find x when w = 20. x = … [3]

3 marks

Mark scheme: 12 24 3 B2 for w = 4 x + 1 20 8 or M2 for = x + 1 3 + 1 or M1 for 8 = k 3 + 1

This question in 0607/43 May/June 2025

10 y is inversely proportional to x3. When x = 2, y = 2. 5 . (a) Find y in terms of x. y = … [2] (b) Find the value of x when y = 20000 . x = … [2] (c) p is directly proportional to y2. When x = 3 2, p = 1200 . Find p in terms of x. p = … [3]

7 marks

Mark scheme: 10(a) 20 2 k oe final answer M1 for y = oe  y =  3 3 x x 10(b) 0.1 oe 2 their 20 M1 for x3 = or better 20000 10(c) 4800 3 M1 for p = cy2 oe [ p = ] oe final answer x 6  their 20  2 M1 for 1200 = c oe    2 

This question in 0607/41 Oct/Nov 2025