Cambridge IGCSE Mathematics - International 0607 — 2025 Feb/March Paper 2 · Variant 2

1 Write down the reciprocal of 5. ................................................. [1]

Mark scheme: Question Answer Marks Partial Marks 1 5 1

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2 Show the inequality –2 1 x G 3 on the number line. x −4 −3 −2 −1 0 1 2 3 4 [2]

Mark scheme: 2 2 B1 for line going from –2 to 3 or B1 for both ‘circles’ correct and no complete –2 3 line drawn

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4 9 15 12 8 13 14 Find the range of these numbers. ................................................. [1]

Mark scheme: 4 7 1

5 y B (1, 7) NOT TO SCALE A (9, 2) O x Point A is translated to point B. Find vector AB. AB = f p [2]

Mark scheme: 5  −8  2 B1 for each    8   5  If 0 scored SC1 for    −5 

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6 U = {natural numbers G 20 } A = {multiples of 3} B = {triangle numbers} (a) Write down the elements of set A and the elements of set B. Set A ....................................................... Set B ....................................................... [2] (b) Write down the elements of A + B . ................................................. [1]

Mark scheme: 6(a) [A] 3 6 9 12 15 18 2 B1 for each [B] 1 3 6 10 15 6(b) 3 6 15 1 FT their sets

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7 Write 84 as a product of its prime factors. ................................................. [2]

Mark scheme: 7 2 2 3 7 oe 2 B1 for 2, 3, 7 only as list of factors not multiplied or M1 for 2 correct stages in factor tree or factor ladder

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8 By writing each number correct to 1 significant figure, estimate the value of 5923 - 2198 . 0.5461 # 39 .43 ................................................. [2]

Mark scheme: 8 6000 − 2000 M1 Three correct values [0].5  40 200 A1

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9 These are the equations of two lines. 4y = x + 7 y + 4x = 6 (a) Find the coordinates of the point where these two lines intersect. (.............. , ..............) [3] (b) Are the two lines perpendicular? Give a reason for your answer. ................................. because ........................................................................................................... ..................................................................................................................................................... [2]

Mark scheme: 9(a) (1, 2) final answer 3 M1 for correctly equating one set of coefficients A1 for x = 1 A1 for y = 2 Correct answers spoiled scores SC2 If 0 scored SC1 for their solutions satisfying one equation 9(b) 1 B1 both gradients seen –4 and 4 1 B1 FT their gradients and correct conclusion Yes and −4 = −1 4

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10 (a) The probability that an event happens is 0.95 . Write down the probability that the event does not happen. ................................................. [1] (b) An unbiased die is numbered 1, 2, 3, 4, 5, 6. Jamisha rolls the die once. Find the probability that Jamisha rolls (i) an even number ................................................. [1] (ii) a 3 or a 5. ................................................. [1]

Mark scheme: 10(a) 0.05 1 10(b)(i) 3 1 oe 6 10(b)(ii) 2 1 oe 6

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11 These are the first five terms of a sequence. 3 9 19 33 51 Find the nth term of this sequence. ................................................. [2]

Mark scheme: 11 2n 2 + 1 2 M1 for second differences of 4 or any quadratic expression

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12 y 6 5 4 T 3 2 B 1 0 1 2 3 4 5 6 7 8 9 x – 1 A – 2 – 3 (a) Describe fully the single transformation that maps (i) triangle T onto triangle A ............................................................................................................................................. ............................................................................................................................................. [2] (ii) triangle T onto triangle B. ............................................................................................................................................. ............................................................................................................................................. [3] J N – 4 (b) Draw the image of triangle T after a translation of [2] KK OO. 1 L P

Mark scheme: 12(a)(i) Reflection 2 B1 for each y = 1 12(a)(ii) Enlargement 3 B1 for each [Centre] (6, 3) [Scale factor] –2 12(b) Correct triangle 2 k  − 4  B1 for translation or (4, 5) (4, 4) (2, 5)    1  k 

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13 y 5 4 3 2 1 – 5 – 4 – 3 – 2 – 1 0 1 2 3 4 5 6 x – 1 – 2 Find the 3 inequalities that define the shaded region. ............................................................................................................................................................. [4]

Mark scheme: 13 x  5 4 B2 for x [=] 5, x + y 4= oe, y 1= x + 2 oe x + y 4 oe 2 1 y x + 2 oe or B1 for y 1= x + 2 oe or x + y 4= oe 2 2 B1 for one correct inequality

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14 (a) Work out ( 0 .5 ) 2. ................................................. [1] (b) Work out 3 64 # 3 2 . ................................................. [2] (c) 16 n = 2 n – 1 Find the value of n. n = ................................................ [2]

Mark scheme: 14(a) 0.25 oe 1 14(b) 36 2 M1 for  3 64 =  4   14(c) 1 2 M1 for 4n = n − 1 − n 4 n 3 or B1 for [16 = ] 2

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15 Expand and simplify. ( 3a + 4b)( a - 2 b) ................................................. [2]

Mark scheme: 15 3a 2 − 2ab − 8b 2 final answer 2 B1 for 3 out of 4 terms correct in 3a 2 + 4 ab − 6ab − 8b 2

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16 NOT TO SCALE 4 cm 3 cm The diagram shows a solid half cylinder. The radius of the cross-section is 3 cm. The length of the solid is 4 cm. Find the total surface area of the solid. Give your answer in terms of r. ......................................... cm2 [4]

Mark scheme: 16 21π + 24 4 M3 for  1 2   1   2   3  +  4  2 3  + ( 2 3 4 )  2   2  OR 1 2 M1 for  2   3 2 1 M1 for 4  2 3 2

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17 A is the point (–5, 7) and B is the point (5, –2). Find the equation of the line AB. Give your answer in the form ax + by = k where a, b and k are integers. ................................................. [4]

Mark scheme: 17 9 x + 10 y = 25 final answer 4 9 5 B3 for y = − x + oe 10 2 OR 7 −−2 M1 for oe −−5 5 M1 for correct substitution of (–5, 7) or (5, –2) and their gradient into y = mx + c oe

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18 (a) Rearrange the formula to make a the subject. a - b = 2c a = ................................................ [2] (b) Rearrange the formula to make p the subject. p - 5 = q 3p - 2 p = ................................................ [3]

Mark scheme: 18(a) 4 c 2 + b final answer 2 B1 for a − b = ( 2c ) 2 oe 18(b) 5 − 2 q 3 M1 for removing fraction e.g. p −=5 (3 p − 2) q final answer 1 − 3q M1 for isolating terms in p e.g. p − 3 pq = 5 − 2q a ( q ) M1 for factorising and dividing e.g. p = b ( q ) maximum M2 if final answer incorrect

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19 h NOT TO SCALE A large cone has height h. The large cone is cut parallel to its base to make a smaller cone and a frustum. 3 The height of the frustum is h . 5 Find the ratio volume of small cone : volume of frustum. ....................... : ...................... [3]

Mark scheme: 19 8 : 117 3 3  2  M2 for   oe  5  or M1 for 2[ h ] 5

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20 A box contains 2 yellow pencils, 3 blue pencils and 4 green pencils. Micah takes two pencils from the box at random without replacement. Find the probability that the pencils are both blue. ................................................. [2]

Mark scheme: 20 6 2 3 2 oe M1 for  oe 72 9 8

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21 (a) Simplify. 3 12 - 48 + 75 ................................................. [3] (b) Rationalise the denominator and simplify. 6 5 - 2 ................................................. [3]

Mark scheme: 21(a) 7 3 3 B1 for 3  2 3 B1 for 4 3 or 5 3 21(b) 3 2 5 + 2 2 or 2 5 + 2 6 5 + 2 ( ) ( ) M2 for oe 5 − 2 6 5 + 2 or M1 for  5 − 2 5 + 2

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22 The equation of a quadratic function is y = ax 2 + bx + c . The vertex of the quadratic function is at (1, 3). The quadratic function passes through the point (2, 5). Find the equation of the quadratic function. y = ................................................ [4]

Mark scheme: 22 2 x 2 − 4 x + 5 4 B3 for y = 2 x 2 − 4 x + c or y = 2 x 2 + bx + 5 OR M1 for y = a ( x − 1) 2 + 3 M1 for substituting (2, 5) or (1, 3) into a quadratic If 0 scored SC1 for a sketch of suitable quadratic with (1,3) and (2, 5) indicated

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23 (a) f (x) 3 0 90 180 270 360 x – 3 The diagram shows the graph of f ( x°) = a cos bx . Find the value of a and the value of b. a = ....................................................... b = ....................................................... [2] 3 (b) Solve - 1 = 0 for values of x between 0° and 360°. tanx ................................................. [3]

Mark scheme: 23(a) a = 3 2 B1 for each b = 2 23(b) 60, 240 only 3 B2 for 60 highlighted or M1 for tan x = 3 If 0 scored SC1 for their two answers (inside the range), which have a difference of 180

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