Mark-by-Mark · Cambridge IGCSE Additional Mathematics 0606

0606/23 May/Jun 2025: where students lost marks

The 4 questions the June 2025 examiners' report flagged on this paper. Full question, mark scheme and video walkthrough, free. No sign-in needed.

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Chapters

Question 30:34 Question 6(b)(i)2:05 Question 8(a)3:30 Question 10(a)5:04

Question 3

4 marks · Linear Law (After logarithm)

The question

Where students lost marks

A reasonable proportion found m and c, but some square-rooted in the final step instead of squaring, some stopped at √y = 2x³ + 1 thinking it was the answer, and weaker responses used the plotted points as if they were (x, y). Candidates who wrote Y and X for the plotted variables usually avoided stopping early.

The trap: The graph is of √y against x³, so the line gives √y = 2x³ + 1. That is not y yet: square both sides.

Source: Paper 0606/23 (Calculator), Question 3, June 2025 examiners' report (paraphrased)

Mark scheme, mark by mark

M1√y = m x³ + c stated or implied

M1m = (21 − 5)/(10 − 2) = 2

M1c = 5 − 2(2) = 1

A1y = (2x³ + 1)²

Teacher's note: Rename first: Y = √y, X = x³. Then it is a straight-line question you already know, and the last step, undoing the Y, can't be forgotten.

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Question 6(b)(i)

2 marks · Permutation

The question

Where students lost marks

This was the part candidates found most challenging. The most successful listed a pattern like RBRBRBRBRBRBR and then used the product rule, 7! × 6!. Others found 7! and 6! but never multiplied them, or used combinations instead of permutations.

The trap: Seven reds, six browns, no two reds together: the only pattern is R B R B … R. Then the reds can swap among themselves AND the browns can: 7! × 6!, multiplied.

Source: Paper 0606/23 (Calculator), Question 6(b)(i), June 2025 examiners' report (paraphrased)

Mark scheme, mark by mark

M17! × 6!

A13 628 800

Teacher's note: With 7 reds and only 6 browns there is exactly one seating pattern, so there's no extra factor for the pattern itself.

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Question 8(a)

5 marks · Application of dy/dx

The question

Where students lost marks

Most could differentiate e^(3x + 2), but some evaluated the derivative at x = 0.1 with the calculator in degree mode, some wrote the derivative of tan x as sec x or cosec²x, and some did not recall the small-changes relationship or made no use of h at all.

The trap: Calculus with trig means radians. In degree mode the same correct working gives about 10.0h instead of 13.1h, and the accuracy mark is gone.

Source: Paper 0606/23 (Calculator), Question 8(a), June 2025 examiners' report (paraphrased)

Mark scheme, mark by mark

B1d/dx of e^(3x + 2) = 3e^(3x + 2)

M1product rule structure

A1dy/dx = 3e^(3x + 2) tan x + e^(3x + 2) sec²x

M1δy ≈ (dy/dx at x = 0.1) × h

A1δy ≈ 13.1h

Teacher's note: Check the mode before the first trig value: tan 0.1 should be about 0.1003. If you see 0.0017, you're in degrees.

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Question 10(a)

3 marks · Postion and velocity vector

The question

Where students lost marks

Candidates needed the unit vector in the given direction, multiplied by the speed. Some found the magnitude 29 but combined it with the speed wrongly, usually getting 10i + 10.5j; others skipped the unit vector and used 58(20i + 21j). Some final position vectors left out t.

The trap: 20i + 21j is a direction, not a velocity. Its length is 29, so scale it by 58 ÷ 29 = 2: v = 40i + 42j. Then r = −30j + t(40i + 42j).

Source: Paper 0606/23 (Calculator), Question 10(a), June 2025 examiners' report (paraphrased)

Mark scheme, mark by mark

B1|20i + 21j| = √(20² + 21²) = 29

M1v = (58/29)(20i + 21j)

A1r = −30j + t(40i + 42j)

Teacher's note: Sense check: the length of your velocity must equal the speed. √(40² + 42²) = 58. ✓

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