AQA GCSE Mathematics Paper 1 (Higher), June 2025: Question 25

Mark scheme

Show the mark scheme

Q Answer Mark Comments

65 may be seen on a tree diagram

or M1

n n −1

65 1 oe eg n(n – 1) = 240

× = M1dep

n n −1 8

Correctly rearranges their equation, dep on 1st M1

which must be correct or of the 2 2

6 a n – n = 240 or n – n – 240 = 0

form × = , where a and implies M1M1M1

n n + b 8 M1dep

b are integers, into a quadratic

equation with brackets expanded

and no unknowns in denominators

For their three-term quadratic,

correctly factorises eg (n + 15)(n – 16) (= 0)

− −1± (−1)2 − 4 × × −1240

or correctly substitutes into the eg

quadratic formula 2×1

M1

or correctly completes the square eg ± 240 25.

25 2

to the form n = … for their quadratic

or

–15 and 16

16 with at least the first two marks SC1 16 with no other marks awarded

A1

awarded and not from incorrect working

Additional Guidance

Accept the use of any letter throughout

16 from trial and improvement, or without working SC1

65 1 2 M1M0M1

× = , n = 240, n = 4 15

n n 8 M0A0

The second mark may be awarded for work done in stages

65 36 36 1 6 5 1

eg × = then = implies × = M1M1

n n −1 n n( −1) n n( −1) 8 n n −1 8

(36 should be 30, which does not affect these marks but means they cannot

get the third mark)

How to answer it

Conditional Probability & Quadratic Equations

GCSE Higher Tier • 5 Marks • Question 25

What this question tests

Dependent probability without replacement: Writing algebraic fractions for successive events where the denominator and numerator decrease by 1.
Forming algebraic equations: Multiplying algebraic fractions and setting up an equation equal to a given probability.
Rearranging into standard quadratic form: Clearing algebraic denominators to form an² + bn + c = 0 .
Solving quadratics: Factoring or applying the quadratic formula, and interpreting the context to discard negative solutions.

Question Breakdown & Model Solution

Part-by-part guide to securing all 5 marks

📐 Step-by-Step Algebraic Solution

1 Write the probability of picking the first black counter:
Total counters = n, Black counters = 6.
P(1st Black) = 6 / n [M1 for 6/n or 5/(n - 1)]

2 Write the probability of picking the second black counter:
Because there is no replacement, one black counter is gone and the total reduces by 1:
Remaining black = 5, Remaining total = n - 1.
P(2nd Black) = 5 / (n - 1)

3 Set up the product equation:
The probability both are black is given as 1/8:
(6 / n) × (5 / (n - 1)) = 1 / 8 [M1dep]
Multiply the numerators and denominators:
30 / [n(n - 1)] = 1 / 8

4 Eliminate fractions and form a 3-term quadratic:
Cross-multiply:
30 × 8 = 1 × n(n - 1)
240 = n² - n
Rearrange to standard form ax² + bx + c = 0 :
n² - n - 240 = 0 [M1dep: expanding & eliminating denominators]

5 Solve the quadratic equation:
Find two numbers that multiply to -240 and add to -1. The factors are -16 and +15.
(n - 16)(n + 15) = 0 [M1: factorising or quadratic formula]
So, n = 16 or n = -15 .

6 Select the valid solution:
Since n represents a physical count of counters, n cannot be negative ( n > 0 ).
Therefore, n = 16 .

✅ Final Answer

n = 16

A1: Awarded for 16 achieved from valid algebraic working (with at least the first two method marks).

💡 Key Knowledge

Without replacement (conditional): Always decrease both the target count and the total count by 1 for the second pick.
The "AND" rule: P(A and B) = P(A) × P(B|A).
Contextual constraints: In real-world problems, items, lengths, and counts cannot be negative. Always reject negative solutions explicitly if required.

🧠 Exam Technique & Examiner Insight

"Algebraic method" instruction: The question explicitly says "Use an algebraic method". If you guess numbers or use trial and improvement to find 16, the mark scheme restricts you to a maximum of SC1 (only 1 out of 5 marks!).
Tree diagrams help: Sketching just the "Black → Black" branch of a probability tree ensures you don't forget to subtract 1 from the denominator.
Formula check: If factorising 240 is tricky in an exam, immediately switch to the quadratic formula:
n = [-(-1) ± √((-1)² - 4(1)(-240))] / 2 .

❌ Common Traps to Avoid

Treating it as "with replacement": Writing (6/n) × (6/n) or (6/n) × (5/n) leads to n² = 240 , giving n = 4√15 ≈ 15.5 . This scores a maximum of 1 mark!
Numerator arithmetic slips: Calculating 6 × 5 = 36 or forgetting to multiply numerators entirely.
Sign errors in the quadratic: Writing n² - n + 240 = 0 , which gives no real solutions. Remember that moving +240 to the other side makes it -240.
Leaving both answers: Writing n = 16 and n = -15 on the answer line without identifying 16 as the final answer.

Topics

Algebra · Probability · 3.2.1 Notation, vocabulary and manipulation · 3.2.3 Solving equations and inequalities · 3.5 Probability