Full Exam Walkthrough

This is a capstone lesson: a full walkthrough of a representative Edexcel GCSE Mathematics (1MA1) Higher Paper 1 (non-calculator). We focus on Paper 1 because it is the hardest paper for most students — no calculator to rescue arithmetic slips, and surd/recurring-decimal/algebraic-proof questions appear here most often.

Rather than inventing a full 80-mark paper, we have selected 10 representative questions that together cover every major topic area tested on Higher Paper 1 (Number, Algebra, Ratio and Proportion, Geometry, Probability, Statistics). Each question is shown in the style you will see on the exam, followed by:

A worked model solution with every step visible.
A mark-scheme breakdown showing exactly where marks are earned.
An examiner's commentary highlighting the most common errors from real Edexcel examiner reports.

How to Read This Walkthrough

This is not a passive read. Treat each question as a live exam:

Cover the solution and attempt the question on paper, under timed conditions (roughly 1 minute per mark).
Mark your own work against the mark-scheme breakdown — be strict. If a method step is missing, no M mark.
Read the commentary and diagnose the class of error (e.g. sign error, rounded too early, missed the hence instruction). Errors almost always cluster into a handful of repeated classes.
Note weaknesses in your error log and revisit the relevant topic lesson.

A quick navigation aid:

Question	Topic area	Marks	Approx. grade
1	Number — fractions of amounts	3	4
2	Algebra — indices and simplification	3	4
3	Ratio — sharing in a ratio	3	5
4	Number — recurring decimals [H]	3	6
5	Algebra — solving a quadratic by factorising	3	5
6	Geometry — Pythagoras with surds	4	6
7	Probability — tree diagrams without replacement	5	7
8	Algebra — completing the square [H]	4	7
9	Surds — rationalising the denominator [H]	4	8
10	Algebraic proof [H]	5	8–9

Total marks in this walkthrough: 37 marks — roughly the second half of a typical Paper 1.

Before You Start (the first 2 minutes)

On exam day, use the first 2 minutes before writing any working:

Check you have the right paper (Paper 1 Higher non-calculator).
Write your name and candidate number.
Flip through to check no pages are missing.
Glance at the formula sheet — remind yourself what is provided (quadratic formula, cone volume, sphere volume, sine/cosine rules).
Note the total marks (80) and your pace target $(\approx 1$ minute per mark).

These two minutes cost you nothing in the scoring but save you from disasters — missing a page, working in radians, or forgetting that the quadratic formula is provided so you do not need to memorise it.

Question 1 — Number: fractions of amounts (3 marks)

A school trip has 360 students. 2/5 of the students are in Year 10, and 1/4 of the remaining students are in Year 11. How many students are in Year 11?

Model solution

Step 1: Year 10 students $= (2/5) \times 360 = 2 \times 72 = 144$ .

Step 2: Remaining students (Years 7, 8, 9, 11) = 360 − 144 = 216.

Step 3: Year 11 students $= (1/4) \times 216 =$ 54.

Mark scheme

M1 for a correct method to find Year 10 (e.g. $360 \div 5 \times 2)$ .
M1 for subtracting to find the remaining students (216).
A1 for 54.

Examiner's commentary

The single most common error here is interpreting "1/4 of the remaining students" as "1/4 of 360", giving 90 — an error worth two of the three marks. The word remaining is deliberate; when you see a chained fraction problem, always compute the reduced total before applying the next fraction.

A secondary error: students who can do the fraction arithmetic mentally sometimes write only "54" with no working. On a 3-mark question that risks all three marks if the answer is wrong (e.g. 90). Always show the intermediate value of 216.

Question 2 — Algebra: indices and simplification (3 marks)

Simplify fully $(2x^{3}y)^{2} \times 3x^{-4}y^{5}$ .

Model solution

Step 1: Expand the bracket. Square each factor inside the bracket: $(2x^{3}y)^{2} = 2^{2} \times (x^{3})^{2} \times y^{2} = 4x^{6}y^{2}$ .

Step 2: Multiply: $4x^{6}y^{2} \times 3x^{-4}y^{5} = (4 \times 3) \times x^{6 + (-4)} \times y^{2 + 5}$ $= 12x^{2}y^{7}$ .

Mark scheme

M1 for squaring the bracket correctly $(4x^{6}y^{2} seen)$ .
M1 for combining powers of $x (x^{2}$ or equivalent) or $y (y^{7})$ .
A1 for $12x^{2}y^{7}$ .

Examiner's commentary

Two classic slips:

Forgetting to square the coefficient: writing $(2x^{3}y)^{2} = 2x^{6}y^{2}$ instead of $4x^{6}y^{2}$ . The squaring must apply to every factor.
Sign error in the exponent: 6 + (−4) = 2, not −2. A double-negative mistake cascades and wipes out the A1.

Also watch out for students writing $x^{6-4} = x^{2}$ . This is correct because of the negative exponent becoming a subtraction, but the intermediate notation "6 + (−4)" makes the reasoning visible to the examiner and the student.

Question 3 — Ratio: sharing in a ratio (3 marks)

Ayesha, Ben and Carl share £840 in the ratio 5 : 3 : 4. Work out how much more Ayesha receives than Carl.

Model solution

Total parts = 5 + 3 + 4 = 12.

One part = £ $840 \div 12 =$ £70.

Ayesha $= 5 \times$ £70 = £350. Carl $= 4 \times$ £70 = £280.

Difference = £350 − £280 = £70.

Mark scheme

M1 for dividing 840 by the total number of parts (12) to get £70 per part.
M1 for finding both Ayesha's and Carl's shares.
A1 for £70.

Examiner's commentary

This question often traps the student who reads too quickly: it asks for the difference, not Ayesha's individual share. An answer of £350 scores two out of three marks at best (M1 M1 A0).

Sensible sanity check: Ayesha has 5 parts, Carl has 4 parts — one more part than Carl. Since one part = £70, the difference must be £70. If your final calculation gives a different number, the arithmetic is wrong. This kind of cross-check takes 5 seconds and catches the most common error.

Question 4 — Number: recurring decimals [H] (3 marks)

Using algebra, prove that 0.1̇3̇ = 13/99.

Model solution

Let x = 0.131313...

Multiply by 100 (since the repeating block has 2 digits): 100x = 13.131313...

Subtract x from 100x (to eliminate the recurring part): 100x − x = 13.131313... − 0.131313... 99x = 13 x = 13/99.

Therefore 0.1̇3̇ = 13/99. ∎

Mark scheme

M1 for setting x = the recurring decimal and multiplying by 100 (or equivalent).
M1 for the subtraction 100x − x = 13.
A1 for x = 13/99 with a concluding statement.

Examiner's commentary

"Using algebra" is a deliberate command — inspection answers score zero, even if correct. The full method is mandatory.

The most common error is multiplying by 10 rather than 100. The rule: multiply by $10^{n}$ where n is the length of the repeating block. A block of "13" is 2 digits, so multiply by 100. Students who multiply by 10 get 10x − x = 9x = 1.2 (approximately), which doesn't simplify cleanly — a warning sign that you've chosen the wrong power.

Remember to write the concluding statement ("Therefore 0.1̇3̇ = 13/99"). On a "prove" question, the final A1 is partly awarded for communicating that the proof is complete.

Question 5 — Algebra: solving a quadratic by factorising (3 marks)

Solve $x^{2} - 5x - 14 = 0$ .

Model solution

Find two numbers that multiply to give −14 and add to give −5.

Factors of −14: (1, −14), (−1, 14), (2, −7), (−2, 7).

The pair 2 and −7: $2 \times (-7) = -14$ ✓; 2 + (−7) = −5 ✓.

So $x^{2} - 5x - 14 = (x + 2)(x - 7)$ .

Setting each factor to zero: $x + 2 = 0 \to x = -2$ . $x - 7 = 0 \to x = 7$ .

Answer: x = −2 or x = 7.

Mark scheme

M1 for a correct factorisation or use of the quadratic formula.
A1 for x = −2.
A1 for x = 7.

Examiner's commentary

The classic error here is the sign flip: writing "(x − 2)(x + 7)" instead of "(x + 2)(x − 7)". Always verify by quickly expanding your factorisation — does it give back the original?

$(x + 2)(x - 7) = x^{2} - 7x + 2x - 14 = x^{2} - 5x - 14$ ✓.

A second error: forgetting to give both roots. A single-line answer "x = 7" on a factorised quadratic loses one of the two A marks automatically.

Because this is Paper 1 (non-calculator), the quadratic formula is possible but unnecessarily cumbersome — always try factorising first on this paper.

Question 6 — Geometry: Pythagoras with surds (4 marks)

A right-angled triangle has legs of length $2\sqrt{3} cm$ and $\sqrt{6} cm$ . Find the length of the hypotenuse, giving your answer as a simplified surd.

Model solution

By Pythagoras' theorem: $h^{2} = (2\sqrt{3})^{2} + (\sqrt{6})^{2}$ .

$(2\sqrt{3})^{2} = 4 \times 3 = 12$ . $(\sqrt{6})^{2} = 6$ .

So $h^{2} = 12 + 6 = 18$ .

$h = \sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}$ .

Answer: $3\sqrt{2} cm$ .

Mark scheme

M1 for applying Pythagoras' theorem.
M1 for squaring both surds correctly.
M1 for reaching $h^{2} = 18$ .
A1 for $h = 3\sqrt{2} cm ($ must be simplified).

Examiner's commentary

Three traps:

Squaring $(2\sqrt{3})$ — students sometimes write $(2\sqrt{3})^{2} = 2^{2} \times \sqrt{3} = 4\sqrt{3}$ , forgetting to square the $\sqrt{3}$ as well. The correct expansion is $2^{2} \times (\sqrt{3})^{2} = 4 \times 3 = 12$ .
Leaving the answer as $\sqrt{18}$ — this earns M1 M1 M1 but loses the final A1 because "simplify fully" is implicit in "simplified surd". Always extract the largest square factor.
Forgetting units — because the problem gave centimetres, the answer needs centimetres. Edexcel is usually lenient on units in surd questions, but it's a habit worth forming.

Question 7 — Probability: tree diagrams without replacement (5 marks)

A bag contains 4 red balls and 6 blue balls. Two balls are taken out at random, one after the other, without replacement.

Find the probability that the two balls are the same colour.

Model solution

On the first draw: P(red) = 4/10, P(blue) = 6/10.

After one red is removed, the bag has 3 red + 6 blue = 9 balls:

P(red | first was red) = 3/9.

After one blue is removed, the bag has 4 red + 5 blue = 9 balls:

P(blue | first was blue) = 5/9.

Same colour = (red then red) OR (blue then blue):

P(both $red) = (4/10) \times (3/9) = 12/90$ . P(both $blue) = (6/10) \times (5/9) = 30/90$ .

P(same colour) = 12/90 + 30/90 = 42/90 = 7/15.

Mark scheme

M1 for identifying that the second probability depends on the outcome of the first (without replacement).
M1 for computing P(both red) correctly $(4/10 \times 3/9)$ .
M1 for computing P(both blue) correctly $(6/10 \times 5/9)$ .
M1 for adding the two branch probabilities.
A1 for 7/15 (or equivalent, e.g. 42/90).

Examiner's commentary

The single biggest error: using $4/10 \times 3/10 ($ replacing the ball mentally). "Without replacement" means the denominator drops by 1 for the second draw and the relevant colour count drops by 1 for the matching colour. Draw a tree diagram with the updated probabilities written on each branch — this one visual aid eliminates almost all of the errors we see in this question type.

The second common error: computing each branch correctly but then multiplying the final result rather than adding. Remember: AND is multiply, OR is add. "Same colour" means "red then red" OR "blue then blue" — an addition.

Finally, students often stop at 42/90 without simplifying. "Give a fraction" problems always imply "in simplest form" — divide by 6 to get 7/15.

Question 8 — Algebra: completing the square [H] (4 marks)

Write $x^{2} - 6x + 2$ in the form $(x + a)^{2} + b$ , where a and b are integers. Hence, state the minimum value of the expression.

Model solution

Halve the coefficient of x: −6 / 2 = −3. So a = −3.

$(x - 3)^{2} = x^{2} - 6x + 9$ .

To make this equal $x^{2} - 6x + 2$ , subtract 7: $x^{2} - 6x + 2 = (x - 3)^{2} - 7$ .

So a = −3 and b = −7.

For the minimum value: $(x - 3)^{2} \geq 0$ for all x, with equality when x = 3. Minimum value of $(x - 3)^{2} - 7$ is −7, occurring at x = 3.

Mark scheme

M1 for halving the coefficient of x correctly (−3).
M1 for adjusting the constant (writing ...−7 or equivalent).
A1 for $(x - 3)^{2} - 7$ .
A1 for the minimum value of −7.

Examiner's commentary

Completing the square is probably the single most-tested Higher algebra technique. Common errors:

Sign error on a: writing $(x + 3)^{2}$ instead of $(x - 3)^{2}$ . The coefficient of x in the original is −6, so a must be negative.
Not adjusting the constant: writing $(x - 3)^{2} + 2$ — forgetting that $(x - 3)^{2}$ already contains a +9 term that has to be cancelled.
For the minimum value, writing x = −7 instead of "minimum = −7". The question asks for the minimum value of the expression, which is the y-coordinate of the vertex, not the x-coordinate.

Quick check: expand your answer back out. $(x - 3)^{2} - 7 = x^{2} - 6x + 9 - 7 = x^{2} - 6x + 2$ ✓.

Question 9 — Surds: rationalising the denominator [H] (4 marks)

Rationalise the denominator of $(2 + \sqrt{3}) / (5 - \sqrt{3})$ . Give your answer in the form $(p + q\sqrt{3}) / r$ where p, q and r are integers.

Full Exam Walkthrough

Full Exam Walkthrough

How to Read This Walkthrough

Before You Start (the first 2 minutes)

Question 1 — Number: fractions of amounts (3 marks)

Model solution

Mark scheme

Examiner's commentary

Question 2 — Algebra: indices and simplification (3 marks)

Model solution

Mark scheme

Examiner's commentary

Question 3 — Ratio: sharing in a ratio (3 marks)

Model solution

Mark scheme

Examiner's commentary

Question 4 — Number: recurring decimals [H] (3 marks)

Model solution

Mark scheme

Examiner's commentary

Question 5 — Algebra: solving a quadratic by factorising (3 marks)

Model solution

Mark scheme

Examiner's commentary

Question 6 — Geometry: Pythagoras with surds (4 marks)

Model solution

Mark scheme

Examiner's commentary

Question 7 — Probability: tree diagrams without replacement (5 marks)

Model solution

Mark scheme

Examiner's commentary

Question 8 — Algebra: completing the square [H] (4 marks)

Model solution

Mark scheme

Examiner's commentary

Question 9 — Surds: rationalising the denominator [H] (4 marks)

Model solution

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