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Real questions, real marking

2 real questions this paper has actually set, drawn only from genuine Pearson question papers. Each one links to its real mark scheme, its real examiner report where one has been published, and our own level-by-level breakdown of how marks are actually earned — write your own answer first, then compare.

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June 2023 · Q2

6 marks

Forming and solving simultaneous equations from a word problem, then applying a stated constraint to choose between two root-pairings

A rectangular sports pitch has length x metres and width y metres, where x > y. Given that the perimeter of the pitch is 350 m, (a) write down an equation linking x and y. Given also that the area of the pitch is 7350 m², (b) write down a second equation linking x and y. (c) hence find the value of x and the value of y.

Our analysis

A rectangular sports pitch has length x metres and width y metres, where x > y. Perimeter 350 m, area 7350 m². Find the value of x and the value of y.

6 marks available

a — 1 mark

01
2x + 2y = 350
Correct equation linking x and y from the given perimeter — any correct equivalent form is accepted.
B1

b — 1 mark

101
xy = 7350
Correct equation linking x and y from the given area — an independent mark, not dependent on part (a).
B1

c — 4 marks

201
x(175 − x) = 7350
Substitutes for x or y into either equation from (a)/(b) to produce a 2- or 3-term quadratic equation in one variable. The mark scheme does not require the terms to be collected yet — reaching this unsimplified form is enough.
M1
202
x² − 175x + 7350 = 0 ⟹ (x − 70)(x − 105) = 0 ⟹ x = 70 or x = 105
Dependent on the first M1 — solves the resulting 3-term quadratic by any correct method (factorisation, the formula, or a calculator).
dM1
203
x = 70 or x = 105
Correct simplified roots of the quadratic. The mark scheme awards this for the two values themselves, not yet for assigning which is x and which is y.
A1
204
Since x > y: x = 105, y = 70
Both x and y correctly assigned using the stated restriction x > y, and all previous marks scored. This is the specific mark the real examiner report names as commonly lost.
A1

June 2022 · Q5(a)

4 marks

Reconstructing a quadratic f(x) from its turning point and a root — the reverse of sketching a quadratic from its equation

The curve C has equation y = f(x). Given that f(x) is a quadratic expression, the maximum turning point on C has coordinates (−2, 12), and C cuts the negative x-axis at −5, (a) find f(x).

Our analysis

The curve C has equation y = f(x), where f(x) is a quadratic expression, the maximum turning point on C has coordinates (−2, 12), and C cuts the negative x-axis at −5. Find f(x).

4 marks available

01
f(x) = 12 − a(x + 2)² [using the maximum turning point (−2, 12)]
Uses a quadratic form built from two pieces of the given information. Equally credited here: the factorised form a(x−1)(x+5), using the given root x=−5 together with the second root x=1 found by symmetry about the turning point's x-coordinate (−2); or the general form ax²+bx+c substituted with two known points.
M1
02
f(x) = 12 − a(x + 2)² [equation set up with the unknown constant a still to be found]
Correctly set up equation/expression with the correct unknown constants included, reaching a position where only the value of a (or one further constant, if using the general form) remains to be found.
A1
03
Substitute the root (−5, 0): 0 = 12 − a(−5 + 2)² = 12 − 9a ⟹ a = 4/3
Dependent on the first M1 — a full method using the remaining given information (the root at x = −5) to solve for the multiplier a. This is the step the real examiner report identifies as the one most candidates never completed, stopping instead at a correctly-shaped but under-determined expression.
dM1
04
f(x) = 12 − (4/3)(x + 2)² = −(4/3)(x − 1)(x + 5) = −(4/3)x² − (16/3)x + 20/3
Correct, fully-determined equation for f(x), accepted in any equivalent form.
A1