June 2025 Paper 2 Q11
11. A company is trying to determine the most profitable selling price for a new toy.
Given that
- if the selling price of each toy is £30, the company expects to sell 1500 toys in one year
- if the selling price of each toy is £50, the company expects to sell 300 toys in one year
Using a linear model, with \(y\) being the expected number of toys sold in one year and \(x\) pounds being the selling price of the toy,
Given that
- the cost of making each toy is £10
- the company has additional costs of £8 000 per year
Use the model given in part (b) to answer parts (c), (d) and (e).
Given that the company wishes to make a profit on sales of the toy,
In one particular year, the company sold the toy for £35 and made £21 750 profit.
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{300 - 1500}{50 - 30} = (-60)\) | M1 | 3.1b |
| \(y - 1500 = \text{``}{-}60\text{''}(x - 30)\) | dM1 | 1.1b |
| \(y = 3300 - 60x\) | A1 | 3.3 |
| (3) |
Notes
M1: Attempts the gradient of the straight line either way round.
Look for \(\dfrac{300 - 1500}{50 - 30} = (-60)\) o.e. or \(\dfrac{50 - 30}{300 - 1500} = \left(-\dfrac{1}{60}\right)\) o.e.
Condone 1 sign/copying slip only if a correct formula is seen or implied e.g. \(\dfrac{y_2 - y_1}{x_2 - x_1}\)
Alternatively, sets up simultaneous equations \(1500 = 30a + b\) and \(300 = 50a + b\) (or \(30 = 1500a + b\) and \(50 = 300a + b\)) and attempts to solve, leading to a value for \(a\) and/or \(b\).
If using simultaneous equations, condone slips.
dM1: Attempts the equation relating \(x\) and \(y\) with either coordinate (50, 300) or (30, 1500) used correctly. Look for \(y - 1500 = \text{``}{-}60\text{''}(x - 30)\) or \(x - 30 = \text{``}{-}\dfrac{1}{60}\text{''}(y - 1500)\) or using (50, 300). Note that e.g. \(y - 30 = \text{``}{-}60\text{''}(x - 1500)\) scores dM0 as the values have been used incorrectly. If using \(y = mx + c\) they must proceed to \(c = \ldots\)
If using simultaneous equations, they must proceed to values for \(a\) and \(b\)
(Note for reference \(x\) in terms of \(y\) is \(x = -\dfrac{1}{60}y + 55\))
If they use e.g. \(\dfrac{y - 300}{1500 - 300} = \dfrac{x - 50}{30 - 50}\) both M’s can be scored together.
Condone 1 sign/copying slip only if a correct formula is seen or implied.
A1: cao Either \(y = 3300 - 60x\) or \(y = -60x + 3300\) or equivalent with \(y\) in terms of \(x\)
The correct equation with no working scores full marks.
| Scheme | Marks | AO |
|---|---|---|
| Either \(x\left(\text{``}3300 - 60x\text{''}\right)\) or \(-10\left(\text{``}3300 - 60x\text{''}\right)\) seen | M1 | 1.1b |
| \((P =)\ \dfrac{x\left(\text{``}3300 - 60x\text{''}\right) - 10\left(\text{``}3300 - 60x\text{''}\right) - 8000}{1000}\) | dM1 | 3.3 |
| \(P = 3.3x - 0.06x^2 - 33 + 0.6x - 8\) \(P = -0.06x^2 + 3.9x - 41\) * | A1* | 2.1 |
| (3) |
Notes
M1: Attempts either \(\pm x\left(\text{``}3300 - 60x\text{''}\right)\) or \(\pm 10\left(\text{``}3300 - 60x\text{''}\right)\) with or without dividing by 1000.
May see both in one step, i.e., \(\pm(x \pm 10)\left(\text{``}3300 - 60x\text{''}\right)\)
Alternatively, allow \((P =)\ \pm xy \pm 10y \pm 8000\) with or without dividing by 1000.
Allow their model for \(y\) to be any expression in \(x\) (i.e. it does not need to be linear) for M1.
dM1: Full attempt to find \(P\) in terms of \(x\) only. There is no need to see a LHS for this mark.
\((P =)\ \dfrac{x\left(\text{``}3300 - 60x\text{''}\right) - 10\left(\text{``}3300 - 60x\text{''}\right) - 8000}{1000}\) or \((P =)\ \dfrac{(x - 10)\left(\text{``}3300 - 60x\text{''}\right) - 8000}{1000}\)
Condone the consistent absence of division by 1000 e.g.
\((P =)\ x\left(\text{``}3300 - 60x\text{''}\right) - 10\left(\text{``}3300 - 60x\text{''}\right) - 8000\) or \((P =)\ (x - 10)\left(\text{``}3300 - 60x\text{''}\right) - 8000\)
Condone invisible brackets if recovered. Dependent on the previous method mark.
Their model for \(y\) must now be a linear expression in \(x\) for dM1.
A1*: Achieves the given answer through rigorous argument with all elements pulled together clearly. Requires \(P =\) (not just “profit =”) at some point which may be on a previous line.
Any brackets seen must have been expanded in an intermediate line before the given answer.
Allow otherwise correct work leading to \(P = -60x^2 + 3900x - 41000\) followed by \(P = -0.06x^2 + 3.9x - 41\) (without justifying the division by 1000).
Allow recovery if they write e.g. 800 but it is recovered before the given answer.
Note: Attempts by verification are unlikely to score any marks. If unsure, send to review.
e.g. \(x = 10 \rightarrow P = -0.06(10)^2 + 3.9(10) - 41 = -8\) without an accompanying argument involving the \((x, y)\) values (50, 300) and (30, 1500) scores M0dM0A0*.
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{-3.9 \pm \sqrt{3.9^2 - 4(-0.06)(-41)}}{2 \times -0.06}\) | M1 | 3.1b |
| \(13.18 \lt x \lt 51.82\) (or \(13.19 \leqslant x \leqslant 51.81\)) | A1 | 3.2a |
| (2) |
Notes
M1: Attempts to find either end point of the interval which may come from a calculator.
The exact values are \(\dfrac{195 \pm 5\sqrt{537}}{6}\) but may just be seen in the quadratic formula, e.g., \(\dfrac{-3.9 \pm \sqrt{3.9^2 - 4(-0.06)(-41)}}{2 \times -0.06}\) or using completing the square – see general marking principles.
Accept awrt 13.2 or awrt 51.8 for this mark.
A1: Either \(13.18 \lt x \lt 51.82\) or \(13.19 \leqslant x \leqslant 51.81\) No units are required but 2d.p. are required.
Correct answer scores both marks. Allow e.g. \(13.18 \lt x \leqslant 51.81\) or \(13.19 \leqslant x \lt 51.82\)
Allow e.g. “\(x \gt 13.18\) and \(x \lt 51.82\)” or “\(x \gt 13.18 \cap x \lt 51.82\)” but not use of exact values and not “\(x \gt 13.18\), \(x \lt 51.82\)” or “\(x \gt 13.18 \cup x \lt 51.82\)” or “\(x \gt 13.18\) or \(x \lt 51.82\)”
Must be \(x\) or e.g. “selling price” for the range, not \(y\) or \(P\).
| Scheme | Marks | AO |
|---|---|---|
| £32.50 | B1ft | 3.4 |
| (1) |
Notes
B1ft: £32.50. Units and 2dp required. Not £32.50p. Ignore attempts to find maximum profit.
Allow FT on the mean of their end points from (c) (their end points must be > 0, their answer requires £ and 2dp).
May come directly from the given answer to (b) using \(x = -\dfrac{b}{2a} = -\dfrac{3.9}{2 \times -0.06} = \text{£}32.50\) or from \(\dfrac{\mathrm{d}P}{\mathrm{d}x} = -0.12x + 3.9 = 0\) or directly from a calculator. No working required.
Note: Following use of 13.2 and/or 51.8 (not e.g. 51.80) with either type of inequality in part (c), £32.5 will score B1ft so that the missing decimal place(s) is only penalised once.
| Scheme | Marks | AO |
|---|---|---|
| \((P =)\ -0.06(35)^2 + 3.9(35) - 41 = \ldots\) | M1 | 3.4 |
| £22 000 which is close to £21 750 so the model is suitable. | A1 | 3.5a |
| (2) | ||
| (11 marks) |
Notes
M1: Attempts to substitute 35 (and not any other value) into the given equation for \(P\) to find a value for \(P\).
May be implied by 22 or 22000. Condone slips.
Alternatively, substitutes \(P = 21.75\) into the model, and solves via any acceptable method, including by calculator, to find a value for \(x\). If using 21750 it must be substituted into their \((P =)\ -60x^2 + 3900x - 41000\). Note \(x = 35.73\) or 29.27
A1: Requires:
- A correct value. Usually (£)22 000 but accept, e.g., 22 thousand (pound). If using 22 they must compare with 21.75. The alternative requires \(x =\) awrt 35.7 to be compared to 35.
- A correct comparison. Some examples below:
- “Close to (£)21 750” o.e. e.g. “the model predicts 22000 and they make close to this amount”
- “Agree to 2sf”. Do not condone “it rounds to 22 000” without mention of the degree of accuracy.
- 1.1% error (accept approx. 1%) Do not be concerned about the mechanics of any percentage error calculation seen.
- “The difference is small”
- Condone e.g. \(22 \approx 21.75\) o.e.
- “Only £250 off”. However, just stating “£250 off” without suggesting this is a (relatively) small difference is not sufficient for this component. Similarly, “£350 off is a small difference” is incorrect (requires correct £250).
- A conclusion. e.g. Concludes it (the model) is suitable / reliable / good / fairly accurate / accurate.