A2 June 2024 Paper 1 Q17
17 In an industrial process, a container initially contains 1000 litres of liquid. Liquid is drawn from the bottom of the container at a rate of 5 litres per minute. At the same time, salt is added to the top of the container at a constant rate of 10 grams per minute. After \(t\) minutes the mass of salt in the container is \(x\) grams, and you are given that \(x = 0\) when \(t = 0\).
In modelling the situation, it is assumed that the salt dissolves instantly and uniformly in the liquid, and that adding the salt does not change the volume of the liquid.
\(\dfrac{\mathrm{d}x}{\mathrm{d}t} + \dfrac{x}{200 - t} = 10\). [3]
Suggest a reason for this. [1]
| Scheme | Marks | AO |
|---|---|---|
| (i) After \(t\) minutes volume of liquid \(= 1000 - 5t\) [litres] so concentration \(= \dfrac{x}{1000 - 5t}\) | B1 | 3.1b |
| [1] | ||
| (ii) \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = 10 \ldots\) | B1* | 3.3 |
| \(\ldots [-]5\left(\dfrac{x}{1000 - 5t}\right)\) | B1* | 3.3 |
| \(\Rightarrow \dfrac{\mathrm{d}x}{\mathrm{d}t} + \dfrac{x}{200 - t} = 10\) | B1dep | 3.3 |
| [3] |
Notes
(a)(i)
B1: AG Volume of liquid (allow “volume”, “amount of liquid”, “capacity” or “liquid”) in the container must be given in first step. Do not allow “volume leaving”.
(a)(ii)
Second B1*: Not oe
B1dep: AG successful completion
| Scheme | Marks | AO |
|---|---|---|
| IF is \(\mathrm{e}^{\int \frac{1}{200 - t}\,\mathrm{d}t}\) | M1 | 3.1a |
| \(= \mathrm{e}^{-\ln(200 - t)} = \dfrac{1}{200 - t}\) | A1 | 2.1 |
| \(\Rightarrow \dfrac{1}{200 - t}\dfrac{\mathrm{d}x}{\mathrm{d}t} + \dfrac{x}{(200 - t)^2} = \dfrac{10}{200 - t}\) | M1 | 2.1 |
| \(\Rightarrow \dfrac{\mathrm{d}}{\mathrm{d}t}\left(\dfrac{x}{200 - t}\right) = \dfrac{10}{200 - t}\) or \(\dfrac{x}{200 - t} = \displaystyle\int \frac{10}{200 - t}\,\mathrm{d}t\) | M1* | 2.1 |
| \(\dfrac{x}{200 - t} = -10\ln(200 - t)\ [+c]\) | A1 | 2.1 |
| \(t = 0, x = 0 \Rightarrow c = 10\ln 200\) | B1FT | 3.1b |
| \(\dfrac{x}{200 - t} = -10\ln(200 - t) + 10\ln 200 = 10\ln\left(\dfrac{200}{200 - t}\right)\) | M1dep | 2.1 |
| \(x = 10(200 - t)\ln\left(\dfrac{200}{200 - t}\right)\) | A1 | 2.1 |
| [8] |
Notes
M1: Finding integrating factor
A1: Allow \(A\frac{1}{200 - t}\) where \(A\) is clearly an unevaluated constant of integration and not from incorrect integration
M1: Multiplying equation by their IF, soi by next line
M1*: Either, ft their IF (i.e. their IF \(\times\ x\))
A1: First A1 must have been scored. Condone missing \(+c\) and bracketing errors.
B1FT: Correct \(c\) for their integral
M1dep: Combining their log terms correctly, do not condone bracketing errors without clear recovery
A1: AG www
| Scheme | Marks | AO |
|---|---|---|
| (i) Half drawn off when \(t = 100\) | M1 | 3.4 |
| \(\Rightarrow x = 10 \times 100 \times \ln 2 = 693\) [g] | A1 | 3.4 |
| [2] | ||
| (ii) [\(x\) is maximised when] \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = 0\) | M1 | 3.1a |
| \(10(200 - t)\ln\left(\dfrac{200}{200 - t}\right) = 10(200 - t)\) | M1* | 1.1 |
| \(\Rightarrow \ln\left(\dfrac{200}{200 - t}\right) = 1\) | A1 | 1.1 |
| \(\Rightarrow \dfrac{200}{200 - t} = \mathrm{e}\) | M1dep | 1.1 |
| \(\Rightarrow t = 126.42\ldots\) | A1 | 3.4 |
| [5] |
Notes
(c)(i)
M1: May be embedded but must be seen
A1: Accept 690 or better, or \(1000\ln 2\)
(c)(ii)
M1: soi. Implied by \(x = 10(200 - t)\).
M1*: Or finding \(\frac{\mathrm{d}x}{\mathrm{d}t} = 10\left(1 - \ln\left(\frac{200}{200 - t}\right)\right)\) correctly
M1dep: Eliminating logarithms correctly
A1: 130 (2sf) or better. Do not accept exact value. Ignore units.
| Scheme | Marks | AO |
|---|---|---|
| In reality the salt would not dissolve/mix/spread/diffuse/disperse/distribute instantly | B1 | 3.5b |
| [1] |
Notes
B1: Or “the salt does not dissolve/mix/spread/diffuse/disperse/distribute uniformly”. Must be explicit that modelling assumption is unrealistic.
Ignore subsequent comments
Do not accept reference to volume or exogenous factors, e.g. evaporation/temperature