A2 June 2019 Paper 1 Q5
5. A tank at a chemical plant has a capacity of 250 litres. The tank initially contains 100 litres of pure water.
Salt water enters the tank at a rate of 3 litres every minute. Each litre of salt water entering the tank contains 1 gram of salt.
It is assumed that the salt water mixes instantly with the contents of the tank upon entry.
At the instant when the salt water begins to enter the tank, a valve is opened at the bottom of the tank and the solution in the tank flows out at a rate of 2 litres per minute.
Given that there are \(S\) grams of salt in the tank after \(t\) minutes,
When the concentration of salt in the tank reaches 0.9 grams per litre, the valve at the bottom of the tank must be closed.
| Scheme | Marks | AO |
|---|---|---|
| The tank initially contains 100L. 3 L are entering every minute and 2 L are leaving every minute so overall 1 L increase in volume each minute so the tank contains \(100 + t\) litres after \(t\) minutes | M1 | 3.3 |
| 2 L leave the tank each minute and if there are \(S\)g of salt in the tank, the concentration will be \(\dfrac{S}{100 + t}\) g/L so salt leaves the tank at a rate of \(2 \times \dfrac{S}{100 + t}\) g per minute | M1 | 3.3 |
| Salt enters the tank at a rate of \(3 \times 1\) g per minute | B1 | 2.2a |
| \(\therefore \dfrac{\mathrm{d}S}{\mathrm{d}t} = 3 - \dfrac{2S}{100 + t}\) * cso | A1* | 1.1b |
| (4) |
Notes
M1: A suitable explanation for the “\(100 + t\)” e.g. as a minimum \((v) = 100 + 3t - 2t = 100 + t\)
M1: A suitable explanation for the \(\dfrac{2S}{100 + t}\)
There need to be some explanation (words) for this part of the formula.
e.g. the concentration of (salt) \(= \dfrac{S}{100 + t}\) therefore (salt) out \(= 2 \times \dfrac{S}{100 + t} = \dfrac{2S}{100 + t}\)
e.g. salt out \(= \dfrac{2S}{\text{volume of water}} = \dfrac{2S}{100 + t}\)
Note: M0 for \(2 \times \dfrac{S}{100 + t} = \dfrac{2S}{100 + t}\) only with no explanation
B1: Correct interpretation for the “3” e.g. salt in = 3 or \(\dfrac{\mathrm{d}S}{\mathrm{d}t}\) in \(= 3\)
Note: Salt water in = 3 is B0
A1*: Puts all the components together to form the given differential equation cso
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}S}{\mathrm{d}t} + \dfrac{2S}{100 + t} = 3\) | ||
| \(I = \mathrm{e}^{\int \frac{2}{100 + t}\,\mathrm{d}t} = (100 + t)^2 \Rightarrow S(100 + t)^2 = \displaystyle\int 3(100 + t)^2\,\mathrm{d}t\) | M1 | 3.1b |
| \(S(100 + t)^2 = (100 + t)^3\,(+c)\) OR \(S(100 + t)^2 = 30\,000t + 300t^2 + t^3\,(+c)\) | A1 | 1.1b |
| \(t = 0,\ S = 0 \Rightarrow c = -10^6\) | M1 | 3.4 |
| \(t = 10 \Rightarrow S = 100 + 10 - \dfrac{10^6}{(100 + 10)^2}\) OR \(S(100 + 10)^2 = (100 + 10)^3\,(+c) \Rightarrow S = \ldots\) | dM1 | 1.1b |
| \(=\) awrt 27 (g) or \(\dfrac{3310}{121}\) (g) | A1 | 2.2b |
| (5) |
Notes
M1: Uses the model to find the integrating factor and attempts the solution of the differential equation. Look for \(I.F. = \mathrm{e}^{\int \frac{2}{100 + t}\,\mathrm{d}t} \Rightarrow S \times \text{‘their } I.F.\text{’} = \displaystyle\int 3 \times \text{‘their } I.F.\text{’}\,\mathrm{d}t\)
A1: Correct solution condone missing \(+ c\)
For the next three mark there must be a constant of integration
M1: Interprets the initial conditions, \(t = 0\ \ S = 0\), and uses in their equation to find the constant of integration.
dM1: Dependent on having a constant of integration. Uses their solution to the problem to find the amount of salt after 10 minutes.
A1: Awrt 27 or \(\dfrac{3310}{121}\). (If the units are stated they must be correct)
Note: If achieves \(S(100 + t)^2 = 30\,000t + 300t^2 + t^3 + c\) the constant of integration \(c = 0\) and the correct amount of salt can be achieved. If there is no \(+ c\) the maximum they can score is M1A1M0M0A0
| Scheme | Marks | AO |
|---|---|---|
| Concentration is \(\left(100 + t - \dfrac{10^6}{(100 + t)^2}\right) \div (100 + t) = 0.9\) OR \(S = 0.9(100 + t) \Rightarrow 0.9(100 + t) = (100 + t) - \dfrac{10^6}{(100 + t)^2}\) OR \(S = 0.9(100 + t) \Rightarrow 0.9(100 + t)^3 = (100 + t)^3 - 10^6\) | M1 | 3.4 |
| \((100 + t)^3 = 10^7 \Rightarrow t = \ldots\) OR \(t^3 + 300t^2 + 30\,000t - 9\,000\,000 = 0 \Rightarrow t = \ldots\) | dM1 | 1.1b |
| \(t =\) awrt 115 (minutes) | A1 | 2.2b |
| (3) |
Notes
Note: Look out for setting \(S = 0.9\) in this part, which scores no marks.
M1: Uses their solution to the model and divides by \(100 + t\) as an interpretation of the concentration and sets \(= 0.9\).
Alternatively recognises that the amount of salt \(= 0.9(100 + t)\) and substitutes for \(S\) in their solution to the model.
dM1: Dependent on previous method mark. Solves their equation to obtain a value for \(t\). May use a calculator.
A1: Awrt 115 (If the units are stated they must be correct) or 1hr 45 mins with units
(The second line of the scheme is corrected from the printed mark scheme: it is printed with \((100 + 10)^2\) in the denominator, which should be \((100 + t)^2\).)
| Scheme | Marks | AO |
|---|---|---|
E.g.
| B1 | 3.5a |
| (1) | ||
| (13 marks) |
Notes
B1: Evaluates the model by making a suitable comment – see scheme for examples.