# Question 7:

## Part (a)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int\frac{1}{5000-N}\,dN = \int\frac{(kt-1)}{t}\,dt \left\{\text{or } \int\left(k-\frac{1}{t}\right)dt\right\}$ | B1 | Separates variables correctly |
| $-\ln(5000-N) = kt - \ln t + c$ | M1 A1; A1 | See notes |
| Leading to $N = 5000 - Ate^{-kt}$ with no incorrect working/statements | A1* cso | See notes |

## Part (b)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\{t=1, N=1200\Rightarrow\}\ 1200=5000-Ae^{-k}$ and $\{t=2, N=1800\Rightarrow\}\ 1800=5000-2Ae^{-2k}$ | B1 | At least one correct statement written down using the boundary conditions |
| $Ae^{-k}=3800$ and $2Ae^{-2k}=3200$ or $Ae^{-2k}=1600$ | | |
| $\frac{e^{-k}}{2e^{-2k}}=\frac{3800}{3200}$ or $\frac{2e^{-2k}}{e^{-k}}=\frac{3200}{3800}$ giving $\frac{1}{2}e^k=\frac{3800}{3200}$ or $2e^{-k}=\frac{3200}{3800}$ | M1 | An attempt to eliminate $A$ by producing an equation in only $k$ |
| $k=\ln\left(\frac{7600}{3200}\right)$ or equivalent $\left\{eg\ k=\ln\left(\frac{19}{8}\right)\right\}$ | A1 | At least one of $A=9025$ cao or $k=\ln\left(\frac{7600}{3200}\right)$ or exact equivalent |
| $A=3800(e^k)=3800\left(\frac{19}{8}\right)\Rightarrow A=9025$ | A1 | Both $A=9025$ cao or $k=\ln\left(\frac{7600}{3200}\right)$ or exact equivalent |

### Alternative Method for M1 in (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $e^{-k}=\frac{3800}{A}$, then $2A\left(\frac{3800}{A}\right)^2=3200$ | M1 | An attempt to eliminate $k$ by producing an equation in only $A$ |

## Part (c)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $t=5,\ N=5000-9025(5)e^{-5\ln\left(\frac{19}{8}\right)}$ | | |
| $N=4402.828401... = 4400$ (fish) (nearest 100) | B1 | Anything that rounds to 4400 |

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