## Question 10(a)(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| When $t = 0$, $L = 2800$. She invests £2800 | B1 | cao |

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## Question 10(a)(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Each year the amount is multiplied by $1.023$, which is $2.3\%$ annual interest | B1 | cao |

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## Question 10(b):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $A = 3000 \times 1.02^t$, so $a = 3000$ | B1 | Allow for $a$ and $b$ given explicitly or embedded in an exponential expression |
| And $b = 1.02$ | B1 | |

## Question 10(c):

**Equal amounts when $3000 \times 1.02^t = 2800 \times 1.023^t$**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\ln 3000 + t\log 1.02 = \ln 2800 + t\ln 1.023$ | M1 | 3.1b — Use of laws of logarithms leading to a linear equation in $t$ using their values of $a$ and $b$ |
| $t = \dfrac{\ln 3000 - \ln 2800}{\ln 1.023 - \ln 1.02} = 23.5$ | M1 | 1.1a — Collecting terms |
| So they have equal amounts after 23.5 years | A1 | 1.1b — Cao must be 1 d.p. |

**Alternative method:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{3}{2.8} = \left(\dfrac{1.023}{1.02}\right)^t$ so $t = \dfrac{\log\frac{3}{2.8}}{\log\frac{1.023}{1.02}} = 23.5$ | M1 | Equating and attempt to collect terms using their values of $a$ and $b$ leading to an equation in which $t$ appears only once |
| | M1 | Uses logarithms leading to a value for $t$; allow $\log_{1.003}1.07$ or $\dfrac{\log 1.07}{\log 1.003}$ or better for method mark |
| So they have equal amounts after 23.5 years | A1 | Cao must be 1 d.p. Note: obtained from exact values or using 1.00294 and 1.0714 or better. Allow full credit for trial and improvement giving 23.5 and £4778 to nearest pound. If M0M0: allow SC2 for 23.5 seen without £4778; allow SC1 for at least 2 trials clearly seen even if root not found |

**Total: [3]**

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