Question

On the 1st of January 2014, Carol invested some money in a bank account. The account pays 2.5% compound interest per year. On the 1st of January 2015, Carol withdrew £1000 from the account. On the 1st of January 2016, she had £23 517.60 in the account. Work out how much Carol originally invested in the account?

Answer 1

She started with x.
After 1 year, she had 1.025x.
She withdrew £1000, so now she has 1.025x - 1000.
Then it earned interest for 1 year and ended up as 1.025(1.025x - 1000).
The actual amount of money was £23 517.60.
Therefore,

1.025(1.025x - 1000) = 23517.60

1.025x - 1000 = 22 944

1.025x = 23 944

x = 23 360

Her original deposit was £23 360

Answer 2

The compound interest paid by the account is applied to the account

balance.

• The amount she originally invested is £23,360.

Reasons:

The compound interest paid by the account per year = 2.5%

Amount carol withdrew from the account on 1st of January 2015 = £1,000

Amount she had on the 1st of January 2015 = £23,517.60

Therefore, we have;

$A = P \cdot \left(1 + \dfrac{r}{n} \right)^{n \cdot t}$

Which gives;

$A = P \cdot \left(1 +0.025 \right)^{ t}$

Amount in the account on the of January is $A_1 = P \cdot \left(1 +0.025 \right)^{ 1} = 1.025 \cdot P$

The principal from January 2015 = 1.025·P

After withdrawing 1,000, we get;

The principal = 1.025·P - 1000

The amount in the account in 2016 is given as follow;

£23,517.60 = (1.025·P - 1000)×(1.025)

$1.025 \cdot P - 1000 = \dfrac{23,517.60}{1.025} = 22,944$

$P = \dfrac{22944 + 1000}{1.025} = 23,360$

• The amount Carol originally invested, P = £23,360

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