In which year will a sum of Rs. 800 at 20% compounded half-yearly exceed Rs.1000?

Let the time taken for this amount to reach Rs.1000 be X.

The important thing to note is that this sum is compounded half-yearly. Hence, we use the formula:

$(A = P{\left( {1 + \frac{r}{{m \times 100}}} \right)^{mT}})$

Where, A= Amount,

P = Principal

r = interest rate

m = no. of periods within a year

T = no. of years

We need to obtain T such that the RHS should be greater than the LHS:

$(\Rightarrow A < P{\left( {1 + \frac{r}{{m \times 100}}} \right)^{mT}})$

In this case,

A = 1000

P = 800

r = 20%

m = 2 (since it is half-yearly)

Substituting these values, we have

$(\begin{array}{l} 1000 < 800{\left( {1 + \frac{{20}}{{2 \times 100}}} \right)^{2T}} = 800{\left( {\frac{{11}}{{10}}} \right)^{2T}}\\ \Rightarrow \;\frac{{1000}}{{800}} < {\left( {\frac{{121}}{{100}}} \right)^T} \end{array})$

⇒ 1.25 < (1.21)T

Now, we need to use trial and error to check for the values of T.

For T = 1, 1.25 > 1.21, hence the condition is not satisfied.

For T = 2, 1.25 < 1.4641, hence the condition is met.

Therefore, it is the second year in which the amount would be greater than Rs.1000.