Question

Calculate the compound interest for the second year on ₹ 8000 invested for 3 years at 10% p.a. also find the sum due at the end of third year

Answer 1

We know,
$P(1 + \frac{r}{100})^{n} = C.I \\ = 8000(1 + \frac{10}{100})^{3} \\ = 8000 \times \frac{1331}{1000} \\ = 10648 Rs \\ Sum \: due \: at \: the \: end \: of \: three \: years = C.I - P = (10648-8000) Rs \\ = Rs. 2648$

Answer 2

It is given that

Principal = ₹ 8000

Rate of interest = 10% p.a.

We know that

Interest for the first year = Prt/100

Substituting the values

= (8000 × 10 × 1)/ 100

= ₹ 800

So the amount after the first year or principal for the second year = 8000 + 800 = ₹ 8800

(i) Interest for the second year = (8800 × 10 × 1)/ 100

= ₹ 880

So the amount after second year or principal for the third year = 8800 + 880 = ₹ 9680

Interest for the third year = (9680 × 10 × 1)/ 100

= ₹ 968

(ii) Amount due at the end of the third year = 9680 + 968

= ₹ 10648