Question

Sneha borrows 10000 rupees
for 2 years at the rate of 4% interest
while the interest is compounded
annually. How much compound interest
will she pay after 2 years?
a) 760
b) 660
c) 812
d) 816

​

Answer 1

Step-by-step explanation:

Principal (P) = ₹10000

Time (n) = 2 years

Rate (R) = 4%

$\sf \dashrightarrow Amount = P\bigg(1 + \dfrac{r}{100}\bigg)^{n}$

$\sf \dashrightarrow A= 10000\times \bigg(1 + \dfrac{4}{100}\bigg)^{2}$

$\sf \dashrightarrow A= 10000\times \bigg(\dfrac{100 + 4}{100}\bigg)^{2}$

$\sf \dashrightarrow A= 10000\times \bigg( \dfrac{104}{100}\bigg)^{2}$

$\sf \dashrightarrow A= 10000\times \bigg(\dfrac{26}{25}\bigg)^{2}$

$\sf \dashrightarrow A= 10000\times \dfrac{26 \times 26 }{25 \times 25}$

$\sf \dashrightarrow A= 10000\times \dfrac{676 }{625}$

$\sf \dashrightarrow A= 10816$

$\sf \therefore Amount = Rs.10816$

$\sf Compound \: Interest = Amount - Principal$

$\sf \dashrightarrow CI = 10816 - 10000$

$\sf \dashrightarrow CI = 816$

$\underline{\boxed{\sf \therefore Compound \: Interest = Rs.816}}$