Question

2. What would be the compound interest obtained on an amount of Rs. 12,000 at the rate of 9
p.c.p.a. for 3 years ?(rounded off to two digits after decimal)
(1) RS.3840
(2) Rs. 3740.75
(3) RS.3640
(4) Rs. 3540.35
(5) None of these​

Answer 1

$\red{\textbf{Answer}}$- (4) 3540.35

Step-by-step explanation:

Given :-

p = 12000

r = 9%

f = n = 3 years

$amount \: = p(1 + \large\frac{r}{100} {)}^{n}$

$amount \: = 12000(1 + \large\frac{9}{100} {)}^{3}$

$= 12000( \large\frac{109}{100} {)}^{3}$

$= 12000 \times \large\frac{109}{100}\times \large\frac{109}{100} \times \large\frac{109}{100}$

$Amount \: = 15540.348$

$C.I= A - P$

$= 15540.348 - 12000$

$C.I = 3540.348$

$\red{\textbf{More \: information}}$-

compound interest = $p(1 + \large\frac{r}{n} {)}^{nt}$

where ,

P = initial principal balance

r = interest rate

n = number of times interest applied per time period

t = number of time periods elapsed