Question

hey friends! what's up please answer my question fast
the compound interest on Rs. 10000 at at 12% per annumm for 1.5 years compounded annually is​

Answer 1

Hey friend!!!

Here is your answer..

⬇️⬇️⬇️⬇️⬇️⬇️

P= Rs.10000

R=12%

t=1.5 year

$a = p {(1 + \frac{r}{100} })^{t} \\ a = 10000( {1 + \frac{12}{100} )}^{1.5}$

Solve the eq.

Get the value of A

Then,

Interest =A-P

HOPE that it will help you ✌️ ✌️

Answer 2

Answer:

1872

Step-by-step explanation:

Given, Principal = 10000, R = 12%, Time n = 1.5 years.

(i) First we take n = 1 year.

A = P(1 + r/100)ⁿ

  = 10000(1 + 12/100)

  = 10000(112/100)

  = 10000 * 1.12

  = 11200

So, Amount after 1 year = 11200.

Now, Principal = 11200.

(ii) For last 1/2 year:

Simple Interest = PRT/100

                          = (11200 * 12 * 1)/200

                          = 672

So, Amount after 1 (1/2) years = 11200 + 672

                                                 = 11872

Now,

Compound Interest = Amount - Principal

                                 = 11872 - 10000

                                 = 1872.

Therefore, Compound Interest = 1872.

Hope it helps!