Question

find CI on rupees 30000 at 12% per annum for 2 years

Answer 1

Solution ;-

Given ;-
• Principal = ₹30000
• Time = 2 years
• Rate = 12% p.a.

Now ,
Finding the the Compound Interest ;-

We know that ;-
A = P ( 1 + R /100 )n

So ,

$= > 30000(1 + \frac{12}{100})2$

$= > 30000 \times \frac{112}{100} \times \frac{112}{100}$

$= > 37632$

Then ,

CI = A - P

$= > 37632 - 30000 = 7632$

Hence ,
$<b>Compound Interest = ₹7632</b>$

Answer 2

Answer:

=> The compound interest will be Rs.7632.

Step-by-step explanation:

In context to the question asked,

We have to find the compound interest,

As per data given in the question,

We have,

Principal ( P ) = Rs. 30,000

Rate ( R ) = 12%

Time ( T ) = 2 years

We know that,

$C.I = A - P$

Now we have to calculate the amount,

For that the formula to be used,

$A=P (1+ \frac{R}{100})^t$

Substitute the given values,

$A=30,000 (1+ \frac{12}{100})^2$

$A=30000(1+0.12)^2$

$A=30000(1.12)^2$

$A=30000\times 1.12\times1.12$

$A = Rs.37632$

Now let us calculate C.I,

$C.I = A - P$

$C.I = 37632 - 30000$

$C.I = Rs.7632$

Hence, the required compound is Rs.7632