Question

Rahul deposited a sum of rupees 8000 at compound interest in the bank at the rate of 18% per annum for 3 years. what interest will he get after the end of 3 years​

Answer 1

Rahul deposited a sum of rupees 8000 at compound interest in the bank at the rate of 18% per annum for 3 years.

Here -

• Principle (P) = Rs. 8000

• Rate (R) = 18%

• Time (t) = 3 years

We know that..

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

Substitute the known values above

=> $\sf{A\:=\:8000\:(1\:+\:\frac{18}{100})^3}$

=> $\sf{A\:=\:8000\:(\frac{100\:+\:18}{100})^3}$

=> $\sf{A\:=\:800\:(\frac{118}{100})^3}$

=> $\sf{A\:=\:800\:\times\:\frac{118}{100}\:\times\:\frac{118}{100}\:\times\:\frac{118}{100}}$

=> $\sf{A\:=\:8000\:\times\:1.18\:\times\:1.18\:\times\:1.18}$

=> $\sf{A\:=\:8000\:\times\:1.64303}$

=> $\sf{A\:=\:Rs.\:13144.24}$

So,

Total interest in 3 years = Rs. 13144.24 - Rs. 8000

=> Rs. 5144.24

Answer 2

${\boxed{\tt{Given\::-}}}$

Here P = Rs 8000, R = 5% Per annum

and, n = 3.

Amount after 3 Year

${\boxed{\tt{Formula\:Applied:\:-}}}$

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

$⇒ P {(1 + \frac{R}{100}) }^{n}$

$⇒8000(1 + \frac{18}{100})^{3}$

$⇒8000 ( \frac{100 + 18}{100} )^{3}$

$⇒8000 ( \frac{118}{100} ) ^{3}$

$⇒ \: 8000( \frac{118}{100} \times \frac{118}{100} \times \frac{118}{100} )$

$⇒8000 \: \: (1.18 \times 1.18 \times 1.18)$

$⇒ \: 8000 \: (1.643032)$

$⇒ \: 13144.25$

Now, The Total interest in Total 3 Years = A—P

i.e., 13144.2—8000 = ₹5144.256