Question

What is the compounded annual interest for a loan amount of800000 at 10% per annum for a period of 2 year

Answer 1

Given:

• Principal,P = 80,000

• Rate,r = 10%

• Time, n = 2 years

To find out:

Find the Amount and Compound interest.

Formula used:

$\boxed { \red{A = P \times (1 + \frac{r}{100} ) {}^{n} }}$

$\boxed{ \red{Compound \: interest = A - P}}$

Solution:

$A =800000 \times (1 + \frac{10}{100}) {}^{2}$

$= 800000 \times ( \frac{100 + 10 }{100} ) {}^{2}$

$= 800000 \times ( \frac{110}{100} ) {}^{2}$

$= 800000 \times \frac{11 0}{100} \times \frac{110}{100}$

$= 800000 \times \frac{12100}{10000}$

$= 8000 \times 121$

$= 968000$

Now,

$Compound \: interest = A - P$

$= 968000 - 800000$

$= 168000$

Answer 2

Given →

Principal Amount (P) = 800000

Rate (r) = 10 %

Time (n) = 2 years

To Find →

The compound Interest.

$\rule{200}{1}$

Answer →

Step by step Explanation →

We know that →

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

So ,

$a = 800000( {1 + \frac{10}{100} )}^{2} \\ \\ a = 800000( {1 + 0.1)}^{2} \\ a = 800000( {1.1)}^{2}$

$a = 800000 \times 1.21$

$a = 968000$

Now ,

• Compound Interest = A - P

Compound Interest = 968000 - 800000

$\boxed{Compound \:Interest = 168000 }$

$\rule{200}{1}$

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