Question

What is the compound interest 2000 and the rate is 10%per annum compounded annually for 2years​

Answer 1

Given:

Principal (p) = Rs. 2000

Time (t) = 2 years

Rate (r) = 10%

Find:

Compound Interest (C.I)

Solution:

We have,

The Amount (A) = p(1 + r/100)n

We know that,

=> A = p(1 + r/100)n

=> A = 2000(1 + 10/100)²

=> A = 2000(100 + 10/100)²

=> A = 2000(110/100)²

=> A = 2000 × (11/10)²

=> A = 2000 × 11/10 × 11/10

=> A = 20 × 11 × 11

=> A = 2420

The Compound Interest = C.I = Amount - Principal

=> C.I = 2420 - 2000

=> C.I = 420

Hence, the Compound Interest is Rs. 420.

Important Information:

Simple Interest = (P × R × T) ÷ 100

Amount = SI + P

A = {(P × R × T) ÷ 100} + P

Where,

• SI = Simple Interest
• A = Amount/Future Value
• P = Principal Amount
• R = Rate of Interest per annum
• T = Time in years

I hope it will help you.

Regards.

Answer 2

GIVEN :-

• Principal ( P ) = Rs. 2000.
• Time ( n ) = 2 years.
• Rate ( R ) = 10 %.

TO FIND :-

• The compound interest ( CI ).

SOLUTION :-

As we know that,

$\\ : \implies \displaystyle \sf \: CI = P\bigg[ \bigg(1 + \dfrac{R}{100}\bigg)^{n} - 1\bigg]$

$: \implies \displaystyle \sf \: CI = 2000\bigg[ \bigg(1 + \dfrac{10}{100}\bigg)^{2} - 1\bigg]$

$: \implies \displaystyle \sf \: CI = 2000\bigg[ \bigg( \dfrac{100 + 10}{100}\bigg)^{2} - 1\bigg]$

$: \implies \displaystyle \sf \: CI = 2000\bigg[ \bigg( \dfrac{110}{100}\bigg)^{2} - 1\bigg]$

$: \implies \displaystyle \sf \: CI = 2000\bigg[ \bigg( \dfrac{110 \times 110}{100 \times 100}\bigg) - 1\bigg]$

$: \implies \displaystyle \sf \: CI = 2000\bigg[ \dfrac{12100}{10000} - 1\bigg]$

$: \implies \displaystyle \sf \: CI = 2000\bigg[ \dfrac{12100 - 10000}{10000} \bigg]$

$: \implies \displaystyle \sf \: CI = 2000\bigg[ \dfrac{2100 }{10000} \bigg]$

$: \implies \displaystyle \sf \: CI =2000 \times \frac{2100}{10000}$

$: \implies \displaystyle \sf \: CI =2 \times 210$

$: \implies \underline{ \boxed{ \displaystyle \sf \: CI =420}}$