Question

Find the amount and the compound interest on rs.2500 for 3 years at 6% per annum compounded annually

Answer 1

Given :-

• Rs.2500 for 3 years at 6% per annum compounded annually

To find :-

• The amount and the compound interest

Solution :-

As we know that

→ Amount = P(1 + R/100)ⁿ

Where " P " is Principal, " R " is rate of interest and " n " is time

According to question

• Principal = Rs.2500

• Time = 3 years

• Rate of interest = 6 %

Put the values

→ Amount = P(1 + R/100)ⁿ

→ A = 2500(1 + 6/100)³

→ A = 2500(1 + 3/50)³

→ A = 2500(50 + 3/50)³

→ A = 2500(53/50)³

→ A = 2500 × 53/50 × 53/50 × 53/50

→ A = 50 × 53 × 53/50 × 53/50

→ A = 53 × 53 × 53/50

→ A = 148877/50

→ A = Rs.2977.50

Hence,

• Amount is Rs.2977.5

Now,

→ Compound Interest = Amount - Principal

→ C.I = 2977.5 - 2500

→ C.I = Rs.477.5

Therefore,

• Compound interest is Rs.477.5

Answer 2

Answer:

• Amount = Rs. 2997.54
• Compound interest = Rs. 477.54

Step-by-step explanation:

Given that,

• Principal (P) = Rs. 2500
• Time (t) = 3 years
• Rate of interest (R) = 6%

As we know that,

$\red \bigstar \boxed{ \rm\orange {A = P(1 + \frac{R}{100})^t}}$

[ Putting values ]

$\leadsto \rm \: A = 2500(1 + \frac{6}{100} ) {}^{3} \\ \\ \leadsto \rm \: A =2500( \frac{100 + 6}{100} ) {}^{3} \\ \\ \leadsto \rm \: A =2500( \frac{106}{100} ) {}^{3} \\ \\ \leadsto\rm \: A =2500 \times \frac{106}{100} \times \frac{106}{100} \times \frac{106}{100} \\ \\ \leadsto \rm \: A =25 \times 106 \times 1.06 \times 1.06 \\ \\ \large \leadsto \rm \underbrace{ Amount = Rs. \: 2997.54 }\: \purple \bigstar$

_________________

Now, we know that,

$\pink\bigstar \boxed{ \rm \green{Compound \: interest = Amount - Principal }}$

[ Putting values ]

$\large\leadsto\rm\: Compound \: interest = Rs.(2997.54 - 2500) \\ \\ \large\leadsto\rm\: \underbrace{Compound \: interest = Rs.477.54} \: \: \blue \bigstar</p><p></p><p>$