Question

A sum of Rs. 1000 is invested for 5 years
at rate of 12% per annum. Find
compound interest and due amount​

Answer 1

Given:

• Principal = Rs. 1000
• Rate = 12 %
• Time = 5 years

What To Find:

We have to find -

• Compound Interest and amount.

Formulas Needed:

$\bf \to A = P \bigg[ 1 + \dfrac{R}{100} \bigg]^{T}$

$\bf \to CI = A - P$

Solution:

• Finding the A.

Use the formula,

$\bf \to A = P \bigg[ 1 + \dfrac{R}{100} \bigg]^{T}$

Substitute the values,

$\bf \to A = 1000 \bigg[ 1 + \dfrac{12}{100} \bigg]^{5}$

Solve the brackets,

$\bf \to A = 1000 \bigg[ \dfrac{112}{100} \bigg]^{5}$

Remove the brackets,

$\bf \to A = 1000 \times \dfrac{112}{100} \times \dfrac{112}{100} \times \dfrac{112}{100} \times \dfrac{112}{100} \times \dfrac{112}{100}$

Solve the fractions,

$\bf \to A = 1000 \times \dfrac{17623416832}{10000000000}$

Cancel the zeros.

$\bf \to A = \dfrac{17623416832}{10000000}$

Divide the numbers,

$\bf \to A = Rs. \: 1762.3 - [approx.\!]$

• Finding the CI.

Use the formula,

$\bf \to CI = A - P$

Substitute the values,

$\bf \to CI = 1762.3 - 1000$

Subtract the numbers,

$\bf \to CI = Rs. \: 762.3$

Final Answer:

∴ Thus, the amount is Rs. 1762.3 and the compound interest is Rs. 762.3.

Answer 2

Given :

• Principal (P) = ₹1000
• Time or period (n) = 5years
• Rate (R) = 12%

Find :

• The amount and compound interest.

Solution :

A = P (1 +R/100)^n

$= > 1000 (\frac{100 + 12}{100}) ^{5}$

$= > 1000 (\frac{112}{100}) ^{5}$

$= > 1000 \times \frac{112}{100} \times \frac{112}{100} \times \frac{112}{100} \times \frac{112}{100} \times \frac{112}{100}$

$= > 1762.32$

CI = A - P

=> ₹1762.32 - ₹1000

=> ₹762.32

Hence :

The amount is ₹1762.32 and compound interest is ₹762.32.