Wednesday, November 29, 2023
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### EteacherG

a educational group

# Compound Interest Formula | Compound Interest Examples and Compound Interest Definition

Compound Interest Today this Online tutoring Class in “EteacherG” we are study about Compound Interest. In this chapter we also read about Compound Interest Formula and Compound Interest Example, Annual Compound Interest Formula. This is a important chapter and many question comes in it.

# Compound Interest

### Definition of Compound Interest –

“When interest to be taken is never not a simple. We calculate interest on last year Principal then this combination called Compound Interest.”

## Compound Interest Formula

$$\displaystyle CI=P{{\left( {1+\frac{R}{N}} \right)}^{{-NT}}}-P$$
Here : CI = Compound Interest
P = Principal
R = Compound Interest Rate
T = Time
N = Compounding Frequency Per Annum
Note : If the compounding frequency per annum is 1 i.e. if the interest is compounding annually, the Compound Interest formula is given below.
$$\displaystyle C.I=P{{\left( {1+\frac{R}{{100}}} \right)}^{T}}-P$$

$$\displaystyle A=P{{\left( {1+\frac{R}{N}} \right)}^{{-NT}}}$$
Here : A = Total Accumulated Amount
P = Principal Amount
R = Compound Interest Rate
T = Time
N = Compounding Frequency Per Annum

$$\displaystyle A=P{{\left( {1+\frac{R}{{100}}} \right)}^{T}}$$

## Compound Interest Example

Above we learn about compound interest Definition and compound interest formula. To see the compound interest formula we know that we need Principal, compound interest rate and time to calculate it. So lets start to do solve some important Compound Interest Example –

Q.1 Find the amount if $10,000 is invested at 10% p.a for 2 years. Solution : First we write value which are given in question Principal (P) =$ 10000
Rate of Compound Interest (R) = 10% (P.A.)
Time (T) = 2 years
Amount (A) = To be known
$$\displaystyle A=P{{\left( {1+\frac{R}{{100}}} \right)}^{T}}$$
Put the value in the compound interest formula
$$\displaystyle \begin{array}{l}A=10000{{\left( {1+\frac{{10}}{{100}}} \right)}^{2}}\\=10000{{\left( {\frac{{110}}{{100}}} \right)}^{2}}\\=10000\times \frac{{110}}{{100}}\times \frac{{110}}{{100}}\end{array}$$
A = 110 × 110
A = 12100
so the total Amount is $12100. Compound Interest (C.I) = Amount (A) – Principal (P) C.I = 12100 – 10000 C.I =$ 2,100

Q.2 Find the Compound Interest (CI), if $2000 was invested for 1.5 years at 20% p.a. compounded half yearly. Solution : First we write value which are given in question Principal (P) =$ 2000
Rate of Compound Interest (R) = 20% (H.Y)
Time (T) = 1.5 years or $$\displaystyle \frac{1}{2}$$
Compound Interest (CI) = To be known
$$\displaystyle C.I=P{{\left( {1+\frac{R}{{100}}} \right)}^{T}}-P$$
because here Compound Interest Rate is compounded half yearly so Time will be doubled and Interest Rate will be half. lets do it.
so now Time (T) = 3 years
and Compound Interest Rate (R) = 10%
$$\displaystyle C.I=P{{\left( {1+\frac{R}{{100}}} \right)}^{T}}-P$$
Put the value in the compound interest formula
$$\displaystyle \begin{array}{l}CI=2000{{\left( {1+\frac{{10}}{{100}}} \right)}^{3}}-2000\\=2000{{\left( {\frac{{110}}{{100}}} \right)}^{3}}-2000\\=2000\times \frac{{11}}{{10}}\times \frac{{11}}{{10}}\times \frac{{11}}{{10}}-2000\end{array}$$
CI = (2 × 11 × 11 × 11) – 2000
CI = 2662 – 2000
CI = $662 Amount = Principal Amount (P) +Compound Interest (CI) = 2000 + 662 =$ 2000

Q.3 Mr. Steve purchased a refrigerator for $2500. Five year later he sold this refrigerator for$1500. If the refrigerator is viewed as an investment, what is rate did he loss?

Solution :
If we view this as an investment of P = $2500 and the future value A =$1500
In Time = 5 years
here we will use N = 1 and then
$$\displaystyle A=P{{\left( {1+\frac{R}{N}} \right)}^{{-NT}}}$$
put the value in the formula
$$\displaystyle \begin{array}{l}1500=2500{{\left( {1+\frac{R}{1}} \right)}^{{1\times 5}}}\\1500=2500{{\left( {1+R} \right)}^{5}}\\\frac{{1500}}{{2500}}={{\left( {1+R} \right)}^{5}}\\\frac{3}{5}={{\left( {1+R} \right)}^{5}}\end{array}$$
Take the left-hand side to 1/5th power to clear the power of 5 on the right side.
$$\displaystyle \begin{array}{l}\frac{3}{5}={{\left( {1+R} \right)}^{5}}\\{{\left( {\frac{3}{5}} \right)}^{{\frac{1}{5}}}}=\left( {1+R} \right)\\\sqrt{{\frac{3}{5}}}=\left( {1+R} \right)\\\frac{{\sqrt{3}}}{{\sqrt{5}}}=1+R\end{array}$$
Rationalize the denominator
$$\displaystyle \begin{array}{l}\frac{{\sqrt{{3\times {{5}^{4}}}}}}{5}=1+R\\\frac{{\sqrt{{3\times 625}}}}{5}=1+R\\\frac{{\sqrt{{1875}}}}{5}=1+R\end{array}$$
0.9028805 = 1 + R
Transfer the variable
R = -1 + 0.9028805
R = – 0.0971195% (Approximately)
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