Principal is 10,000 rupess at a compound intreast of rate 10 percent . Find tbe difference between compound intreast and simple intreast at tbe same rate for 2 years ??
harvgupta
Answers
Answer:
Principal is 10,000 rupess at a compound interest of rate 10 percent per annum. Find tbe difference between compound interest and simple interest at the same rate for 2 years.
\large\underline{\sf{Solution-}}
Solution−
Case :- 1 Simple interest
Principal, P = Rs 10000
Rate of interest, r = 10 % per annum
Time, n = 2 years
We know,
Simple interest ( SI ) received on a certain sum of money of Rs P invested at the rate of r % per annum for n years is given by
\begin{gathered}\boxed{\sf{ \: \: SI \: = \: \frac{P \: \times \: r \times \: n}{100} \: \: }} \\ \end{gathered}
SI=
100
P×r×n
So, on substituting the values, we get
\begin{gathered}\rm \: SI \: = \: \frac{10000 \times 10 \times 2}{100} \: \\ \end{gathered}
SI=
100
10000×10×2
\begin{gathered}\rm\implies \:\boxed{\sf{ \: \: SI \: = \: 2000 \: }} \\ \end{gathered}
⟹
SI=2000
Case :- 2 Compound interest
Principal, P = Rs 10000
Rate of interest, r = 10 % per annum
Time, n = 2 years
We know,
C8 interest ( CI ) received on a certain sum of money of Rs P invested at the rate of r % per annum compounded annually for n years is given by
\begin{gathered}\boxed{\sf{ \:CI \: = \: P \: {\bigg[1 + \dfrac{r}{100} \bigg]}^{n} \: - \: P \: }} \\ \end{gathered}
CI=P[1+
100
r
]
n
−P
So, on substituting the values, we get
\begin{gathered}\rm \: CI \: = \: 10000 {\bigg[1 + \dfrac{10}{100} \bigg]}^{2} - 10000 \\ \end{gathered}
CI=10000[1+
100
10
]
2
−10000
\begin{gathered}\rm \: CI \: = \: 10000 {\bigg[1 + \dfrac{1}{10} \bigg]}^{2} - 10000 \\ \end{gathered}
CI=10000[1+
10
1
]
2
−10000
\begin{gathered}\rm \: CI \: = \: 10000 {\bigg[ \dfrac{10 + 1}{10} \bigg]}^{2} - 10000 \\ \end{gathered}
CI=10000[
10
10+1
]
2
−10000
\begin{gathered}\rm \: CI \: = \: 10000 {\bigg[ \dfrac{11}{10} \bigg]}^{2} - 10000 \\ \end{gathered}
CI=10000[
10
11
]
2
−10000
\begin{gathered}\rm \: CI \: = \: 10000 \times \frac{121}{100} - 10000 \\ \end{gathered}
CI=10000×
100
121
−10000
\begin{gathered}\rm \: CI \: = \: 12100 - 10000 \\ \end{gathered}
CI=12100−10000
\begin{gathered}\rm\implies \:\boxed{\sf{ \: \: CI \: = \: 2100 \: \: }} \\ \end{gathered}
⟹
CI=2100
Hence,
\begin{gathered}\rm \: CI \: - \: SI \\ \end{gathered}
CI−SI
\begin{gathered}\rm \: = \: 2100 - 2000 \\ \end{gathered}
=2100−2000
\begin{gathered}\rm \: = \: 100 \\ \end{gathered}
=100
Thus,
\begin{gathered}\rm\implies \:\boxed{\sf{ \:CI \: - \: SI \: = \: Rs \: 100 \: }} \\ \end{gathered}
⟹
CI−SI=Rs100
▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
Additional Information
1. Amount received on a certain sum of money of Rs P invested at the rate of r % per annum compounded annually for n years is given by
\begin{gathered}\boxed{\sf{ \:Amount\: = \: P \: {\bigg[1 + \dfrac{r}{100} \bigg]}^{n}\: \: }} \\ \end{gathered}
Amount=P[1+
100
r
]
n
2. Amount received on a certain sum of money of Rs P invested at the rate of r % per annum compounded semi - annually for n years is given by
\begin{gathered}\boxed{\sf{ \:Amount\: = \: P \: {\bigg[1 + \dfrac{r}{200} \bigg]}^{2n}\: \: }} \\ \end{gathered}
Amount=P[1+
200
r
]
2n
3. Amount received on a certain sum of money of Rs P invested at the rate of r % per annum compounded quarterly for n years is given by
\begin{gathered}\boxed{\sf{ \:Amount\: = \: P \: {\bigg[1 + \dfrac{r}{400} \bigg]}^{4n}\: \: }} \\ \end{gathered}
Amount=P[1+
400
r
]
4n
4. Amount received on a certain sum of money of Rs P invested at the rate of r % per annum compounded monthly for n years is given by
\begin{gathered}\boxed{\sf{ \:Amount\: = \: P \: {\bigg[1 + \dfrac{r}{1200} \bigg]}^{12n}\: \: }} \\ \end{gathered}
Amount=P[1+
1200
r
]
12n
Explanation:
how are you isha di
hope that helps you
Mark my answer in brainlist please
Answer:
compound intreast of rate 10 percent . Find tbe difference between compound intreast and
