Question

If rs.1000 amounts to rs.2197 at compound intrest in 3 years,then rate of intrest oer annum is

Answer 1

The  rate of interest per annum 30%.

Step-by-step explanation:

Given:

If rs.1000 amounts to rs.2197 at compound interest in 3 years.

To Find:

The  rate of interest per annum.

Formula used:

$X=Y(1+\frac{Z}{100} )^{T}$   -------------- formula no.01.

X= represents the new principal sum or the total amount of money after compounding period

Y= represents the original amount or initial amount

Z= is the annual interest rate(in percentage)

T= represents the number of years

Solution:

As given,if rs.1000 amounts to rs.2197 at compound interest in 3 years.

X= Rs.2197,    Y=  Rs.1000   and 3 years.

Applying the formula no.01.

$2197=1000(1+\frac{Z}{100} )^{3}$

$\frac{2197}{1000} =(1+\frac{Z}{100} )^{3}$

$(\frac{13}{10})^{3} =(1+\frac{Z}{100} )^{3}$

$\frac{13}{10}=1+\frac{Z}{100}$

$\frac{13}{10}-1=\frac{Z}{100}$

$\frac{13-10}{10}=\frac{Z}{100}$

$\frac{3}{10}=\frac{Z}{100}$

$Z=\frac{3\times100}{10}$

$Z=30\%$

Thus,the  rate of interest per annum 30%.

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Answer 2

Given principal is $Rs. 1000$

Principal+interest = 2197

Time = 3 years

What is compound interest?

The formula for finding compound interest

$A=P(1+\frac{r}{100})^n$, where

A = 2197

P = 1000

n = 3

We need to find the rate of interest $r\%$

$2197=1000(1+\frac{r}{100})^3\\\Rightarrow (1+\frac{r}{100})^3=\frac{2197}{1000}\\\Rightarrow (1+\frac{r}{100})=\sqrt[3]{\frac{2197}{1000}}\\\Rightarrow (1+\frac{r}{100})=\sqrt[3]{(\frac{13}{10})^3}\\\Rightarrow 1+\frac{r}{100}=\sqrt[3]{(\frac{13}{10})^3}\\\Rightarrow (1+\frac{r}{100})=\frac{13}{10}\\\Rightarrow \frac{r}{100}=\frac{13}{10}-1\\\Rightarrow \frac{r}{100}=\frac{13-10}{10}\\\Rightarrow \frac{r}{100}=\frac{3}{10}\\\Rightarrow r=\frac{3}{10} \times 100\\\Rightarrow r=30$

Therefore, the rate of interest is 30%

To learn more about compound interest from the given link

https://brainly.in/question/1077196

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