Question

If A = P {1 + ( R/100)} power n find R when. A= 1102.50, P= 1000 n= 2


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Answer 1

Amount = ₹ 1102.50

Principal = ₹ 1000

Time = 2 yrs

Let Rate of Interest be R

$a \: = \: p \: {( \: 1 + \frac{r}{100})}^{n}$

$1102.50 \: = \: 1000 \: {(1 + \frac{r}{100} )}^{2} \\ \\ \frac{1102.50}{1000} = {(1 + \frac{r}{100} )}^{2} \\ \\ \frac{441}{400} = {(1 + \frac{r}{100} )}^{2} \\ \\ \frac{ \sqrt{441} }{\sqrt{400} } = {(1 + \frac{r}{100} )}^{2} \\ \\ {(\frac{21}{20})}^{2} = {(1 + \frac{r}{100} )}^{2} \\ \\ \frac{21}{20} = 1 + \frac{r}{100} \\ \\ \frac{21}{20} - 1 = \frac{r}{100} \\ \\ \frac{21 - 20}{20} = \frac{r}{100} \\ \\ \frac{1}{20} = \frac{r}{100} \\ \\ \frac{100}{20} = r \: \\ \\ r = 5\%$

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