Question

A person invests rupees 10000 for the two years at a certain rate of interest, compounded annually. At the end of one year this sum amounts to rupees 11200. Calculate:(i) the rate of interest per annum. (ii) the amount at the end of second year.​

Answer 1

$\huge{\underline{\purple{\rm{Solution:}}}}$

$\large{\underline{\red{\rm{Given\:in\:1st\:case:}}}}$

• $\sf{Principal=Rs.10000}$
• $\sf{Amount=Rs.11200}$
• $\sf{Time=1\: year}$

$\large{\underline{\red{\rm{To\: Find:}}}}$

• $\rm\green{Rate\:of\: Interest}$
• $\rm\red{Amount\: after\:2\: year's}$

$\large{\underline{\red{\rm{Using\: formula\:Now:}}}}$

$\rm{\implies A=P(1+\dfrac{r}{100})^n}$

$\rm{\implies 11200=10000(1+\dfrac{r}{100})^1}$

$\rm{\implies \dfrac{28}{25}=\dfrac{100+r}{100}}$

$\rm{\implies 2500+25r=2800}$

$\rm{\implies 25r=2800-2800}$

$\rm{\implies \cancel{25}r=\cancel{300}}$

$\rm\red{\implies r=12\%}$

$\sf\purple{Now\:calculate\:amount\: for\:2\: year's}$

$\rm{\implies A=P(1+\dfrac{r}{100})^n}$

$\rm{\implies A=10000(1+\dfrac{12}{100})^2}$

$\rm{\implies A=10000(\dfrac{112}{100})^2}$

$\rm{\implies A=10000*\dfrac{112}{100}*\dfrac{112}{100}}$

$\rm{\implies A=112*112}$

$\rm\red{\implies A=Rs.12544}$

Hence,

$\sf\purple{Amount\:at\:the\:of\:2\: year's=Rs.12544}$

$\rm\green{Rate\:of\: Interest=12\%}$