Question

the compound interest on rupees 36,000 lent on 10% per annum for 5\2 years, if the interest is compounded annually is?​

Answer 1

$\sf\large\underline{Given:}$

• $\rm{\implies Principal=Rs.36000}$
• $\rm{\implies Time=5/2=2\dfrac{1}{2}\: years}$
• $\rm{\implies Rate=10\%}$

$\sf\large\underline{To\: Find:}$

$\rm{\implies Compound\: interest=?}$

$\sf\large\underline{Solution:}$

• By applying formula here to calculate CI]

$\sf\large\underline{Formula\:used:}$

Note:-If time is given in fraction than we will use this formula here]

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

$\tt{\implies A=36000\bigg(1+\dfrac{10}{100}\bigg)^2\bigg(1+\dfrac{10}{100*2}\bigg)}$

$\tt{\implies A=36000\bigg(1+\dfrac{10}{100}\bigg)^2\bigg(1+\dfrac{10}{200}\bigg)}$

$\tt{\implies A=36000\bigg(\dfrac{100+10}{100}\bigg)^2\bigg(\dfrac{200+10}{200}\bigg)}$

$\tt{\implies A=36000\bigg(\dfrac{110}{100}\bigg)^2\bigg(\dfrac{210}{200}\bigg)}$

$\tt{\implies A=36000\bigg(\dfrac{11}{10}\bigg)^2\bigg(\dfrac{21}{20}\bigg)}$

$\tt{\implies A=36000\bigg(\dfrac{121}{100}\bigg)\bigg(\dfrac{21}{20}\bigg)}$

$\tt{\implies A=36000\times\dfrac{121}{100}\times\dfrac{21}{20}}$

$\tt{\implies A=\dfrac{360\times\:121\times\:21}{20}}$

$\tt{\implies A=18\times\:121\times\:21}$

$\tt{\implies A=Rs.45738}$

• Now calculate CI here]

$\tt{\implies CI=A-P}$

$\tt{\implies CI=45738-36000}$

$\tt{\implies CI=Rs.9738}$

$\sf\large{Hence,}$

$\tt{\implies Compound\: interest=Rs.9738}$