3. A sum of money is lent at 8% per annumcompound interest. If the int

1.	3. A sum of money is lent at 8% per annumcompound interest. If the interest for thesecond year exceeds that for the first year by*96, find the sum of money.
Answer» $\large\underline{\sf{Solution-}}$ Let the sum of MONEY invested be Rs P. CASE :- 1 Principal = Rs P Time, N = 1 YEAR Rate, r = 8 % per annum So, Compound interest is given by $\tt{\implies CI=P\bigg(1+\dfrac{r}{100}\bigg)^{n}-P}$ $\tt{\longmapsto CI=P\bigg(1+\dfrac{8}{100}\bigg)^{1}-P}$ $\tt{\longmapsto CI=P\bigg(\dfrac{100 + 8}{100}\bigg)-P}$ $\tt{\longmapsto CI=P\bigg(\dfrac{108}{100}\bigg)-P}$ $\tt{\longmapsto CI=P(1.08)-P}$ $\tt{\longmapsto CI=0.08P} - - (1)$ Case :- 2 Principal = Rs P Time, n = 3 year Rate, r = 8 % per annum So, Compound interest is given by $\tt{\implies CI=P\bigg(1+\dfrac{r}{100}\bigg)^{n}-P}$ $\tt{\longmapsto CI=P\bigg(1+\dfrac{8}{100}\bigg)^{2}-P}$ $\tt{\longmapsto CI=P\bigg(\dfrac{100 + 8}{100}\bigg)^{2}-P}$ $\tt{\longmapsto CI=P\bigg(\dfrac{108}{100}\bigg)^{2}-P}$ $\tt{\longmapsto CI=P\bigg(1.08\bigg)^{2}-P}$ $\tt{\longmapsto CI=P\bigg(1.1664\bigg)-P}$ $\tt\longmapsto CI=0.1664P - - - (2)$ Now, According to statement, Cimpound interest for the second year exceeds that for the first year by Rs 96. $\tt\longmapsto \: (0.1664P \: - \: 0.08 P) - 0.08 P= 96$ $\tt\longmapsto \: 0.0064P \: = 96$ $\bf\implies \:P = 15000$ Hence, Sum invested is Rs 15000 Additional Information :- Let us assume that certain sum of money Rs P is invested at the rate of r % per annum compounded annually for n years then Amount is given by $\tt{\implies Amount=P\bigg(1+\dfrac{r}{100}\bigg)^{n}}$ Let us assume that certain sum of money Rs P is invested at the rate of r % per annum compounded half yearly for n years then Amount is given by $\tt{\implies Amount=P\bigg(1+\dfrac{r}{200}\bigg)^{2n}}$ Let us assume that certain sum of money Rs P is invested at the rate of r % per annum compounded QUARTERLY for n years then Amount is given by $\tt{\implies Amount=P\bigg(1+\dfrac{r}{400}\bigg)^{4n}}$