Practice MCQ Questions and Answer on Interest

Solution: Let the sum at 8% be Rs. x Let the sum at 5% be 1550 - x According to the given condition, $$\eqalign{ & \frac{{{\text{x}} \times 8 \times 3}}{{100}} + \frac{{\left( {1550 - {\text{x}}} \right) \times 5 \times 3}}{{100}} = 300 \cr & \Rightarrow \frac{{24{\text{x}} + \left( {1550 - {\text{x}}} \right) \times 15}}{{100}} = 300 \cr & \Rightarrow \frac{{24{\text{x}} + 23250 - 15{\text{x}}}}{{100}} = 300 \cr & \Rightarrow 24{\text{x}} - 15{\text{x}} = 30000 - 23250 \cr & \Rightarrow 24{\text{x}} - 15{\text{x}} = 6750 \cr & \Rightarrow 9{\text{x}} = 6750 \cr & \Rightarrow {\text{x}} = \frac{{6750}}{9} = 750 \cr & \frac{{{\text{Money lent at 5}}\% }}{{{\text{Money lent at 8}}\% }} \cr & = \frac{{1550 - 750}}{{750}} \cr & = \frac{{800}}{{750}} \cr & = \frac{{16}}{{15}} \cr & = 16:15 \cr} $$

Solution: Let Asif lent Rs. X at 14% per year. Hence, Money lent at 12% = (1500 - x); Given, total interest = Rs. 186 $$ {\frac{{\left( {x \times 14 \times 1} \right)}}{{100}}} \, + $$    $$ {\frac{{\left[ {\left( {1500 - x} \right) \times 12 \times 1} \right]}}{{100}}} $$      = 186 $$\eqalign{ & \frac{{14x}}{{100}} + \frac{{ {18000 - 12x} }}{{100}} = 186 \cr & 14x + 18000 - 12x = 186 \times 100 \cr & 2x = 18600 - 18000 \cr & x = \frac{{600}}{2} = {\text{Rs}}{\text{. }}300 \cr & {\text{Hence, money lent}}{\kern 1pt} {\text{at }}12\% \cr & = 1500 - 300 \cr & = {\text{Rs}}{\text{.}}\,1200 \cr} $$

Solution: $$\eqalign{ & {\text{Let the principal}} \cr & {\text{ = 9 units }} \cr & {\text{Hence simple interest}} \cr & {\text{ = }}\frac{4}{9} \times {\text{9 = 4 units}} \cr & {\text{Let,}} \cr & {\text{Rate of interest = R% }} \cr & {\text{R = T (given)}} \cr & {\text{By using formula}} \cr & \boxed{{\text{SI = }}\frac{{{\text{P}} \times {\text{T}} \times {\text{R}}}}{{100}}} \cr & 4 = \frac{{9 \times {\text{R}} \times {\text{R}}}}{{100}} \cr & {{\text{R}}^2} = \frac{{400}}{9} \cr & {\text{R = }}\frac{{20}}{3} = 6\frac{2}{3}\% \cr} $$

Solution: Let the rates of interest in the former and latter cases be R% and (R + 1) % p.a. Then, $$\eqalign{ & 5000 \times {\text{R}} \times 4 = 4000 \times \left( {{\text{R}} + 1} \right) \times 4 \cr & \Rightarrow \frac{{{\text{R}} + 1}}{{\text{R}}} = \frac{{5000 \times 4}}{{4000 \times 4}} \cr & \Rightarrow 1 + \frac{1}{{\text{R}}} = 1 + \frac{1}{4} \cr & \Rightarrow {\text{R}} = 4 \cr & {\text{Hence,}} \cr & {\text{Required rate}} = 4\% \,{\text{p}}{\text{.a}}{\text{.}} \cr} $$

Solution: $$\eqalign{ & {\text{Let}}, \cr & {\text{Principal}} = Rs.\,100 \cr & {\text{Amount}} = Rs.\,200 \cr & {\text{Rate}} = r\% \cr & {\text{Time}} = 4\,{\text{years}} \cr & {\text{Now}}, \cr & A = P \times {\left[ {1 + \left( {\frac{r}{{100}}} \right)} \right]^n} \cr & 200 = 100 \times {\left[ {1 + \left( {\frac{r}{{100}}} \right)} \right]^4} \cr & 2 = {\left[ {1 + \left( {\frac{r}{{100}}} \right)} \right]^4} - - - - - - \left( i \right) \cr & {\text{If}}\,{\text{sum}}\,{\text{become}}\,{\text{8}}\,{\text{times}}\,{\text{in}}\,{\text{the}}\,{\text{time}}\,n\,{\text{years}} \cr & {\text{then,}} \cr & 8 = {\left( {1 + \left( {\frac{r}{{100}}} \right)} \right)^n} \cr & {2^3} = {\left( {1 + \left( {\frac{r}{{100}}} \right)} \right)^n} - - - - - - \left( {ii} \right) \cr & {\text{Using}}\,{\text{eqn}}\,\left( i \right)in\left( {ii} \right),\,{\text{we}}\,{\text{get}} \cr & {\left( {{{\left[ {1 + \left( {\frac{r}{{100}}} \right)} \right]}^4}} \right)^3} = {\left( {1 + \left( {\frac{r}{{100}}} \right)} \right)^n} \cr & {\left[ {1 + \left( {\frac{r}{{100}}} \right)} \right]^{12}} = {\left( {1 + \left( {\frac{r}{{100}}} \right)} \right)^n} \cr & {\text{Thus}},\,n = 12\,{\text{years}}. \cr} $$

Solution: $$\eqalign{ & {\text{P}} = {\text{Rs}}{\text{. }}16800 \cr & {\text{R}} = 6\frac{1}{4}\% = \frac{{25}}{4}\% \cr & {\text{T}} = 9\,{\text{mths}} = \frac{3}{4}{\text{yr}}{\text{.}} \cr & \therefore {\text{S}}{\text{.I}}{\text{.}} = {\text{Rs}}{\text{.}}\left( {16800 \times \frac{{25}}{4} \times \frac{3}{4} \times \frac{1}{{100}}} \right) \cr & = {\text{Rs}}{\text{.}}\,787.50 \cr} $$

Solution: \[\frac{1}{4}\begin{array}{*{20}{c}} {\,\,\,\, \leftarrow {\text{S}}{\text{.I}}{\text{.}}} \\ { \leftarrow P} \end{array}\] $$\eqalign{ & {\text{Let }}P = 4x{\text{ and S}}{\text{.I}}{\text{.}} = x \cr & {\text{Time}} = t{\text{ years}} \cr & {\text{Rate}} = 4t\% \cr & \therefore {\text{S}}{\text{.I}}{\text{.}} = \frac{{P \times R \times T}}{{100}} \cr & x = \frac{{4x \times 4t \times t}}{{100}} \cr & {t^2} = \frac{{100}}{{16}} \cr & {t^2} = \frac{{25}}{4} \cr & \therefore t = \frac{5}{2} = 2.5{\text{ years}} \cr & \therefore R = 4 \times 2.5{\text{ years}} = 10\% \cr & {\text{New rate}} = 10\% + 2\% \cr & P = {\text{Rs}}{\text{. }}5000 \cr & T = 3{\text{ years}} \cr & {\text{S}}{\text{.I}}{\text{.}} = \frac{{P \times R \times T}}{{100}} \cr & = \frac{{5000 \times 12 \times 3}}{{100}} \cr & = {\text{Rs}}{\text{. }}1800 \cr} $$

Solution: $$\eqalign{ & {\text{Let sum}} = x \cr & {\text{Then,}} \cr & {\text{S}}{\text{.I}}{\text{.}} = \frac{x}{9}. \cr & {\text{Let rate}} = {\text{R}}\% \,{\text{and}} \cr & {\text{time}} = {\text{R}}\,{\text{years}}{\text{.}} \cr & \therefore \left( {\frac{{x \times {\text{R}} \times {\text{R}}}}{{100}}} \right) = \frac{x}{9} \cr & \Rightarrow {{\text{R}}^2} = \frac{{100}}{9} \cr & \Rightarrow {\text{R}} = \frac{{10}}{3} = 3\frac{1}{3} \cr & {\text{Hence, rate}} = 3\frac{1}{3}\% \cr} $$

Solution: $$\eqalign{ & {\bf{Given:}} \cr & {P_1} = 7000,\,{T_1} = 3{\text{ years, }}{R_1} = 5\% \cr & {P_1} = 7000,\,{T_2} = 3{\text{ years, }}{R_2} = 6\frac{1}{3}\% \cr & {\bf{Formula}}\,{\bf{used:}} \cr & {\text{S}}{\text{.I}}{\text{.}} = \frac{{{\text{Principal amout}} \times {\text{Rate of interest}} \times {\text{Time}}}}{{100}} \cr & {\bf{Calculation:}} \cr & {\text{S}}{\text{.I}}{{\text{.}}_1} = \frac{{{P_1} \times {R_1} \times {T_1}}}{{100}} \cr & = \frac{{7000 \times 5 \times 3}}{{100}} \cr & = 1050 \cr & {\text{S}}{\text{.I}}{{\text{.}}_1}{\text{ for one year}} = \frac{{1050}}{3} = 350 \cr & {\text{S}}{\text{.I}}{{\text{.}}_2} = \frac{{{P_1} \times {R_2} \times {T_2}}}{{100}} \cr & = \frac{{7000 \times \frac{{19}}{3} \times 3}}{{100}} \cr & = 1330 \cr & {\text{S}}{\text{.I}}{{\text{.}}_2}{\text{ for one year}} = \frac{{1330}}{3} = 443.33 \cr & {\text{Gain for one year}} = 443.33 - 350 = 93.33 \cr & \therefore {\text{He gain in the transaction per year 93}}{\text{.33}} \cr} $$

Solution: Let the sum invested at 9% be Rs. x and that invested at 11% be Rs. (100000 - x). Then, $$\eqalign{ & = \left( {\frac{{x \times 9 \times 1}}{{100}}} \right) + \left[ {\frac{{\left( {100000 - x} \right) \times 11 \times 1}}{{100}}} \right] \cr & = \left( {100000 \times \frac{{39}}{4} \times \frac{1}{{100}}} \right) \cr & \Leftrightarrow \frac{{9x + 1100000 - 11x}}{{100}} = \frac{{39000}}{4} = 9750 \cr & \Leftrightarrow 2x = \left( {1100000 - 975000} \right) = 125000 \cr & \Leftrightarrow x = 62500 \cr & \therefore {\text{Sum invested at 9}}\% \cr & = {\text{Rs}}{\text{. }}62500 \cr & {\text{Sum invested at 11% }} \cr & {\text{ = Rs}}{\text{.}}\left( {100000 - 62500} \right) \cr & = {\text{Rs}}{\text{. }}37500 \cr} $$