Practice MCQ Questions and Answer on Compound Interest

Solution: $$\eqalign{ & {\text{Rate = 12}}\% \cr & {\text{Time = 2 years}} \cr & {\text{Effective rate of CI for 2 years}} \cr & {\text{ = 12 + 12 + }}\frac{{12 \times 12}}{{100}} \cr & = 25.44\,\% \cr & {\text{Effective rate of SI for 2 years}} \cr & {\text{ = 12}} \times 2{\text{ = 24}}\,\% \cr & {\text{According to question,}} \cr & {\text{Required SI}} \cr & {\text{ = }}\frac{{2544}}{{25.44}} \times {\text{24}} = {\text{ Rs. 2400}} \cr & {\text{Required sum}} \cr & {\text{ = Rs. 2400}} \cr} $$

Solution: $$\eqalign{ & 2{\text{ year}} \to {\text{CI}} \cr & = 10 + 10 + \frac{{10 \times 10}}{{100}} = 21\% \cr & {\text{In 2}}\frac{1}{2}{\text{ year CI}} \cr & = 21 + 5 + \frac{{21 \times 5}}{{100}} = 27.05\% \cr & P \times \frac{{27.05}}{{100}} = 1623 \cr & P = {\text{Rs}}{\text{. 6000}} \cr} $$

Solution: \[\left. \begin{gathered} {\text{Rate of interest}} = 6\% {\text{ year}} \hfill \\ {\text{Rate of interest}} = 8\% {\text{ year}} \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4\% {\text{ half year}} \hfill \\ {\text{Net rate}} = 4 + 4 + \frac{{16}}{{100}} = 8.16\% \hfill \\ \end{gathered} \right)\begin{array}{*{20}{c}} {2.16\mu \to 10800} \\ {1\mu \to 50} \\ {100\mu \to {\mathbf{5000}}} \end{array}\]

Solution: \[P = 10500\xrightarrow{{\,\,\,\,\,2{\text{ year}}\,\,\,\,\,}}13650\] $$\eqalign{ & 2{\text{ year}} \to 3150 \cr & 3150 = \frac{{10500 \times R \times 2}}{{100}} \cr & R = 15\% \cr & A = 10500{\left( {1 + \frac{{15}}{{200}}} \right)^2} \cr & = 10500 \times \frac{{215 \times 215}}{{200 \times 200}} \cr & = \frac{{105 \times 43 \times 43}}{{16}} \cr & = 12134 \cr} $$

Solution: $$\eqalign{ & 12\,{\text{months}} \to 12\% \cr & 8\,{\text{months}} \to 8\% \cr & 8\% \to \frac{2}{{25}} \cr} $$

Solution: 10% = $$\frac{1}{{10}}$$ \[\begin{array}{*{20}{c}} {}&{{\text{Principal}}\,\,\,\,\,\,\,\,\,{\text{Interest}}} \\ {{\text{I}} \to }&{{{10}_{ \times 11 = 110}}\,\,\,\,\,{{11}_{ \times 11 = 121}}} \\ {{\text{II}} \to }&{{{10}^2}_{ = 100}\,\,\,\,\,\,\,\,\,\,{{11}^2}_{ = 121}} \\ {{\text{Total}} \to }&{\overline {\underline {\,210\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,242\,\,\,} } } \end{array}\] Total interest on sum = 242 - 210 = 32 units 121 units → 5808 1 unit → 48 32 units → 48 × 32 = 1536 Total interest of 60% = 1536 × $$\frac{{60}}{{100}}$$ = Rs. 921.6 ∼ Rs. 922

Solution: $$\eqalign{ & {\text{A = P }}{\left( {1 + \frac{{\text{R}}}{{100}}} \right)^n} \cr & \Rightarrow 1348.32 = 1200{\left( {1 + \frac{{\text{R}}}{{100}}} \right)^2} \cr & \Rightarrow \frac{{134832}}{{120000}} = {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^2} \cr & \Rightarrow \frac{{231525}}{{200000}} = {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^2} \cr & \Rightarrow \frac{{2809}}{{2500}} = {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^2} \cr & \Rightarrow {\left( {\frac{{53}}{{50}}} \right)^2} = {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^2} \cr & \Rightarrow \frac{{53}}{{50}} = 1 + \frac{{\text{R}}}{{100}} \cr & \Rightarrow {\text{R}} = {\text{ 6% }} \cr} $$

Solution: $$\eqalign{ & {\text{Rate}} = 10\% ,\, \cr & {\text{Let}}\,{\text{Principal}} = P \cr & {\text{S}}{\text{.I}}{\text{.}} = \frac{{P \times 10 \times 3}}{{100}} = \frac{{3P}}{{10}} \cr & {\text{C}}{\text{.I}}{\text{.}} = P\left\{ {{{\left( {1 + \frac{1}{{10}}} \right)}^3} - 1} \right\} \cr & \Rightarrow {\text{C}}{\text{.I}}{\text{.}}\,\, - \,\,{\text{S}}{\text{.I}}{\text{.}} = 992 \cr & \Rightarrow P\left\{ {{{\left( {1 + \frac{1}{{10}}} \right)}^3} - 1} \right\} - \frac{{3P}}{{10}} = 992 \cr & \Rightarrow P\left\{ {{{\left( {\frac{{11}}{{10}}} \right)}^3} - 1 - \frac{3}{{10}}} \right\} = 992 \cr & \Rightarrow P\left\{ {\frac{{\left( {1331 - 1000 - 300} \right)}}{{1000}}} \right\} = 992 \cr & \Rightarrow P\left( {\frac{{31}}{{1000}}} \right) = 992 \cr & \Rightarrow P = 32000 \cr} $$ Alternate: Rate = 10% = $$\frac{1}{{10}}$$ Let principal ⇒ (10)3 = 1000 Interest 1st year → 100 Interest 2nd year → 100 + 10 Interest 3rd year = 100 + 10 + 10 + 1 C.I – S.I = 331 – 300 = 31 31 → 992 1 → 32 P → 32000

Solution: $$\eqalign{ & {\text{S}}{\text{.I}}{\text{. = Rs}}{\text{. }}\left( {15000 \times \frac{{25}}{2} \times 2 \times \frac{1}{{100}}} \right) \cr & \,\,\,\,\,\,\,\,\, = {\text{Rs}}{\text{. }}3750 \cr & {\text{C}}{\text{.I}}{\text{. = Rs}}{\text{. }}\left[ {15000{{\left( {1 + \frac{{25}}{{2 \times 100}}} \right)}^2} - 15000} \right] \cr & = {\text{Rs}}{\text{. }}\left( {15000 \times \frac{9}{8} \times \frac{9}{8} - 15000} \right) \cr & = {\text{Rs}}{\text{. }}\left( {18948.375 - 15000} \right) \cr & = {\text{Rs}}{\text{. }}3984.375 \cr & {\text{Difference }} \cr & {\text{ = Rs}}{\text{. }}\left( {3984.375 - 3750} \right) \cr & = {\text{Rs}}{\text{. }}234.375 \cr} $$

Solution: 8% → yearly 4% → half yearly 4 → $$\frac{1}{{25}}$$