Practice MCQ Questions and Answer on Compound Interest

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: Let the R% p.a. Then, $$\left[ {26000 \times {{\left( {1 + \frac{{\text{R}}}{{100}}} \right)}^3} - 26000} \right] - $$       $$\left( {\frac{{26000 \times {\text{R}} \times 3}}{{100}}} \right) = $$     $$2994.134$$ $$ \Rightarrow 26000\left[ {{{\left( {1 + \frac{{\text{R}}}{{100}}} \right)}^3} - 1 - \frac{{3{\text{R}}}}{{100}}} \right] = $$        $$2994.134$$ $$ \Rightarrow 26000$$ $$\left[ {\frac{{{{\left( {100 - {\text{R}}} \right)}^3} - 1000000 - 30000{\text{R}}}}{{1000000}}} \right] = $$        $$2994.134$$ $$ \Rightarrow 26\left[ {\left\{ {1000000 + {{\text{R}}^3} + 300{\text{R}}\left( {100 + {\text{R}}} \right) - 1000000 - 30000{\text{R}}} \right\}} \right] = 2994134$$ $$ \Rightarrow {{\text{R}}^3} + 300{{\text{R}}^2} = \frac{{2994134}}{{26}} = $$      $$115159$$ $$ \Rightarrow {{\text{R}}^2}\left( {{\text{R}} + 300} \right) = 115159$$ $$ \Rightarrow {\text{R = 19}}\% $$

Solution: $$\eqalign{ & 28.90 = x{\left( {\frac{{8.5}}{{100}}} \right)^2} \cr & 28.90 = \frac{{x \times 8.5 \times 8.5}}{{10000}} \cr & 1.7 = \frac{{x \times 0.5 \times 8.5}}{{10000}} \cr & 0.1 \times 10000 = x \times 0.5 \times 0.5 \cr & x = {\text{Rs}}{\text{. }}4000 \cr} $$

Solution: $$\eqalign{ & {\text{Effective rate of CI for 2 years}} \cr & {\text{= 5 + 5 + }}\frac{{5 \times 5}}{{100}} \cr & = 10.25\% \cr & {\text{Effective rate of SI for 3 years}} \cr & {\text{= 6}} \times {\text{3 = 18% }} \cr & {\text{According to question}} \cr & {\text{Required SI}} \cr & {\text{= }}\frac{{246}}{{10.25}} \times 18 \cr & = {\text{Rs. 432}} \cr} $$

Solution: Given, Amount = 12,100; r = 10%, t = 2 yrs $$\eqalign{ & {\text{Amount}} = P{\left[ {1 + \frac{r}{{100}}} \right]^t} \cr & 12100 = P{\left[ {1 + \frac{{10}}{{100}}} \right]^2} \cr & \Rightarrow 12100 = P{\left[ {\frac{{11}}{{10}}} \right]^2} \cr & \Rightarrow 12100 = P \times \frac{{11}}{{10}} \times \frac{{11}}{{10}} \cr & \Rightarrow P = \frac{{12100 \times 10 \times 10}}{{11 \times 11}} \cr & \Rightarrow P = 10000 \cr} $$

Solution: $$\eqalign{ & {\text{Amount}} = Rs.\,\left[ {25000 \times {{\left( {1 + \frac{{12}}{{100}}} \right)}^3}} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = Rs.\,\left( {25000 \times \frac{{28}}{{25}} \times \frac{{28}}{{25}} \times \frac{{28}}{{25}}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = Rs.\,35123.20 \cr & \therefore {\text{C}}{\text{.I}}{\text{.}} = Rs.\left( {35123.20 - 25000} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = Rs.\,10123.20 \cr} $$

Solution: Let the share of the younger and elder sons be Rs. x and Rs. (16400 - x) Then, amount of Rs. x after 3 years = Amount of Rs. (16400 - x) after 2 years $$\eqalign{ & \Rightarrow x{\left( {1 + \frac{5}{{100}}} \right)^3} = \left( {16400 - x} \right){\left( {1 + \frac{5}{{100}}} \right)^2} \cr & \Rightarrow x\left( {1 + \frac{5}{{100}}} \right) = \left( {16400 - x} \right) \cr & \Rightarrow \frac{{21x}}{{20}} + x = 16400 \cr & \Rightarrow \frac{{41x}}{{20}} = 16400 \cr & \Rightarrow x = \left( {\frac{{16400 \times 20}}{{41}}} \right) \cr & \Rightarrow x = 8000 \cr} $$

Solution: $$\eqalign{ & {\text{Amount}} = Rs.\,\left( {30000 + 4347} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = Rs.\,34347 \cr & {\text{Let}}\,{\text{the}}\,{\text{time}}\,{\text{be}}\,n\,{\text{years}} \cr & {\text{Then}},30000\,{\left( {1 + \frac{7}{{100}}} \right)^n} = 34347 \cr & \Rightarrow {\left( {\frac{{107}}{{100}}} \right)^n} = \frac{{34347}}{{30000}} = \frac{{11449}}{{10000}} = {\left( {\frac{{107}}{{100}}} \right)^2} \cr & \therefore n = 2\,{\text{years}} \cr} $$

Solution: $$\eqalign{ & {\text{Rate of interest (r)}} \cr & {\text{ = 8}}\frac{3}{4}\% = \frac{7}{{80}} = \frac{{87 \to {\text{ Installment}}}}{{80 \to {\text{Principal}}}} \cr} $$ ⇒     I     80×87     →     87×87   ......(i) ⇒     II     6400     →     7569   ......(ii) Since, installment is equal hence multiply equation (i) by 87 ⇒ Total principal = 6960 + 6400 = 13360 ⇒ 13360 units = Rs. 13360 ⇒ 1 units = Rs. 1 ⇒ 7569 units = Rs. 7569 ∴ Each installment = Rs. 7569

Solution: $$\eqalign{ & \Rightarrow P\left[ {{{\left( {\frac{{21}}{{20}}} \right)}^3} - 1 - \frac{3}{{20}}} \right] = 122 \cr & \Rightarrow P\left[ {\frac{{{{21}^3} - {{20}^3} - 3 \times {{20}^2}}}{{{{20}^3}}}} \right] = 122 \cr & \Rightarrow P\left[ {\frac{{9261 - 8000 - 1200}}{{8000}}} \right] = 122 \cr & \Rightarrow P \times \frac{{61}}{{8000}} = 122 \cr & \Rightarrow P = \frac{{8000 \times 122}}{{61}} \cr & \Rightarrow P = {\text{Rs}}{\text{.}}\,16000 \cr} $$