Practice MCQ Questions and Answer on Compound Interest

Solution: Time (t) = 1 years Rate % = 4% Amount = Rs. 7803 When interest is compounded half yearly New Rate = $$\frac{4}{2}$$ = 2% Time = 1 × 2 = 2 years Required rate% for 2 years CI $${\text{ = 2}} + {\text{2}} + \frac{{2 \times 2}}{{100}} = 4.04\% $$ According to question, (100 + 4.04)% of sum = Rs. 7803 $$\eqalign{ & \therefore {\text{Sum = }}\frac{{7803}}{{104.04}} \times 100 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\text{Rs}}{\text{. }}7500 \cr} $$

Solution: $$\eqalign{ & {\text{According to the question}} \cr & {\text{Principal = Rs S}} \cr & {\text{Rate }}\% {\text{ = 2r}}\,\% {\text{ p}}{\text{.a}}{\text{.}} \cr & {\text{Time = 3 years}} \cr & \therefore {\text{A = P}}{\left( {1 + \frac{{\text{r}}}{{100}}} \right)^T} \cr & \Leftrightarrow {\text{A = S}}{\left( {1 + \frac{{{\text{2r}}}}{{100}}} \right)^3} \cr & \Leftrightarrow {\text{A = S}}{\left( {1 + \frac{{\text{r}}}{{50}}} \right)^3} \cr} $$

Solution: $$\eqalign{ & 12000 \times {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^5} = 24000 \cr & \Rightarrow {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^5} = 2 \cr & \therefore {\left[ {{{\left( {1 + \frac{{\text{R}}}{{100}}} \right)}^5}} \right]^4} = {2^4} = 16 \cr & \Rightarrow {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^{20}} = 16 \cr & \Rightarrow {\text{P}}{\left( {1 + \frac{{\text{R}}}{{100}}} \right)^{20}}{\text{ = 16P}} \cr & \Rightarrow 12000{\left( {1 + \frac{{\text{R}}}{{100}}} \right)^{20}} = 16 \times 12000 \cr & \Rightarrow 12000{\left( {1 + \frac{{\text{R}}}{{100}}} \right)^{20}} = 192000 \cr} $$

Solution: Total cost of the flat = Rs. 55000 Down payment = Rs. 4275 Balance = Rs. (55000 - 4275) = Rs. 50725 Rate of interest = 8% per half year Let the value of each instalment be Rs. x P.W. of Rs. x due 6 months hence + P.W. of Rs. x due 1 year hence + P.W. of Rs. x due $$1\frac{1}{2}$$ years hence = 50725   $$ \Rightarrow \frac{x}{{\left( {1 + \frac{8}{{100}}} \right)}} + $$    $$\frac{x}{{{{\left( {1 + \frac{8}{{100}}} \right)}^2}}} + $$   $$\frac{x}{{{{\left( {1 + \frac{8}{{100}}} \right)}^3}}} = $$    $$50725$$ $$\eqalign{ & \Rightarrow \frac{{25x}}{{27}} + \frac{{625x}}{{729}} + \frac{{15625x}}{{19683}} = 50725 \cr & \Rightarrow \frac{{50725x}}{{19683}} = 50725 \cr & \Rightarrow x = \left( {\frac{{50725 \times 19683}}{{50725}}} \right) = 19683 \cr} $$

Solution: $$\eqalign{ & {\text{Principal (}}{{\text{P}}_1}{\text{) = Rs. 6000}} \cr & {\text{Time (t) = 2 years}} \cr & {\text{Rate % = 5% }} \cr & {\text{Simple interest}} \cr & {\text{ = }}\frac{{6000 \times 5 \times 2}}{{100}}{\text{ = Rs. 600}} \cr & {\text{Principal (}}{{\text{P}}_2}{\text{) = Rs. 5000}} \cr & {\text{Time (t) = 2 years}} \cr & {\text{Rate % = 8% }} \cr} $$ 2 year effective rate for Compound interest $$\eqalign{ & = 8 + 8 + \frac{{8 \times 8}}{{100}} = 16.64\% \cr & {\text{Compound}}\,{\text{Interest}} \cr & {\text{ = 5000}} \times \frac{{16.64}}{{100}} = {\text{Rs}}{\text{. 832}} \cr & {\text{Difference}} \cr & {\text{ = Rs}}{\text{. }}\left( {832 - 600} \right) \cr & = {\text{Rs}}{\text{.}}\,232 \cr & {\text{ }} \cr} $$

Solution: $$\eqalign{ & {\text{Let the rate be R}}\% {\text{ p}}{\text{.a}}{\text{. }} \cr & {\text{then,}} \cr & {\text{25500}}{\left( {1 + \frac{{\text{R}}}{{100}}} \right)^3} \cr & = 25500 + 8440.50 \cr & = 33940.50 \cr} $$   $$ \Rightarrow {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^3} = \frac{{33940.50}}{{25500}} = $$       $$\frac{{1331}}{{1000}} = $$  $${\left( {\frac{{11}}{{10}}} \right)^3}$$ $$\eqalign{ & \Rightarrow 1 + \frac{{\text{R}}}{{100}} = \frac{{11}}{{10}} \cr & \Rightarrow \frac{{\text{R}}}{{100}} = \frac{1}{{10}} \cr & \Rightarrow {\text{R}} = 10\,\% \cr & S.I. = {\text{R}}s.\left( {\frac{{25500 \times 10 \times 3}}{{100}}} \right) \cr & = {\text{Rs}}{\text{.}}\,7650 \cr} $$

Solution: S.I. on Rs. 4624 for 1 year $$\eqalign{ & {\text{ = Rs. }}\left( {4913 - 4624} \right) \cr & {\text{ = Rs. 289}} \cr & \therefore {\text{Rate}} = \left( {\frac{{100 \times 289}}{{4624 \times 1}}} \right)\% \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 6\frac{1}{4}\% \cr & {\text{Now,}} \cr & x{\left( {1 + \frac{{25}}{{400}}} \right)^2} = 4624 \cr & \Rightarrow x \times \frac{{17}}{{16}} \times \frac{{17}}{{16}} = 4624 \cr & \Rightarrow x = \left( {4624 \times \frac{{16}}{{17}} \times \frac{{16}}{{17}}} \right) \cr & \Rightarrow x = 4096 \cr} $$

Solution: Let the sum be Rs. x $$\eqalign{ & \therefore 1352 = x{\left( {1 + \frac{4}{{100}}} \right)^2} \cr & \Rightarrow 1352 = x{\left( {1 + \frac{1}{{25}}} \right)^2} \cr & \Rightarrow 1352 = x{\left( {\frac{{26}}{{25}}} \right)^2} \cr & \Rightarrow x = \frac{{1352 \times 25 \times 25}}{{26 \times 26}} \cr & \Rightarrow x = {\text{Rs}}{\text{.}}\,1250 \cr} $$

Solution: Answer & Solution Answer: Option C No explanation is given for this question Let's Discuss on Board

Solution: Let the principal = 1000, r = 10%, T = 3Year Rate r = 10% = $$\frac{1}{10}$$ Amount = 300 + 30 + 1 + 1000(Principal) = 1331 ⇒1331 unit → 5324 ⇒ 1unit → $$\frac{5324}{1331}$$ ∴ 1000unit = $$1000 \times \frac{5324}{1331}$$     = 4000