Compound Interest - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & {\text{Rate of interest}} \Rightarrow {\text{ 10% = }}\frac{1}{{10}} \cr & {\text{Each installment of 2 years}} \cr & \Rightarrow \frac{{10}}{{11}} \times \frac{{\left( {10 + 11} \right)}}{{11}} \times {\text{ Installment = P}}{\text{.A}} \cr & \Rightarrow \frac{{10}}{{11}} \times \frac{{\left( {10 + 11} \right)}}{{11}} \times {\text{ Installment = 210}} \cr & \Rightarrow {\text{Installment = 121}} \cr} $$

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

$$\eqalign{ & {\text{According to the question,}} \cr & {\text{compounded half yearly}} \cr & {\text{Rate = }}\frac{{\text{R}}}{2} \cr & {\text{Time = }}\frac{{{\text{2T}}}}{3} \cr & {\text{Amount = P}}{\left( {1 + \frac{{\text{R}}}{{2 \times 100}}} \right)^3} \cr & \Rightarrow 2315.25 = 2000{\left( {1 + \frac{{\text{R}}}{{2 \times 100}}} \right)^3} \cr & \Rightarrow \frac{{2315.25}}{{2000}} = {\left( {1 + \frac{{\text{R}}}{{2 \times 100}}} \right)^3} \cr & \Rightarrow \frac{{231525}}{{200000}} = {\left( {1 + \frac{{\text{R}}}{{2 \times 100}}} \right)^3} \cr & \Rightarrow \frac{{9261}}{{8000}} = {\left( {1 + \frac{{\text{R}}}{{2 \times 100}}} \right)^3} \cr & \Rightarrow {\left( {\frac{{21}}{{20}}} \right)^3} = {\left( {1 + \frac{{\text{R}}}{{2 \times 100}}} \right)^3} \cr & \Rightarrow 1 + \frac{{\text{R}}}{{200}} = \frac{{21}}{{20}} \cr & \Rightarrow {\text{R = 10}}\% \cr} $$

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
  $$ \Rightarrow \frac{x}{{\left( {1 + \frac{8}{{100}}} \right)}} + $$    $$\frac{x}{{{{\left( {1 + \frac{8}{{100}}} \right)}^2}}} + $$   $$\frac{x}{{{{\left( {1 + \frac{8}{{100}}} \right)}^3}}} = $$    $$50725$$
$$\eqalign{ & \Rightarrow \frac{{25x}}{{27}} + \frac{{625x}}{{729}} + \frac{{15625x}}{{19683}} = 50725 \cr & \Rightarrow \frac{{50725x}}{{19683}} = 50725 \cr & \Rightarrow x = \left( {\frac{{50725 \times 19683}}{{50725}}} \right) = 19683 \cr} $$

Rate of interest = 18%
Time = 2 year
When the interest is payable half yearly
Then, rate of interest = 9%
Time = 4 half - years
Let the principal be Rs. x
$$\eqalign{ & {\text{C}}{\text{.I}}{\text{. = }}x\left[ {{{\left( {1 + \frac{R}{{100}}} \right)}^T} - 1} \right]{\text{ }} \cr & = x\left[ {{{\left( {1 + \frac{9}{{100}}} \right)}^4} - 1} \right] \cr & = x\left[ {{{\left( {\frac{{109}}{{100}}} \right)}^4} - 1} \right] \cr & = x\left[ {1.4116 - 1} \right] \cr & = Rs.\,0.4116x \cr & {\text{According to question}} \cr & = x\left[ {{{\left( {1 + \frac{{18}}{{100}}} \right)}^2} - 1} \right] \cr & = x\left[ {{{\left( {\frac{{118}}{{100}}} \right)}^2} - 1} \right] \cr & = x\left[ {{{\left( {1.18} \right)}^2} - 1} \right] \cr & = x\left[ {1.3924 - 1} \right] \cr & = Rs.\,0.3924x \cr & {\text{According to question,}} \cr & 0.4116x - 0.3924x = 960 \cr & \Rightarrow x = \frac{{960}}{{0.0192}} \cr & \Rightarrow x = \frac{{960 \times 10000}}{{192}} \cr & \Rightarrow x = 50000 \cr & {\text{Hence, sum of money}} \cr & {\text{ = Rs. 50000}} \cr} $$

$$\eqalign{ & {\text{According to the question,}} \cr & {\text{Amount}} = {\text{ }}{\left( {1 + \frac{{\text{R}}}{{2 \times 100}}} \right)^{2 \times {\text{t}}}} \cr & \Rightarrow 68921 = 64000{\left( {1 + \frac{5}{{2 \times 100}}} \right)^{2 \times {\text{t}}}} \cr & \Rightarrow \frac{{68921}}{{64000}} = {\left( {1 + \frac{1}{{40}}} \right)^{2 \times {\text{t}}}} \cr & \Rightarrow {\left( {\frac{{41}}{{40}}} \right)^3} = {\left( {\frac{{41}}{{40}}} \right)^{2 \times {\text{t}}}} \cr & \Rightarrow 2{\text{t = 3}} \cr & \Rightarrow {\text{t = }}\frac{3}{2} \cr & \Rightarrow {\text{t = 1}}\frac{1}{2}{\text{ years}} \cr} $$

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

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

$$\eqalign{ & {\text{Rate of interest = 5}}\% {\text{ per annum}} \cr & {\text{Time = 2 year}} \cr & {\text{Accroding to question,}} \cr & \Rightarrow P\left[ {{{\left( {1 + \frac{r}{{100}}} \right)}^n} - 1} \right] - \frac{{P \times r \times t}}{{100}}{\text{ = 63}} \cr & \Rightarrow P\left[ {{{\left( {1 + \frac{5}{{100}}} \right)}^2} - 1} \right] - \frac{{P \times 5 \times 2}}{{100}}{\text{ = 63}} \cr & \Rightarrow P\left[ {{{\left( {1 + \frac{5}{{100}}} \right)}^2} - 1} \right] - \frac{{10P}}{{100}}{\text{ = 63}} \cr & \Rightarrow P\left[ {{{\left( {\frac{{105}}{{100}}} \right)}^2} - 1} \right] - \frac{{10P}}{{100}}{\text{ = 63}} \cr & \Rightarrow P\left( {\frac{{11025 - 10000}}{{10000}}} \right) - \frac{{10P}}{{100}} = 63 \cr & \Rightarrow \frac{{1025P}}{{10000}} - \frac{{10P}}{{100}} = 63 \cr & \Rightarrow \frac{{1025P - 1000P}}{{10000}} = 63 \cr & \Rightarrow 25P = Rs.630000 \cr & \Rightarrow P = \frac{{630000}}{{25}} \cr & \Rightarrow P = Rs. 25200 \cr & {\text{Hence}},\,{\text{sum Rs}}{\text{. 25200}} \cr} $$

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