Practice MCQ Questions and Answer on Interest

Solution: $$\eqalign{ & \frac{{{\text{Decrease in second year}}}}{{{\text{Decrease in third year}}}} \cr & = \frac{{100}}{{\left( {100 - r} \right)}} \cr & = \frac{{10}}{9} \to r = 10\% \cr} $$ Let the population of vultures 3 years ago be P, then $$\eqalign{ & P \times {\left[ {1 - \left( {\frac{{10}}{{100}}} \right)} \right]^3} = 29160 \cr & \to P = 40000 \cr} $$

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: $$\eqalign{ & {\text{Principal = Rs}}{\text{. 1750}} \cr & {\text{Let the first part = }}x \cr & {\text{Hence second part}} \cr & {\text{ = }}\left( {1750 - x} \right) \cr & {\text{According to the question,}} \cr & \Rightarrow x \times \frac{8}{{100}} \times 1 = \left( {1750 - x} \right) \times \frac{6}{{100}} \times 1 \cr & \Rightarrow 4x = 5250 - 3x \cr & \Rightarrow 7x = 5250 \cr & \Rightarrow x = 750 \cr & {\text{First part = Rs}}{\text{. 750}} \cr & \therefore {\text{ Second part}} \cr & {\text{ = Rs}}{\text{. }}\left( {1750 - 750} \right) \cr & {\text{ = Rs}}{\text{. 1000}} \cr & {\text{Required interest}} \cr & {\text{ = 750}} \times \frac{8}{{100}} \cr & {\text{ = Rs}}{\text{. 60}} \cr} $$ Alternate Note : In such type of questions to save your valuable time follow the given below method. Let, Principal = 100 units in both cases       1st part           2nd part       Total     Principal     →     100×3     :     100×4     +→     700 units     Interest     →     8×3     :     6×4       Note : Interest is same in both cases According to the question, $$\eqalign{ & {\text{700 units = Rs}}{\text{. 1750}} \cr & {\text{1 unit = Rs}}{\text{. }}\frac{{1750}}{{700}} \cr & {\text{24 units = Rs}}{\text{. }}\frac{{1750}}{{700}} \times 24 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\text{Rs}}{\text{. 60}} \cr & {\text{Hence, required interest}} \cr & {\text{ = Rs}}{\text{. 60}} \cr} $$

Solution: $$\eqalign{ & {\text{Let the sum be Rs}}{\text{. }}x{\text{ and}} \cr & {\text{original rate be R}}\% \cr & {\text{Then,}} \cr & \Rightarrow \frac{{x \times \left( {{\text{R}} + 1} \right) \times 3}}{{100}} - \frac{{x \times {\text{R}} \times 3}}{{100}} = 5100 \cr & \Rightarrow 3{\text{R}}x + 3x - 3{\text{R}}x = 510000 \cr & \Rightarrow 3x = 510000 \cr & \Rightarrow x = 170000. \cr & {\text{Hence,}} \cr & {\text{Sum}} = {\text{Rs}}.170000 \cr} $$

Solution: $$\eqalign{ & {\text{Rate of interest}} = {\text{5}}\% {\text{ p}}{\text{.a}}{\text{.}} \cr & {\text{sum}} = {\text{Rs}}{\text{. 3200}} \cr & {\text{Time}} = {\text{2 years 6 months}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2\frac{1}{2}{\text{years}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{5}{2}{\text{years}} \cr & {\text{S}}{\text{.I}}{\text{.}} = \frac{{{\text{Prinicipal}} \times {\text{Time}} \times {\text{Rate}}}}{{100}} \cr & \,\,\,\,\,\,\,\,\,\, = \frac{{3200 \times 5 \times 5}}{{2 \times 100}} \cr & \,\,\,\,\,\,\,\,\,\, = {\text{Rs}}{\text{. }}400 \cr & \therefore {\text{Amount}} = {\text{Rs}}{\text{. sum}} + {\text{S}}{\text{.I}}. \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {3200 + 400} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\text{Rs}}{\text{. 3600}}{\text{}} \cr} $$

Solution: Principal = Rs. 12000 Rate % = 10% Interest paid by the person in 5 years $$\eqalign{ & = \frac{{12000 \times 10 \times 5}}{{100}} \cr & = {\text{Rs}}{\text{. 6000}} \cr} $$ Interest received by the person after 3 years $$\eqalign{ & = {\text{Rs}}{\text{. }}\left( {6000 - 3320} \right) \cr & = {\text{Rs}}{\text{. 2680}} \cr & {\text{By using formula,}} \cr & {\text{Rate}}\% \cr & {\text{ = }}\frac{{2680}}{{12000}} \times \frac{{100}}{3} \cr & = \frac{{67}}{9} \cr & = 7\frac{4}{9}\% \cr & {\text{Hence required rate}}\% \cr & {\text{ = 7}}\frac{4}{9}\% \cr} $$

Solution: Let, Principal, P = Rs. 100; Amount, A = Rs. 200; Time = 12 years; Interest = Rs. 100; Rate of interest $$\eqalign{ & = \frac{{{\text{Total Interest}}}}{{{\text{Given Time}}}} \cr & = \frac{{100}}{{12}} \cr & = 8\frac{1}{3}\% \cr} $$

Solution: $$\eqalign{ & {\text{Let sum}} = {\text{Rs}}{\text{. }}x \cr & {\text{Then,}} \cr & {\text{S}}{\text{.I}}{\text{.}} = {\text{Rs}}{\text{.}}\,x \cr & \therefore \text{Rate}\,\% \cr & = \left( {\frac{{100 \times x}}{{x \times 7}}} \right)\% \cr & = \frac{{100}}{7}\% \cr & {\text{Now, sum}} = {\text{Rs}}{\text{. }}x \cr & {\text{S}}{\text{.I}}. = {\text{Rs}}{\text{. }}3x \cr & \text{Rate} = \frac{{100}}{7}\% \cr & \therefore {\text{Total Time}} \cr & = \left( {\frac{{100 \times 3x}}{{x \times \frac{{100}}{7}}}} \right){\text{years}} \cr & = 21\,{\text{years}} \cr} $$

Solution: $$\eqalign{ & {\text{P}} = {\text{Rs}}.\,520, \cr & {\text{R}} = 13\% \cr & {\text{T}} = \frac{1}{2}yr. \cr & \therefore {\text{S}}{\text{.I}}{\text{.}} = {\text{Rs}}{\text{.}}\left( {\frac{{520 \times 13}}{{100 \times 2}}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\text{Rs}}{\text{. }}33.80 \cr & {\text{Hence, amount after 6 months}} \cr & = {\text{Rs}}{\text{. }}\left( {520 + 33.80} \right) \cr & = {\text{Rs}}{\text{. }}553.80 \cr} $$

Solution: Rate of interest $$ = \frac{{100(x - 1)}}{t}\% $$ Where x is the no. of times the sum becomes of itself. Here the sum is getting 3 times. Therefore, x = 3 Where t is the time taken by sum to become x times of itself. Here, t = 12 years By the short trick approach, we get $$\eqalign{ & = \frac{{100(3 - 1)}}{{12}} \cr & = \frac{{200}}{{12}} \cr & = \frac{{50}}{3} = 16\frac{2}{3}\% \cr} $$ Alternative Method : Let the principal be P and in the 2nd scenario, SI = 2P $$\eqalign{ & {\text{Rate}} = \frac{{{\text{SI}} \times 100}}{{{\text{Principal}} \times {\text{Time}}}} \cr & = \frac{{2{\text{P}} \times 100}}{{{\text{P}} \times 12}} \cr & = \frac{{50}}{3} \cr & = 16\frac{2}{3}\% \cr} $$