Practice MCQ Questions and Answer on Interest

Solution: Answer & Solution Answer: Option E Solution: Let the original rate be R%. Then, new rate = (2R)%. Note: Here, original rate is for 1 year(s); the new rate is for only 4 months i.e. $$\frac{1}{3}$$ year(s). $$\eqalign{ & \therefore {\frac{{725 \times R \times 1}}{{100}}} + {\frac{{362.50 \times 2R \times 1}}{{100 \times 3}}} \cr & = 33.50 \cr & \Rightarrow \left( {2175 + 725} \right)R = 33.50 \times 100 \times 3 \cr & \Rightarrow \left( {2175 + 725} \right)R = 10050 \cr & \Rightarrow \left( {2900} \right)R = 10050 \cr & \Rightarrow R = \frac{{10050}}{{2900}} = 3.46 \cr & \therefore \text{Original rate} = 3.46\% \cr} $$

Solution: $$\eqalign{ & \frac{{P \times 5.4 \times 9.25}}{{100}} = \left( {14395.20 - P} \right) \cr & P = 9600 \cr & \frac{{9600 \times 8.6 \times 4.5}}{{100}} = {\text{SI}} \cr & {\text{SI}} = 3715.20 \cr} $$

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: $$\eqalign{ & \frac{{{\text{Principal}}}}{{{\text{Amount}}}} = \frac{{4 \times 5}}{{5 \times 5}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{20}}{{25}} \cr & {\text{After three year}} \cr & \frac{{\text{P}}}{{\text{A}}} = \frac{{5 \times 4}}{{7 \times 4}} \cr & \,\,\,\,\,\,\,\, = \frac{{20}}{{28}} \cr & {\text{In three year S}}{\text{.I}}{\text{.}} \cr & = 28x - 25x \cr & = 3x \cr & \text{So, the required interest will be} \cr & 3x = \frac{{20x \times {\text{R}} \times 3}}{{100}} \cr & {\text{R}} = 5\% \cr} $$

Solution: Answer & Solution Answer: Option E Solution: $$\eqalign{ & {\text{Total}}\,{\text{cow}}\,{\text{milksold}}\,{\text{in}}\,{\text{five}}\,{\text{days}} \cr & = 5 \times 8 = 40\,{\text{litres}} \cr & {\text{Total}}\,{\text{buffalo}}\,{\text{milk}}\,{\text{sold}}\,{\text{in}}\,{\text{five}}\,{\text{days}} \cr & = 5 \times 10 = 50\,{\text{litres}} \cr & {\text{Hence}}, \cr & {\text{SP}}\,{\text{of}}\,{\text{40}}\,{\text{litres}}\,{\text{cow}}\,{\text{milk}} \cr & = 40 \times 55 = Rs.\,2200 \cr & {\text{Hence}}, \cr & {\text{Profit}}\,{\text{on}}\,{\text{cow}}\,{\text{milk}} \cr & = \frac{{ {2200 \times 12} }}{{112}} = Rs.\,235.71 \cr & {\text{Profit}}\,{\text{on}}\,{\text{buffalo}}\,{\text{milk}} \cr & = \frac{{ {50 \times 36 \times 20} }}{{120}} = Rs.\,300 \cr & {\text{Thus}}, \cr & {\text{Total}}\,{\text{Profit}} \cr & = 235.71 + 300 = Rs.\,535.71 \cr} $$

Solution: $$\eqalign{ & \frac{{\text{P}}}{{{\text{S}}{\text{.I}}{\text{.}}}} = \frac{{10}}{3} \cr & {\text{Let Principal = 10}} \cr & {\text{S}}{\text{.I}}{\text{. for 5 years = 3}} \cr & {\text{S}}{\text{.I}}{\text{. for 1 year = 0}}{\text{.6}} \cr & {\text{Rate = }}\frac{{{\text{S}}{\text{.I}}{\text{.}}}}{{{\text{Principal}}}} \times 100 \cr & {\text{Rate = }}\frac{{0.6}}{{10}} \times 100 \cr & \,\,\,\,\,\,\,\,\,\,\,\, = 6\% \cr} $$

Solution: $$\eqalign{ & {\text{Time}} = \left( {\frac{{100 \times 81}}{{450 \times 4.5}}} \right){\text{years}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4\,{\text{years}} \cr} $$

Solution: $$\eqalign{ & {\text{Let}}\,{\text{each}}\,{\text{installment}}\,{\text{be}}\,P \cr & {\text{Hence}}, \cr & {\frac{x}{{ {\left( {\frac{{100}}{{100 + r}}} \right) + {{\left( {\frac{{100}}{{100 + r}}} \right)}^2}} }}} \cr & Or,\,\frac{x}{{ {1 + {\frac{{15}}{{400}}} } }} + \frac{x}{{ {1 + {{\left( {\frac{{15}}{{400}}} \right)}^2}} }} = Rs.\,13040 \cr & {\text{On}}\,{\text{solving,}}\,{\text{it}}\,{\text{gives}}\,x = Rs.\,6889 \cr} $$

Solution: $$\eqalign{ & {\text{Let sum}} = x. \cr & {\text{Then,}} \cr & {\text{S}}{\text{.I}}{\text{.}} = 0.125x = \frac{1}{8}x \cr & {\text{R}} = 10\% \cr & \therefore \text{Rate} \cr & = \left( {\frac{{100 \times x}}{{x \times 8 \times 10}}} \right){\text{years}} \cr & = \frac{5}{4}{\text{years}} \cr & = {\text{1}}\frac{1}{4}{\text{years}} \cr} $$

Solution: $$\eqalign{ & {\text{Present Population}} \cr & = {\text{P}}{\left( {\frac{{1 - {\text{R}}}}{{100}}} \right)^n} \cr & = 10000{\left( {\frac{{1 - 20}}{{100}}} \right)^2} \cr & = 10000{\left( {\frac{{100 - 20}}{{100}}} \right)^2} \cr & = 10000{\left( {\frac{{80}}{{100}}} \right)^2} \cr & = 10000{\left( {\frac{4}{5}} \right)^2} \cr & = 10000 \times \frac{{16}}{{25}} \cr & = 400 \times 16 \cr & = 6400 \cr} $$