Speed Time And Distance - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & {\text{Let the length of train}} = l{\text{ km}} \cr & \frac{{l + 0.2}}{x} = \frac{{30}}{{3600}} \cr & 120\left( {l + 0.2} \right) = x \cr & 120\left( {l + 24} \right) = x{\text{ }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {\text{i}} \right) \cr & \frac{l}{{x - 6}} = \frac{{20}}{{3600}} \cr & 180l = x - 6 \cr & 180l + 6 = x{\text{ }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {{\text{ii}}} \right) \cr & {\text{From equation }}\left( {\text{i}} \right)\,\& \,\left( {{\text{ii}}} \right) \cr & 120l + 24 = 180l + 6 \cr & 60l = 18 \cr & l = \frac{3}{{10}} \cr & {\text{Now from equation }}\left( {\text{i}} \right) \cr & 120 \times \frac{3}{{10}} + 24 = x \cr & x = 60{\text{ km/h}} \cr} $$

$$\eqalign{ & {\text{Required distance}} = \left( {75 + 85} \right) \times 3\min \cr & = 160 \times \frac{3}{{60}} \cr & = 8{\text{ km}} \cr} $$

$$\eqalign{ & A + 600 = \left( {81 - s} \right) \times \frac{5}{{18}} \times 72{\text{ }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {\text{i}} \right) \cr & A + 600 = \left( {81 + s} \right) \times \frac{5}{{18}} \times 36{\text{ }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {{\text{ii}}} \right) \cr & {\text{From equation }}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right) \cr & \left( {81 - s} \right) \times \frac{5}{{18}} \times 72 = \left( {81 + s} \right) \times \frac{5}{{18}} \times 36 \cr & \left( {81 - s} \right) \times 2 = \left( {81 + s} \right) \times 1 \cr & 162 - 2s = 81 + s \cr & 3s = 81 \cr & s = 27 \cr & {\text{Now, put }}s = 27{\text{ in equation }}\left( {\text{i}} \right) \cr & A + 600 = \left( {81 - 27} \right) \times \frac{5}{{18}} \times 72 \cr & A + 600 = 54 \times 20 \cr & A = 1080 - 600 \cr & A = 480{\text{ m}} \cr} $$

Let the distance be 400 km
Then time taken by him = $$\frac{{400}}{{130}}$$
75% of distance is = 300 km
Time taken by him to complete the distance of 300 km $$ = \frac{{400}}{{130}} \times \frac{2}{3} = \frac{{80}}{{39}}$$
Rest distance = 100 km
Time taken by him $$ = \frac{{400}}{{130}} - \frac{{80}}{{39}} = \frac{{40}}{{39}}{\text{ h}}$$
According to the question,
$$\eqalign{ & \frac{{100}}{{\frac{{40}}{{39}}}} = x \cr & x = \frac{{390}}{4} \cr & \therefore \frac{x}{2} = \frac{{390}}{8} = 48.75 \cr} $$

$$\eqalign{ & {\text{Average speed}} = \frac{{2 \times 90 \times 126}}{{216}} \cr & = \frac{{2 \times 5 \times 126}}{{12}} \cr & = 105{\text{ km/h}} \cr & {\text{Time}} = \frac{{525}}{{105}} = 5{\text{ hours}} \cr} $$

$$\eqalign{ & \frac{{300}}{{x + 10}} - \frac{{300}}{x} = 5 \cr & 60\left[ {\frac{{10}}{{x\left( {x + 10} \right)}}} \right] = 1 \cr & 600 = x\left( {x + 10} \right) \cr & {\text{Put }}x = 20 \cr & {\text{Average speed}} = \frac{{2 \times 20 \times 30}}{{20 + 30}} \cr & = \frac{{2 \times 600}}{{50}} \cr & = 24{\text{ km/hr}}{\text{.}} \cr} $$

$$\eqalign{ & {\text{Average speed}} = \frac{{{\text{Total Journey}}}}{{{\text{Time Taken}}}} \cr & \Rightarrow {\text{Remaining Distance}} \cr & = 1 - \left( {\frac{1}{3} + \frac{1}{4}} \right) \cr & = \frac{5}{{12}}{\text{ km}} \cr & \therefore {\text{Average speed}} \cr & = \frac{1}{{\frac{1}{{3 \times 80}} + \frac{1}{{50 \times 4}} + \frac{{1 \times 5}}{{100 \times 12}}}} \cr & = \frac{1}{{\frac{{5 + 6 + 5}}{{1200}}}} \cr & = \frac{{1200}}{6} \cr & = 75{\text{ km/hr}} \cr} $$

$$\eqalign{ & {\text{Average speed}} = \frac{{{\text{Total Distance}}}}{{{\text{Total Time}}}} \cr & = \frac{{210 + 198}}{{\frac{{210}}{{60}} + \frac{{198}}{{66}}}} \cr & = 62.769 \cr & = 62.8{\text{ km/hr}} \cr} $$

$$\eqalign{ & {\text{Let total distance}} = 100 \cr & {\text{Average speed}} = \frac{{{\text{Total Distance}}}}{{{\text{Total Time}}}} \cr & = \frac{{100}}{{\frac{{40}}{8} + \frac{{24}}{9} + \frac{{36}}{{12}}}} \cr & = \frac{{100}}{{5 + \frac{8}{3} + 3}} \cr & = \frac{{100 \times 3}}{{32}} \cr & = \frac{{75}}{8} \cr & = 9\frac{3}{8} \cr} $$

$$\eqalign{ & {\text{Time}} = 1{\text{ hr }}20{\text{ m}} = \frac{4}{3}{\text{ hr}} \cr & {\text{Speed}} = 51{\text{ km/hr}} \cr & \therefore {\text{Distance}} = \frac{4}{3} \times 51 = 68 \cr & {\text{Average speed}} = \frac{{{\text{Total Distance}}}}{{{\text{Total Time}}}} \cr & = \frac{{30 + 68}}{{\frac{{30}}{{45}} + \frac{4}{3}}} \cr & = \frac{{98}}{{\frac{2}{3} + \frac{4}{3}}} \cr & = 49{\text{ km/hr}} \cr} $$