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

$$\eqalign{ & {\text{Distance}} = {\text{Speed}} \times {\text{Time}} \cr & {\text{Distance}} = \left( {6 \times \frac{{50}}{{60}}} \right){\text{km}} \cr & {\text{Distance}} = 5{\text{ km}} \cr & \therefore {\text{ Required time}} \cr & = \frac{{{\text{Distance}}}}{{{\text{Speed}}}} \cr & = \frac{5}{{10}}{\text{ hrs}} \cr & = \frac{1}{2}{\text{ hrs}} \cr & = {\text{30 min}} \cr} $$

$$\eqalign{ & {\text{Distance}} = {\text{Speed}} \times {\text{Time}} \cr & {\text{Distance}} = \left( {6 \times \frac{{100}}{{60}}} \right){\text{km}} \cr & {\text{Distance}} = 10{\text{ km}} \cr} $$
$$\eqalign{ & \therefore {\text{ Required time}} = \frac{{{\text{Distance}}}}{{{\text{Speed}}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{10}{{10}}{\text{ hrs}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\text{1 hr}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\text{60 min}} \cr} $$

Let the total journey be x km
Then,
$$\eqalign{ & \Leftrightarrow \frac{{3x}}{5} + \frac{{7x}}{{20}} + 6.5 = x \cr & \Leftrightarrow \frac{{12x + 7x + 20 \times 6.5}}{{20}} = x \cr & \Leftrightarrow 12x + 7x + 20 \times 6.5 = 20x \cr & \Leftrightarrow x = 130{\text{ km}} \cr} $$

Total distance travelled :
= (10 + 12) km
= 22 km
Total time taken :
$$\eqalign{ & = \left( {\frac{{10}}{{12}} + \frac{{12}}{{10}}} \right){\text{ hrs}} \cr & = \frac{{61}}{{30}}{\text{ hrs}} \cr} $$
∴ Average speed :
$$\eqalign{ & = \left( {22 \times \frac{{30}}{{61}}} \right){\text{ km/hr}} \cr & = 10.8{\text{ km/hr}} \cr} $$

Due to stoppages, it covers 9 k less.
Time taken to cover 9 km :
$$\eqalign{ & = \left( {\frac{9}{{54}} \times 60} \right){\text{ min}} \cr & = 10\min \cr} $$

Distance covered by first person in time t = $$\frac{2}{8}$$ round = $$\frac{1}{4}$$ round
Distance covered by second person in time t = $$\frac{3}{8}$$ round
Speed of the first person = $$\frac{1}{4t}$$
Speed of the second person = $$\frac{3}{8t}$$
Since the two persons start from A and B respectively, so they shall meet each other when there is a difference of $$\frac{7}{8}$$ round between the two.
Relative speed of A and B :
$$\eqalign{ & = \left( {\frac{3}{{8t}} - \frac{1}{{4t}}} \right) \cr & = \frac{1}{{8t}} \cr} $$
Time taken to cover $$\frac{7}{8}$$ round at this speed :
$$\eqalign{ & = \left( {\frac{7}{8} \times 8t} \right) \cr & = 7t \cr} $$

Let the speed of train Z be z mph
Distance travelled by train X in 1 hr = x miles
Relative speed of train Z w.r.t. train X = (z - x) mph
To catch train X at 5.30 am, train Z will have to cover x miles at relative speed in 3 hr 30 min, i.e., $$\frac{7}{2}$$ hrs
$$\eqalign{ & \therefore \left( {z - x} \right) \times \frac{7}{2} = x \cr & \Rightarrow \frac{7}{2}z = \frac{9}{2}x \cr & \Rightarrow z = \left( {\frac{9}{2} \times \frac{2}{7}} \right)x \cr & \Rightarrow z = \frac{9}{7}x \cr} $$

1 m/sec = $$\frac{18}{5}$$ km/hr
Car cover 20 metres in a second
∴ 20 m/sec = $$\frac{20 ×18}{5}$$  = 72 km/hr

Ratio of speeds :
$$\eqalign{ & = \left( {\frac{{550}}{{60}} \times \frac{{18}}{5}} \right):\left( {\frac{{33}}{{45}} \times 60} \right) \cr & = 33:44 \cr & = 3:4 \cr} $$

Time taken to travel 45 km :
$$\eqalign{ & = \left( {\frac{{45}}{{20}}} \right){\text{ hrs}} \cr & = \frac{9}{4}{\text{ hrs}} \cr & = 2\frac{1}{4}{\text{ hrs}} \cr & = 2{\text{ hrs }}15\min \cr} $$
Remaining time = (3 hrs - 2 hrs 15 min) = 45 min
Hence, required speed :
$$\eqalign{ & = \left( {\frac{{45}}{{45}}} \right){\text{ km/min}} \cr & = 1{\text{ km/min}} \cr} $$