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

Speed :
$$\eqalign{ & = \left( {\frac{{600}}{{5 \times 60}}} \right){\text{ m/sec}} \cr & = {\text{ 2 m/sec}} \cr & = \left( {2 \times \frac{{18}}{5}} \right){\text{ km/hr}} \cr & = {\text{ 7}}{\text{.2 km/hr}} \cr} $$

Let the distance travelled by train be x km
Then, distance travelled by bus = (285 - x) km
$$\eqalign{ & \therefore \left( {\frac{{285 - x}}{{40}}} \right) + \frac{x}{{55}} = 6 \cr & \Leftrightarrow \frac{{\left( {285 - x} \right)}}{8} + \frac{x}{{11}} = 30 \cr & \Leftrightarrow \frac{{11\left( {285 - x} \right) + 8x}}{{88}} = 30 \cr & \Leftrightarrow 3135 - 11x + 8x = 2640 \cr & \Leftrightarrow 3x = 495 \cr & \Leftrightarrow x = 165 \cr} $$
Hence, distance travelled by train = 165 km

Time taken :
$$\eqalign{ & = 1\,{\text{hr}}\,40\,{\text{min}}\,48\,{\text{sec}} \cr & = 1\,{\text{hr}}\,40\frac{4}{5}{\text{min}} \cr & = 1\frac{{51}}{{75}}{\text{hrs}} \cr & = \frac{{126}}{{75}}{\text{hrs}} \cr} $$
Let the actual speed be x km/hr
Then,
$$\eqalign{ & \frac{5}{7}x \times \frac{{126}}{{75}} = 42 \cr & {\text{Or }}x = \left( {\frac{{42 \times 7 \times 75}}{{5 \times 126}}} \right) \cr & x = 35{\text{ km/hr}} \cr} $$

Let Abhay's speed be x km/hr
Then,
$$\eqalign{ & \Rightarrow \frac{{30}}{x} - \frac{{30}}{{2x}} = 3 \cr & \Rightarrow 6x = 30 \cr & \Rightarrow x = 5{\text{ km/hr}} \cr} $$

Let the thief be caught x metres from the place where the policeman started running.
Let the speed of the policeman and the thief be 5y m/s and 4y m/s respectively.
Then, time taken by the policeman to cover x metres = time taken by the thief to cover (x - 100) m
$$\eqalign{ & \Rightarrow \frac{x}{{5y}} = \frac{{\left( {x - 100} \right)}}{{4y}} \cr & \Rightarrow 4x = 5\left( {x - 100} \right) \cr & \Rightarrow x = 500 \cr} $$
So, the thief ran (500 - 100) i.e., 400 m before being caught.

At the time of meeting, let the distance travelled by the first train be x km.
Then, discount covered by the second train = (x + 120) km
$$\eqalign{ & \therefore \frac{x}{{50}} = \frac{{x + 120}}{{60}} \cr & \Leftrightarrow 60x = 50x + 6000 \cr & \Leftrightarrow 10x = 6000 \cr & \Leftrightarrow x = 600 \cr} $$
So, distance between A and B
= (x + x + 12) km
= 1320 km

Speed of vehicle = 80 kmph
$$\eqalign{ & = \left( {\frac{{80 \times 1000}}{{60}}} \right){\text{m/min}} \cr & = \left( {\frac{{4000}}{3}} \right){\text{m/min}} \cr} $$
∴ Required distance :
$$\eqalign{ & = \frac{{4000}}{3} \times 15 \cr & = 20000{\text{ metres}} \cr} $$

Speed of the bike :
$$\eqalign{ & = \left( {\frac{{180}}{4}} \right){\text{km/hr}} \cr & = {\text{45 km/hr}} \cr} $$
Speed of the bicycle :
$$\eqalign{ & = \left( {\frac{1}{6} \times 45} \right){\text{km/hr}} \cr & = 7.{\text{5 km/hr}} \cr} $$
∴ Required distance :
= (7.5 × 8) km
= 60 km

Total time taken :
$$\eqalign{ & = \left[ {\frac{x}{y} + \frac{x}{{\left( {\frac{{3y}}{5}} \right)}} + \frac{x}{{\left( {\frac{{2y}}{5}} \right)}}} \right]{\text{hours}} \cr & {\text{ = }}\left( {\frac{x}{y} + \frac{{5x}}{{3y}} + \frac{{5x}}{{2y}}} \right){\text{hours}} \cr & = \left( {\frac{{6x + 10x + 15x}}{{6y}}} \right){\text{hours}} \cr & = \left( {\frac{{31x}}{{6y}}} \right){\text{hours}} \cr & \therefore \frac{{31x}}{{6y}} = 1 \cr & \Leftrightarrow \frac{x}{y} = \frac{6}{{31}} \cr & \Leftrightarrow x:y = 6:31 \cr} $$

Let A's speed = x km/hr
Then, B's speed = (7 - x) km/hr
So,
$$\eqalign{ & \Leftrightarrow \frac{{24}}{x} + \frac{{24}}{{\left( {7 - x} \right)}} = 14 \cr & \Leftrightarrow 24\left( {7 - x} \right) + 24x = 14x\left( {7 - x} \right) \cr & \Leftrightarrow 14{x^2} - 98x + 168 = 0 \cr & \Leftrightarrow {x^2} - 7x + 12 = 0 \cr & \Leftrightarrow \left( {x - 3} \right)\left( {x - 4} \right) = 0 \cr & \Leftrightarrow x = 3{\text{ or }}x = 4 \cr} $$
Since A is faster than B,
So, A's speed = 4 km/hr
And B's speed = 3 km/hr