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

Distance covered by thief in (2 pm - 1.30 pm)
= $$\frac{1}{2}$$ hr at speed of 40 km/hr
= 40 × $$\frac{1}{2}$$
= 20 kms
Their relative speed in same direction :
= (50 - 40) km/hr
= 10 km/hr
According to the question,
20 km, is the distance that has to be covered by owner to catch the thief.
$$\eqalign{ & {\text{Required time}} = \frac{{20{\text{ km}}}}{{10{\text{ km/hr}}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2{\text{ hours}} \cr} $$
Therefore, he will over take the thief at :
= 2 pm + 2 hr
= 4 pm

In these type of questions go through options to save your valuable time :
Option (A)
Abhay's speed = 5 km/hr
Abhay's time = $$\frac{30}{5}$$ = 6 hr
Sameer's time = 6 - 2 = 4 hr
Abhay's new time :
= $$\frac{30}{5 × 2}$$
= 3 hr
Hence option (A) is correct as it satisfies all the conditions.

Time taken by Kamal :
$$ = \frac{{100}}{{18 \times \frac{5}{{18}}}} = 20\sec $$
Time taken by Bimal :
$$\eqalign{ & = 20 + 5 \cr & = 25\sec \cr} $$
Speed of Bimal :
$$\eqalign{ & = \frac{{100}}{{25}} \times \frac{{18}}{5} \cr & = 14.4{\text{ kmph}} \cr} $$

D = S × T
D = 80 × 7
D = 560 km

$$\eqalign{ & {{\text{V}}_{{\text{rel}}{\text{.}}}} = 77 + 67 = 144{\text{ km/hr}} \cr & {\text{ = 144}} \times \frac{5}{{18}}{\text{ m/sec}} \cr & = 40{\text{ m/sec}} \cr & \therefore {\text{ T}} = \frac{{\text{D}}}{{{{\text{V}}_{{\text{rel}}{\text{.}}}}}} \cr & \Rightarrow {\text{T}} = \frac{{140 + 160}}{{40}} \cr & \Rightarrow {\text{T}} = \frac{{300}}{{40}} \cr & \Rightarrow {\text{T}} = 7.5\sec {\text{or 7}}\frac{1}{2}\,\sec \cr} $$

Since Raju and Prem walk in opposite directions.
Distance covered per hour = Relative speed × Time
= (3 + 2) × 1 = 5 km   [opposite direction]
∴ Distance covered in 2 hours = 5 × 2 = 10 km

$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{A}}\,{\text{:}}\,{\text{B}}\,\,\,\,\,\,\,{\text{length}} \cr & {\text{Ratio of A's and }} \to 5:3\,\,\,\left( {5x:3x} \right) \cr & {\text{ B's length}} \cr & {\text{Ratio of A's and }} \to 6:5\,\,\,\left( {6y:5y} \right) \cr & {\text{ B's speed}} \cr} $$
We know that,
When a train crosses a pole, i.e., it covers the distance equal to its length
Time taken by train A to cross the pole :
$$ = \frac{{{\text{Total distance}}}}{{{\text{Speed}}}} = \frac{{5x}}{{6y}}$$
Time taken by train B to cross the pole :
$$ = \frac{{{\text{Total distance}}}}{{{\text{Speed}}}} = \frac{{3x}}{{5y}}$$
Ratio of the their time :
$$\eqalign{ & = {\text{A}}:{\text{B}} \cr & = \frac{{5x}}{{6y}}:\frac{{3x}}{{5y}} \cr & = 25:18 \cr} $$

Time taken by man if he did not stop :
$$\eqalign{ & = \frac{{5{\text{ km}}}}{{10{\text{ kmph}}}} \cr & = \frac{1}{2}{\text{hr}} \cr & = 30{\text{ min}} \cr} $$
$$\because $$ Man takes rest for 5 minutes on each km
Total rest time = 5 × 4 = 20 min
Total travelling time :
= 30 min + 20 min
= 50 min

Total distance covered by the length of both trains :
= 50 m + 75 m
And, their relative speed in same direction :
= 68 - 50 = 18 km/hr
$$\because $$ (Speed subtracted in same direction)
Then, the time to cross each other will be :
$$\eqalign{ & = \frac{{125{\text{ m/s}}}}{{18{\text{ kms}}}} \cr & = \frac{{125{\text{ }} \times {\text{18}}}}{{18 \times 5{\text{ }}}}{\text{m/s}} \cr & \left\{ {\because 1{\text{ km/hr = }}\frac{5}{{18}}{\text{ m/s}}} \right\} \cr & = 25\sec \cr} $$

Let 10% of journey's = 40 km
Then, total journey = 400 kms
And, $${\text{Average speed }} = \frac{{{\text{Total distance }}}}{{{\text{Total time}}}}$$
$$\eqalign{ & 30\% {\text{ of journey}} = 400 \times \frac{{30}}{{100}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 120{\text{ km}} \cr & 60\% {\text{ of journey}} = 400 \times \frac{{60}}{{100}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 240{\text{ km}} \cr & 10\% {\text{ of journey}} = 400 \times \frac{{10}}{{100}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 40{\text{ km}} \cr & {\text{Average speed}} = \frac{{400}}{{\frac{{120}}{{20}} + \frac{{240}}{{40}} + \frac{{40}}{{10}}}} \cr & {\text{Average speed}} = \frac{{400}}{{\left( {6 + 6 + 4} \right)}} \cr & {\text{Average speed}} = \frac{{400}}{{16}} \cr & {\text{Average speed}} = 25{\text{ km/hr}} \cr} $$