1. Two trains of equal length, running in opposite directions, pass a pole in 18 and 12 seconds. The trains will cross each other in:
a) 15.5 seconds
b) 14.4 seconds
c) 20.2 seconds
d) 18.8 seconds
Explanation: Let length of each train be x meter.
Then, speed of 1st train = $$\frac{x}{{18}}$$ m/sec
Speed of 2nd train = $$\frac{x}{{12}}$$ m/sec
When both trains cross each other, time taken
$$\eqalign{ & = {\frac{{2x}}{{ { {\frac{x}{{18}}} + {\frac{x}{{12}}} } }}} \cr & = \frac{{2x}}{{ {\frac{{ {2x + 3x} }}{{36}}} }} \cr & = \frac{{2x \times 36}}{{5x}} \cr & = \frac{{72}}{5} \cr & = 14.4\,\,{\text{seconds}} \cr} $$
2. A train passes two persons walking in the same direction at a speed of 3 kmph and 5 kmph respectively in 10 seconds and 11 seconds respectively. The speed of the train is
a) 28 kmph
b) 27 kmph
c) 25 kmph
d) 24 kmph
Explanation: Let the speed of the train be S. And length of the train be x.
When a train crosses a man, its travels its own distance.
$$\eqalign{ & {\text{According to question}}; \cr & \frac{x}{{ {\left( {s - 3} \right) \times {\frac{5}{{18}}} } }} = 10 \cr & {\text{or}},\,18x = 50 \times s - 150.....({\text{i}}) \cr & {\text{and}} \cr & \frac{x}{{ {\left( {x - 5} \right) \times {\frac{5}{{18}}} } }} = 11 \cr & 18x = 55 \times s - 275......({\text{ii}}) \cr & {\text{Equating equation }}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right) \cr & 50 \times s - 150 = 55 \times s - 275 \cr & {\text{or}},\,5 \times s = 125 \cr & {\text{or}},\,s = 25\,{\text{kmph}} \cr} $$
3. A plane left half an hour later than the scheduled time and in order to reach its destination 1500 kilometers away in time, it had to increase its speed by 33.33 percent over its usual speed. Find its increased speed.
a) 500 kmph
b) 250 kmph
c) 750 kmph
d) 1000 kmph
Explanation: By increasing the speed by 33.33%, it would be able to reduce the time taken for traveling by 25%.
But since this is able to overcome the time delay of 30 minutes, 30 minutes must be equivalent to 25% of the time originally taken.
Hence, the original time must have been 2 hours and the original speed would be 750 kmph.
Hence, the new speed would be 1000 kmph.
4. Two rifles are fired from the same place at a difference of 11 minutes 45 seconds. But a man who is coming towards the place in a train hears the second sound after 11 minutes. Find the speed of train. (speed of sound = 330 m/s)
a) 72 kmph
b) 36 kmph
c) 81 kmph
d) 108 kmph
Explanation: Distance travelled by man-train in 11 minute = distance traveled by sound in (11 min 45 sec - 11 min) = 45 sec.
Let the speed of train be x kmph.
$$x \times \frac{{11}}{{60}} = \frac{{45}}{{3600}} \times 330 \times \frac{{18}}{5}$$
Or, x = 81 kmph.
5. In a 100m race, Kamal defeats Bimal by 5 seconds. If the speed of Kamal is 18 kmph, then the speed of Bimal is:
a) 15.4 kmph
b) 14 kmph
c) 14.5 kmph
d) 14.4 kmph
Explanation:
$$\eqalign{ & {\text{Time taken by Kamal}} \cr & = \frac{{100}}{{ {\frac{{18 \times 5}}{{18}}} }} \cr & = 20\,{\text{seconds}} \cr & {\text{Hence, }} \cr & {\text{Time taken by Bimal}} \cr & 20 + 5 = 25\,{\text{seconds}} \cr & {\text{So, Bimal's speed}} \cr & = \frac{{100}}{{25}} \cr & = 4\ \cr & = \frac{{4 \times 18}}{5} \cr & = 14.4\,{\text{kmph}} \cr} $$
6. Two trains 105 meters and 90 meters long, run at the speeds of 45 kmph and 72 kmph respectively, in opposite directions on parallel tracks. The time which they take to cross each other, is:
a) 8 seconds
b) 6 seconds
c) 7 seconds
d) 5 seconds
Explanation: Length of the 1st train = 105 m
Length of the 2nd train = 90 m
Relative speed of the trains,
= 45 + 72 = 117 kmph
= $$\frac{{117 \times 5}}{{18}}$$ = 32.5 m/sec
Time taken to cross each other,
= $$\frac{{{\text{Length}}\,{\text{of}}\,{1^{{\text{st}}}}\,{\text{train}} + {\text{length of}}\,{2^{{\text{nd}}}}\,{\text{train}}}}{{{\text{relative }}\,{\text{speed }}\,{\text{of}}\,{\text{the }}\,{\text{trains}}}}$$
Time taken = $$\frac{{195}}{{32.5}}$$ = 6 seconds.
7. A train travelling with a speed of 60 kmph catches another train travelling in the same direction and then leaves it 120 m behind in 18 seconds. The speed of the second train is
a) 36 kmph
b) 26 kmph
c) 63 kmph
d) 35 kmph
Explanation: Let speed of the 2nd train is S m/sec.
60 kmph $$ = \frac{{60 \times 5}}{{18}} = \frac{{50}}{3}$$ m/sec.
As trains are traveling in same distance, Then Relative distance,
$$\eqalign{ & = \frac{{60 \times 5}}{{18}} = \frac{{50}}{3} \cr & \frac{{50}}{3} - S = \frac{{120}}{{18}} \cr & \Rightarrow \frac{{50 - 3S}}{3} = \frac{{20}}{3} \cr & \Rightarrow 50 - 3S = 20 \cr & \Rightarrow 3S = 50 - 20 \cr & \Rightarrow 3S = 30 \cr & \therefore S = 10\,\,{\text{m/sec}} \cr} $$
Or, Speed of the 2nd train = $${\text{10}} \times \frac{{18}}{5}$$ = 36 kmph.
8. A car driver, driving in a fog, passes a pedestrian who was walking at the rate of 2 km/h in the same direction. The pedestrian could see the car for 6 minutes and it was visible to him up to distance of 0.6 km. what was speed of the car?
a) 30 kmph
b) 15 kmph
c) 20 kmph
d) 8 kmph
Explanation: In 6 minutes, the car goes ahead by 0.6 km.
Hence, the relative speed of the car with respect to the pedestrian is equal to 6 kmph.
Hence, Net speed of the car is 8 kmph.
9. Between 5 am and 5 pm of a particular day for how many times are the minute and the hour hands together?
a) 22
b) 11
c) 44
d) 33
Explanation: It will come 11 times together.
10. Two boats A and B start towards each other from two places, 108 km apart. Speed of the boats A and B in still water are 12 km/h and 15 km/h respectively. If A proceeds down and B up the stream, they will meet after:
a) 4.5 hours
b) 4 hours
c) 5.4 hours
d) 6 hours
Explanation: Let the speed of the stream be x kmph and both the boats meet after t hour.
According to the question,
(12 + x) × t + (15 - x) × t = 108
12t + 15t = 108
27t = 108
∴ t = $$\frac{{108}}{{27}}$$ = 4 hours