1. A train passes two bridges of length 800 m and 400 m in 100 seconds and 60 seconds respectively. The length of the train is:

a) 80 m

b) 90 m

c) 200 m

d) 150 m

Explanation: Let length of the train be x m and speed of the train is s kmph.

Speed, s = $$\frac{{x + 800}}{{100}}$$ . . . . . (i)

Speed, s = $$\frac{{x + 400}}{{60}}$$ . . . . . (ii)

Equating equation (i) and (ii), we get,

$$\frac{{x + 800}}{{100}}$$ = $$\frac{{x + 400}}{{60}}$$

5x + 200 = 3x + 2400

2x = 400

x = 200m

2. A cyclist moving on a circular track of radius 100 meters completes one revolution in 2 minutes. What is the average speed of cyclist (approx.)?

a) 314 m/min

b) 200 m/min

c) 300 m/min

d) 900 m/min

Explanation: Distance covered in one complete revolution

$$\eqalign{ & = 2\pi r \cr & = \frac{{2 \times 100 \times 22}}{7} \cr & \approx 628\,m \cr & {\text{Average}}\,{\text{speed}} \cr & = \frac{{628}}{2} \cr & = 314\,m/\min \cr} $$

3. A train, 300m long, passed a man, walking along the line in the same direction at the rate of 3 kmph in 33 seconds. The speed of the train is:

a) 30 kmph

b) $$32\frac{8}{{11}}$$ kmph

c) $$35\frac{8}{{11}}$$ kmph

d) 32 kmph

Explanation: Let the speed of train = x km/ hr

Length of train = 300 metres

Their relative speed in same direction

= (x - 3) km/hr

According to the question,

$$\frac{{\left( {300 + 0} \right)m}}{{\left( {x - 3} \right) \times \frac{5}{{18}}m/s}} = 33$$

[Here man's length is 0 metre]

$$\eqalign{ & \frac{{100 \times 18}}{{5x - 15}} = 11 \cr & 1800 = 55x - 165 \cr & 55x = 1965 \cr} $$

∴ Speed of the train = $$\frac{{1965}}{{55}}$$ = $$35\frac{8}{{11}}$$ kmph

4. In a race of 200 meters, B can gives a start of 10 meters to A, and C can gives a start of 20 meters to B. The starts that C can gives to A, in the same race is:

a) 30 meters

b) 25 meters

c) 29 meters

d) 27 meters

Explanation: When B runs 200 meters, A runs 190 meters;

Hence, when B runs 180 meters,

A runs = $$\frac{{190 \times 180}}{{200}}$$ = 171 meters;

When C runs 200m, B runs 180 meters.

C will give a start to A by = 200 - 171

= 29 meters

5. A person can row $$7\frac{1}{2}$$ km an hour in still water. Finds that it takes twice the time to row upstream than the time to row downstream. The speed of the stream is:

a) 2 kmph

b) 2.5 kmph

c) 3 kmph

d) 4 kmph

Explanation: Let the distance covered be x km and speed of stream = y kmph.

Speed downstream = $$\frac{{15}}{2} + y$$ kmph

Speed upstream = $$\frac{{15}}{2} - y$$ kmph

$$\eqalign{ & {\text{According}}\,{\text{to}}\,{\text{question,}} \cr & {\frac{{2x}}{{ { {\frac{{15}}{2}} + y} }}} = {\frac{x}{{ { {\frac{{15}}{2}} - y} }}} \cr & or,\,15 - 2y = {\frac{{15}}{2}} + y \cr & 3y = 15 - {\frac{{15}}{2}} = \frac{{15}}{2} \cr & y = \frac{{15}}{6} = 2.5\,{\text{kmph}} \cr} $$

6. A train travelling at 48 kmph crosses another train, having half of its length and travelling in opposite direction at 42 kmph, in 12 seconds. It also passes a railway platform in 45 seconds. The length of railway platform is:

a) 200 m

b) 300 m

c) 350 m

d) 400 m

Explanation: Let the length of the train traveling at 48 kmph be 2x meters.

And length of the platform is y meters.

$$\eqalign{ & {\text{Relative speed of train}} \cr & = \left( {48 + 42} \right)\,{\text{kmph}} \cr & = {\frac{{90 \times 5}}{{18}}} \cr & = 25\,m/\sec \cr & {\text{And}}\,{\text{48}}\,{\text{kmph}} \cr & = \frac{{48 \times 5}}{{18}} \cr & = \frac{{40}}{3}\,m/\sec \cr & \cr & {\text{According to the question}}, \cr & \frac{{ {2x + x} }}{{25}} = 12; \cr & \,3x = 12 \times 25 = 300 \cr & \,x = \frac{{300}}{3} = 100m \cr & {\text{Then, length of the train}} \cr & = 2x \cr & = 100 \times 2 = 200m \cr & \frac{{200 + y}}{{ {\frac{{40}}{3}} }} = 45 \cr & 600 + 3y = 40 \times 45 \cr & {\text{Or}},\,3y = 1800 - 600 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1200 \cr & {\text{Or}},\,y = \frac{{1200}}{3} = 400m \cr & {\text{Length of the platform}} \cr & = 400m \cr} $$

7. A man goes downstream with a boat to some destination and returns upstream to his original place in 5 hours. If the speed of the boat in still water and the stream are 10 kmph and 4 kmph respectively, the distance of the destination from the starting place is:

a) 16 km

b) 18 km

c) 21 km

d) 25 km

Explanation: Let the distance of the destination from the starting point = x km.

Speed downstream = (10 + 4) = 14 kmph

Speed upstream = (10 - 4) = 6 kmph

Total time taken = 5 hours

$$\eqalign{ & {\frac{x}{{14}}} + {\frac{x}{6}} = 5 \cr & \frac{{ {3x + 7x} }}{{42}} = 5 \cr & {\text{or}},\,10x = 42 \times 5 \cr & {\text{or}},\,x = \frac{{42 \times 5}}{{10}} \cr & {\text{So}},\,x = 21\,{\text{km}} \cr} $$

8. The speed of a motor-boat is that of the current of water as 36:5. The boat goes along with the current in 5 hours 10 minutes. It will come back in:

a) 5 hours 50 minutes

b) 6 hours 50 minutes

c) 6 hours

d) 12 hours 10 minutes

Explanation: Let the speed of the motor boat = 36x kmph and speed of current = 5x kmph.

The boat goes along with the current in 5 hours 10 minutes = $$\frac{{31}}{6}$$ hour

$$\eqalign{ & {\text{Hence, Distance}} \cr & = {\frac{{31}}{6}} \times \left( {36x + 5x} \right) \cr & = \frac{{41x + 31}}{6}\,km \cr & {\text{Speed of boat upstream}} \cr & = 36x + 5x = 41x\,{\text{kmph}} \cr & {\text{Hence, time taken to come back}} \cr & = {\frac{{ {41x \times {\frac{{31}}{6}} } }}{{31x}}} \cr & = \frac{{41}}{6}\,{\text{hours}} \cr & = 6\,{\text{hours}}\,\,50\,{\text{minutes}} \cr} $$

9. Two trains started at the same time, one from A to B and other from B to A. if they arrived at B and A respectively in 4 hours and 9 hours after they passed each other, the ratio of the speeds of the two trains was:

a) 2 : 1

b) 3 : 2

c) 4 : 3

d) 5 : 4

Explanation: A →____________________← B

Ratio of the speed of trains is given by;

$$\eqalign{ & \frac{{{\text{Speed}}\,{\text{of}}\,{\text{train}}\,A}}{{{\text{Speed}}\,{\text{train}}\,{\text{of}}\,B}} = \frac{{\sqrt B }}{{\sqrt A }} \cr & = \frac{{\sqrt 9 }}{{\sqrt 4 }} \cr & = \frac{3}{2} \cr & = 3:2 \cr} $$

10. In a 1-kilometre race, A can beat B by 30 meters, while in a 500-meter race B can beat C by 25 meters. By how many meters will A beats C in a 100-meter race?

a) 7.25

b) 7.15

c) 7.85

d) 7.03

Explanation: When A runs 100 meters, B runs 970 meters.

Hence, when A runs 100 meter, b runs 97 meter.

When B runs 500 meter, C runs 475 meter.

When B runs 97 meter, C runs, $$\frac{{475 \times 97}}{{500}}$$ = 92.15 meter.

A will beat C by (100 - 92.15) = 7.85 meters.