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21. 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 kmph

C. $32 \frac{8}{11}$ kmph

D. $35 \frac{8}{11}$ kmph

Discuss

Solution:

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

[Here man's length is 0 metre]

\begin{aligned} \frac{100 \times 18}{5 x - 15} = 11 \\ 1800 = 55 x - 165 \\ 55 x = 1965 \end{aligned}

∴ Speed of the train =

\frac{1965}{55}

35 \frac{8}{11}

kmph

22. 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

Discuss

Solution:

A →____________________← B
Ratio of the speed of trains is given by;

\begin{aligned} \frac{Speed of train A}{Speed train of B} = \frac{\sqrt{B}}{\sqrt{A}} \\ = \frac{\sqrt{9}}{\sqrt{4}} \\ = \frac{3}{2} \\ = 3 : 2 \end{aligned}

23. 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. 14.4 seconds

B. 15.5 seconds

C. 18.8 seconds

D. 20.2 seconds

Discuss

Solution:

Let length of each train be x meter.
Then, speed of 1^st train =

\frac{x}{18}

m/sec
Speed of 2^nd train =

\frac{x}{12}

m/sec
Now,
When both trains cross each other, time taken

\begin{aligned} = \frac{2 x}{\frac{x}{18} + \frac{x}{12}} \\ = \frac{2 x}{\frac{2 x + 3 x}{36}} \\ = \frac{2 x \times 36}{5 x} \\ = \frac{72}{5} \\ = 14.4 seconds \end{aligned}

24. 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. 200m

B. 300m

C. 350m

D. 400m

Discuss

Solution:

Let the length of the train traveling at 48 kmph be 2x meters.
And length of the platform is y meters.

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

25. 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

Discuss

Solution:

Length of the 1^st train = 105 m
Length of the 2^nd 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{Length of 1^{st} train + length of 2^{nd} train}{relative speed of the trains}

∴ Time taken =

\frac{195}{32.5}

= 6 secs.

26. 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

Discuss

Solution:

1^st method:
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.

\begin{aligned} According to question; \\ \frac{x}{(s - 3) \times \frac{5}{18}} = 10 \\ or, 18 x = 50 \times s - 150..... (i) \\ and \\ \frac{x}{(x - 5) \times \frac{5}{18}} = 11 \\ 18 x = 55 \times s - 275...... (ii) \\ Equating equation (i) and (ii) \\ 50 \times s - 150 = 55 \times s - 275 \\ or, 5 \times s = 125 \\ or, s = 25 kmph \end{aligned}

27. 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. 80m

B. 90m

C. 200m

D. 150m

Discuss

Solution:

1^st Method:
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,
Or,

\frac{x + 800}{100}

\frac{x + 400}{60}

Or, 5x + 200 = 3x + 2400
Or, 2x = 400
Or, x = 200m

2^nd Method:
As in both cases, the speed of the train is constant, and then we have;
Time α distance

\begin{aligned} \frac{100}{60} = \frac{x + 800}{x + 400} \\ or, x = 200 m \end{aligned}

28. 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. 26 kmph

B. 35 kmph

C. 36 kmph

D. 63 kmph

Discuss

Solution:

Let speed of the 2^nd train is S m/sec.
And,
60 kmph

= \frac{60 \times 5}{18} = \frac{50}{3}

m/sec.
As trains are traveling in same distance, Then Relative distance,

\begin{aligned} = \frac{60 \times 5}{18} = \frac{50}{3} \\ \frac{50}{3} - S = \frac{120}{18} \\ \Rightarrow \frac{50 - 3 S}{3} = \frac{20}{3} \\ \Rightarrow 50 - 3 S = 20 \\ \Rightarrow 3 S = 50 - 20 \\ \Rightarrow 3 S = 30 \\ ∴ S = 10 m/sec \end{aligned}

Or, Speed of the 2^nd train =

10 \times \frac{18}{5}

= 36 kmph.

29. 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

Discuss

Solution:

Let the distance of the destination from the starting point = x km.
Speed downstream = (10 + 4) = 14 kmph
Speed upstream = (10 - 4) = 6 kmph
According to the question,
Total time taken = 5 hours

\begin{aligned} \frac{x}{14} + \frac{x}{6} = 5 \\ \frac{3 x + 7 x}{42} = 5 \\ or, 10 x = 42 \times 5 \\ or, x = \frac{42 \times 5}{10} \\ So, x = 21 km \end{aligned}

30. 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

Discuss

Solution:

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

\begin{aligned} According to question, \\ \frac{2 x}{\frac{15}{2} + y} = \frac{x}{\frac{15}{2} - y} \\ o r, 15 - 2 y = \frac{15}{2} + y \\ o r, 3 y = 15 - \frac{15}{2} = \frac{15}{2} \\ o r, y = \frac{15}{6} = 2.5 kmph \end{aligned}

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