Time & Distance Questions and Answers

Q: How many seconds will a train 100 meters long take to cross a bridge 150 meters long if the speed of the train is 36 kmph?

\[ 1 \text{ kmph} = \frac{5}{18} \text{ m/s} \]\[ 36 \text{ kmph} = 10 \text{ m/s} \]\[ t = \frac{100 + 150}{10} = 25 \text{ seconds} \]

Q: A person crosses a 600 m long street in 5 minutes. What is his speed in km per hour?

Convert both quantities to the units the answer needs.\[ 600 \text{ m} = 0.6 \text{ km}, \qquad 5 \text{ min} = \frac{1}{12} \text{ h} \]\[ v = \frac{0.6}{1/12} = 7.2 \text{ km/hr} \]Equivalently, 2 m/s multiplied by \( \tfrac{18}{5} \) gives the same 7.2 km/hr.

Q: P, Q and R are in a cycle race of 4500 meters. P cycles twice as fast as Q. R cycles 1/3rd as fast as Q. R completes the race in 45 minutes. Then where was Q from the finishing line when P finished the race?

Start from R, the only cyclist whose time is given.\[ v_{R} = \frac{4500}{45} = 100 \text{ m/min} \]Since \( v_{R} = \tfrac{1}{3}v_{Q} \) and \( v_{P} = 2v_{Q} \):\[ v_{Q} = 300 \text{ m/min}, \qquad v_{P} = 600 \text{ m/min} \]\[ t_{P} = \frac{4500}{600} = 7.5 \text{ min} \]In that time Q has covered \( 300 \times 7.5 = 2250 \) m, so\[ 4500 - 2250 = 2250 \text{ m from the finish} \]P is twice as fast as Q, so Q is exactly halfway when P finishes - a check that needs no arithmetic at all.

Q: A train of 24 carriages, each of 60 meters length, when an engine also of 60 meters length is running at a speed of 60 kmph. In what time will the train cross a bridge 1.5 km long?

\[ 1 \text{ kmph} = \frac{5}{18} \text{ m/s} \]The whole train is the engine plus 24 carriages, each 60 m:\[ 25 \times 60 = 1500 \text{ m} \]\[ 60 \text{ kmph} = \frac{50}{3} \text{ m/s} \]\[ t = \frac{1500 + 1500}{50/3} = \frac{3000 \times 3}{50} = 180 \text{ s} = 3 \text{ minutes} \]The engine counts as a twenty-fifth 60 m unit, and the bridge's 1.5 km must be converted to 1500 m.

Q: A trained covered x km at 40 kmph and another 2x km at 20 kmph. Find the average speed of the train in covering the entire 3x km.

Add the times, not the speeds.\[ t = \frac{x}{40} + \frac{2x}{20} = \frac{x}{40} + \frac{4x}{40} = \frac{5x}{40} = \frac{x}{8} \]\[ \bar{v} = \frac{3x}{x/8} = 24 \text{ kmph} \]Averaging the two speeds gives 30 kmph, too high - two thirds of the journey is at the slower rate.

Q: Two trains each 250 m in length are running on the same parallel lines in opposite directions with the speed of 80 kmph and 70 kmph respectively. In what time will they cross each other completely?

Trains in the same direction close at the difference of their speeds; in opposite directions, at the sum.\[ 1 \text{ kmph} = \frac{5}{18} \text{ m/s} \]\[ \text{relative speed} = 80 + 70 = 150 \text{ kmph} = \frac{125}{3} \text{ m/s} \]\[ t = \frac{250 + 250}{125/3} = \frac{500 \times 3}{125} = 12 \text{ sec} \]Both lengths are covered, since the trains must clear each other completely.

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81.

A motorcyclist goes from Bombay to Pune, a distance of 192 kms at an average of 32 kmph speed. Another man starts from Bombay by car 2 ½ hours after the first, and reaches Pune ½ hour earlier. What is the ratio of the speed of the motorcycle and the car?

A.1:2
B.1:3
C.10:27
D.5:4

Answer: A

\[ t_{\text{motorcycle}} = \frac{192}{32} = 6 \text{ hours} \]

The car starts \( 2\tfrac{1}{2} \) hours later and arrives half an hour earlier:

\[ t_{\text{car}} = 6 - 2.5 - 0.5 = 3 \text{ hours} \]

\[ v_{\text{car}} = \frac{192}{3} = 64 \text{ kmph} \]

\[ 32 : 64 = 1 : 2 \]

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82.

A man on tour travels first 160 km at 64 km/hr and the next 160 km at 80 km/hr. The average speed for the first 320 km of the tour is:

A.35.55 km/hr
B.36 km/hr
C.71.11 km/hr
D.71 km/hr

Answer: C

Average speed is total distance over total time, which for two equal legs is the harmonic mean, not the arithmetic one:

\[ \bar{v} = \frac{2v_{1}v_{2}}{v_{1} + v_{2}} \]

The two legs are equal in distance, so

\[ \bar{v} = \frac{2 \times 64 \times 80}{64 + 80} = \frac{10240}{144} = 71.11 \text{ km/hr} \]

The arithmetic mean of 72 is close but slightly high. The 35.55 option is exactly half the right answer - the result of dividing by the total distance instead of one leg.

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83.

How many seconds will a train 100 meters long take to cross a bridge 150 meters long if the speed of the train is 36 kmph?

A.18
B.22
C.25
D.28

Answer: C

\[ 1 \text{ kmph} = \frac{5}{18} \text{ m/s} \]

\[ 36 \text{ kmph} = 10 \text{ m/s} \]

\[ t = \frac{100 + 150}{10} = 25 \text{ seconds} \]

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84.

A person crosses a 600 m long street in 5 minutes. What is his speed in km per hour?

A.3.6
B.7.2
C.8.4
D.10

Answer: B

Convert both quantities to the units the answer needs.

\[ 600 \text{ m} = 0.6 \text{ km}, \qquad 5 \text{ min} = \frac{1}{12} \text{ h} \]

\[ v = \frac{0.6}{1/12} = 7.2 \text{ km/hr} \]

Equivalently, 2 m/s multiplied by \( \tfrac{18}{5} \) gives the same 7.2 km/hr.

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85.

P, Q and R are in a cycle race of 4500 meters. P cycles twice as fast as Q. R cycles 1/3rd as fast as Q. R completes the race in 45 minutes. Then where was Q from the finishing line when P finished the race?

A.300 m
B.1500 m
C.2250 m
D.3000 m

Answer: C

Start from R, the only cyclist whose time is given.

\[ v_{R} = \frac{4500}{45} = 100 \text{ m/min} \]

Since \( v_{R} = \tfrac{1}{3}v_{Q} \) and \( v_{P} = 2v_{Q} \):

\[ v_{Q} = 300 \text{ m/min}, \qquad v_{P} = 600 \text{ m/min} \]

\[ t_{P} = \frac{4500}{600} = 7.5 \text{ min} \]

In that time Q has covered \( 300 \times 7.5 = 2250 \) m, so

\[ 4500 - 2250 = 2250 \text{ m from the finish} \]

P is twice as fast as Q, so Q is exactly halfway when P finishes - a check that needs no arithmetic at all.

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86.

Surendra travels from home to office by car. With an average speed of 50 km/hr, he is late by 30 minutes. But when he comes with a speed of 60 km/h, he reaches his office 10 minutes earlier. How far is his office from his home?

A.60 km
B.80 km
C.200 km
D.100 km

Answer: C

One trip is 30 minutes late and the other 10 minutes early, so the travel times differ by 40 minutes.

\[ \frac{D}{50} - \frac{D}{60} = \frac{40}{60} = \frac{2}{3} \text{ h} \]

\[ D\left(\frac{6 - 5}{300}\right) = \frac{2}{3} \quad\Longrightarrow\quad \frac{D}{300} = \frac{2}{3} \]

\[ D = 200 \text{ km} \]

Late and early must be added to get the gap: 30 + 10 = 40 minutes, not their difference.

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87.

A train of 24 carriages, each of 60 meters length, when an engine also of 60 meters length is running at a speed of 60 kmph. In what time will the train cross a bridge 1.5 km long?

A.1 ½ mins
B.3 mins
C.3 ½ mins
D.7 mins

Answer: B

\[ 1 \text{ kmph} = \frac{5}{18} \text{ m/s} \]

The whole train is the engine plus 24 carriages, each 60 m:

\[ 25 \times 60 = 1500 \text{ m} \]

\[ 60 \text{ kmph} = \frac{50}{3} \text{ m/s} \]

\[ t = \frac{1500 + 1500}{50/3} = \frac{3000 \times 3}{50} = 180 \text{ s} = 3 \text{ minutes} \]

The engine counts as a twenty-fifth 60 m unit, and the bridge's 1.5 km must be converted to 1500 m.

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88.

In what time will a railway train 60 m long moving at the rate of 36 kmph pass a telegraph post on its way?

A.9 sec
B.8 sec
C.7 sec
D.6 sec

Answer: D

A telegraph post has no length, so the train covers only its own 60 m.

\[ 36 \text{ kmph} = 36 \times \frac{5}{18} = 10 \text{ m/s} \]

\[ t = \frac{60}{10} = 6 \text{ seconds} \]

Passing a platform or bridge would require adding its length as well; a post adds nothing.

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89.

A trained covered x km at 40 kmph and another 2x km at 20 kmph. Find the average speed of the train in covering the entire 3x km.

A.30 kmph
B.25 kmph
C.24 kmph
D.28 kmph

Answer: C

Add the times, not the speeds.

\[ t = \frac{x}{40} + \frac{2x}{20} = \frac{x}{40} + \frac{4x}{40} = \frac{5x}{40} = \frac{x}{8} \]

\[ \bar{v} = \frac{3x}{x/8} = 24 \text{ kmph} \]

Averaging the two speeds gives 30 kmph, too high - two thirds of the journey is at the slower rate.

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90.

Two trains each 250 m in length are running on the same parallel lines in opposite directions with the speed of 80 kmph and 70 kmph respectively. In what time will they cross each other completely?

A.10 sec
B.11 sec
C.12 sec
D.14 sec

Answer: C

Trains in the same direction close at the difference of their speeds; in opposite directions, at the sum.

\[ 1 \text{ kmph} = \frac{5}{18} \text{ m/s} \]

\[ \text{relative speed} = 80 + 70 = 150 \text{ kmph} = \frac{125}{3} \text{ m/s} \]

\[ t = \frac{250 + 250}{125/3} = \frac{500 \times 3}{125} = 12 \text{ sec} \]

Both lengths are covered, since the trains must clear each other completely.

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