Time & Distance Questions and Answers

Q: A train sets off at 2 p.m. at the speed of 70 kmph. Another train starts at 3:30 p.m. in the same direction at the rate of 85 kmph. At what time the trains will meet?

The first train has a \( 1\tfrac{1}{2} \) hour head start:\[ 70 \times 1.5 = 105 \text{ km} \]The gap closes at \( 85 - 70 = 15 \) kmph:\[ t = \frac{105}{15} = 7 \text{ hours} \]\[ 3{:}30 \text{ p.m.} + 7 \text{ h} = 10{:}30 \text{ p.m.} \]

Q: A and B go cycling in the same direction with speeds of 6 km/hr and 12 km/hr. A car from behind passes them in 9 and 10 seconds respectively. What is the speed of the car?

Overtaking means the car's own length is covered at the relative speed. Let the car do \( c \) km/hr and be \( L \) metres long.\[ L = (c - 6) \times 9 \times \frac{5}{18} = (c - 12) \times 10 \times \frac{5}{18} \]The factor \( \tfrac{5}{18} \) is common, so\[ 9(c - 6) = 10(c - 12) \quad\Longrightarrow\quad 9c - 54 = 10c - 120 \]\[ c = 66 \text{ km/hr} \]The car's length never has to be found - it cancels because the same car passes both cyclists.

Q: If John covers a certain distance in 1 hr. 24 min. by covering two third of the distance at 4 kmph and the rest at 5 kmph, then find the total distance.

Let the total distance be \( D \). The two legs take\[ \frac{2D/3}{4} + \frac{D/3}{5} = 1 \text{ h } 24 \text{ min} = 1.4 \text{ h} \]\[ \frac{D}{6} + \frac{D}{15} = \frac{5D + 2D}{30} = \frac{7D}{30} \]\[ \frac{7D}{30} = 1.4 \quad\Longrightarrow\quad D = 6 \text{ km} \]Checking: 4 km at 4 kmph is 1 hour and 2 km at 5 kmph is 24 minutes.

Q: Two trains 121 meters and 165 meters in length respectively are running in opposite directions, one at the rate of 80 km and the other at the rate of 65 kmph. In what time will they be completely clear of each other from the moment they meet?

Trains in the same direction close at the difference of their speeds; in opposite directions, at the sum.\[ 1 \text{ kmph} = \frac{5}{18} \text{ m/s} \]\[ \text{relative speed} = 80 + 65 = 145 \text{ kmph} = \frac{725}{18} = 40.28 \text{ m/s} \]\[ t = \frac{121 + 165}{40.28} = \frac{286}{40.28} \approx 7.1 \text{ sec} \]"Clear of each other" means both lengths must pass, which is why 286 m and not 165 m is the distance.

Q: In covering a distance of 30 km, Abhay takes 2 hours more than Sameer. If Abhay doubles his speed, then he would take 1 hour less than Sameer. Abhay's speed is:

Let Abhay's speed be \( a \) and Sameer's be \( s \).\[ \frac{30}{a} - \frac{30}{s} = 2 \qquad \text{(i)} \]\[ \frac{30}{2a} = \frac{30}{s} - 1 \qquad \text{(ii)} \]From (i), \( \tfrac{30}{s} = \tfrac{30}{a} - 2 \). Substituting into (ii):\[ \frac{15}{a} = \frac{30}{a} - 2 - 1 \quad\Longrightarrow\quad \frac{15}{a} = 3 \]\[ a = 5 \text{ kmph} \]Checking: Abhay takes 6 hours, Sameer 4; at 10 kmph Abhay would take 3 hours - one less than Sameer, as required.

Q: If a train, travelling at a speed of 90 kmph, crosses a pole in 5 sec, then the length of train is?

\[ 1 \text{ kmph} = \frac{5}{18} \text{ m/s} \]\[ 90 \text{ kmph} = 25 \text{ m/s} \]A pole has no length, so the train covers only itself:\[ L = 25 \times 5 = 125 \text{ m} \]

Q: How much time will a train of length 200 m moving at a speed of 72 kmph take to cross another train of length 300 m, moving at 36 kmph in the same direction?

Trains in the same direction close at the difference of their speeds; in opposite directions, at the sum.\[ 1 \text{ kmph} = \frac{5}{18} \text{ m/s} \]\[ \text{relative speed} = 72 - 36 = 36 \text{ kmph} = 10 \text{ m/s} \]\[ t = \frac{200 + 300}{10} = 50 \text{ sec} \]Both lengths count, because the whole of one train must pass the whole of the other.

Q: A man walks at a speed of 3 km/hr and runs at a speed of 7 km/hr. How much time will the man require to cover a distance of 10 1/2 km, if he completes half of the distance, i.e., (5 1/4) km on foot and the other half by running?

Each half is \( 5\tfrac{1}{4} = 5.25 \) km.\[ t = \frac{5.25}{3} + \frac{5.25}{7} = 1.75 + 0.75 = 2.5 \text{ hours} \]That is \( 2\tfrac{1}{2} \) hours, which appears nowhere on the list, so the answer is none of these. The \( 2\tfrac{1}{4} \) offered is close enough to catch a rounded calculation.

Q: Rohit covers one-fourth of the total distance at 20 km/hr, one-fourth at 10 km/h and rest of his journey at 80 km/h. Find Rohit’s average speed for the whole distance?

Take the total distance as \( D \) and add the three times.\[ t = \frac{D/4}{20} + \frac{D/4}{10} + \frac{D/2}{80} = \frac{D}{80} + \frac{D}{40} + \frac{D}{160} \]\[ = \frac{2D + 4D + D}{160} = \frac{7D}{160} \]\[ \bar{v} = \frac{D}{7D/160} = \frac{160}{7} = 22.85 \text{ km/hr} \]Averaging the three speeds would give \( \tfrac{110}{3} = 36.7 \) km/hr, far too high - the slow 10 km/hr leg dominates the total time even though it covers only a quarter of the distance.

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61.

A train sets off at 2 p.m. at the speed of 70 kmph. Another train starts at 3:30 p.m. in the same direction at the rate of 85 kmph. At what time the trains will meet?

A.9.30 p.m.
B.8.30 p.m.
C.10.30 p.m.
D.10.45 p.m.

Answer: C

The first train has a \( 1\tfrac{1}{2} \) hour head start:

\[ 70 \times 1.5 = 105 \text{ km} \]

The gap closes at \( 85 - 70 = 15 \) kmph:

\[ t = \frac{105}{15} = 7 \text{ hours} \]

\[ 3{:}30 \text{ p.m.} + 7 \text{ h} = 10{:}30 \text{ p.m.} \]

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62.

A and B go cycling in the same direction with speeds of 6 km/hr and 12 km/hr. A car from behind passes them in 9 and 10 seconds respectively. What is the speed of the car?

A.22 km/hr
B.33 km/hr
C.66 km/hr
D.44 km/hr

Answer: C

Overtaking means the car's own length is covered at the relative speed. Let the car do \( c \) km/hr and be \( L \) metres long.

\[ L = (c - 6) \times 9 \times \frac{5}{18} = (c - 12) \times 10 \times \frac{5}{18} \]

The factor \( \tfrac{5}{18} \) is common, so

\[ 9(c - 6) = 10(c - 12) \quad\Longrightarrow\quad 9c - 54 = 10c - 120 \]

\[ c = 66 \text{ km/hr} \]

The car's length never has to be found - it cancels because the same car passes both cyclists.

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63.

If John covers a certain distance in 1 hr. 24 min. by covering two third of the distance at 4 kmph and the rest at 5 kmph, then find the total distance.

A.4 km
B.6 km
C.6.8 km
D.7.2 km

Answer: B

Let the total distance be \( D \). The two legs take

\[ \frac{2D/3}{4} + \frac{D/3}{5} = 1 \text{ h } 24 \text{ min} = 1.4 \text{ h} \]

\[ \frac{D}{6} + \frac{D}{15} = \frac{5D + 2D}{30} = \frac{7D}{30} \]

\[ \frac{7D}{30} = 1.4 \quad\Longrightarrow\quad D = 6 \text{ km} \]

Checking: 4 km at 4 kmph is 1 hour and 2 km at 5 kmph is 24 minutes.

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64.

Two trains 121 meters and 165 meters in length respectively are running in opposite directions, one at the rate of 80 km and the other at the rate of 65 kmph. In what time will they be completely clear of each other from the moment they meet?

A.7 sec
B.8.12 sec
C.7.25 sec
D.7.15 sec

Answer: D

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

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

\[ \text{relative speed} = 80 + 65 = 145 \text{ kmph} = \frac{725}{18} = 40.28 \text{ m/s} \]

\[ t = \frac{121 + 165}{40.28} = \frac{286}{40.28} \approx 7.1 \text{ sec} \]

"Clear of each other" means both lengths must pass, which is why 286 m and not 165 m is the distance.

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65.

In covering a distance of 30 km, Abhay takes 2 hours more than Sameer. If Abhay doubles his speed, then he would take 1 hour less than Sameer. Abhay's speed is:

A.5 kmph
B.6 kmph
C.6.25 kmph
D.7.5 kmph

Answer: A

Let Abhay's speed be \( a \) and Sameer's be \( s \).

\[ \frac{30}{a} - \frac{30}{s} = 2 \qquad \text{(i)} \]

\[ \frac{30}{2a} = \frac{30}{s} - 1 \qquad \text{(ii)} \]

From (i), \( \tfrac{30}{s} = \tfrac{30}{a} - 2 \). Substituting into (ii):

\[ \frac{15}{a} = \frac{30}{a} - 2 - 1 \quad\Longrightarrow\quad \frac{15}{a} = 3 \]

\[ a = 5 \text{ kmph} \]

Checking: Abhay takes 6 hours, Sameer 4; at 10 kmph Abhay would take 3 hours - one less than Sameer, as required.

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66.

If a train, travelling at a speed of 90 kmph, crosses a pole in 5 sec, then the length of train is?

A.104 m
B.125 m
C.140 m
D.152 m

Answer: B

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

\[ 90 \text{ kmph} = 25 \text{ m/s} \]

A pole has no length, so the train covers only itself:

\[ L = 25 \times 5 = 125 \text{ m} \]

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67.

How much time will a train of length 200 m moving at a speed of 72 kmph take to cross another train of length 300 m, moving at 36 kmph in the same direction?

A.50 sec
B.25 sec
C.40 sec
D.65 sec

Answer: A

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

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

\[ \text{relative speed} = 72 - 36 = 36 \text{ kmph} = 10 \text{ m/s} \]

\[ t = \frac{200 + 300}{10} = 50 \text{ sec} \]

Both lengths count, because the whole of one train must pass the whole of the other.

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68.

A man walks at a speed of 3 km/hr and runs at a speed of 7 km/hr. How much time will the man require to cover a distance of 10 1/2 km, if he completes half of the distance, i.e., (5 1/4) km on foot and the other half by running?

A.1 3/4 hrs
B.2 1/4 hrs
C.3 1/2 hrs
D.None of these

Answer: D

Each half is \( 5\tfrac{1}{4} = 5.25 \) km.

\[ t = \frac{5.25}{3} + \frac{5.25}{7} = 1.75 + 0.75 = 2.5 \text{ hours} \]

That is \( 2\tfrac{1}{2} \) hours, which appears nowhere on the list, so the answer is none of these. The \( 2\tfrac{1}{4} \) offered is close enough to catch a rounded calculation.

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69.

How much time does a train 125 metres long running at 60 km/hr take to pass a pole?

A.6s
B.2.08s
C.7.5s
D.8s

Answer: C

A pole has no length, so the train covers only its own length in passing it.

\[ 60 \text{ km/hr} = 60 \times \frac{5}{18} = \frac{50}{3} \text{ m/s} \]

\[ t = \frac{125}{50/3} = \frac{375}{50} = 7.5 \text{ s} \]

Passing a platform would require adding the platform's length; a pole adds nothing.

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70.

Rohit covers one-fourth of the total distance at 20 km/hr, one-fourth at 10 km/h and rest of his journey at 80 km/h. Find Rohit’s average speed for the whole distance?

A.22.85 km/hr
B.25.15 km/hr
C.50 km/hr
D.40 km/hr

Answer: A

Take the total distance as \( D \) and add the three times.

\[ t = \frac{D/4}{20} + \frac{D/4}{10} + \frac{D/2}{80} = \frac{D}{80} + \frac{D}{40} + \frac{D}{160} \]

\[ = \frac{2D + 4D + D}{160} = \frac{7D}{160} \]

\[ \bar{v} = \frac{D}{7D/160} = \frac{160}{7} = 22.85 \text{ km/hr} \]

Averaging the three speeds would give \( \tfrac{110}{3} = 36.7 \) km/hr, far too high - the slow 10 km/hr leg dominates the total time even though it covers only a quarter of the distance.

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