Time & Distance Questions and Answers

Q: Two trains of length 120 m and 280 m are running towards each other on parallel lines at 42 kmph and 30 kmph respectively. In what time will they be 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.\[ 1 \text{ kmph} = \frac{5}{18} \text{ m/s} \]\[ \text{relative speed} = 42 + 30 = 72 \text{ kmph} = 20 \text{ m/s} \]\[ t = \frac{120 + 280}{20} = \frac{400}{20} = 20 \text{ sec} \]

Q: Convert the 13/36 m/s into kilometers per hour?

\[ 1 \text{ m/s} = \frac{18}{5} \text{ kmph} \]\[ \frac{13}{36} \times \frac{18}{5} = \frac{13 \times 18}{36 \times 5} = \frac{13}{10} = 1.3 \text{ kmph} \]The 18 cancels half of the 36, which is what makes the arithmetic so clean.

Q: If a man can cover 12 metres in one second, how many kilometres can he cover in 3 hours 45 minutes?

Convert the time to seconds and multiply.\[ 3 \text{ h } 45 \text{ min} = 3.75 \times 3600 = 13{,}500 \text{ s} \]\[ d = 12 \times 13{,}500 = 162{,}000 \text{ m} = 162 \text{ km} \]Equivalently, 12 m/s is \( 12 \times \tfrac{18}{5} = 43.2 \) kmph, and \( 43.2 \times 3.75 = 162 \).

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: Pavan travelled for 11 hours. He covered the first half of the distance at 30 kmph and remaining half of the distance at 25 kmph. Find the distance travelled by Pavan.

Each half of the distance is \( \tfrac{d}{2} \).\[ \frac{d/2}{30} + \frac{d/2}{25} = 11 \]\[ \frac{d}{60} + \frac{d}{50} = \frac{5d + 6d}{300} = \frac{11d}{300} \]\[ \frac{11d}{300} = 11 \quad\Longrightarrow\quad d = 300 \text{ km} \]Equivalently, the average speed is the harmonic mean \( \tfrac{2 \times 30 \times 25}{55} = 27.27 \) kmph, and \( 27.27 \times 11 = 300 \).

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: A person covered one-fourth of the total distance at 26 kmph and remaining distance at 24 kmph. What is the average speed for the total distance?

Add the times for the two portions.\[ t = \frac{d/4}{26} + \frac{3d/4}{24} = \frac{d}{104} + \frac{d}{32} \]Over the common denominator 416:\[ t = \frac{4d + 13d}{416} = \frac{17d}{416} \]\[ \bar{v} = \frac{d}{17d/416} = \frac{416}{17} = 24\tfrac{8}{17} \text{ kmph} \]The average must lie between 24 and 26, and nearer 24 since three quarters of the distance is covered at that speed - a check that rules out anything below 24.

Important Instructions

131.

Two trains of length 120 m and 280 m are running towards each other on parallel lines at 42 kmph and 30 kmph respectively. In what time will they be clear of each other from the moment they meet?

A.14 sec
B.21 sec
C.17 sec
D.20 sec

Answer: D

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

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

\[ \text{relative speed} = 42 + 30 = 72 \text{ kmph} = 20 \text{ m/s} \]

\[ t = \frac{120 + 280}{20} = \frac{400}{20} = 20 \text{ sec} \]

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132.

Convert the 13/36 m/s into kilometers per hour?

A.1.5
B.1.2
C.1.3
D.1.4

Answer: C

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

\[ \frac{13}{36} \times \frac{18}{5} = \frac{13 \times 18}{36 \times 5} = \frac{13}{10} = 1.3 \text{ kmph} \]

The 18 cancels half of the 36, which is what makes the arithmetic so clean.

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133.

If a man can cover 12 metres in one second, how many kilometres can he cover in 3 hours 45 minutes?

A.168
B.162
C.150
D.156

Answer: B

Convert the time to seconds and multiply.

\[ 3 \text{ h } 45 \text{ min} = 3.75 \times 3600 = 13{,}500 \text{ s} \]

\[ d = 12 \times 13{,}500 = 162{,}000 \text{ m} = 162 \text{ km} \]

Equivalently, 12 m/s is \( 12 \times \tfrac{18}{5} = 43.2 \) kmph, and \( 43.2 \times 3.75 = 162 \).

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134.

A person takes 20 minutes more to cover a certain distance by decreasing his speed by 20%. What is the time taken to cover the distance at his original speed?

A.1 hour 30 minutes
B.1 hour 15 minutes
C.1 hour 20 minutes
D.2 hours

Answer: C

Let the distance and original speed be d km and k kmph respectively.

d/0.8k - d/k = 20/60 => 5d/4k - d/k = 1/3

=> (5d - 4d)/4k = 1/3 => d = 4/3 k

Time taken to cover the distance at original speed

= d/k = 4/3 hours = 1 hour 20 minutes.

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135.

Kiran travels from A to B by car and returns from B to A by cycle in 7 hours. If he travels both ways by car he saves 3 hours. What is the time taken to cover both ways by cycle?

A.10 hours
B.13 hours
C.14 hours
D.12 hours

Answer: A

Let the time taken to cover from A to B in car and cycle be x hours and y hours respectively.

x + y = 7 --- (1) ; 2x = 4 --- (2)

solving both the equations, we get y = 5

So, time taken to cover both ways by cycle = 2y hours = 10 hours.

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136.

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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137.

Pavan travelled for 11 hours. He covered the first half of the distance at 30 kmph and remaining half of the distance at 25 kmph. Find the distance travelled by Pavan.

A.240 km
B.280 km
C.260 km
D.300 km

Answer: D

Each half of the distance is \( \tfrac{d}{2} \).

\[ \frac{d/2}{30} + \frac{d/2}{25} = 11 \]

\[ \frac{d}{60} + \frac{d}{50} = \frac{5d + 6d}{300} = \frac{11d}{300} \]

\[ \frac{11d}{300} = 11 \quad\Longrightarrow\quad d = 300 \text{ km} \]

Equivalently, the average speed is the harmonic mean \( \tfrac{2 \times 30 \times 25}{55} = 27.27 \) kmph, and \( 27.27 \times 11 = 300 \).

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138.

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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139.

Murali travelled from city A to city B at a speed of 40 kmph and from city B to city C at 60 kmph. What is the average speed of Murali from A to C given that the ratio of distances between A to B and B to C is 2 : 3?

A.48 kmph
B.50 kmph
C.52 kmph
D.56 kmph

Answer: B

Let the distances between city A to B and B to C be 2x km and 3x km respectively.

Total time taken to cover from A to C

= (2x)/40 + (3x)/60 = (6x + 6x)/120 = 12x/120 = x/10 Average speed = (2x + 3x)/(x/10) = 50 kmph.

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140.

A person covered one-fourth of the total distance at 26 kmph and remaining distance at 24 kmph. What is the average speed for the total distance?

A.21(2/3) kmph
B.\( 24\tfrac{8}{17} \) kmph
C.22 (1/3) kmph
D.22 (2/3) kmph

Answer: B

Add the times for the two portions.

\[ t = \frac{d/4}{26} + \frac{3d/4}{24} = \frac{d}{104} + \frac{d}{32} \]

Over the common denominator 416:

\[ t = \frac{4d + 13d}{416} = \frac{17d}{416} \]

\[ \bar{v} = \frac{d}{17d/416} = \frac{416}{17} = 24\tfrac{8}{17} \text{ kmph} \]

The average must lie between 24 and 26, and nearer 24 since three quarters of the distance is covered at that speed - a check that rules out anything below 24.

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