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# Time and Distance Questions

Home > Quantitative Aptitude > Time and Distance > General Questions
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A person crosses a 600 m long street in 5 minutes. What is his speed in km per hour?

A3.6

B7.2

C8.4

D10

Explanation:

Speed = 600/(5*60) m/sec. = 600/300 m/sec = 2 m/sec
Now convert m/sec to km/hr
= 2 x 18/5 km/hr = 7.2 km/hr.

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How Many minutes does shyam take to cover a distance of 500 m, if he runs at a speed of 20km/h?

A1 min 30 sec.

B1 min 25 sec.

C1 min.

D1 min 10 sec

Explanation:

Ram's speed = 20 km/h = (20 * 5/18) m/sec. =50/9 m/sec.

Formula Used: Time = Distance/Speed.

Time taken to cover 500m = 500/(50/9) sec

= (500 *9)/50 sec = 90 sec = 1 min 30 sec..

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Virat travelled 75 kms in 7 hours. He went some distance at the rate of 12 km/hr and the rest at 10 km/hr. How far did he travel at the rate of 12 km/hr?

A30 kms

B25 kms

C40 kms

D35 kms.

Explanation:

Let the distance travelled at the rate of 12 kmph be x km
Time taken to travel this distance = x/12 .......(1)
Then the distance travelled at the rate of 10 km/hr will be (75 - x)
Time taken to travel this distance = (75 - x)/10 .......(2)
Total time taken to cover the total distance = 7 hrs
=>(x/12) + (75 - x)/10 = 7 [sum of (1) and (2)]
=> (10x + 900 - 12x) = 840
=> -2x = -60
=> x = 30.
Therefore, the distance travelled at the rate of 12 km/hr = 30km.

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A boy rides his bicycle 10 kms at an average speed of 12 kmph and again travels 12 kms at an average speed of 10 kmph. His average speed for the entire trip is approximately:

A10.4 kmph

B10.8 kmph

C11.0 kmph

D12.2 kmph

Explanation:

Average speed is represented by total distance divided by the time taken.
Total distance = 10 kms + 12 kms = 22 kms
Total Time = (10/12) + (12/10)
? (10/12) + (12/10) = (100 + 144)/120 = 244/120 hrs
Average speed = 22/(244/120) = 10.8 kmph.
Average speed for the entire trip is 10.8 kmph.

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An athlete runs half of the distance at the speed of 10 km/hr and remaining 15 km at the speed of 20 km/hr, how much time will be taken to cover the total distance?

A95 mins

B135 mins

C120 mins

D115 mins.

Explanation:

We know that time = distance/speed
Time taken by the athlete for the first half = 15/10
And time taken by the athlete for the second half = 15/20
Total time = 15/10 + 15/20 = (45/20) x 60 = 135 mins
Time taken to cover the total distance is 135 mins.

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A man covered a certain distance at some speed. Had he moved 3 kmph faster, he would have taken 40 minutes less. If he had moved 2 kmph slower, he would have taken 40 minutes more. The distance (in km) is:

A35

B36 2/3

C37 1/2

D40

Explanation:

Let distance = x km and usual rate = y kmph.

Then, x/y - x/(y+3) = 40/60
2y(y + 3) = 9x ....(i)

And, x/(y-2) - x/y = 40/60
y(y - 2) = 3x ....(ii)
By Solving (i) by (ii), we get: x = 40

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Sanu travelled a distance of 20 kms at the speed of 5 km/hr, he reached 40 minutes late. If he had walked at the speed of 8 km/hr, how early from fixed time will he reach?

A15 mins

B50 mins

C25 mins

D1.5 hrs.

Explanation:

Time taken to travel 20 kms at the speed of 5 kmph.

=> t = 20/5 = 4 hrs.

Actual time of reaching = (4 hrs - 40 mins) = 3 hrs 20 mins ......(1)

Time taken to travel 20 kms at the speed of 8 kmph.

=> t = 20/8 = 2.5 hrs. ......(2)

Difference between (1) and (2) => (3 hrs 20 mins - 2 hrs 30 mins)

(200 mins - 150 mins) = 50 mins.

Sanu would have reached 50 mins earlier.

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Excluding stoppages, the speed of a bus is 54 kmph and including stoppages, it is 45 kmph. For how many minutes does the bus stop per hour?

A9

B10

C12

D20

ENone of these

Explanation:

Speed of the bus excluding stoppages =54 kmph

Speed of the bus including stoppages =45 kmph

Loss in speed when including stoppages =54âˆ’45=9 kmph

â‡’ In 1 hour, bus covers 9 km less due to stoppages.

Hence, time in which the bus stops per hour

= Time taken to cover 9 km

=distancespeed=954 hour=16 hour  =606 min=10 min

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If an object travels at five feet per second, how many feet does it travel in one hour?

A30

B300

C720

D1800

E18000

Explanation:

5 ft per second so in 1 minute 300 ft

In 60 minutes = 60 * 300= 18000 ft.

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Hare in the other. The hare starts after the tortoise has covered 1/5 of its distance and that too leisurely3. A hare and a tortoise have a race along a circle of 100 yards diameter. The tortoise goes in one direction and the. The hare and tortoise meet when the hare has covered only 1/8 of the distance. By what factor should the hare increase its speed so as to tie the race?

A37.8

B8

C40

D5

Explanation:

Assume the circumference of the circle is 200 meters.  Hare and tortoise started at the same point but moves in the opposite direction. It is given that by that time tortoise covered 40 m (1/5th of the distance), Hare started and both met after hare has covered 25. This implies, in the time hare has covered 25m, hare has covered 200 - 40 - 25 = 135 meters.
So Hare : tortoise speeds = 25 : 135 = 5 : 27
Now Hare and tortoise has to reach the starting point means, Hare has to cover 175 meters and Tortoise has to cover only 25 meters in the same time.
As time =$\frac{\mathrm{D}\mathrm{i}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{c}\mathrm{e}}{\mathrm{S}\mathrm{p}\mathrm{e}\mathrm{e}\mathrm{d}}=\frac{25}{27}=\frac{175}{5×K}$
Ie., Hare has to increase its speed by a factor K. Solving we get K = 37.8

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