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# Time and Distance Questions

Home > Quantitative Aptitude > Time and Distance > General Questions
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A person rides a bicycle round a circular path of radius 50 m. The radius of the wheel of the bicycle is 50 cm. The cycle comes to the starting point for the first time in 1 hour. What is the number of revolutions of the wheel in 15 minutes?

A20

B25

C30

D35

Explanation:

Circumference of the circular path = 2/\50 = 100 /\ m
This means 100? m is the distance covered in 60 minutes
Therefore distance covered in 15 minutes = 1/4 x 100 /\ m = 25/\ m
Now distance covered in one revolution of the wheel = 2 /\(1/2) m = /\ m
Therefore number of revolutions in 15 minutes = 25/\ //\ = 25

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A scooterist covers a certain distance at 36 kmph. How many meters does he cover in 2 min?

A800 meters.

B1100 meters.

C700 meters.

D1200 meters.

Explanation:

Speed = 36 kmph = 36 * 5/18 = 10mps
So, Distance covered in 2 min = (10 * 2 * 60)m = 1200 meters.

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The diameter of the driving wheel of a bus is 140cm. How many revolutions per minute must the wheel make in order to keep a speed of 66 kmph?

A250

B150

C300

D200

Explanation:

Distance to be covered in 1 min=(66*1000)/60 m=1100m
Circumference of the wheel =(2*22/7*0.70)m=4.4m.
So, Number of revolutions per min=1100/4.4=250.

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By walking at 3/4th of his usual speed, a man reaches office 20 minutes later than usual. What is his usual time?

A60 minutes.

B50 minutes.

C30 minutes.

DNone of these

Explanation:

3/4 of a man's usual speed means, he takes 4/3 of his usual time to cover the same distance,
i.e. he takes 4/3 â€“ 1 = 1/3 time extra.
Given 1/3 time is 20 minutes
Usual time = 20 * 3 = 60 minutes.

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If a man walks at the rate of 5kmph, he misses a train by only 7min. However if he walks at the rate of 6 kmph he reaches the station 5 minutes before the arrival of the train.Find the distance covered by him to reach the station.

A10 km

B8 km

C6 km

D5 km

Explanation:

Lets assume the required distance = x km.
Difference in the times taken at two speeds=12mins=1/5 hr.
Therefore (x/5-x/6)=1/5 or (6x-5x) = 6 or x = 6km.
So required distance = 6 km

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Yana and Gupta leave points x and y towards y and x respectively simultaneously and travel in the same route. After meeting each other on the way, Yana takes 4 hours to reach her destination, while Gupta takes 9 hours to reach his destination. If the speed of Yana is 48 km/hr, what is the speed of Gupta?

A30 mph

B50 mph

C35 mph

D20 mph

Explanation:

Let the speed of Yana be v1 and Gupta be v2 and time taken by them be t1 and t2 respectively.
There is a shortcut for such questions as v1^2/v2^2= t2/t1

Putting the respective values we get v2= 32km/hr

1.6 km =1 miles

So 32/1.6= 20mph.

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Walking 5/6 of its usual speed, a train is 10min late. Find the usual time to cover the journey?

A55 min.

B45 min.

C35 min.

D50 min.

Explanation:

New speed = 5/6 of usual speed
New time = 6/5 of usual time
Therefore, (6/5 of usual time) - usual time = 10min
Therefore Usual time = 50min

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Vivek travelled 1200km by air which formed 2/5 of his trip.One third of the whole trip , he travelled by car and the rest of the journey he performed by train. The distance travelled by tain was:

A1000km

B900km

C700km

D800km

Explanation:

Let the total trip be x km.
Then 2x/5=1200
x=1200*5/2=3000km
Distance travelled by car =1/3*3000=1000km
Journey by train =[3000-(1200+1000)]=800km.

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A girl rode her bicycle from home to school, a distance of 15 miles, at an average speed of 15 miles per hour. She returned home from school by walking at an average speed of 5 miles per hour. What was her average speed for the round trip if she took the same route in both directions?

A8.5 miles/hr.

B7 miles/hr.

C7.5 miles/hr.

DNone of these

Explanation:

Avg speed=(total disance)/(total time)
Total distance= 15+15 = 30 miles.
Total time= 1+3=4 hrs.
So average Spedd = 30/4 = 7.5 miles/hr.

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A ship went on a voyage after 180 miles a plane started with 10 times speed that of the ship. Find the distance when they meet from starting point.

A200

B202

C120

D100

Explanation:

Lets assume u is the speed of ship and 10u is of plane.
So, relative speed = 9u
Distance covered = (time to meet * speed of plane)
=180*10u/9u =200 miles

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