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# Time and Distance Questions

Home > Quantitative Aptitude > Time and Distance > General Questions
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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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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 thief steals a car at 2.30 p.m and drives it at 60 kmph. The theft is discovered at 3 p.m and the owner sets off in another car at 75 kmph. When will be overtake the thief.

A4.30 p.m

B5.15 p.m

C4.45 p.m

D5:00 p.m

Explanation:

As Theft is discovered at 3:00pm but Thief stole the car at 2:30.
This means thief covered some distance in this 30 min gap.
Distance travelled by thief in 30 min = 60 * 1/2 = 30 km

Owner Discovered Car at 3:00pm

Now relative speed = (75-60)km/hr = 15km/hr
Time needed to travel 30km by the speed of 15km/hr.
Time at which owner meets thief = 30/15 = 2 hrs
So After 2 hrs (i.e,at 5:00 pm) the owner will catch/overtake the thief

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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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Kartik rides a bicycle 8 km/hr and reaches his school 2.5 minutes late. The next day he increases his speed to 10 km/hr and reaches the school 5 minutes early. How far is the school from his house?

A5/8 km

B8 km

C5 km

D10 km.

Explanation:

Let the total distance be d km. We have speed = 8km/hr and 10km/hr.
We know, time (t) = distance (d)/speed (s).
= > t = d/8 (for the first day)
=> t = d/10 (for the second day)
Difference between the time taken on first day and second day is 2.5 + 5 = 7.5 mins.
Therefore, (d/8) - (d/10) = 7.5/60. (7.5 mins converted to hours)
=> (10d - 8d)/80 = 75/600
=> d = (75 x 80)/1200.
=> d = 5 kms.
Therefore, the distance between his school and house 5 km.

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If I walk with 30 miles/hr i reach 1 hour before and if i walk with 20 miles/hr i reach 1 hour late. Find the distance between 2 points and the exact time of reaching destination is 11 am then find the speed with which it walks.

A125 miles and 24 miles/hr

B120 miles and 30 miles/hr

C120 miles and 20 miles/hr

D120 miles and 24 miles/hr

Explanation:

the exact time is 11 am
then 30miles/ph reach it on 10am
20 miles/ph reach it on 12am
if they start walking in 6am
then 30m/ph-(10am-6am) 4h*30=120miles
20m/ph-(12am-6am) 6h*20=120miles

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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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A thief is noticed by a policeman from a distance of 200 m. The thief starts running and the policeman chases him. The thief and the policeman run at the rate of 10 km and 11 km per hour respectively. What is the distance between them after 6 minutes ?

A50 meter

B100 meter

C110 meter

DNone of above

Explanation:

find the relative speed of the thief and policeman = (11-10)km/hr = 1 km/hr

Distance covered in 6 minutes = (1/60)*6=1/10km => 100meters

So distance between them after 6 minutes = (200-100)=100 meters

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