Time and Distance Questions

Answer: Option D

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

Answer: Option B

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.

Answer: Option B

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

Answer: Option D

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

Answer: Option B

Speed of the bus excluding stoppages =54 kmph

Speed of the bus including stoppages =45 kmph



Loss in speed when including stoppages =54−45=9 kmph

⇒ 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

Answer: Option E

5 ft per second so in 1 minute 300 ft

In 60 minutes = 60 * 300= 18000 ft.

Answer: Option C

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.

Answer: Option D

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

Answer: Option A

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 =${\displaystyle \frac{\mathrm{D}\mathrm{i}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{c}\mathrm{e}}{\mathrm{S}\mathrm{p}\mathrm{e}\mathrm{e}\mathrm{d}}=\frac{25}{27}=\frac{175}{5\times K}}$ 
Ie., Hare has to increase its speed by a factor K. Solving we get K = 37.8

Answer: Option B

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