Speed Problem

[gmat009]

Mary passed a certain gas station on a highway while traveling west at a constant speed of 50 miles per hour. Then, 15 minutes later, Paul passed the same gas station while traveling west at a constant speed of 60 miles per hour. If both drivers maintained their speeds and both remained on the highway for at least 2 hours, how long after he passed the gas station did Paul catch up with Mary?
A. 30 min
B. 45 min
C. 1 hr
D. 1 hr 15 min
E. 1 hr 30 min

OA is D

[amitansu]

Since both are travelling at same direction relative spedd = 60-50=10 mph

In 15 min. Mary would travel = 50/4 miles (since her speed is 50 mph)

So for Paul to catch up her , has to travel 50/4 miles with the relative speed which is 10 mph.

T=Distance/speed=(50/4)/10=50/40 hrs=75 min=1 hr 15 min.


Amit

[andreasonlinegr]

In 15 min. Mary would travel = 50/4 miles (since her speed is 50 mph)

Why is it 50/4? Should it not be that in 15min Mary would travel
d=r x t => d=50mph x 0.25h= d = 12.5miles

[andreasonlinegr]

In 15 min. Mary would travel = 50/4 miles (since her speed is 50 mph)

Why is it 50/4? Should it not be that in 15min Mary would travel
d=r x t => d=50mph x 0.25h= d = 12.5miles

[andreasonlinegr]

In 15 min. Mary would travel = 50/4 miles (since her speed is 50 mph)

Why is it 50/4? Should it not be that in 15min Mary would travel
d=r x t => d=50mph x 0.25h= d = 12.5miles

[vivek.kapoor83]

in 60 mins she travels - 50miles
so in 15 min she travels - 12.5 miles
relative speed = 10 mph
Time taken to cover distance = 12.5 /10 = 125/100 = 5/4 = 1.25 hrs
not if decimal is 100 then 60 min
if decimal is 1 ...60/100
if decimal .25 = (60/100) *.25 = 15 mins
So, time taken to catch her...1 hr 15 mins.