Let Car A = car that starts at 9 AM
Car B = car that starts at 9:10 AM
Time for which car A travels at speed of 40 m per hour = 1.5 hours
Distance travelled by Car A = 40 *1.5 = 60 miles

Since Car B catches up Car A at 10:30 , time = 80 mins = 4/3 hour
Speed of Car B = 60/(4/3) = 45 miles per hour

Answer A
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Distance covered by the car in 1.5 hours ( 10:30 A.M. - 9:00 A.M.) is 1.5 * 40 = 60 miles
Now car A has covered 60 miles by 10:30 A.M , so car B must also cover the same distance in ( 10:30 A.M - 9:10 A.M.) 80/60 hr

So, speed of car B must be = 60 * 6/8 i.e. 45 miles per hour

So, Answer must be (A)
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Distnace covered in 1.5 hr (90 minutes i.e. between 09:00AM - 10:30AM ) = 60 miles (40 * 1.5)

Second covered 60 miles in 80 minutes (i.e. 10 minutes less than the first car because second car started at 09:10AM).

Speed of the second car = 60miles / (80/60) hour = 45miles/hour

Option A is correct.
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Distance travelled by first car in 10 mins = 40/6 = 20/3 miles

Relative Distance between two cars at the time second car starts = 20/3 miles

Relative speed of two cars = (s-14) where s is the speed of second car

Time taken by second car to catch first car = (9:10 to 10:30) = 1 hour 20 mins = 4/3 hours

Relative speed = Relative Distance / time

(s-40) = (20/3)/(4/3) = 5
i.e. s = 5+40 = 45 miles per hour

Answer: Option A
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Here is one other way to solve the problem.
Car 1 took 90 minutes to cover some distance, whereas the second car took 80 minutes to do the same. We can set up following equation-
Ratio of time = inverse of the ratio of speed
80/90 = 40/x => 2/90=1/x=> x = 45m.

the problem can also be solved by the relative speed equation
distance to be caught up by the second car is equal to the distance that the first car did in first 10 minutes
40/60*10
This is to be caught up in 80 minutes or 8/6 hours
40/60*10/(x-40) = 8/6. Solve this to get x = 45.

We are given that a car left at 9 a.m. traveling at a rate of 40 mph and another car left 10 minutes later. Since the second car caught up to the first car at 10:30 a.m., the time of the first car is 1.5 hours and the distance is 1.5 x 40 = 60 miles.

If we let r = the rate of the second car, the distance of the second car is:

r x (1 hour 20 minutes)

r x (1 1/3 hours)

r x (4/3) = 4r/3

Since the second car catches up to the first car we can set the distances equal and determine r:

4r/3 = 60

4r = 180

r = 45

Answer: A
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Very little calculation here.
The second car travelled for 80 minutes and covered 60 miles. The question asks the speed of the second car in miles per hour. OR
How many miles did the second car travel in 60mins?

80mins------60miles
60mins------?
\(\frac{(60*60)}{80}\) = \(\frac{(3600)}{80}\) = slightly over 40 but less than 50

Answer is A

Hi GMATPrepNow

COuld you please help me with this solution . I got the realtive distance of the gap as 20/3 miles. I am not sure how to proceed next

Concept: when the 2 cars depart from the same place, and the 2nd Car catches up to the 1st Car —-

Distance car 1 travels = Distance car 2 travels


Car 1 travels for 3/2 hours (from 9 AM to 10:30 AM) at a Speed of 40 mph



Car 2 drives for 10 minutes less, so the travel Time for Car 2 is:

(3/2) - (1/6) = 8/6 = 4/3 hours

Car 2’s Unknown Speed = R


Distance = Speed * Time


Again, when the 2 cars Meet—-

D of Car 1 = D of Car 2

40 * (3/2) = R * (4/3)

R = (3 * 40 * 3) / (2 * 4)

R = 3 * 5 * 3 = 45 mph

-A-

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