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by karthikpandian19 » Thu May 10, 2012 6:32 pm
A car going at 40 miles per hour set out on an 80-mile trip at 9:00 A.M. Exactly 10 minutes later, a second car left from the same place and followed the same route. How fast, in miles per hour, was the second car going if it caught up with the first car at 10:30 A.M.?

(A) 45

(B) 50

(C) 53

(D) 55

(E) 60
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by mathbyvemuri » Thu May 10, 2012 9:16 pm
Whatever the first car covers in 1 hour 30 min (90 sec), that is covered by the second car in a 10 min lesser time.
=>The time taken by second car = 1hr20min = 80 sec

The key required is here=> speed = distance/time

As the distance covered is same,the ratio of speeds is inversely proportional to the ratio of times taken.
=>(speed of second car)/(speed of first car) = (time taken by first car)/(time taken by second car)
=>speed of second car = 40*(90/80) = 45 m/h

Anser "A"
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by GMATGuruNY » Fri May 11, 2012 7:17 am
karthikpandian19 wrote:A car going at 40 miles per hour set out on an 80-mile trip at 9:00 A.M. Exactly 10 minutes later, a second car left from the same place and followed the same route. How fast, in miles per hour, was the second car going if it caught up with the first car at 10:30 A.M.?

(A) 45

(B) 50

(C) 53

(D) 55

(E) 60
10 minutes = 1/6 of an hour.
The distance traveled by the first car in 1/6 of an hour = r*t = 40(1/6) = 20/3 miles.

10:30am - 9:10am = 80 minutes = 4/3 of an hour.
Required rate for the second car to catch up by 20/3 miles in 4/3 of an hour = d/t = (20/3)/(4/3) = 5 miles per hour.
Thus, the second car must travel 5 miles per hour FASTER than the first car.

Rate of the second car = rate of the first car + 5 = 40+5 = 45 miles per hour.

The correct answer is A.
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by Jeff@TargetTestPrep » Thu Dec 14, 2017 5:49 pm
karthikpandian19 wrote:A car going at 40 miles per hour set out on an 80-mile trip at 9:00 A.M. Exactly 10 minutes later, a second car left from the same place and followed the same route. How fast, in miles per hour, was the second car going if it caught up with the first car at 10:30 A.M.?

(A) 45

(B) 50

(C) 53

(D) 55

(E) 60

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 ca,r we can set the distances equal and determine r:

4r/3 = 60

4r = 180

r = 45

Answer: A

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