Distance Rate Problem - Help!

Hi TxGMAT,

This question can be approached in a couple of different ways. Since it's essentially about basic arithmetic (re: two 'pairs' of numbers that have a product of 400), you would likely find it fastest to TEST THE ANSWERS.

To start, the difference in the time that the two cars traveled is exactly 2 hours and we know that Car B was going exactly 10 mph slower than Car A. Based on the prompt and the answer choices, we're clearly dealing with nice 'round' numbers, so let's start with an answer that divides evenly into 400...

Let's TEST Answer C: 40 miles/hour

IF.... Car B is traveling 40 mph, then it takes 400/40 = 10 hours to complete that drive.
We're told that Car A's speed is 10 mph greater than Car B's speed, so....
Car A is traveling 50 mph... and that would take 400/50 = 8 hours to complete that drive.
Here, the difference in travel time is exactly 2 hours - and this matches what we were told - so this MUST be the answer.

Final Answer: C

GMAT assassins aren't born, they're made,
Rich

TxGMAT wrote:The time it took Car A to travel 400 miles was 2 hours less than the time it took Car B to travel the same distance. If Car A's average speed was 10 mph greater than that of Car B, what was Car B's average speed in miles/hr.

A)20
B)30
C)40
D)50
E)60

I have a way to get to the answer I believe, but I'm definitely not sure about it.

Hi TxGMAT,

Say Car B takes x hours, thus Car A takes (x-2) hours to complete 400 miles.

Speed of Car A = 400 / (x-2) miles/hour

Speed of Car B = 400 / x miles/hour

Thus, 400 / (x-2) - 400 / x = 10

=> 400[1/(x-2) -1/x] = 10

=> x = 10 hours

Speed of Car B = 400 / 10 = 40 miles/hour

The correct answer: C

Hope this helps!

Relevant book: Manhattan Review GMAT Word Problems Guide

-Jay
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TxGMAT wrote:The time it took Car A to travel 400 miles was 2 hours less than the time it took Car B to travel the same distance. If Car A's average speed was 10 mph greater than that of Car B, what was Car B's average speed in miles/hr.

A)20
B)30
C)40
D)50
E)60

We are given that Car A and B both traveled 400 miles and that Car A’s average speed was 10 mph greater than that of Car B. We can let the rate of Car B = r and the rate of Car A = r + 10.

Since both cars traveled 400 miles and time = distance/rate, the time of Car A is 400/(r+10).

We are also given that it took Car A 2 hours less than it took Car B to travel the 400 miles. We can set up the following equation:

400/(r+10) + 2 = 400/r

Multiplying the entire equation by r(r+10), we have:

400r + 2(r)(r+10) = 400(r+10)

400r + 2r^2 + 20r = 400r + 4000

2r^2 + 20r - 4000 = 0

r^2 + 10r - 2000 = 0

(r + 50)(r - 40) = 0

We see that r = -50 or r = 40, but r must be positive, so r = 40.

Answer:C