Bunuel wrote:

Madelaine88 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 miles per hour greater than that of car B, what was car B's average speed in miles per hour?

A. 20

B. 30

C. 40

D. 50

E. 80

Let the rate of B be \(x\) miles per hour, then the rate of A would be \(x+10\) miles per hour.

We also know that the time of B equals to time of A + 2: \(\frac{400}{x}=\frac{400}{x+10}+2\), from this point it's better to plug the answer choices rather than to solve the quadratics. You can quickly find that \(\frac{400}{40}=10=\frac{400}{40+10}+2=8+2\)

Answer: C.

I'm struggling with this one for some odd reason:

The setup for my work was as follows

Car B: Rate = R + 10, Time = T + 2, Distance = 400

Car A: Rate = R, Time = T, Distance = 400

I then substituted B's Rate (400/T) into Ys equation --> giving me --> (400 / T) + 10 = (400 / T - 2)

After solving this out, I got T = 20, therefore Car B's average speed is 20mph.

Can either of you help me out with where I went wrong here. Much appreciated. Just over a month from my first GMAT experience!