GMAT problem (possible typo?) involving relative rates.

Question

So here's the question. It's on page 35 of a popular GMAT book.

Car X is 40 miles west of Car Y. Both cars are traveling east and Car X is going 50% faster than Car Y. If both cars travel at a constant rate and it takes Car X 2 hours and 40 minutes to catch up to Car Y, how fast is Car Y going?

IT seems like the solution can be this:

The two cars are getting closer to each other at a rate of 0.5r (1.5r - 1r). The time it takes is 8/3 of an hour. They need to travel 40 miles.

So...

0.5r * 8/3 = 40

0.5r = 15

r = 30

Car Y is going at 30 mph.

BUT

the book sets up an RTD chart like this:

__R | T | D

---

X: 1.5r|8/3|d

---

Y: r|8/3|d+40

Why is the distance for Y d + 40? Shouldn't X be traveling farther?

Answer 1

Yes, your solution is correct and the book has a typo, it should be d-40 in the Y distance it it is d in the X distance.

Answer 2

Y has to travel 40 miles less than X. Therefore X travel more miles in a given time. To get an equality you have to add 40 miles to the travelling distance of Y.

$S_X=S_Y+40$

Therefore X is travelling farther, because Y travel the distance of Y plus 40 miles.

And the travelled distance of X is $S_X=1.5\cdot V_Y\cdot 2\frac{2}{3}=1.5\cdot V_Y\cdot \frac{8}{3}$.

The travelled distance of X is $S_Y= V_Y\cdot 2\frac{2}{3}= V_Y\cdot \frac{8}{3}$.

It looks like that this equation are equivalent to the table of the book, if d is the travelled distance of X and Y, respectively.