A question from Applerouth’s Guide to the New SAT in the chapter “Solving Systems” of equations (p. 503):

a = 4800 – 6t

b = 5400 – 8tIn the system of equations above, a and b represent the distance, in meters, two marathon runners are from the finish line after running for four hours and t seconds. How far will runner a be from the finish line when runner b passes her?

A) 200 meters

B) 300 meters

C) 1000 meters

D) 3000 metersAnswer is supposed to be D.

(No idea how to solve!)

I guess you’re supposed to assume that, despite the fact that *a* is currently closer to the finish line and they’ve been running for the same amount of time, *b* is currently running faster than *a*. I guess you’re also meant to assume that *a* is now running at a constant rate of 6 meters per second and *b* is now running at a constant rate of 8 meters per second, and that both runners will continue at those paces until *b* passes *a*. That’s a lot of assumptions that you wouldn’t have to make on a real SAT question.

*b* will pass *a* when *b* = *a*, so you can set the equations equal to solve for *t*:

*b* = *a*

5400 – 8*t = *4800 – 6*t*

600 = 2*t*300 =

*t*

That tells you that *b* will catch up to *a* in 300 seconds. To figure out how far they’ll be from the finish line at that point, plug 300 into either equation:

*a* = 4800 – 6(300)*a* = 4800 – 1800*a* = 3000