Question

Harold left town driving east at 55 mph. Carol left 2 hours later driving the same route at 65 mph. How many hours after Harold left will it take Carol to catch up to Harold?
A. 10
B. 11
C. 12
D. 13

Answer 1

Formula for distance traveled at a constant speed is \[d = st\] where \(d\) is the distance, \(s\) is the speed, and \(t\) is the time.

Harold starts driving at \(t=0\). He drives at 55 mph. The distance he has traveled is given by \[d_{harold} = 55t\]

Carol starts driving at \(t = 2\). She drives at 65 mph. The distance she has traveled is given by \[d_{carol} = 65(t-2)\]

Carol catches up with Harold when \[d_{carol} = d_{harold}\]or\[65(t-2) = 55t\]

Answer 2

Another way to think of this problem: Harold drives for 2 hours at 55 mph for 2*55=110 miles before Carol starts driving. Carol drives 65-55=10 mph faster than Harold, so every hour that they both drive, she erases 10 miles from Harold's head start of 110 miles.

Answer 3

so 10 is how long it will take?

Answer 4

If you do the second approach, notice that it is NOT the answer for the problem directly, but how many hours Carol has to drive.

No, 10 is not the answer. How did you get that?

Answer 5

After 10 hours, Harold has driven 55*10 = 550 miles. After 10 hours, Carol has driven 65*(10-2) = 520 miles. Are they equal distances?

Answer 6

NO

Answer 7

So its 13, cause 55X13=715 and 65 (13-2)=715

Answer 8

Yes!