Question

Mike leaves school on his bike at 1 PM, traveling at 12 mph. Janis leaves the same school one quarter of an hour later, traveling at 16 mph in the same direction. At what time will Janis catch up with Mike?

Answer 1

Distance = Rate x Time

Answer 2

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Answer 3

@kaytru

Answer 4

2 PM
Let t be the elapsed time for Janis to \[1\text{ PM} + \frac{1}{4}+\frac{3}{4}=2\text{ PM} \]catch up with Mike.\[12 \left(\frac{1}{4}\right)+12t=16t\]\[t=\frac{3}{4} \]

Answer 5

Correct...Janice in my solution is t - 1/4...so t = 1, t - 1/4 = 3/4

I stand corrected.

Answer 6

3/4 hr after 1:15 is 2PM.

Answer 7

Sorry for the discombobulated posting above. Tried to correct it but openstudy.com decided to freeze up and/or shutdown.