Question 1163872
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Since it is arbitrary when A took the 18 minute break, simplify the problem by assuming that s/he took the break before s/he left and so didn't start the trip until 7:18.  Meanwhile, B left at 7:00 and was driving for 18 minutes at 50 km/h.  18 minutes is 0.3 hours, so by the time A left at 7:18, B had traveled 50 times 0.3 or 15 km.  So when the two cars start moving toward each other they are 300 minus 15 or 285 km apart approaching each other at the sum of their speeds, namely 95 km/h.  285 divided by 95 is 3, so they meet 3 hours after A started at 7:18, or 10:18 AM.


																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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