At 7 a.m. John leaves his home riding his bicycle due west, at a speed

Distance travelled by John in 5 hours = 5 * 11 = 55 miles.

Time after 5 hours = 12 pm

Distance remaining to Mark's house = 260 - 55 = 205 miles

Time taken for Mark and John to meet = 205/(11+9) = 205/20 hours = 10.25 hours = 10 hours 15 mins.

Time after 10 hours 15 mins = 10:15 PM

Answer D. 10:15 p.m.

At 7 a.m. John leaves his home riding his bicycle due west, at a speed of 11 miles per hour, toward Mark's house, which is 260 miles away. Five hours later Mark leaves his home along the same route to John's house on his bicycle traveling at 9 miles per hour. At what time do they meet each other on the way?

A. 1:00 p.m.
B. 5:00 p.m.
C. 8:15 p.m.
D. 10:15 p.m.
E. 10:25 p.m.

Total Distance = 260 miles
John started his journey at 7 am.
Mark starts his journey 5 hours later, ie; at 12 noon.
Mark Speed = 9 miles/hr
John Speed = 11miles/hr
Distance traveled by John in 5 hours = 5 x 11 = 55 miles
Distance between Mark and John when Mark started his journey = 260 - 55 = 205 mile
[Relative speed = When 2 objects are moving in opposite directions with speeds x and y miles/hr, then the relative speed = (x + y) miles/ hr]
Relative speed of John and Mark = 11 + 9 = 20 miles/hr
Required time taken for Mark and John to meet = Time = Distance / Relative Speed = \(\frac{205}{20}\) = 10.25 hr or 10 hr 15 mins.
Therefore Required Time = 10:15 pm
Answer D...

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We are given that John travels at a rate of 11 mph and that Mark leaves 5 hours later traveling at a rate of 9 mph. If we let t = Mark’s travel time and (t + 5) = John’s travel time, then Mark’s distance = 9t and John’s distance = 11(t + 5) = 11t + 55.

Since we have a converging rate problem, we can use the following formula:

John’s distance + Mark’s distance = 260

11t + 55 + 9t = 260

20t = 205

t = 205/20 = 10.25 hours = 10 hours and 15 minutes

Since Mark left at noon, they will meet at 12 p.m. + 10 hours and 15 minutes = 10:15 p.m.

Answer: D

When Mark starts cycling (at 9 mph) towards John's home at 12 noon, there is only 205 miles separating him from John who had started from his home toward Mark's at 7 am moving at 11 mph thus covering 55 miles in the interim 5 hours. Let us say they meet after 't' hrs. So in this 't' hrs, John covers 11*t miles and Mark covers 9*t miles of the 205 miles which had separated them when Mark started at 12 noon. Thus:
9*t + 11*t = 205 or t = 10.25 = 10 hrs 15 mins.
So they meet at 10:15 PM. Ans: D

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