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

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

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07 May 2017, 13:45
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Difficulty:

45% (medium)

Question Stats:

70% (02:11) correct 30% (02:51) wrong based on 151 sessions

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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.

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

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Updated on: 20 May 2017, 01:48
5
3
Before solving this question lets revise one concept

RELATIVE SPEED

“Relative” means “in comparison to”. We use the concept of relative Speed when we have two or more bodies moving with some Speeds. To make things simpler, we make one body stationary (Speed =zero) and take the Speed of the other body with respect to the stationary body, which is the sum of the Speeds if the bodies are moving in the opposite direction and the difference if they are moving in the same direction. This Speed of the moving body with respect to the stationary body is called Relative Speed.

Relative Speed of 2 bodies = Sum of their individual Speeds if they are moving in the opposite direction = Difference of their individual Speeds if they are moving in the same direction

Now coming back to question.

Distance traveled by John in 5 hours = 5 * 11 = 55 miles.
using distance = speed * time

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.
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Originally posted by guptarahul on 19 May 2017, 03:47.
Last edited by guptarahul on 20 May 2017, 01:48, edited 1 time in total.
##### General Discussion
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Joined: 06 May 2017
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Re: At 7 a.m. John leaves his home riding his bicycle due west, at a speed  [#permalink]

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07 May 2017, 14:33
after 5 hours john already run 55miles, so he is 260-55=205 miles away from mark.
at the time Tj=Tm, VjTj + VmTm = 205 ---> Tj=Tm=205/(Vj+Vm)
= 205/20 --> 10,25 hours
we are at 12.00am + 10,25 hours = they will meet at 10:15pm so [D]
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Re: At 7 a.m. John leaves his home riding his bicycle due west, at a speed  [#permalink]

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07 May 2017, 14:51
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

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

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19 May 2017, 04:12
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

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Re: At 7 a.m. John leaves his home riding his bicycle due west, at a speed  [#permalink]

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22 May 2017, 18:45
1
Bunuel wrote:
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.

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.

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Re: At 7 a.m. John leaves his home riding his bicycle due west, at a speed  [#permalink]

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01 Apr 2019, 00:46
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
Re: At 7 a.m. John leaves his home riding his bicycle due west, at a speed   [#permalink] 01 Apr 2019, 00:46
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