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John and Jacob set out together on bicycle traveling at 15 and 12 mile

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John and Jacob set out together on bicycle traveling at 15 and 12 mile  [#permalink]

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New post 06 Oct 2014, 07:20
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A
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Question Stats:

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Tough and Tricky questions: Distance/Rate .



John and Jacob set out together on bicycle traveling at 15 and 12 miles per hour, respectively. After 40 minutes, John stops to fix a flat tire. If it takes John one hour to fix the flat tire and Jacob continues to ride during this time, how many hours will it take John to catch up to Jacob assuming he resumes his ride at 15 miles per hour? (consider John's deceleration/acceleration before/after the flat to be negligible)

A. 3
B. 3 1/3
C. 3 1/2
D. 4
E. 4 1/2

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John and Jacob set out together on bicycle traveling at 15 and 12 mile  [#permalink]

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New post Updated on: 22 Oct 2014, 08:22
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John's speed - 15 miles/hr
Jacob's speed - 12 miles/hr

After 40min (i.e 2/3hr), distance covered by John = 15x2/3 = 10 miles.
Jacob continues to ride for a total of 1hour and 40min (until John's bike is repaired). Distance covered in 1 hour 40min (i.e 5/3hr) = 12x5/3 = 20 miles.

Now, when John starts riding back, the distance between them is 10 miles. Jacob and John are moving in the same direction. For John to catch Jacob, the effective relative speed will be 15-12 = 3 miles/hr.

Thus, to cover 10 miles at 3 miles/hr, John will take 10/3 = 3 1/3 hours

Answer B
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Originally posted by swanidhi on 22 Oct 2014, 07:55.
Last edited by swanidhi on 22 Oct 2014, 08:22, edited 1 time in total.
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Re: John and Jacob set out together on bicycle traveling at 15 and 12 mile  [#permalink]

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New post 06 Oct 2014, 07:34
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Bunuel wrote:

Tough and Tricky questions: Distance/Rate .



John and Jacob set out together on bicycle traveling at 15 and 12 miles per hour, respectively. After 40 minutes, John stops to fix a flat tire. If it takes John one hour to fix the flat tire and Jacob continues to ride during this time, how many hours will it take John to catch up to Jacob assuming he resumes his ride at 15 miles per hour? (consider John's deceleration/acceleration before/after the flat to be negligible)

A. 3
B. 3 1/3
C. 3 1/2
D. 4
E. 4 1/2


distance travelled by john in 40 minutes = 15(40/60) = 10 km
distance travelled by jacob in 40 minutes = 12(40/60) =8 km
after 40 minutes john leads jacob by 2 km.
after 40 minutes jacob continues to ride for 1 hour, while john stops to fix a flat tire. during this time jacob covers 12km. but since initially john was leading the jacob by 2km thus, distance now between john and jacob =12-2 =10km

thus time taken by john to catch up with jacob = 10/(15-12)
=3 1/3

hence B
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Re: John and Jacob set out together on bicycle traveling at 15 and 12 mile  [#permalink]

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New post 22 Oct 2014, 03:23
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Distance travelled by John in 40 Minutes \(= \frac{15}{60} * 40 = 10\) Miles (At this point, his car gets punctured)

Distance travelled by Jacob in 40 Minutes\(= \frac{12}{60}* 40 = 8\)Miles (He is still moving)

John is idle for 1hr, means he's still at 10 Miles from initial start

Jacob is still moving for 1hr, means he has travelled another 12 Miles; total distance travelled from initial start = 12+8 = 20 Miles

Relative distance between Jacob & John = 10 Miles

Refer diagram below:
Attachment:
meet.png
meet.png [ 3.37 KiB | Viewed 11164 times ]



Time taken by John to reach the meeting point would be same as time taken by Jacod

Setting up the equation

\(\frac{10+x}{15} = \frac{x}{12}\)

x = 40

Total catch up time taken \(= \frac{10+40}{15} = \frac{31}{3} Hrs\)

Answer = B
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Re: John and Jacob set out together on bicycle traveling at 15 and 12 mile  [#permalink]

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New post 22 Oct 2014, 08:08
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PareshGmat wrote:

Total catch up time taken \(= \frac{10+40}{15} = \frac{31}{3} Hrs\)

Answer = B


I agree with Paresh. Just wanted to make a slight modification since it looks like there is a typo in the last line:

Total catch up time taken \(= \frac{10+40}{15} = 3\frac{1}{3}\) Hrs
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Re: John and Jacob set out together on bicycle traveling at 15 and 12 mile  [#permalink]

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New post 11 Oct 2017, 14:58
John and Jacob set out together on bicycle traveling at 15 and 12 miles per hour, respectively. After 40 minutes, John stops to fix a flat tire. If it takes John one hour to fix the flat tire and Jacob continues to ride during this time, how many hours will it take John to catch up to Jacob assuming he resumes his ride at 15 miles per hour? (consider John's deceleration/acceleration before/after the flat to be negligible)?

At 1 hour and 40 minutes (time at which John resumes after his 1 hour halt) = John's distance covered is 15*40/60 = 10 miles and distance covered by Jacob is 20 (12*40/60+12).

Difference in their speeds = 3
Time that will be taken to catch up = difference in distances / differences in speeds = (20-10)/3 = 10/3 (B)
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Re: John and Jacob set out together on bicycle traveling at 15 and 12 mile  [#permalink]

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New post 11 Oct 2017, 19:21
Poorvasha wrote:
John and Jacob set out together on bicycle traveling at 15 and 12 miles per hour, respectively. After 40 minutes, John stops to fix a flat tire. If it takes John one hour to fix the flat tire and Jacob continues to ride during this time, how many hours will it take John to catch up to Jacob assuming he resumes his ride at 15 miles per hour? (consider John's deceleration/acceleration before/after the flat to be negligible)?

At 1 hour and 40 minutes (time at which John resumes after his 1 hour halt) = John's distance covered is 15*40/60 = 10 miles and distance covered by Jacob is 20 (12*40/60+12).

Difference in their speeds = 3
Time that will be taken to catch up = difference in distances / differences in speeds = (20-10)/3 = 10/3 (B)



10 miles and distance covered by Jacob is 20 (12*40/60+12). Why do you add +12 here?
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Re: John and Jacob set out together on bicycle traveling at 15 and 12 mile  [#permalink]

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