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# Ben and Sam set out together on a bicycle travelling at 15

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Intern
Joined: 20 Jun 2013
Posts: 5
Ben and Sam set out together on a bicycle travelling at 15  [#permalink]

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21 Jul 2013, 21:41
14
19
00:00

Difficulty:

45% (medium)

Question Stats:

72% (02:24) correct 28% (02:20) wrong based on 570 sessions

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Ben and Sam set out together on a bicycle traveling at 15 and 12 miles per hour, respectively. After 40 minutes, Ben stops to fix a flat tire. If it takes Ben one hour to fix the flat tire and Sam continues to ride during this time, how many hours will it take Ben to catch up Sam and assuming he resumes his ride at 15 miles per hour?

A. 3
B. 3.33
C. 3.5
D. 4
E. 4.5

Don't forget Kudos if you like the question
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Joined: 02 Sep 2009
Posts: 64101
Re: Ben and Sam set out together on a bicycle travelling at 15  [#permalink]

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21 Jul 2013, 21:53
7
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KalEl wrote:
Ben and Sam set out together on a bicycle traveling at 15 and 12 miles per hour, respectively. After 40 minutes, Ben stops to fix a flat tire. If it takes Ben one hour to fix the flat tire and Sam continues to ride during this time, how many hours will it take Ben to catch up Sam and assuming he resumes his ride at 15 miles per hour?

A. 3
B. 3.33
C. 3.5
D. 4
E. 4.5

Don't forget Kudos if you like the question

The distance between Ben and Sam after 40 minutes (2/3 hours) is (distance) = (time)(speed) = 2/3*(15-12) = 2 miles (Ben will be 2 miles ahead of Sam).

In one hour Sam covers 12 miles, so after an hour Sam will be 12 - 2 = 10 miles ahead of Ben.

To catch up Ben will need (time) = (distance)/(speed) = 10/(15-12) = 10/3 hours.

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Re: Ben and Sam set out together on a bicycle travelling at 15  [#permalink]

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21 Jul 2013, 22:09
Thanks Bunuel!!

What difficulty would you rate this question as?
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Joined: 02 Sep 2009
Posts: 64101
Re: Ben and Sam set out together on a bicycle travelling at 15  [#permalink]

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21 Jul 2013, 22:11
KalEl wrote:
Thanks Bunuel!!

What difficulty would you rate this question as?

_________
Around 650.
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Re: Ben and Sam set out together on a bicycle travelling at 15  [#permalink]

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21 Jul 2013, 22:38
1
1. Let the total distance traveled be x.
2. Total time taken by sam = x/12 = 2/3 +1+ t where t is the time Ben needs to catch up after restarting.
3. Time gained by Ben is due to the difference in speed i.e., x/12 - x/15
4. Time lost by Ben is the repair time = 1
5. (3) and (4) are equal when Ben catches up. i.,e., x/12 - x/15 =1. x=60
6.Substituting x in (2) we have t= 3.33 hrs.
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Re: Ben and Sam set out together on a bicycle travelling at 15  [#permalink]

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02 Aug 2013, 10:37
Ben and Sam set out together on a bicycle traveling at 15 and 12 miles per hour, respectively. After 40 minutes, Ben stops to fix a flat tire. If it takes Ben one hour to fix the flat tire and Sam continues to ride during this time, how many hours will it take Ben to catch up Sam and assuming he resumes his ride at 15 miles per hour?

We need to find how long Ben and Sam travel until Ben needs to change his tire. Then, we need to get their respective rates to see how long it will take Ben to catch up to Sam.

Rate = 15/60
Rate = 1/4 miles/minute

Rate = 12/60
Rate = 1/5 miles/minute

Distance (Ben) = 1/4*40
Distance (Ben) = 10 miles

Distance (Sam) = 1/5*40
Distance (Sam) = 8 miles

Ben stops for one hour. Sam Travels for 1/5*60 = 12 miles. Because Ben was two miles ahead of Sam when he stopped, Sam traveled 10 miles ahead of Ben.
Ben's rate of gain on Sam is 3 miles/hour (15-12). Therefore, the time it will take for Ben to catch up to Sam: Time = distance/rate ==> Time = 10/3 ==> Time = 3.33 hours.

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Re: Ben and Sam set out together on a bicycle travelling at 15  [#permalink]

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02 Aug 2013, 12:09
1
Diff style of thinking...

T = Time sam cycles after fixing tyre to catch up with ben

Total time Sam Cycled at 15mph = 40mins+t = t + 40/60 hrs
Total Distance travelled = 15 * (t+ 40/60) ------- (1)

Total time Ben cycled at 12mph = 40mins + 60mins (When sam was fixing tyre) + t = t + 100/60 hrs
Total Distance traveled = 12 * (t+100/60) --------- (2)

(1) = (2) and solving t = 3.33
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Re: Ben and Sam set out together on a bicycle travelling at 15  [#permalink]

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15 Aug 2013, 12:59
KalEl wrote:
Ben and Sam set out together on a bicycle traveling at 15 and 12 miles per hour, respectively. After 40 minutes, Ben stops to fix a flat tire. If it takes Ben one hour to fix the flat tire and Sam continues to ride during this time, how many hours will it take Ben to catch up Sam and assuming he resumes his ride at 15 miles per hour?

A. 3
B. 3.33
C. 3.5
D. 4
E. 4.5

Don't forget Kudos if you like the question

Sb = 15 * 40/60 = 10
Ss = 12 * 40/60 = 8 . 2 mile behind.
then 1 hour more, Ss = 12 * 1 = 12
after 1 hour Sam will go x more mile in t hours. so, x = 12t ..................(2)
and same has to go, 10+x = 15*t ..............(1)

from (1) and (2) , t = 10/3 = 3.333333 = B
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Re: Ben and Sam set out together on a bicycle travelling at 15  [#permalink]

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23 Nov 2014, 03:23
1
Ben after 40 min = 15*2/3=10 miles
Sam after 40 min=12*2/3=8 miles

Sam goes 8+12=20 miles, so Ben should cover 20-10=10 miles with speed difference in 3m/h

10/3=3.33

B
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Ben and Sam set out together on a bicycle travelling at 15  [#permalink]

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30 Jan 2015, 04:11
Bunuel wrote:
KalEl wrote:
Ben and Sam set out together on a bicycle traveling at 15 and 12 miles per hour, respectively. After 40 minutes, Ben stops to fix a flat tire. If it takes Ben one hour to fix the flat tire and Sam continues to ride during this time, how many hours will it take Ben to catch up Sam and assuming he resumes his ride at 15 miles per hour?

A. 3
B. 3.33
C. 3.5
D. 4
E. 4.5

Don't forget Kudos if you like the question

The distance between Ben and Sam after 40 minutes (2/3 hours) is (distance) = (time)(speed) = 2/3*(15-12) = 2 miles (Ben will be 2 miles ahead of Sam).

In one hour Sam covers 12 miles, so after an hour Sam will be 12 - 2 = 10 miles ahead of Ben.

To catch up Ben will need (time) = (distance)/(speed) = 10/(15-12) = 10/3 hours.

Great Bunuel, if I may ask, do we subtract Sam's time in the eqution above because he keeps moving?
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Re: Ben and Sam set out together on a bicycle travelling at 15  [#permalink]

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30 Jan 2015, 04:46
1
pacifist85 wrote:
Bunuel wrote:
KalEl wrote:
Ben and Sam set out together on a bicycle traveling at 15 and 12 miles per hour, respectively. After 40 minutes, Ben stops to fix a flat tire. If it takes Ben one hour to fix the flat tire and Sam continues to ride during this time, how many hours will it take Ben to catch up Sam and assuming he resumes his ride at 15 miles per hour?

A. 3
B. 3.33
C. 3.5
D. 4
E. 4.5

Don't forget Kudos if you like the question

The distance between Ben and Sam after 40 minutes (2/3 hours) is (distance) = (time)(speed) = 2/3*(15-12) = 2 miles (Ben will be 2 miles ahead of Sam).

In one hour Sam covers 12 miles, so after an hour Sam will be 12 - 2 = 10 miles ahead of Ben.

To catch up Ben will need (time) = (distance)/(speed) = 10/(15-12) = 10/3 hours.

Great Bunuel, if I may ask, do we subtract Sam's time in the eqution above because he keeps moving?

Yes.

This is the concept of Relative Speed.

When two objects move in same direction, their speeds get subtracted.

When two objects (speeds V1 and V2) move in opposite directions (towards each other or away from each other), they cover the distance between them (or create distance between them) at the rate of (V1 + V2).
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Re: Ben and Sam set out together on a bicycle travelling at 15  [#permalink]

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06 Jul 2015, 05:03
Distance taken by Ben and Same is equal to each other. Since distance is equal to speed and time,
15 *(40min/60min + t running time after the fix)+ 0*(1 hr of pause time)= 12 * (40/60+1hr + t)
hence, 10+15t=8+12+12t
3t=10
t=3.33 hr
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Re: Ben and Sam set out together on a bicycle travelling at 15  [#permalink]

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31 Mar 2017, 04:53
My solution:

Before the tire break: Ben did 10 miles & Sam did 8 miles
During the repair: Ben did 0 mile & Sam did 12 miles

When Ben resumed, he had 10 miles to catch-up at a relative speed of 3 miles per hour.

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Re: Ben and Sam set out together on a bicycle travelling at 15  [#permalink]

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04 Apr 2017, 15:07
KalEl wrote:
Ben and Sam set out together on a bicycle traveling at 15 and 12 miles per hour, respectively. After 40 minutes, Ben stops to fix a flat tire. If it takes Ben one hour to fix the flat tire and Sam continues to ride during this time, how many hours will it take Ben to catch up Sam and assuming he resumes his ride at 15 miles per hour?

A. 3
B. 3.33
C. 3.5
D. 4
E. 4.5

Ben and Sam both travel for 40 minutes or 2/3 of an hour. During that time, Ben rides 15 x 2/3 = 10 miles and Sam rides 12 x 2/3 = 8 miles. During the one hour that Ben is fixing his flat tire, Sam rides another 12 miles. So, after that hour, Sam has traveled 8 + 12 = 20 miles and Ben has traveled 10 miles.

We can use the following formula to determine how long it will take Ben to catch Sam:

difference in distance/difference in rate

(20 - 10)/(15 - 12) = 10/3 = 3.33

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Re: Ben and Sam set out together on a bicycle travelling at 15  [#permalink]

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06 Aug 2019, 03:48
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Re: Ben and Sam set out together on a bicycle travelling at 15   [#permalink] 06 Aug 2019, 03:48