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At 1:00, Jack starts to bicycle along a 60 mile road at a constant spe

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At 1:00, Jack starts to bicycle along a 60 mile road at a constant spe [#permalink]

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At 1:00, Jack starts to bicycle along a 60 mile road at a constant speed of 15 miles per hour. Thirty minutes earlier, Scott started bicycling towards Jack on the same road at a constant speed of 12 miles per hour. At what time will they meet?

A. 1:30
B. 3:00
C. 5:00
D. 6:00
E. 9:00

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[Reveal] Spoiler: OA

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Re: At 1:00, Jack starts to bicycle along a 60 mile road at a constant spe [#permalink]

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Bunuel wrote:
At 1:00, Jack starts to bicycle along a 60 mile road at a constant speed of 15 miles per hour. Thirty minutes earlier, Scott started bicycling towards Jack on the same road at a constant speed of 12 miles per hour. At what time will they meet?

A. 1:30
B. 3:00
C. 5:00
D. 6:00
E. 9:00

Kudos for a correct solution.



13:30 Scott 12m
14:30 Scott 24m
15:30 Scott 36m -> 15:00 Scott 36 - 6m = 30m

14:00 Jack 15m
15:00 Jack 30m

Answer B

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Bunuel wrote:
At 1:00, Jack starts to bicycle along a 60 mile road at a constant speed of 15 miles per hour. Thirty minutes earlier, Scott started bicycling towards Jack on the same road at a constant speed of 12 miles per hour. At what time will they meet?

A. 1:30
B. 3:00
C. 5:00
D. 6:00
E. 9:00

Kudos for a correct solution.


IN THESE QUESTIONS WE FIND THE DISTANCE BETWEEN TWO WHEN BOTH ARE MOVING AND THEN FIND THE TIME IT TAKES THE FASTER ONE TO COVER THAT DISTANCE...
Scott has already moved for half an hour so he is 12/2 miles ahead.. =6 miles
jack covers 3 miles extra each hour when compared to scott...
so he will cover 6 miles in 6/3=2 hours..
2 hours after 1:00 is 3:00...ans B
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Re: At 1:00, Jack starts to bicycle along a 60 mile road at a constant spe [#permalink]

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New post 14 Feb 2015, 21:36
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Hi All,

This is an example of the 'Combined Rate' question; on Test Day, they're relatively rare (you'll probably see just 1) and they're usually written in a way that requires a certain amount of real "math" work to get to the correct answer.

By using the answer choices in this question, you can actually AVOID most of the work though.

In the first sentence, we know Jack's starting time (1pm), speed (15 mph) and distance (60 miles), so we can quickly figure out how long it would take Jack to travel the entire distance (4 hours). This tells us that, at 5pm, Jack would be at the other end of the road. We know that Scott is cycling TOWARDS him though, so some time BEFORE 5PM, they'll meet. Eliminate Answers C, D and E.

With the remaining two answers, 1:30pm is much TOO early to be the meeting time. Neither Jack nor Scott is traveling fast enough to meet up that early. Eliminate A. There's only one answer left..

Final Answer:
[Reveal] Spoiler:
B


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Re: At 1:00, Jack starts to bicycle along a 60 mile road at a constant spe [#permalink]

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New post 15 Feb 2015, 12:01
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let x be time.

15x+12(x+1/2)=60;
27x=54;
x=2;

=>

1:00 + 2 hr = 3:00

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Re: At 1:00, Jack starts to bicycle along a 60 mile road at a constant spe [#permalink]

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Bunuel wrote:
At 1:00, Jack starts to bicycle along a 60 mile road at a constant speed of 15 miles per hour. Thirty minutes earlier, Scott started bicycling towards Jack on the same road at a constant speed of 12 miles per hour. At what time will they meet?

A. 1:30
B. 3:00
C. 5:00
D. 6:00
E. 9:00

Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION

Correct Answer: B

Explanation: Solution: B The relevant formula is Rate x Time = Distance. If Jack's time is t, then Scott's time is t + .5, since he started one-half hour earlier. Jack's distance is therefore 15t, and Scott's distance is 12 (t + .5). The sum of their distances will be the distance apart that they started. Therefore we can say that 15t + 12 (t + .5) = 60. This simplifies to 15t + 12t + 6 = 60, which means that 27t = 54, and t = 2. Since Jack left at 1:00, and his time was t, he met Scott at 3:00.
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Re: At 1:00, Jack starts to bicycle along a 60 mile road at a constant spe [#permalink]

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New post 19 Feb 2015, 01:59
Answer = B = 3:00

At 01:00, Scott has already travelled 6 kms, so remaining distance = 60-4 = 54

Say Jack meets Scott at "x" kms from his own starting point

\(\frac{x}{15} = \frac{54-x}{12}\)

x = 30

Time required by Jack to travel 30kms = 2 Hrs

They will meet at 03:00
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New post 23 Jan 2017, 18:01
Given : Distance is 60 miles

Scott has a 30 mins lead and hence covers 6 miles.

So when Jack starts he and Scott have to cover a distance of 54miles ( 60-6) to reach each other, with a speed of 27 miles/hour ( 15+12 ) => 2 Hours,

1:00 + 2 hours = 3:00...
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Re: At 1:00, Jack starts to bicycle along a 60 mile road at a constant spe [#permalink]

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New post 08 Feb 2017, 11:12
Scott has already moved for half an hour so he is 12/2 miles ahead.. =6 miles
jack covers 3 miles extra each hour when compared to scott...
so he will cover 6 miles in 6/3=2 hours..
2 hours after 1:00 is 3:00...ans B
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Re: At 1:00, Jack starts to bicycle along a 60 mile road at a constant spe [#permalink]

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New post 15 Feb 2017, 15:47
Scott is ahead by 30min x 12 miles per 60 min = 6 miles.

Find the relative speed of Jack to Scott, 15 miles/hr - 12 miles/hr = 3 miles/hr

For Jack to catch up and meet Scott is 6 / 3 = 2 hrs

1:00 + 2 hrs = 3:00 = Answer

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Re: At 1:00, Jack starts to bicycle along a 60 mile road at a constant spe [#permalink]

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New post 18 Mar 2017, 20:40
Distance Traveled by Jockey = 6 + Distance Traveled by Scott
Lets Say they will meet after T hours.
15T = 12T + 6
T=2 Hours
therefore, They Will meet @ 03:00 pm

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Re: At 1:00, Jack starts to bicycle along a 60 mile road at a constant spe [#permalink]

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New post 13 Aug 2017, 20:57
Bunuel wrote:
At 1:00, Jack starts to bicycle along a 60 mile road at a constant speed of 15 miles per hour. Thirty minutes earlier, Scott started bicycling towards Jack on the same road at a constant speed of 12 miles per hour. At what time will they meet?

A. 1:30
B. 3:00
C. 5:00
D. 6:00
E. 9:00

Kudos for a correct solution.


12(x + 1/2) +15x =60
27x =54
x=2

2 hours from 1 or
2 hours and 30 mins from 12 :30

B

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Re: At 1:00, Jack starts to bicycle along a 60 mile road at a constant spe   [#permalink] 13 Aug 2017, 20:57
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