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Two cyclists start biking from a trail's start 3 hours apart. The seco

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Two cyclists start biking from a trail's start 3 hours apart. The second cyclist travels at 10 miles per hour and starts 3 hours after the first cyclist who is traveling at 6 miles per hour. How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking?

A. 2 hours
B. 4 ½ hours
C. 5 ¾ hours
D. 6 hours
E. 7 ½ hours
[Reveal] Spoiler: OA

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Two cyclists start biking from a trail's start 3 hours apart. The seco [#permalink]

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New post 05 Oct 2017, 07:33
Bunuel wrote:
Two cyclists start biking from a trail's start 3 hours apart. The second cyclist travels at 10 miles per hour and starts 3 hours after the first cyclist who is traveling at 6 miles per hour. How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking?

A. 2 hours
B. 4 ½ hours
C. 5 ¾ hours
D. 6 hours
E. 7 ½ hours

I use rate and speed interchangeably in these problems. Speed \(s\) is the same as rate \(r\)

Short version, no explanation
Cyclist 1, at a speed of 6 mph * 3 hrs, travels 18 miles

Cyclist 2 rides at 10 mph, and will cover that distance at relative speed (=relative rate) of 10 - 6 = 4 mph

18 miles/ 4 mph = 4.5 hours

Second cyclist catches first cyclist in 4.5 hours from the time second cyclist starts.

Answer B

Answer with explanation

One way to think about this question: it's a gap and chase problem.

1) Find the distance between the cyclists, which is the "gap"

The first cyclist creates the distance between the two cyclists. At a rate of 6 mph, he rides for 3 hours before the other cyclist starts.

D = r*t. Cyclist 1, riding alone, creates

\(D: (6 mph * 3 hrs)\) =18 miles between them

At the moment the second cyclist starts from the same place, the first cyclist is exactly 18 miles ahead. That's the "gap's" distance.

2) Find the rate at which the gap is closed in a "chase"

Because they start from the same place, and ride in the same direction, this is a chase* problem. She chases him to close the gap.*

She rides faster, at 10 mph.

The rate* at which she closes the gap is the difference of their speeds, called "relative rate" or "relative speed."

Relative rate: 10 - 6 = 4 mph

3) Find the time it takes for Cyclist 2 to catch up to Cyclist 1 while both travel at the same time*

How much time will pass before the second cyclist catches up with the first, from the time the second cyclist started biking?

Time = D/r

Distance = 18 miles
(Relative) Rate = (10 - 6) = 4 mph

Time: \(\frac{18miles}{4mph}\) = \(4.5\) hours

Cyclist 2 will catch Cyclist 1 in 4.5 hours after Cyclist 2 begins.

Answer B

*"Chase" means two travelers travel in the same direction, where one travels faster than the other. One traveler typically starts behind the other, either in time, or in distance. The person behind has to catch the person ahead. She is "chasing" him.

*rate = "relative rate." How this relative rate is calculated depends on travelers' directions, original positions, and speeds. Same direction, same starting point: subtract slower rate from faster rate

*"closing the gap" = "catching up to" the person ahead

*this question has nothing to do with the first cyclist's first three hours, except for its role in creating the distance of the gap between them. After that, the question asks only about the time segment that begins when Cyclist 2 starts to ride.

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Re: Two cyclists start biking from a trail's start 3 hours apart. The seco [#permalink]

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New post 06 Oct 2017, 00:16
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Relative distance at the instant when both cyclists are underway = 18m (6m/h x 3hrs)

Relative speed= 10-6= 4mph

So, Time to catch up= 18/4= 4.5hrs

Hence, Ans B

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Re: Two cyclists start biking from a trail's start 3 hours apart. The seco [#permalink]

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New post 09 Oct 2017, 16:42
Bunuel wrote:
Two cyclists start biking from a trail's start 3 hours apart. The second cyclist travels at 10 miles per hour and starts 3 hours after the first cyclist who is traveling at 6 miles per hour. How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking?

A. 2 hours
B. 4 ½ hours
C. 5 ¾ hours
D. 6 hours
E. 7 ½ hours


We can let the time of the first cyclist = t + 3 and the time of the second cyclist = t, and thus:

6(t + 3) = 10t

6t + 18 = 10t

18 = 4t

18/4 = t

4.5 = t

Thus, 4.5 hours from the time the second cyclist started biking will pass before the second cyclist catches the first.

Answer: B
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Re: Two cyclists start biking from a trail's start 3 hours apart. The seco [#permalink]

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New post 10 Oct 2017, 07:46
The second cyclist starts 3 hours after the first cyclist.
Distance travelled by the first cyclist in 3 hours =3 hrs x 6 miles/hr=18 miles
When 2nd cyclist starts 1st cyclist is already at a distance of 18 miles and is travelling at 6mph.
Relative speed between the two cyclists=10mph - 6mph = 4 mph
Time to cover the distance of 18 miles = \(\frac{18}{4}\) = 4\(\frac{1}{2}\) hrs.
Answer B.
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Two cyclists start biking from a trail's start 3 hours apart. The seco [#permalink]

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New post 10 Oct 2017, 09:10
Distance from start point at time 0:
C1 (travels at 6 mph) = 0
C2 (travels at 10 mph) = 0

In 3 hours, C1 will be at 18 miles marker, and C2 will be just starting and will be at mile marker 0. Since they are now traveling in same direction, their relative speed is given as: 10 - 6 = 4 mph

Distance between them = 18 - 0 = 18 miles.

Time for cyclist C2 to catch up with C1 = Distance/Relative Speed = 18/4 = 4.5 hours

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Two cyclists start biking from a trail's start 3 hours apart. The seco   [#permalink] 10 Oct 2017, 09:10
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