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Pam and Cathy begin running at the same time on a straight

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Pam and Cathy begin running at the same time on a straight  [#permalink]

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New post Updated on: 02 Aug 2014, 00:05
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Pam and Cathy begin running at the same time on a straight path. Pam runs at 10 miles per hour, and Cathy runs at 8 miles per hour. After 45 minutes, Pam stops to stretch. If it takes Pam 30 minutes to stretch and Cathy continues to run during this time, how many minutes will it take Pam to catch up to Cathy assuming Pam resumes running at 10 miles per hour?

A. 30 minutes
B. 40 minutes
C. 45 minutes
D. 60 minutes
E. 75 minutes

Originally posted by neeti1813 on 01 Aug 2014, 17:31.
Last edited by Bunuel on 02 Aug 2014, 00:05, edited 1 time in total.
Renamed the topic and edited the question.
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Pam and Cathy begin running at the same time on a straight  [#permalink]

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New post 02 Aug 2014, 00:20
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neeti1813 wrote:
Pam and Cathy begin running at the same time on a straight path. Pam runs at 10 miles per hour, and Cathy runs at 8 miles per hour. After 45 minutes, Pam stops to stretch. If it takes Pam 30 minutes to stretch and Cathy continues to run during this time, how many minutes will it take Pam to catch up to Cathy assuming Pam resumes running at 10 miles per hour?

A. 30 minutes
B. 40 minutes
C. 45 minutes
D. 60 minutes
E. 75 minutes


Pam ran for 45 minutes and was stretching for 30 minutes. The distance covered in 45 minutes, or in 3/4 hours = 3/4*10 = 30/4.
Cathy ran for 45 + 30 = 75 minutes. The distance covered in 75 minutes, or in 5/4 hours = 5/4*8 = 40/4.

The distance between them = 40/4 - 30/4 = 10/4 = 5/2 miles.
Their relative speed = 10 - 8 = 2 miles per hour.

Therefore, it will take (time) = (distance)/(rate) = (5/2)/(2) = 5/4 hours or 75 minutes Pam to catch up to Cathy.

Answer: E.

P.S. Please name topics properly. Check rule 3 here: rules-for-posting-please-read-this-before-posting-133935.html Thank you.


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Re: Pam and Cathy begin running at the same time on a straight  [#permalink]

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New post 01 Aug 2014, 20:09
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Since at the time that they stop, the distance they both ran are the same, therefore \(r_{Pam} * t_{Pam} = r_{Cathy} * t_{Cathy}\). Since Pam stopped running for 30 minutes, then \(t_{Cathy}=t_{Pam}+0.5\). Therefore \(10 * t_{Pam} = 8 *( t_{Pam}+0.5)\) yields \(t_{Pam} = 2\). 2 hours or 120 minutes is how long Cathy was running. Since she already ran for 45 minutes, after stretching, she ran for 75 minutes .

This corresponds to E.

Furthermore, \(t_{Cathy}=2+0.5=2.5\) or 150 minutes of total running time. She ran for 45 minutes + 30 minutes, so the difference is again 75 minutes. Corresponding to answer E.
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Re: Pam and Cathy begin running at the same time on a straight  [#permalink]

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New post 01 Aug 2014, 20:52
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Here is the simple solution

After 45 mins

Pam would have covered 45/60 * 10 mph = 7.5 miles
Cathy would have covered 45/60 * 8 mph = 6 miles

Pam does 30 minutes of stretching .. so after 75 mins

Pam would still be at 7.5 miles
Cathy would have covered 75/60 * 8 = 10 miles.

So after 75 mins Cathy is 2.5 miles ahead.


Now to catch up with Cathy, Pam must cover 2.5 miles more than Cathy in the same time period. With simple mental calculation based on their speeds, if Pam covers 2 miles more than Cathy per hour, how much time would she take to cover 2.5 miles more than Cathy. An hour and a quarter ( 0.5 is one quarter of 2)

which is 75 mins.
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Re: Pam and Cathy begin running at the same time on a straight  [#permalink]

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New post 02 Aug 2014, 01:52
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neeti1813 wrote:
Pam and Cathy begin running at the same time on a straight path. Pam runs at 10 miles per hour, and Cathy runs at 8 miles per hour. After 45 minutes, Pam stops to stretch. If it takes Pam 30 minutes to stretch and Cathy continues to run during this time, how many minutes will it take Pam to catch up to Cathy assuming Pam resumes running at 10 miles per hour?

A. 30 minutes
B. 40 minutes
C. 45 minutes
D. 60 minutes
E. 75 minutes


Time to catch up= Distance to catch up / relative speed
Distance to catch up= (8*0.75 - 10*0.75) + 8*0.5 = 2.5 miles
Time to catch up= 2.5/(10-8) = 75 minutes
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Re: Pam and Cathy begin running at the same time on a straight  [#permalink]

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New post 03 Sep 2014, 23:14
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Fig I

Race starts with Pam @ 10 Miles/Hr & Cathy @ 8 Miles/Hr

Fig II

After 45 Minutes,

Distance travelled by Pam = 7.5 Miles
Distance travelled by Cathy = 6 Miles

Pam starts taking rest for 30 Minutes

Fig III

Pam completes resting for 30 Minutes. At this point, location of Pam remains same (We'll make this the starting point now)

Distance travelled by Cathy = 2.5 (Further; with respect to Pam)

Lets say they meet at distance x from the current location of Pam,

means Cathy has to travel x-2.5 to reach the same location

Setting up the equation

\(\frac{x}{10} = \frac{x-2.5}{8}\)

x = 12.5 = Distance travelled by Pam (After resting) to meet Cathy

\(Time taken = \frac{12.5}{\frac{10}{60}} = 75 Minutes\)

Answer = E
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Re: Pam and Cathy begin running at the same time on a straight  [#permalink]

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New post 05 Aug 2016, 13:53
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neeti1813 wrote:
Pam and Cathy begin running at the same time on a straight path. Pam runs at 10 miles per hour, and Cathy runs at 8 miles per hour. After 45 minutes, Pam stops to stretch. If it takes Pam 30 minutes to stretch and Cathy continues to run during this time, how many minutes will it take Pam to catch up to Cathy assuming Pam resumes running at 10 miles per hour?


\(P = 10\\
C = 8\)

Obtain the distance travelled for both after the break.
\(P \times \frac{3}{4} + 0 \times \frac{1}{2}= \frac{30}{4}\\
C \times (\frac{3}{4} + \frac{1}{2}) = 8 \times \frac{5}{4} = 10\)

Distance difference = \(10 - \frac{30}{4} = \frac{10}{4}\)

Relative speed = \(P - C = 2\)

\(\text{Time for them to meet }= \frac{\text{Distance difference }}{\text{Relative speed}} = \frac{\frac{10}{4}}{2} = \frac{5}{4} = 75 \text{ minutes}\)

E. 75 minutes

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Re: Pam and Cathy begin running at the same time on a straight  [#permalink]

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New post 13 Feb 2017, 04:28
to answe r this question we need to calculate the distance after 45 mins cathy has covered that in 30 mins which equal to 4 miles (in 45mins cathy has covered 6miles and pan has covered 7.5miles) now pam will cover 2.5km(as cathy has cover total 10 miles in 45+30 mins so pam has to calculate the difference between his and cathy's distance that i equal to 2.5miles)with relative speed of 2 such as 2.5/2=75mins
hence option E must be the answer
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Pam and Cathy begin running at the same time on a straight  [#permalink]

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New post 13 Feb 2017, 12:41
neeti1813 wrote:
Pam and Cathy begin running at the same time on a straight path. Pam runs at 10 miles per hour, and Cathy runs at 8 miles per hour. After 45 minutes, Pam stops to stretch. If it takes Pam 30 minutes to stretch and Cathy continues to run during this time, how many minutes will it take Pam to catch up to Cathy assuming Pam resumes running at 10 miles per hour?

A. 30 minutes
B. 40 minutes
C. 45 minutes
D. 60 minutes
E. 75 minutes



after 75 minutes,
P is (5/4)*8-(3/4)*10=2.5 miles behind C
P gains 10-8=2 mph on C
2.5 miles/2 mph=1.25 hours=75 minutes for P to catch up
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Re: Pam and Cathy begin running at the same time on a straight  [#permalink]

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New post 14 Feb 2017, 15:53
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neeti1813 wrote:
Pam and Cathy begin running at the same time on a straight path. Pam runs at 10 miles per hour, and Cathy runs at 8 miles per hour. After 45 minutes, Pam stops to stretch. If it takes Pam 30 minutes to stretch and Cathy continues to run during this time, how many minutes will it take Pam to catch up to Cathy assuming Pam resumes running at 10 miles per hour?

A. 30 minutes
B. 40 minutes
C. 45 minutes
D. 60 minutes
E. 75 minutes



We are given that Pam and Cathy start running at the same time in a straight path. Pam has a rate of 10 mph and Cathy has a rate of 8 mph. Pam stops after 45 minutes and stretches for 30 minutes as Cathy continues to run for the entire 75 minutes.

Since distance = rate x time, we can determine how far Pam runs in 45 minutes (or 3/4 of an hour) and how far Cathy runs in 75 minutes (or 5/4 hours).

Pam’s distance = 10 x 3/4 = 30/4 = 15/2 = 7.5 miles

Cathy’s distance = 8 x 5/4 = 40/4 = 10 miles

Thus, Cathy has run 10 - 7.5 = 2.5 miles farther than Pam.

We can use the formula (change in distance)/(change in rate) to determine how long it will take Pam to catch Cathy.

time = 2.5/(10-8) = 2.5/2 = 1.25 hours = 75 minutes

Answer: E
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Re: Pam and Cathy begin running at the same time on a straight  [#permalink]

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New post 16 Feb 2017, 12:44
Hi All,

While this question is a bit 'wordy', solving it simply requires that you stay organized and do the necessary math work step-by-step.

To start, we're given the rates that two women run at and a length of time that they each run for. We can use that data to calculate the distance they both travel...

Pam: runs 10 miles/hour for 45 minutes, then STOPS for 30 minutes = (10 mi/hr)(3/4 hour) = 7.5 miles run
Cathy: runs 8 miles/hour for 1 hour 15 minutes = (8 mi/hr)(5/4 hour) = 40/4 = 10 miles run during that same time

Thus, after 1 hour 15 minutes, Cathy is 2.5 miles AHEAD of Pam

We're then asked how long it will take Pam to 'catch up' to Cathy if Pam runs at 10 miles/hour. Since Pam is now running 2 miles/hour FASTER than Cathy, then Pam will 'catch up' 2 miles every hour. Here, we can take advantage of how the answer choices are written. Cathy is 2.5 miles AHEAD of Pam, so we know that it will take MORE than 1 hour for Pam to catch Cathy. There's only one answer that makes sense...

Final Answer:

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Re: Pam and Cathy begin running at the same time on a straight  [#permalink]

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