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

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.

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

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

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

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 = \(P - C = 2\)

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

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

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
E

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

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:

GMAT assassins aren't born, they're made,
Rich

Pam

R= 10 mph
Time= 45 min or 3/4 hours
Distance= 10 * 3/4
Distance= 30/4

Cathy

R= 8 mph
Time= 75 min or 5/4 hours
Distance= 8 * 5/4
Distance= 40/4

(D) Distance between Cathy and Pam = 40/4 - 30/4 = 10/4 = 5/2
(R) Relative speed= 10mph-8mph =2mph
T= D/R

T= (5/2)/2
T= 5/4
T= 75 min. = E

Hi scott,

but in ttp wasnt it mentioned not to use this formula if the total distance traveled was the same?

To begin with, Pam runs for 45 mins and Cathy runs for 75 mins (while Pam stretches for the last 30 mins).
Distance covered by Pam in 3/4th of an hour = 3/4th of 10 = 7.5 miles
Distance covered by Cathy in 1 and 1/4th of an hour = 8 miles + 1/4th of 8 miles = 8 + 2 miles = 10 miles

At the end of 75 mins, Pam is at 7.5 km point and Cathy is at 10 miles point so distance between them is 2.5 miles. 
Relative speed of Pam wrt to Cathy = 10 - 8 = 2 mph
Time taken for Pam to catch up ­= 2.5/2 hours (which is more than 1 hence answer is more than 60 mins)

Answer (E)

Check this video: Confused about when to use Relative Speed on GMAT Questions?