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Teneka and Francis leave Town A at 9:00 a.m. driving...

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by Vincen » Tue Sep 12, 2017 10:42 am
Teneka and Francis leave Town A at 9:00 a.m. driving north along Highway B, in seperate cars. Teneka drives 30 miles per hour and Francis drives 40 miles per hour. If after traveling 240 miles Francis' car breaks down and stops, at what time will Teneka's car reach Francis' car?

A. 11 p.m.
B. 10 p.m.
C. 6 p.m.
D. 5 p.m.
E. 3 p.m.

The OA is D.

Can someone make a graph of the vehicles so I can understand it better. Please.
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by Brent@GMATPrepNow » Wed Sep 13, 2017 11:15 am
Vincen wrote:Teneka and Francis leave Town A at 9:00 a.m. driving north along Highway B, in seperate cars. Teneka drives 30 miles per hour and Francis drives 40 miles per hour. If after traveling 240 miles Francis' car breaks down and stops, at what time will Teneka's car reach Francis' car?

A. 11 p.m.
B. 10 p.m.
C. 6 p.m.
D. 5 p.m.
E. 3 p.m.
Given: Francis drives 240 miles and then stops (breaks down)

So, Teneka will reach Francis once Teneka drives 240 miles.
How long will it take Teneka to drive 240 miles?
Time = distance/speed = (240 miles)/(30 miles per hour) = 8 hours

Teneka started driving at 9:00am
So, after 8 hours, the time will be 5pm

Answer: D

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Brent
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by Jeff@TargetTestPrep » Tue Sep 19, 2017 3:04 pm
Vincen wrote:Teneka and Francis leave Town A at 9:00 a.m. driving north along Highway B, in seperate cars. Teneka drives 30 miles per hour and Francis drives 40 miles per hour. If after traveling 240 miles Francis' car breaks down and stops, at what time will Teneka's car reach Francis' car?

A. 11 p.m.
B. 10 p.m.
C. 6 p.m.
D. 5 p.m.
E. 3 p.m.

The OA is D.
Since distance/rate = time and 240/30 = 8, Taneka had to drive 8 hours to reach Francis.

So, she reached Francis at 9 a.m. + 8 hours = 5 p.m.

Answer: D

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