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

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Bunuel
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Teneka and Francis leave Town A at 9:00 a.m. driving north along Highw [#permalink] New post  26 Apr 2017, 08:16
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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.
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BrentGMATPrepNow
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Re: Teneka and Francis leave Town A at 9:00 a.m. driving north along Highw [#permalink] New post  03 Apr 2018, 09:51
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Bunuel 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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Re: Teneka and Francis leave Town A at 9:00 a.m. driving north along Highw [#permalink] New post  26 Apr 2017, 10:25
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Hopefully, my solution is correct
If Francis drives 40 miles per hour. After 240 miles Francis' car breaks down then

40 x t = 240
t = 6 hours
Then Francis' car broke down by 3 pm.

Teneka reaches Francis car:
30 x t = 240
t = 8 hours
Then Teneka reaches Francis car by 5 pm

Which is answer D
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Re: Teneka and Francis leave Town A at 9:00 a.m. driving north along Highw [#permalink] New post  29 Aug 2017, 14:45
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Distance traveled = 240/40, which means Francis has driven 6 hours. At 6 hours, Teneka has driven 180 miles (30 * 6), so it will take her (240-180) 60 miles/30 MPH = 2 additional hours.

6 hours after 9:00 is 3:00 P.M.
2 Additional hours makes it 5:00 P.M.
Answer: D
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Re: Teneka and Francis leave Town A at 9:00 a.m. driving north along Highw [#permalink] New post  01 Sep 2017, 12:00
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Bunuel 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.



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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Re: Teneka and Francis leave Town A at 9:00 a.m. driving north along Highw [#permalink] New post  14 Apr 2018, 23:03
Bunuel 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.


Distance = 240 miles

Francis' travel time when car broke down = 240/40 = 6 hours

Means, 9 a.m. + 6 hrs = 3 pm

Teneka's speed = 30 m/hr

Teneka's time when joined Francis = 240/30 = 8hrs

Means, 9 a.m. + 8 hrs = 5 pm

Hence (D)
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Re: Teneka and Francis leave Town A at 9:00 a.m. driving north along Highw [#permalink] New post  08 Nov 2022, 01:55
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Re: Teneka and Francis leave Town A at 9:00 a.m. driving north along Highw [#permalink]
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