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Bunuel
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Andrew passed the point x at 1:00 pm at the average speed 5 mile per 14 May 2025, 00:30
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E
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55% (hard)

Question Stats:

61% (02:15) correct 39% (02:52) wrong based on 168 sessions

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Andrew passed the point x at 1:00 pm at the average speed 5 mile per hour and Ben passed the same point at 2:20 pm with the average speed 6 mile per hour. If Ben followed the same route passed by Andrew, when Ben overtook Andrew?

(A) 7:00pm
(B) 7:30pm
(C) 8:00pm
(D) 8:30pm
(E) 9:00pm

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Andrew passed the point x at 1:00 pm at the average speed 5 mile per 14 May 2025, 01:28
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Distance covered by Andrew by 2:20 = 5 * 80/60 = 20/3 miles
Relative speed = 1m/h


Time = Distance/ Speed
Time = 20/3 hours = 6 hours 40 mins

Adding this to 2:20PM, we get 9:00 PM

Answer E

Bunuel
Andrew passed the point x at 1:00 pm at the average speed 5 mile per hour and Ben passed the same point at 2:20 pm with the average speed 6 mile per hour. If Ben followed the same route passed by Andrew, when Ben overtook Andrew?

(A) 7:00pm
(B) 7:30pm
(C) 8:00pm
(D) 8:30pm
(E) 9:00pm

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Gentle note to all experts and tutors: Please refrain from replying to this question until the Official Answer (OA) is revealed. Let students attempt to solve it first. You are all welcome to contribute posts after the OA is posted. Thank you all for your cooperation!
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Andrew passed the point x at 1:00 pm at the average speed 5 mile per 14 May 2025, 02:01
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Hello ,
As ben will travel distance from x in 1:20 will be 6.65 miles AT 2:20
Time of meet will be (6.65/(6-5)
6:65 hours after 2:20 that will be 9PM

Hence option E is correct
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Andrew passed the point x at 1:00 pm at the average speed 5 mile per 14 May 2025, 04:52
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Andrew passes point x at 1:00 pm at 5 mph, Ben at 2:20 pm at 6 mph, following the same route. We need the time Ben overtakes Andrew.

Point x is the reference. Let Andrew travel for \(t\) hours from 1:00 pm. Ben starts 80 minutes (or \(\frac{4}{3}\) hours) later, so Ben travels for \(t - \frac{4}{3}\) hours. We equate their distances to find \(t\).
PersonSpeed (mph)Time (hours)Distance (miles)
Andrew5\(t\)\(5t\)
Ben6\(t - \frac{4}{3}\)\(6 \left(t - \frac{4}{3}\right)\)

Setting the distances equal when Ben overtakes Andrew, we have:
\(5t = 6 \left(t - \frac{4}{3}\right)\)
\(5t = 6t - \frac{24}{3}\)
\(5t = 6t - 8\)
\(8 = t\)

Andrew’s time \(t = 8\) hours from 1:00 pm is 9:00 pm (1:00 pm + 8 hours).

Verify: Ben’s time is \(8 - \frac{4}{3} = \frac{24}{3} - \frac{4}{3} = \frac{20}{3}\) hours from 2:20 pm. Distances match: Andrew’s \(5 \times 8 = 40\) miles, Ben’s \(6 \times \frac{20}{3} = 40\) miles.

Answer: E
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Andrew passed the point x at 1:00 pm at the average speed 5 mile per 26 Sep 2025, 22:55
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A B C
|<------1hr 20 min-------------->|-----------------x
6m/s-> 5m/s-> (meeting point)
distance b/w A to B= 6*1hr 20 min = 8 mile
let dist b/w B & C be x
distance travelled by Ben A-C be d= 6.t (t=time say).........eq(1)
distance travelled by andrew B-C is d-8=5t.........eq(2)
subtituting 1 in 2
we get t=8 hrs
need to add 8 hrs from when andrew passed point B at 01:00 pm...= 9 pm
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