Andrew passed the point x at 1:00 pm at the average speed 5 mile per

Question

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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ManifestDreamMBA · Answer

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

ravi1522 · Answer

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

kingbucky · Answer

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 .  
    Person  
  Speed (mph)  
  Time (hours)  
  Distance (miles)  
 
    Andrew   
   5   
    t    
    5t    
 
    Ben   
   6   
    t - \frac{4}{3}    
    6 \left(t - \frac{4}{3}ight)

Setting the distances equal when Ben overtakes Andrew, we have:   
 5t = 6 \left(t - \frac{4}{3}ight) 
 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 	imes 8 = 40  miles, Ben’s  6 	imes \frac{20}{3} = 40  miles.

Answer: E

ektalakra · Answer

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

Andrew passed the point x at 1:00 pm at the average speed 5 mile per

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Person	Speed (mph)	Time (hours)	Distance (miles)
Andrew	5	\(t\)	\(5t\)
Ben	6	\(t - \frac{4}{3}\)	\(6 \left(t - \frac{4}{3}\right)\)

GMAT Problem Solving (PS) Questions

Andrew passed the point x at 1:00 pm at the average speed 5 mile per

Prep Toolkit

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My Rewards