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DS -time and distance

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by guerrero » Wed Feb 20, 2013 7:24 am
Buster leaves the trailer at noon and walks towards the studio at a constant rate of B miles per hour. 20 minutes later, Charlie leaves the same studio and walks towards the same trailer at a constant rate of C miles per hour along the same route as Buster. Will Buster be closer to the trailer than to the studio when he passes Charlie?

(1) Charlie gets to the trailer in 55 minutes.

(2) Buster gets to the studio at the same time as Charlie gets to the trailer.

Need help . Please explain the approach

OA B
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by hemant_rajput » Wed Feb 20, 2013 9:05 am
guerrero wrote:Buster leaves the trailer at noon and walks towards the studio at a constant rate of B miles per hour. 20 minutes later, Charlie leaves the same studio and walks towards the same trailer at a constant rate of C miles per hour along the same route as Buster. Will Buster be closer to the trailer than to the studio when he passes Charlie?

(1) Charlie gets to the trailer in 55 minutes.

(2) Buster gets to the studio at the same time as Charlie gets to the trailer.

Need help . Please explain the approach

OA B

Pictorial representation of situation


B------------------------X------------------------C
Trailer<----------->meeting point<------------->Studio

question is asking when Buster meet Charlie, i.e. point X, then the distance BX<CX.


statement 1:-

We know charlie took 55 min, but we don't know anything about the speed of Buster.
Not sufficient.


statement 2:-
Now we know if buster took t min to reach studio then charlie took t+20 min. Now we don't have ratio of their speed. so we have 2 unknown , distance and time.
so not sufficient.


combining two statement.

we know charlie took 55 min and buster took 35 min. what ever may be the Distance we can calculate their speed ratio so we can solve this problem.
Note:- Although, I'm doing showing the further calculation but knowing this is a DS question I don't need to do it.
TS --> distance between Trailer and Studio
B/C = (TS/35)/(TS/55)=> B:C = 11:7{ratio of speed of Buster and Charlie.}


So time taken by Buster and charlie to meet a point X.

TS/(11a + 7a) = TS/18a

distance traveled by Buster in TS/18a = 11a * TS/18a =11*TS/18.

BX = 11*TS/18
CX = 7*TS/18

BX > CX.

Hope it helps.
Cheers,
Hemant

Answer is C.
I'm no expert, just trying to work on my skills. If I've made any mistakes please bear with me.
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by aditya8062 » Thu Feb 21, 2013 1:59 am
my take wud be B .but i feel that the language of B is lightly not gmat types considering the fact that gmat writes nice unambiguous sentences .but the good part is even with both the interpretation of statement B i am able to solve the question

interpretation 1 : when i say that Buster gets to the studio at the same time as Charlie gets to the trailer it can mean that both take lets say 2 hours if they both were to start at the same time--now for this situation we can conclude that speed of both is same that means hypothetically if they had started at the same time from respective ends then the wud have met somewhere in the middle now because buster has already traveled for some 20 min so its obvious that when they meet buster wud be a point that wud be nearer to the studio

interpretation 2: when i say that Buster gets to the studio at the same time as Charlie gets to the trailer --it can also mean that the problem is considering the time as in both finishes their task of traveling at, lets say ,1300 hrs (coz they had started at 1200 hrs )but under this situation buster has already traveled for 20 min .so that means that in my 2nd interpretation the speed of charlie is more than that of buster.so that means even in this situation buster will meat charlie at a point which will be nearer to studio

i hope my explanation makes sense
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by GMATGuruNY » Thu Feb 21, 2013 5:38 am
guerrero wrote:Buster leaves the trailer at noon and walks towards the studio at a constant rate of B miles per hour. 20 minutes later, Charlie leaves the same studio and walks towards the same trailer at a constant rate of C miles per hour along the same route as Buster. Will Buster be closer to the trailer than to the studio when he passes Charlie?

(1) Charlie gets to the trailer in 55 minutes.

(2) Buster gets to the studio at the same time as Charlie gets to the trailer.

Need help . Please explain the approach

OA B
Many DS rate problems are quickly solved with a little REASON.

Statement 1: Charlie gets to the trailer in 55 minutes.
No information about Buster.
INSUFFICIENT.

Statement 2: Buster gets to the studio at the same time as Charlie gets to the trailer.
If Buster and Charlie meet at the halfway point, then -- for each to finish traveling the remaining half of the distance at the same time -- they must travel AT THE SAME SPEED.
But Buster's total time is 20 MINUTES LONGER than Charlie's total time, implying that Buster is traveling MORE SLOWLY than Charlie.
Thus, when Buster and Charlie meet, Buster must be MORE THAN HALFWAY to the studio, so that he can travel the remaining distance AT A SLOWER SPEED and still finish his trip at the same time as Charlie finishes his trip.
SUFFICIENT.

The correct answer is B.
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by prsnt11 » Tue Oct 08, 2013 3:27 pm
Hey GMATGuruNY,

Can you please advice if my reasoning is correct:
Given that B traveled 20 mins more than C and reached at the same time as C, so B's speed is < C's speed. Therefore, Statement 2 implies that they did not meet till they actually reached the studio. This is because they cannot travel the same remaining distance in the same remaining time (after they met) in two different speeds. Hence, B is SUFFICIENT.
Thanks,
-prsnt11
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by Mo2men » Sun Jan 07, 2018 1:05 am
GMATGuruNY wrote:
Statement 2: Buster gets to the studio at the same time as Charlie gets to the trailer.
If Buster and Charlie meet at the halfway point, then -- for each to finish traveling the remaining half of the distance at the same time -- they must travel AT THE SAME SPEED.
But Buster's total time is 20 MINUTES LONGER than Charlie's total time, implying that Buster is traveling MORE SLOWLY than Charlie.
Thus, when Buster and Charlie meet, Buster must be MORE THAN HALFWAY to the studio, so that he can travel the remaining distance AT A SLOWER SPEED and still finish his trip at the same time as Charlie finishes his trip.
SUFFICIENT.

The correct answer is B.
Dear Mitch,

I do not understand how I can interpret St 2 ? What SAME time means? Does mean both take 1 hr but how this happen if Buster had more 2 hrs? Does it mean same clock hour?

Thanks
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buster and charlie

by GMATGuruNY » Sun Jan 07, 2018 3:32 am
Mo2men wrote: Dear Mitch,

I do not understand how I can interpret St 2 ? What SAME time means? Does mean both take 1 hr but how this happen if Buster had more 2 hrs? Does it mean same clock hour?

Thanks
Statement 2: Buster gets to the studio at the same time as Charlie gets to the trailer.
Conveyed meaning:
If Charlie arrives at the trailer at 1pm, then Buster arrives at the studio at 1pm.
If Charlie arrives at the trailer at 2pm, then Buster arrives at the studio at 2pm.
If Charlie arrives at the trailer at 3pm, then Buster arrives at the studio at 3pm.
For each traveler, the arrival time is THE SAME.

Since Charlie begins his journey 20 minutes AFTER Buster -- but completes his journey AT THE SAME TIME as Buster -- Charlie takes LESS TIME than Buster to travel the distance between the trailer and the studio.
Thus, Charlie's rate must be greater than Buster's rate.
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