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M06-19

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New post 16 Sep 2014, 00:27
3
30
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

58% (02:05) correct 42% (02:21) wrong based on 218 sessions

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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.

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New post 16 Sep 2014, 00:27
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Official Solution:


(1) Charlie gets to the trailer in 55 minutes. No info about Buster. Not sufficient.

(2) Buster gets to the studio at the same time as Charlie gets to the trailer. Charlie needed 20 minutes less than Buster to cover the same distance, which means that the rate of Charlie is higher than that of Buster. Since after they pass each other they need the same time to get to their respective destinations (they get at the same time to their respective destinations) then Buster had less distance to cover ahead (at lower rate) than he had already covered (which would be covered by Charlie at higher rate). Sufficient.


Answer: B
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New post 01 Aug 2015, 23:00
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A very good question. You would just have to draw a straight line and think logically to solve this tricky problem.
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New post 27 Feb 2016, 09:35
Could we please expand on this answer, perhaps drawing a line to see how this works? Thank you
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New post 31 May 2016, 08:58
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VeritasPrepKarishma Bunuel

A-----------------X-----------------------------------------------------------------B

Can you please share your thoughts on this problem.

I have tried to visualize it this way. Say we have 100m distance between the two points. Now, See Buster has started 20 minutes before but still reaches at the same time that means his time is t + 20 if t is the time taken by charlie to reach 100 meters distance.

Now, one thing we know is that : charlie has higher speed than buster. So far so good. But how do we know what happens before they cross, at the time of crossing and after they have crossed is something I need more clarification on.

When Charlie starts Buster has already covered a distance equal to 20 times his speed in miles per hour. When charlie and Buster meet at say X point then we have D-X more than X Since if both we running at equal speed then starting 20 minutes early they would have met beyond the midpoint of the D.

I think, I am getting lost beyond this point.

Can you help me get organized from here?
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New post 31 May 2016, 09:23
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ankushbagwale wrote:
VeritasPrepKarishma Bunuel

A-----------------X-----------------------------------------------------------------B

Can you please share your thoughts on this problem.

I have tried to visualize it this way. Say we have 100m distance between the two points. Now, See Buster has started 20 minutes before but still reaches at the same time that means his time is t + 20 if t is the time taken by charlie to reach 100 meters distance.

Now, one thing we know is that : charlie has higher speed than buster. So far so good. But how do we know what happens before they cross, at the time of crossing and after they have crossed is something I need more clarification on.

When Charlie starts Buster has already covered a distance equal to 20 times his speed in miles per hour. When charlie and Buster meet at say X point then we have D-X more than X Since if both we running at equal speed then starting 20 minutes early they would have met beyond the midpoint of the D.

I think, I am getting lost beyond this point.

Can you help me get organized from here?


Hi ankushbagwale and Avigano,

we are concerned what happens when B reaches half the distance..

B takes \(x\) min and C takes \(x-20\)....
B is at half the distance at \(\frac{x}{2}\).....

so lets see where is C at this time -20 min, as he starts 20 minutes later = \(\frac{x}{2} -20\)..
But when does C reach midway = at \(\frac{x-20}{2} =\frac{x}{2} - 10\)
since C reaches half way ONLY at \(\frac{x}{2} -10\), he has to walk for another 10 minutes after \(\frac{x}{2}-20\) to reach mid-way..
so they meet towards the starting point of C..
suff
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New post 02 Jun 2016, 16:28
Thank you Chetan for your explanation. It all makes sense now. I think, logically also what bunuel said is the correct way. Suppose they were to meet at the middle point. Then in that case C would have reached the destination earlier that B. Suppose they meet at any point before the midpoint and towards B then C would have lesser distance and more rate than B, Hence C still would have reached before B. Thus, the only scenario when Both reaches at the same time is when B has already travelled considerable distance ( at least more than the half ) before C started and then C starts 20 minutes later and covers that at higher rate.
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New post 14 Jul 2016, 15:40
I think this is a high-quality question and I agree with explanation.
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New post 17 Apr 2017, 11:35
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I wasn't satisfied with the explanation given here so i tried to solve it:
Smart numbers worked only after doing some attempts as answer used to vary with the distance between b mph and c mph. Let me know if you think i am doing some/anything wrong...


PLEASE USE C = 19 NOT 20.
as i said, the answer only made sense after a few trials...
>> !!!

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New post 06 Jan 2018, 12:56
Is there a better explanation for this question? It looks like a good questions but the explanation is poorly written.
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New post 13 Feb 2018, 07:55
Beauty!!!
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New post 28 Aug 2018, 20:38
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Given: Speed of B = B miles per hour
Speed of C = C miles per hour and C started the journey after 20 minutes.

from1) C took 55 minutes...not giving us any information regarding time taken by B so we can't compare their speeds, hence not sufficient !

from2) B & C reached at the same time.
Let's assume
distance D= 100 miles and time taken by B = 100 mins, therefore time taken by C = 80 mins
Hence speed of B = 1 mile per min and C =10/8 mile per min
C started the journey after 20 mins.....so in 20 mins B covered 20 miles....remaining distance 80 miles
Now Relative speed concept They both are travelling in opposite directions so relative speed = 1+10/8 =18/8 miles per min
time taken to cover 80 miles with this relative speed
= (80/18)*8 = 35.something (this is the time when they cross each other)
B already covered 20 miles and in 35.something time s/he will cover = 35 miles total = 55 miles , this means C covered 100-55 =45 miles. Hence option B !
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New post 02 May 2019, 09:36
Trailer (X)---------------------------A-----------Studio (Y)


Buster gets to the studio at the same time as Charlie gets to the trailer, given Charlie leaves 20 mins later than Buster
=> Charlie walks faster than Buster
=> C > B

Question: Is XA < YA ?

(1) Statement 1 is clearly insufficent because we don't know at which point Charlie and Buster meet

(2)

Trailer (X)---------------------------A-----------Studio (Y)


Buster passes Charlie at point A
In 20 mins, the distance Buster covers = B/3
Let t denotes the period Charlie and Buster must walk until they meet at point A
The distance Buster has covered when he reaches point A: XA=B/3 + Bt
The distance Charlie has covered when he reaches point A: YA = Ct

=> The distance Buster needs to cover ahead = Ct ; the distance Charlie needs to cover ahead = B/3+Bt

Since they arrive at the destinations at same time, let t1 denotes the period both Buster & Charlie need to walk to their destinations.

=> The distance Buster needs to cover ahead = Ct = B*t1 = YA
The distance Charlie needs to cover ahead = B/3+Bt = C*t1 = XA

We need to compare YA and XA => We need to compare B*t1 and C*t1
Since C>B => C*t1 > B*t1 => XA > YA
=> SF
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New post 27 Jul 2019, 02:28
From question stem, buster started earlier by 20mins and was caught-up by Charlie.
So, Charlie must have traveled faster than Buster


catch-up means distance of buster + distance of Charlie = total distance

When charlie caught up with Buster,

Case 1: If Buster had traveled 1/3 of the distance, with a distance closer to the starting point, then Charlie would have traveled 2/3 of the distance
Buster will have to cover 2/3 more of the journey at a speed slower than Charlie's
Charlie will have to cover 1/3 more of the journey at a speed faster than Buster's
----------------------------------------------------------------------------------------



Case 2: If Buster had traveled 2/3 of the distance, with a distance closer to the end point, then Charlie would have traveled 1/3 of the distance
Buster will have to cover 1/3 more of the journey at a speed slower than Charlie's
Charlie will have to cover 2/3 more of the journey at a speed faster than Buster's
----------------------------------------------------------------------------------------



Statement 2 tells us that, they both reached the same time at their respective destination.
Therefore, as soon as they caught-up each other, Buster must have been closer to the end point (as in case 2), since buster was traveling slower than charlie was.

Had Buster closer to the starting point, Buster would have needed more time to complete the journey, not possibly to reach the same time as Charlie.
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New post 13 Oct 2019, 08:50
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Bunuel

We are told that Buster has a 20 minutes head start than Charlie, but still both Buster and Charlie reach their respective destinations at the same time. The distance travelled for both is the same. That means they will not meet the center. Thus, "Will Buster be closer to the trailer than to the studio when he passes Charlie?" will either be Yes or No. We don't really need to know which is it. Hence Statement 2 is sufficient.

Does this logic work?
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New post 13 Oct 2019, 09:10
Bunuel, can you tag this under TTP. This is a question from them.

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New post 13 Oct 2019, 10:27
It's just better to read the options first, as Buster already has 20 minutes head start and they reach at destinations at same time.

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New post 13 Oct 2019, 21:34
aditya2491 wrote:
Bunuel

We are told that Buster has a 20 minutes head start than Charlie, but still both Buster and Charlie reach their respective destinations at the same time. The distance travelled for both is the same. That means they will not meet the center. Thus, "Will Buster be closer to the trailer than to the studio when he passes Charlie?" will either be Yes or No. We don't really need to know which is it. Hence Statement 2 is sufficient.

Does this logic work?


According to me, this is the best and simple explanation. Thank you :)

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New post 17 Oct 2019, 09:53
Buster gets to the studio at the same time as Charlie gets to the trailer.

Since Charlie started 20 min later, he has to cover the same distance as Buster in lesser time for both of them to reach their respective destinations at the same time.

If you imagine, that buster and charlie met mid-way, then both of them have to cover their respective left over distances.
Now, for buster to match his time with charlie he has to cover less distance than Charlie, hence buster have to be closer to the studio than to the trailer.

Hence, B is sufficient.

Great question. Really rattles up your mind in a crunch situation.
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