M06-19

A very good question. You would just have to draw a straight line and think logically to solve this tricky problem.

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

I think this is a high-quality question and I agree with explanation.

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...

Beauty!!!

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 !

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

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
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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
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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.

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.

S1 talks only about C's speed. Ignore!

S2 states that B and C take the same time to reach their destination. This would mean if they meet somewhere in between then the time taken by them to reach their destination after the meeting point will be equal. So, in the D=ST equation, the time is constant so the speed is proportional to the distance. Since B moves with a relatively slower speed he will cover a smaller distance, which means B is closer to his destination (studio).

I think this is a high-quality question and I agree with explanation.

B takes 20 mins more than C to cover the same distance which means speed of c > speed of b.
Also they reach their destinations at the same time which means, after crossing each other they take the same amount of time to reach their destinations. So let's imagine the point at which they cross each other And turn B around. Then conduct a race.. C would beat B as C has a higher speed. So basically b can reach the other end while racing with c but can't reach its origin while racing with C in the same amount of time. Only possible if distance to the other end is smaller.

Posted from my mobile device

Good Question:

Can also be done as . B takes 20 minutes after that C starts. In 20 minutes B has travelled B/3. Left is 2/3 of B. Now. As, if 1/3 B of B covered in 20 minutes, total B will be covered in 60 minutes. So total distance is 60. 20 is done. Now, C takes the same time to cover the distance as B does. Thus 20B + 40 B = 40 C. Thus B:C = 2:3. So distance travelled by B after 20 minutes to meet at the meeting point will be 2/5(40) = 16. So total distance travelled by B = 20 + 16 = 36. Total distance travelled by C = 40-16 = 24.

Very good question indeed, should be bumped ))

I think this is a high-quality question and I agree with explanation. Surprisingly tricky.

From the question stem and the statement (2) we can establish two facts. Firstly, we know that Charlie walks faster than Buster. Secondly, we know that both reach their destinations at same instant.


Thought Process 1:

Now imagine both walking towards each other and meeting at some point between Trailer and Studio. Now from this point Buster walks towards studio and Charlie walks towards trailer. From this meeting point both are taking the same time in reaching their destinations. Now since Charlie is a faster, he will cover more distance as compared to the distance covered by Buster. Hence. we can say that this meeting point was closer to the studio and farther from trailer. Buster and Charlie being at this point when they met, they were closer to the studio. Hence, question stem + statement 2 sufficiently answer the question.


Thought Process 2:

Imagine both Charlie and Buster started walking, passed each other and have reached their destinations. Charlie is at trailer and Buster is at studio. Now visualize all this action going reverse in time like a video frame being played backward. We will see both of them going backwards. Now, we already know that Charlie was the faster one hence they both will meet at some point upto where Charlie has shifted more distance (going backwards) as compared to that covered by Buster. Hence this meeting point will be closer to studio. And this answers our question that they were closer to studio when they met.