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Re: Buster leaves the trailer at noon and walks towards the studio at a co [#permalink]
it seems that the solution of bb is the only best solution for this question. To visualize the solution, ones just need to draw the distance and the point where the two pass each other.
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Re: Buster leaves the trailer at noon and walks towards the studio at a co [#permalink]
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akshata19 wrote:
skidstorm wrote:
it seems that the solution of bb is the only best solution for this question. To visualize the solution, ones just need to draw the distance and the point where the two pass each other.



I agree Bunuel's solution is precise one.
But whenever I stumble upon this question again , i forget the approach how to solve this and it takes me lot of time to figure out .
Especially when trying to solve it algebraically

So here is my visual solution . Hope it helps.


Given: C starts 20 min late than B
To find the relative position

3 possible meeting point i.e X
1. B -------X-------------------------------C

2. B ------------------X--------------------C

3. B -------------------------------X-------C

Stmnt 1 : Total C time 55 min .
With this info cannot find the meeting point whether 1,2 or 3

Stmnt 2 : Both reach at the same time to respective destinations (total distance is same)

Since we know C starts late still covers same distance .
Hence C is speeder is greater than B

Now lets check if stmnt 2 satisfies any of the meeting point

[When they meet they have same time to cover remaining respective distances]

1.) B -------X-------------------------------C

B has to cover larger distance XC and
C has to cover shorter distance XB

But C is faster than B hence it can cover smaller distance before B could cover large distance
hence this meeting point is wrong

2.) B -----------------X--------------------C
Same remaining distance with same time
but since speed is not same
this scenario is also not possible

3.) B --------------------------X-----------C

B has to cover smaller distance XC and
C has to cover Larger distance XB

C can achieve this it as its speed is more B .
So it can reach , till B slowly reaches the studio . This is the only possibility for relative position.
i.e closer B is closer to studio

B


This is by far the most easy and simplest solution.
Epic visual solution.
Thank you so much. I was beating my head around this so much.
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Re: Buster leaves the trailer at noon and walks towards the studio at a co [#permalink]
gordonf35 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.


Statement 1 just conveys about the time taken no info about Buster we cannot comment upon anything
Clearly insuff

Statement 2 this definitely gives rate of buster and charlie since Charlie started after 20 minutes his speed is faster nad Butler slower ,however since they have to reach the posirion at the same time butler has to cover lesser distance than Charlie therefore we can deterministically determine that Butler is closer to the studio than trailer

Therefore IMO B
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Re: Buster leaves the trailer at noon and walks towards the studio at a co [#permalink]
If total distance is 'x' miles and Charlie traveled for 't' hours.
then equation will be
B(t+1/3) + Ct = x;
t(B+C) + B/3 = x - 1

equation for question stem: Ct< B(t+1/3)

Simplifying further : t*(C-B)<B/3 - 2

Statement 2 put in equation
x/B=x/C+1/3

x = BC/3(C-B) - 3

if we substitute eq.3 in eq.1, then

3t(B+C)(C-B)=B^2 - 4

If we put eq.4 in eq.2, then

B^2/(3*(B+C))< B/3

Solving this and knowing fact that B & C are positive numbers

BC>0

this will we true for any value of B and C,
SO statement 2 is sufficient.
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Re: Buster leaves the trailer at noon and walks towards the studio at a co [#permalink]
For the math geeks out there, let's look at it completely mathematically:
Option 1 is not sufficient because we have no information about Buster nor any correlation between their times. So that's out.

Let's look at option 2 now:

Let the total distance between the studio and trailer be D
Let us assume that Buster travels X distance when Buster and Charlie meet. Through this, we have:
And let the time (in hours) at which they meet be T (w.r.t Buster) which means it's T-1/3 for Charlie (20 mins converted to hours is 1/3; taking care of the units as well :P)

From this, we have:
X = BT
D-X = C(T-1/3)

On solving, we get:
D/C + 1/3 - X/C = X/B [EQUATION A]

Now let's assume that they both reach their destinations at a time T'. Given they reach at the same time and given Charlie started 20 mins late, the equations we have are:
D = C(T'-1/3)
D = BT'

Equating T' from both equations, we get:
D/B = D/C + 1/3 [EQUATION B]

Now look closely, does equation A and B have something in common? Yes! Part of the LHS of A = RHS of B. On substituting B in A, we get:

D/B -X/C = X/B
or
(D-X)/B = X/C

Now we know that C>B (because Charlie is faster than Buster and therefore reaches at the same time as Buster even after starting 20 mins late). For this proportion to hold, X>D-X (otherwise the equality won't hold) -> Distance covered by Buster when he meets Charlie is greater than the remaining distance he has to cover to reach his destination -> Buster is closer to his destination compared to his starting point.

I hope this helps! :)
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Re: Buster leaves the trailer at noon and walks towards the studio at a co [#permalink]
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