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Two friends A and B leave point A and point B simultaneously and trave

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Two friends A and B leave point A and point B simultaneously and trave [#permalink]

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Two friends A and B leave point A and point B simultaneously and travel towards Point B and Point A on the same route at their respective constant speeds. They meet along the route and immediately proceed to their respective destinations in 32 minutes and 50 minutes respectively. How long will B take to cover the entire journey between Point B and point A?

A) 85mins
B) 90 mins
C) 95mins
D) 100mins
E) 60 mins


Source: 4gmat
[Reveal] Spoiler: OA

Last edited by alphonsa on 29 Aug 2014, 19:53, edited 1 time in total.

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Re: Two friends A and B leave point A and point B simultaneously and trave [#permalink]

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New post 29 Aug 2014, 15:42
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alphonsa wrote:
Two friends A and B leave point A and point B simultaneously and travel towards Point B and Point A on the same route at their respective constant speeds. They meet along the route at two points and immediately proceed to their respective destinations in 32 minutes and 50 minutes respectively. How long will B take to cover the entire journey between Point B and point A?

A) 85mins
B) 90 mins
C) 95mins
D) 100mins
E) 60 mins


Source: 4gmat

Dear alphonsa,
I'm happy to respond. :-) Something seems flawed with the question, or at least unclear. First of all, I think the word "respectively" is implied by not stated in the first sentence --- that would clear up a certain amount of the ambiguity. From the first sentence, I picture a line, AB. Traveler A starts at A and moves toward B. Traveler B starts at B and moves toward A. So far so good. The completely unclear part of the question is: how on earth do these travelers "meet along the route at two points"? If one goes from A to B, and the other from B to A, then they will cross paths just once, as they passed each other. For them to meet twice --- the route must be more complicated, or each one goes in both directions, or something of that sort. That one condition throws everything about the entire scenario into doubt.

If each one travels a straight line, A to B, and B to A, and they cross paths only once, at one point, then it's a relatively straightforward problem.

The meeting point m divides the total distance into two lengths, which I will call K and L.
Attachment:
track from A to B.JPG
track from A to B.JPG [ 12.02 KiB | Viewed 2672 times ]

Call the time it takes them to meet at the meeting place T.
In that time T, A covers the distance K, and B covers the distance L. Thus
K = (vA)*T
L = (vB)*T
They meet at m, and keep moving. After the meeting, A covers the distance L is 32 minutes, and B covers the distance K in 50 minutes.
L = (vA)*32
K = (vB)*50
Set the expression for the distance equal.
(vA)*T = (vB)*50 and (vB)*T = (vA)*32
Rewrite each equation to equal the ratio of (vA)/(vB)
(vA)/(vB) = 50/T
(vA)/(vB) = T/32
Set these equal and solve for T
50/T = T/32
T^2 = 50*32 = 100*16 = 1600
T = 40
So B traveled 40 minutes along L and 50 minutes along K for a full time of 90 minutes. Answer = (B).

Notice that I used the doubling & halving trick to multiply 32 times 50. See:
http://magoosh.com/gmat/2012/doubling-a ... gmat-math/

That's what happens if the two travelers meet at one point during this process. If they meet at two points, I have no earthly clue what paths they have taken.

Does all this make sense?
Mike :-)
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Re: Two friends A and B leave point A and point B simultaneously and trave [#permalink]

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New post 29 Aug 2014, 19:56
Dear Mike

Sorry.
I have edited the question now.

And thanks for the explanation :-D

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Re: Two friends A and B leave point A and point B simultaneously and trave [#permalink]

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New post 30 Aug 2014, 03:36
Let x per minute be the speed of A and y per minute be the speed of B.
After meeting at a point, A travels for 32 mins and B travels for 50 mins. So distance covered by each of them post point of crossing
A= 32x and B=50y
The distance covered by A and B before they cross each would be distance covered by B and A post crossing respectively.
Therefore distance covered by B before he meets A= 32x
Time taken by B cover 32x distance= 32x/y mins
Therefore total time taken by B= 32x/y + 50 mins ................. I

We need to find value of x in terms of y to arrive at final answer.
Total distance= 32x+50y
Combined speed of A and B= x+y
Therefore time taken before A and B meet en-route= (32x+50y)/(x+y)
Time taken by B reach destination after meeting A= 50 mins
Total travel time for B= [(32x+50y)/(x+y)]+50 mins ...................II

Equate I and II
32x/y+50= [(32x+50y)/(x+y)]+50
(32x+50y)/y=(82x+100y)/(x+y)
32x^2+50xy+32xy+50y^2=82xy+100y^2
32x^2+82xy-82xy+50y^2-100y^2=0
32x^2-50y^2=0
32x^2=50y^2
16x^2=25y^2
Taking square root.. (since x and y denote speed, square root can't be negative)
4x=5y
y=4x/5 ............ III

substitute in I
=32x/(4x/5) + 50
=32x*5/4x + 50
=40+50
=90 mins

Ans=B

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Two friends A and B leave point A and point B simultaneously and trave [#permalink]

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alphonsa wrote:
Two friends A and B leave point A and point B simultaneously and travel towards Point B and Point A on the same route at their respective constant speeds. They meet along the route and immediately proceed to their respective destinations in 32 minutes and 50 minutes respectively. How long will B take to cover the entire journey between Point B and point A?

A) 85mins
B) 90 mins
C) 95mins
D) 100mins
E) 60 mins


Source: 4gmat


Responding to a pm:

Here are two methods:

First check this post that discusses both methods in detail: http://www.veritasprep.com/blog/2013/04 ... -formulas/

If two objects A and B start simultaneously from opposite points and, after meeting, reach their destinations in ‘a’ and ‘b’ hours respectively (i.e. A takes ‘a hrs’ to travel from the meeting point to his destination and B takes ‘b hrs’ to travel from the meeting point to his destination), then the ratio of their speeds is given by:
Sa/Sb = √(b/a)

\(Sa/Sb = \sqrt{50/32} = \sqrt{25/16} = 5/4\)

We need to find the time taken by B to cover the entire journey. We know that B takes 50 mins to cover the distance after meeting. How much time will he take to cover the distance before meeting A (i.e. distance covered by A in 32 mins)?
Since SpeedA:SpeedB = 5:4,
TimeA:TimeB = 4:5
This means that if A takes 32 (which is 4*8) mins to cover that distance, B will take 5*8 = 40 mins to cover it.

Hence total time taken by B = 50 + 40 = 90 mins

Method 2:

Distance between A and Meeting point /Distance between Meeting point and B = Time taken to go from A to Meeting point/Time taken to go from Meeting point to B = 32/t (in case of A) = t/50 (in case of B)
t^2 = 32*50
t = 40

Total time taken by B = 50+40 mins = 90 mins
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Re: Two friends A and B leave point A and point B simultaneously and trave [#permalink]

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New post 03 Sep 2015, 13:18
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let t=time to meeting
start with equation between two ratios:
t/t+32=50/t+50
t^2=1600
t=40
40+50=90

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Re: Two friends A and B leave point A and point B simultaneously and trave [#permalink]

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Re: Two friends A and B leave point A and point B simultaneously and trave   [#permalink] 21 Nov 2017, 15:59
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