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Allen starts from X, goes to Y. At the same time Bob starts from Y and

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Allen starts from X, goes to Y. At the same time Bob starts from Y and [#permalink]

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New post 25 Aug 2015, 21:32
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Allen starts from X, goes to Y. At the same time Bob starts from Y and goes towards X. Once Allen reaches Y he changes his direction and returns to X. Once Bob reaches X, he changes his direction and returns to Y. Throughout Allen travels at 54 kmph and Bob travels at 72kmph. By the time they meet for the second time, Bob covers 36 km more than Allen. Find the distance between X and Y.

a)144km
b)72 km
c)126km
d)84 km
e)42km
[Reveal] Spoiler: OA

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Re: Allen starts from X, goes to Y. At the same time Bob starts from Y and [#permalink]

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New post 27 Aug 2015, 18:57
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IMO the answer must be C 126

as I tried to back solve this one C perfectly fit the statement above.

total distance 126km
in an hour both Allen and Bob covered 126km
that is 54+72=126

they meet for the first time Bob covered 18 more KM than Allen. 72-54=18.
so when they meet for the second time Bob covered 18*2= 36 more km
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Re: Allen starts from X, goes to Y. At the same time Bob starts from Y and [#permalink]

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New post 27 Aug 2015, 19:29
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This is a good question. Definitely more towards the 700 range than the 600 range of the spectrum.

Rate * Time = Distance

Rate Allen =54
Rate Bob =72

They will be working together to cover the total distance from X to Y, so their combined rate, 54+72=126, will be used.

The key is figuring out how times Allen and Bob will collectively traverse the distance from X to Y.

It helps to conceptualize two people moving along this same pattern without factoring in their respective rates just yet.


X----------N----------Y

If the XY distance is drawn as above, the total distance can be represented as (X+N)+(N+Y). If both Allen and Bob meet at point N each time, during the first meeting one will have traveled X+N and the other will have traveled N+Y, for a total distance of (X+N) + (N+Y), or 1x the total distance.

As they pass the other, one will travel N+Y for the first time as the other travels X+N for the first time. Thus, they collectively make another combined trip of the full distance X to Y. They each turn around and cover the same ground they just traveled, for a third combined trip of the total distance. Thus, combined they total distance of 3x the distance from X to Y.

Set up a standard R*T=D equation with their combined rate and let the distance =3D to figure out the amount of time it takes for the guys to meet twice. Set everything in terms of D as we're trying to find out distance.

126* (3D/126)=3D. [T=D/R]

126*(D/42)=3D; Thus, the time it takes to meet twice can be represented as D/42.

Now use this representation of time and set the equations equal to each other with info given in the question.

72D/42=(54D/42)+36

18D/42=36

D=36(42)/18

D=2*42=84

Answer D
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Allen starts from X, goes to Y. At the same time Bob starts from Y and [#permalink]

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New post 27 Aug 2015, 23:02
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Well Actually OA = D

Here is my explanation
Let d be the distance between X and Y

A--> ---------------------- <---B
X---------------------------------Y

A traveled = 54
B traveled = 72

When they meet for 2nd time B = A + 36
B - A = 36 km

When they meet for the 2nd time then the total distance traveled by A & B combined would be = 3d --(i)

So if we consider total distance traveled by A & B = 54+72 = 126
126 =3d
d = 42

But the difference between B and A = 72 - 54 = 18
Given in the question is that the diff is 36 when they meet for the 2nd time.
In other words we can say double the one we calculated.

Thus Distance between X and Y = 42*2 = 84 kms

Hope you guys like my approach :)
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Re: Allen starts from X, goes to Y. At the same time Bob starts from Y and [#permalink]

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New post 27 Aug 2015, 23:18
Thanks for the explanation.

I got the point where I was wrong.
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Allen starts from X, goes to Y. At the same time Bob starts from Y and [#permalink]

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1. At their first meeting, based on their respective speeds, the ratio of B's distance to A's distance is 4:3.
2. If there are 36 kms difference between their distances at the second meeting, there would have been
12 kms difference at the first, as the second meeting required twice the combined distance of the first.
3. 4x-3x=12➡x=12➡(7)(12)=84 kilometers.

Last edited by gracie on 03 Oct 2015, 15:09, edited 2 times in total.
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Re: Allen starts from X, goes to Y. At the same time Bob starts from Y and [#permalink]

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The rate of difference between two is 18
So, 36 KM, is difference in two hours.
The distance covered in 2 hours is 2*72=144
The distance covered in 2 hours is 2*54=108
The total distance at the time of meeting second time is 3 times total distance i.e 3D
3D=144+108
3D=252
D=84
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Re: Allen starts from X, goes to Y. At the same time Bob starts from Y and [#permalink]

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New post 13 Sep 2015, 17:20
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meneny wrote:
The rate of difference between two is 18
So, 36 KM, is difference in two hours.
The distance covered in 2 hours is 2*72=144
The distance covered in 2 hours is 2*54=108
The total distance at the time of meeting second time is 3 times total distance i.e 3D
3D=144+108
3D=252
D=84


How is it three times total distance (3d)?
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Re: Allen starts from X, goes to Y. At the same time Bob starts from Y and [#permalink]

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dina98 wrote:
meneny wrote:
The rate of difference between two is 18
So, 36 KM, is difference in two hours.
The distance covered in 2 hours is 2*72=144
The distance covered in 2 hours is 2*54=108
The total distance at the time of meeting second time is 3 times total distance i.e 3D
3D=144+108
3D=252
D=84


How is it three times total distance (3d)?



Say distance between X and Y is D. When do they meet for the second time? When do they meet for the first time? When Allen is going to Y and Bob is going to X. They meet for the first time, somewhere in between X and Y. After that, Allen reaches Y and Bob reaches X. So both travel for distance D each.
Then, they meet second time when Allen is coming back from Y toward X and Bob is coming back from X toward Y. When they meet somewhere in the middle of X and Y again, they have together covered another D distance. So total they have covered distance 3D.

Originally:
X (Allen)________________________________________________________Y(Bob)
Distance covered together = 0

First Meet:
X ______________________(Allen)(Bob)__________________________________Y
Distance covered together = D

Reaching destination:
X (Bob)________________________________________________________Y(Allen)
Distance covered together = 2D

Second Meet:
X ______________________(Bob)(Allen)__________________________________Y
Distance covered together = 3D
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Re: Allen starts from X, goes to Y. At the same time Bob starts from Y and [#permalink]

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New post 16 Sep 2015, 02:06
gracie wrote:
1. At their first meeting the ratio of A's distance traveled to B's distance traveled is 3:4.
2. If there are 36 kilometers difference between their distances at the second meeting,
there would have been 12 kilometers difference at the first, as the second meeting
entailed twice the total distance of the first.
3. 7 x 12 = 84



Hi,
My approach is slightly different than that of Gracie's;

1) The ratio between their speeds is A:B = 54:72 = 3:4
2) Given that the time is the "constant" here, the total distance travelled by both of them will have the same ratio as their speed's: 3:4
3) Lets call the total distance travelled by A, "d", and B, "d+36". ==> so, d/(d+36) = 3/4 ==> d=108
4) The total distance travelled by both of them is d + (d+36) = 108 + 108 + 36 = 252. As already noted, this equals three times the distance between XY
==> 252 / 3 = 84
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Re: Allen starts from X, goes to Y. At the same time Bob starts from Y and [#permalink]

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New post 17 Sep 2015, 06:18
Let 't' be the time taken by each of them.

Given, their rates are 72 and 54... and when they meet for the second time distance covered by B is 36 km more than by A.
Hence, 72t-54t = 36
or, t=2 hours

Also, total distance covered = 3*d
Distance covered by A is 54t and covered by B is 72t.
Hence, 72t + 54t = 126t = 126*2 = 3*d
d=84
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Re: Allen starts from X, goes to Y. At the same time Bob starts from Y and [#permalink]

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New post 02 Feb 2018, 10:36
every hour, Bob covers 18 km more than Allen => 2 of them must be walk for 2 hour, and the total distance is 3 times as the distance between X and Y => D is the answer.
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Re: Allen starts from X, goes to Y. At the same time Bob starts from Y and [#permalink]

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New post 03 Feb 2018, 01:25
zurvy wrote:
gracie wrote:
1. At their first meeting the ratio of A's distance traveled to B's distance traveled is 3:4.
2. If there are 36 kilometers difference between their distances at the second meeting,
there would have been 12 kilometers difference at the first, as the second meeting
entailed twice the total distance of the first.
3. 7 x 12 = 84



Hi,
My approach is slightly different than that of Gracie's;

1) The ratio between their speeds is A:B = 54:72 = 3:4
2) Given that the time is the "constant" here, the total distance travelled by both of them will have the same ratio as their speed's: 3:4
3) Lets call the total distance travelled by A, "d", and B, "d+36". ==> so, d/(d+36) = 3/4 ==> d=108
4) The total distance travelled by both of them is d + (d+36) = 108 + 108 + 36 = 252. As already noted, this equals three times the distance between XY
==> 252 / 3 = 84


Nice try but i think gracie's approach is faster, i tend to take time with ratios, i myself used equations and took 6 mins :(
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Re: Allen starts from X, goes to Y. At the same time Bob starts from Y and [#permalink]

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New post 03 Feb 2018, 03:38
VenoMfTw wrote:
Allen starts from X, goes to Y. At the same time Bob starts from Y and goes towards X. Once Allen reaches Y he changes his direction and returns to X. Once Bob reaches X, he changes his direction and returns to Y. Throughout Allen travels at 54 kmph and Bob travels at 72kmph. By the time they meet for the second time, Bob covers 36 km more than Allen. Find the distance between X and Y.

a)144km
b)72 km
c)126km
d)84 km
e)42km


\(D_b = D_a + 36\)

\(\frac{S_b}{t} = \frac{S_a}{t} + 36\)

\(\frac{72}{t} = \frac{54}{t} + 36\)

\(\frac{72}{t} - \frac{54}{t} = 36\)

\(\frac{(72t - 54t)}{t^{2}} = 36\)

\(18=36t\)

\(t=0.5\)

Total distance travelled by Bob and Allen = \(D_{ab} = \frac{S_{ab}}{t}\)
\(S_{ab}\) is the relative speed of Bob and Allen = 72 + 54 = 126 km/h
\(D_{ab}=\frac{126}{0.5}\)
\(D_{ab}=252\)

Bob and Allen together traversed the path XY three times until they met the second time.
So distance between X and Y = \(\frac{252}{3}=84km\)
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Re: Allen starts from X, goes to Y. At the same time Bob starts from Y and [#permalink]

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New post 08 Feb 2018, 13:27
I have a weird way of thinking about these problems that I've developed since Highschool:
The thought process basically goes like this:
Let B be the speed of Bob and A be the speed of Allen, let B be the distance between X and Y
B:A = 72: 54 = 4:3
Therefore the distance that Bob travels will always be 4/3 times the distance Allen travels:
By the time Allen meets Bob for the 2nd time, they have travelled a combined distance of 3D
therefore 3D/(7)*4 is the distance that Bob travelled and 3D/7*3 is the distance Allen travelled
The difference between their distances is: 3D/7 = 36, D = 84
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Re: Allen starts from X, goes to Y. At the same time Bob starts from Y and [#permalink]

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New post 12 Feb 2018, 16:38
VenoMfTw wrote:
Allen starts from X, goes to Y. At the same time Bob starts from Y and goes towards X. Once Allen reaches Y he changes his direction and returns to X. Once Bob reaches X, he changes his direction and returns to Y. Throughout Allen travels at 54 kmph and Bob travels at 72kmph. By the time they meet for the second time, Bob covers 36 km more than Allen. Find the distance between X and Y.

a)144km
b)72 km
c)126km
d)84 km
e)42km


We can let the distance between X and Y = d. In the total trip, we see that Allen covers a distance of d plus some more miles. Similarly, Bob also covers a distance of d, plus some more miles. Moreover, when they meet the second time, the sum of the surplus miles covered by Allen and Bob must also equal d; therefore the total distance that they cover is 3d.
We see that Bob travels 18 kmph faster than Allen, and, in order for Bob to travel 36 km more than Allen, they each must have traveled for 2 hours.

We also see that in 2 hours, Allen travels 54 x 2 = 108 km, and Bob travels 72 x 2 = 144 hm. We have argued that their total distance traveled must be three times the distance between X and Y. So we have:

108 + 144 = 3d

252 = 3d

84 = d

Answer: D
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Allen starts from X, goes to Y. At the same time Bob starts from Y and [#permalink]

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New post 12 Feb 2018, 21:55
VenoMfTw wrote:
Allen starts from X, goes to Y. At the same time Bob starts from Y and goes towards X. Once Allen reaches Y he changes his direction and returns to X. Once Bob reaches X, he changes his direction and returns to Y. Throughout Allen travels at 54 kmph and Bob travels at 72kmph. By the time they meet for the second time, Bob covers 36 km more than Allen. Find the distance between X and Y.

a)144km
b)72 km
c)126km
d)84 km
e)42km


rate of B to A=72/54=4/3
if 7/7=combined distance traveled by A and B,
then 4/7-3/7=1/7=difference between distances traveled by A and B respectively
if d=distance between X and Y,
then 3d=combined distance of A and B at their second meeting
36 km=1/7*3d
d=84 km
Allen starts from X, goes to Y. At the same time Bob starts from Y and   [#permalink] 12 Feb 2018, 21:55
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