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A left P for Q at 10:00 am. At the same time B left Q for P.

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A left P for Q at 10:00 am. At the same time B left Q for P. [#permalink]

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New post 27 Aug 2015, 18:17
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A left P for Q at 10:00 am. At the same time B left Q for P. After they met at a point on the way, A took 24 mins to reach Q and B took 54 mins to reach P. At what time did they meet?

A. 10:35 am
B. 10:36 am
C. 10:37 am
D. 10:38 am
E. 10:39 am
[Reveal] Spoiler: OA

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Re: A left P for Q at 10:00 am. At the same time B left Q for P. [#permalink]

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arhumsid wrote:
A left P for Q at 10:00 am. At the same time B left Q for P. After they met at a point on the way, A took 24 mins to reach Q and B took 54 mins to reach P. At what time did they meet?

A. 10:35 am
B. 10:36 am
C. 10:37 am
D. 10:38 am
E. 10:39 am


Let S be the total distance between P and Q

Let Va and Vb be the speeds of A and B respectively.

Let t be the meeting time.

Per the question, Va*t+Vb*t = s ...(1)

Also, Vb*t+Vb*54 = s ---> Vb = s/(t+54) ....(2)

Similarly, Va = s/(t+24) ....(3)

Substituting values of Va and Vb from 2 and 3 respectively into 1 we get,

t = 36 minutes.

Thus they met at 10:00 AM + 36 minutes = 10:36 AM.

B is the correct answer.
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Re: A left P for Q at 10:00 am. At the same time B left Q for P. [#permalink]

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New post 27 Aug 2015, 18:53
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hi here we can use , in Time and distance if both trains start at the same time
and travelled in opposite directions.
when you want to find time of both trains to meet,
we can multiply both the times and take square root of it.

Here one train reached at 54mins and another trains reached at 24 mins.

Multiply both times 54*24=1296.
square root of 1296 is 36mins.

from 1296 we dont need square root also because unit digit ends in 6 . only the number unit ends in 6 only have that square.

We can use and apply logic to save time in GMAT.

So option B is correct.
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Status: One Last Shot !!!
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Re: A left P for Q at 10:00 am. At the same time B left Q for P. [#permalink]

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New post 27 Aug 2015, 20:53
Psiva00734 wrote:
hi here we can use , in Time and distance if both trains start at the same time
and travelled in opposite directions.
when you want to find time of both trains to meet,
we can multiply both the times and take square root of it.


By Time you mean, The time they took after meeting eachother? Does this hold true always?
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A left P for Q at 10:00 am. At the same time B left Q for P. [#permalink]

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New post 28 Aug 2015, 15:22
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Because they derive from the same distance, the ratio between A's pre-meeting time (t)
and total time (t+24) is equal to the ratio between B's post-meeting time (54) and total
time (t+54).
t/t+24=54/t+54
t^2=1296
t=36
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Re: A left P for Q at 10:00 am. At the same time B left Q for P. [#permalink]

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New post 06 Feb 2016, 09:43
I don't get it...
anyone can explain, maybe with graphs how to solve it?
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Re: A left P for Q at 10:00 am. At the same time B left Q for P. [#permalink]

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New post 07 Feb 2016, 22:39
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mvictor wrote:
I don't get it...
anyone can explain, maybe with graphs how to solve it?


Do you need an explanation for Gracie's solution?

If you need an explanation for Gracie's solution, since I wanted to know how her solution ticked, I quickly analyzed it from the ground up:

The first/long section is x, and the second/narrow section is y. The point about 2/3 through is where A and B met. A spent 24 minutes in section y, and B spent 54 minutes in section x.

P |-----------------------|------------| Q

1. Let's state the central idea Gracie's solution is built around: the distance A has to travel is the same distance B has to travel. Please keep this in the back of your mind for the next few steps.

2. For A, we know the time he spent in section y but not section x. As such, we can say that A spent \(\frac{time_x}{{time_x + 24}}\)% of his travel time covering section x, and \(\frac{24}{{time_x + 24}}\)% of his travel time covering section y.

3. For B, we know the time he spent in section x but not section y. As such, we can say that B spent \(\frac{54}{{time_y + 54}}\)% of his travel time covering section x, and \(\frac{time_y}{{time_y + 54}}\)% of his travel time covering section y.

4. Since A and B are traveling at a constant speed (this is an assumption- be extremely careful with assumptions), we know that both will share the same percent of their time in each section. As such, if we select a section, the amount of time each party spent in that section will be equal. Let's use section x going forward.

5. From (4), we have \(\frac{t}{{t + 24}} = \frac{54}{{t + 54}}\)

6. \(t * (t + 54) = 54 * (t + 24)\)

7. \(t^2 + 54t = 54t + 54*24\)

8. \(t^2 = 1296\)

9. \(t = 36\)

10. Since A spent t minutes traveling before meeting B, and t = 36, our answer must be 10:36, or B.

If anyone sees any errors in my reasoning, please let me know.
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Re: A left P for Q at 10:00 am. At the same time B left Q for P. [#permalink]

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New post 07 Feb 2016, 23:09
Engr2012 wrote:
arhumsid wrote:
A left P for Q at 10:00 am. At the same time B left Q for P. After they met at a point on the way, A took 24 mins to reach Q and B took 54 mins to reach P. At what time did they meet?

A. 10:35 am
B. 10:36 am
C. 10:37 am
D. 10:38 am
E. 10:39 am


Let S be the total distance between P and Q

Let Va and Vb be the speeds of A and B respectively.

Let t be the meeting time.

Per the question, Va*t+Vb*t = s ...(1)

Also, Vb*t+Vb*54 = s ---> Vb = s/(t+54) ....(2)

Similarly, Va = s/(t+24) ....(3)

Substituting values of Va and Vb from 2 and 3 respectively into 1 we get,

t = 36 minutes.

Thus they met at 10:00 AM + 36 minutes = 10:36 AM.

B is the correct answer.



the easiest way of doing this question :

meeting time T= sqrt of 24x36 =36
so, they will meet at 10.36 min form start.
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A left P for Q at 10:00 am. At the same time B left Q for P. [#permalink]

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New post 08 Feb 2016, 08:32
Beixi88 wrote:
mvictor wrote:
I don't get it...
anyone can explain, maybe with graphs how to solve it?


Do you need an explanation for Gracie's solution?

If you need an explanation for Gracie's solution, since I wanted to know how her solution ticked, I quickly analyzed it from the ground up:

The first/long section is x, and the second/narrow section is y. The point about 2/3 through is where A and B met. A spent 24 minutes in section y, and B spent 54 minutes in section x.

P |-----------------------|------------| Q

1. Let's state the central idea Gracie's solution is built around: the distance A has to travel is the same distance B has to travel. Please keep this in the back of your mind for the next few steps.

2. For A, we know the time he spent in section y but not section x. As such, we can say that A spent \(\frac{time_x}{{time_x + 24}}\)% of his travel time covering section x, and \(\frac{24}{{time_x + 24}}\)% of his travel time covering section y.

3. For B, we know the time he spent in section x but not section y. As such, we can say that B spent \(\frac{54}{{time_y + 54}}\)% of his travel time covering section x, and \(\frac{time_y}{{time_y + 54}}\)% of his travel time covering section y.

4. Since A and B are traveling at a constant speed (this is an assumption- be extremely careful with assumptions), we know that both will share the same percent of their time in each section. As such, if we select a section, the amount of time each party spent in that section will be equal. Let's use section x going forward.

5. From (4), we have \(\frac{t}{{t + 24}} = \frac{54}{{t + 54}}\)

6. \(t * (t + 54) = 54 * (t + 24)\)

7. \(t^2 + 54t = 54t + 54*24\)

8. \(t^2 = 1296\)

9. \(t = 36\)

10. Since A spent t minutes traveling before meeting B, and t = 36, our answer must be 10:36, or B.

If anyone sees any errors in my reasoning, please let me know.



Thanks, your explanation does clear things up. I did not understand the reasoning behind the equation written by Gracie.
I understood that:
1. A was travelling faster than B
2. point of intersection is closer to Q

yet, I did not see how to solve further this point.
Thanks one more time..would definitely remember this case. Kudos for you :)
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Re: A left P for Q at 10:00 am. At the same time B left Q for P. [#permalink]

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Re: A left P for Q at 10:00 am. At the same time B left Q for P. [#permalink]

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New post 12 Aug 2017, 21:40
Psiva00734 wrote:
hi here we can use , in Time and distance if both trains start at the same time
and travelled in opposite directions.
when you want to find time of both trains to meet,
we can multiply both the times and take square root of it.

Here one train reached at 54mins and another trains reached at 24 mins.

Multiply both times 54*24=1296.
square root of 1296 is 36mins.

from 1296 we dont need square root also because unit digit ends in 6 . only the number unit ends in 6 only have that square.

We can use and apply logic to save time in GMAT.

So option B is correct.


Would be great if someone can explain how this method works
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Re: A left P for Q at 10:00 am. At the same time B left Q for P. [#permalink]

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New post 12 Aug 2017, 23:36
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theperfectgentleman wrote:
Psiva00734 wrote:
hi here we can use , in Time and distance if both trains start at the same time
and travelled in opposite directions.
when you want to find time of both trains to meet,
we can multiply both the times and take square root of it.

Here one train reached at 54mins and another trains reached at 24 mins.

Multiply both times 54*24=1296.
square root of 1296 is 36mins.

from 1296 we dont need square root also because unit digit ends in 6 . only the number unit ends in 6 only have that square.

We can use and apply logic to save time in GMAT.

So option B is correct.


Would be great if someone can explain how this method works


lets take D traveled by A ....and D1 traveled by B....Time traveled is " T "
D/T = S(a) =D = T*S(a)-----------1
D1/T = S(b) =D1 = T*S(b)----------2

now A need 24 min to travel what B has already covered ....(as they meet in a common point)
So D1/24 = S(a)...
substituting 2 here ...
T*S(b)/S(a) = 24 ------------3

similarly for B ...
D/54 = S(b)
substituting 1 here
T*S(a)/S(b) = 54 ------------4

if we multiply 3 and 4 we will get ..
T^2 = 24 * 54
T = 36
Re: A left P for Q at 10:00 am. At the same time B left Q for P.   [#permalink] 12 Aug 2017, 23:36
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