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Fastest way to solve this speed/time/distance problem?

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by Kevinst » Thu Aug 16, 2012 2:01 pm
A left P for Q at 10:00 A.M and at the same time B left Q for P. Both A & B took the same route. They meet at 11:20 A.M. If both of them start from P, B will reach Q 2 hours earlier than A. What is the time taken by A to reach Q?
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by GMATGuruNY » Thu Aug 16, 2012 5:02 pm
Kevinst wrote:A left P for Q at 10:00 A.M and at the same time B left Q for P. Both A & B took the same route. They meet at 11:20 A.M. If both of them start from P, B will reach Q 2 hours earlier than A. What is the time taken by A to reach Q?
The time for A and B to meet = 4/3 hours.
Since B travels faster than A, in this amount of time B has covered more than 1/2 the distance from Q to P.
Thus, it will take B LESS than 4/3 more hours to reach P.
Thus, the time for B to cover the ENTIRE distance from Q to P must be BETWEEN 4/3 hours and 8/3 hours.
It seems VERY likely that B's time = 2 hours, implying that A's time = 4 hours.

Let the distance from P to Q = 8 miles.
Rate for B = d/t = 8/2 = 4 miles per hour.
Rate for A = d/t = 8/4 = 2 miles per hour.
When elements WORK TOGETHER to cover the distance between them, we ADD their rates.
Time for A and B to meet = d/(combined rate) = 8/(4+2) = 4/3 hours.
Success!

Thus, the time for A to travel from P to Q = 4 hours.
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by natali » Thu Aug 16, 2012 9:17 pm
GMATGuruNY wrote:
Kevinst wrote:A left P for Q at 10:00 A.M and at the same time B left Q for P. Both A & B took the same route. They meet at 11:20 A.M. If both of them start from P, B will reach Q 2 hours earlier than A. What is the time taken by A to reach Q?
The time for A and B to meet = 4/3 hours.
Since B travels faster than A, in this amount of time B has covered more than 1/2 the distance from Q to P.
Thus, it will take B LESS than 4/3 more hours to reach P.
Thus, the time for B to cover the ENTIRE distance from Q to P must be BETWEEN 4/3 hours and 8/3 hours.
It seems VERY likely that B's time = 2 hours, implying that A's time = 4 hours.

Let the distance from P to Q = 8 miles.
Rate for B = d/t = 8/2 = 4 miles per hour.
Rate for A = d/t = 8/4 = 2 miles per hour.
When elements WORK TOGETHER to cover the distance between them, we ADD their rates.
Time for A and B to meet = d/(combined rate) = 8/(4+2) = 4/3 hours.
Success!

Thus, the time for A to travel from P to Q = 4 hours.[/quote
mitch,
i still dont understand much.can you please just use tradiional way to solve this problem?
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by cypherskull » Fri Aug 17, 2012 2:30 am
Am not sure if I'd go on to solve this on actual GMAT. I'd rather take a guess and move than solve this problem the way I did. Here's my solution:

Time taken for A and B to meet = 4/3 hrs
Let s1 and s2 represent the speeds of A & B and t1 & t2 represent the times taken.

Therefore,

4/3 (s1 + s2) = d

or,
4(s1+s2) = 3d ------ (1)

Second part of the problem states that both starting P, B took 2 hrs less.

So,

d/s1 - d/s2 = 2

or

d = 2s1*s2/(s2-s1) ------- (2)

Putting (2) in (1), we'd get -

2(s2/s1) - 2(s1/s2) = 3

Solving the above equation, we get - s2 = 2s1.

Now since s2 = 2s1, surely t2 = t1/2 for the same distance.

Its given that time taken by B was 2 hrs less than the time taken by A.

So,

t1 - t2 = 2

or, t1 - t1/2 = 2 => t1 = 4 Answer.
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by GMATGuruNY » Fri Aug 17, 2012 2:35 am
natali wrote:
GMATGuruNY wrote:
Kevinst wrote:A left P for Q at 10:00 A.M and at the same time B left Q for P. Both A & B took the same route. They meet at 11:20 A.M. If both of them start from P, B will reach Q 2 hours earlier than A. What is the time taken by A to reach Q?
The time for A and B to meet = 4/3 hours.
Since B travels faster than A, in this amount of time B has covered more than 1/2 the distance from Q to P.
Thus, it will take B LESS than 4/3 more hours to reach P.
Thus, the time for B to cover the ENTIRE distance from Q to P must be BETWEEN 4/3 hours and 8/3 hours.
It seems VERY likely that B's time = 2 hours, implying that A's time = 4 hours.

Let the distance from P to Q = 8 miles.
Rate for B = d/t = 8/2 = 4 miles per hour.
Rate for A = d/t = 8/4 = 2 miles per hour.
When elements WORK TOGETHER to cover the distance between them, we ADD their rates.
Time for A and B to meet = d/(combined rate) = 8/(4+2) = 4/3 hours.
Success!

Thus, the time for A to travel from P to Q = 4 hours.[/quote
mitch,
i still dont understand much.can you please just use tradiional way to solve this problem?
When elements work together, ADD THEIR RATES.
A's rate + B's rate = their combined rate.

Since A and B leave at 10am and meet at 11:20am, they take 4/3 hours to cover the distance between them.
To make the algebra easier, we can think of the distance between them as ONE JOB, letting d=1.
Thus, their combined rate = d/t = 1/(4/3) = 3/4.

Let A's time = a.
Then A's rate = 1/a.

Since B's time is 2 hours less than A's time, B's time = a-2.
B's rate = 1/(a-2).

Since A's rate + B's rate = their combined rate, we get:
1/a + 1/(a-2) = 3/4
(a-2 + a)/(a²-2a) = 3/4
8a - 8 = 3a² - 6a
3a² -14a + 8 = 0
(3a - 2)(a - 4) = 0
a = 2/3 or a = 4.
a = 2/3 implies that B's time is negative -- not viable.

Thus, A's time = 4 hours.
Last edited by GMATGuruNY on Fri Aug 17, 2012 6:36 am, edited 1 time in total.
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by cypherskull » Fri Aug 17, 2012 2:40 am
Thanks Mitch! That seems like a much easier approach!
Regards,
Sunit

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by Kevinst » Fri Aug 17, 2012 6:16 am
Nice approach, Mitch!

To avoid confusion it should be noted:

"Thus, their combined rate = d/t = 1/(4/3) = 3/4."
Mitch divides 1 by (4/3) because the calculated rate shows the "number of distances" that can be traveled in a certain time, not how many miles or km.

This basically transforms the problem in a "Machines-Rate-problem":

Machines A and B work at different constant rates.
A and B together get one job done in 80 Minutes.
If A and B work separately, B finishes one job 2 hours earlier than A.
How long does it take A to finish one job on his own?


Looking at the problem from this perspective makes it a lot more feasible for me.
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