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Bob and Wendy planned to walk from their home to a

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Bob and Wendy planned to walk from their home to a [#permalink]

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New post 09 Jul 2009, 10:55
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Bob and Wendy planned to walk from their home to a restaurant for dinner together. However, Bob was delayed at work, and Wendy left for the restaurant before Bob did. If the restaurant is 3 miles from their home and Bob left for the restaurant a half-hour after Wendy did, how long did Wendy have to wait for Bob at the restaurant?

(1) Wendy walked at a constant pace of 4 miles per hour

(2) Bob walked at a constant pace of 1 mile per hour faster than Wendy.

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Re: Late for Dinner [#permalink]

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New post 09 Jul 2009, 13:15
mbaMission wrote:
Bob and Wendy planned to walk from their home to a restaurant for dinner together. However, Bob was delayed at work, and Wendy left for the restaurant before Bob did. If the restaurant is 3 miles from their home and Bob left for the restaurant a half-hour after Wendy did, how long did Wendy have to wait for Bob at the restaurant?

(1) Wendy walked at a constant pace of 4 miles per hour

(2) Bob walked at a constant pace of 1 mile per hour faster than Wendy.

MGMAT/CAT1/DS


(1) from this information we can get wendy's rate, but this is not enough to solve the problem, we don't know Bob's rate.
(2) let's X be Wendy's rate and X+1 Bob's. the distance is the same. We know that Wendy left for a half hour earlier, so 0.5 + 3/x = 3/x+1
By knowing rates of both Wendy and Bob we can determine the time Wendy had to wait for Bob.
So (2) alone is sufficient to solve the problem.
P.s. I am almost sleeping now and hope didn;t write something very stupid above
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Re: Late for Dinner [#permalink]

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New post 09 Jul 2009, 13:58
I don't know if I am missing something but don't we not know how fast Wendy walked without (1)? We can write an equation for how long Wendy waited but without knowing how fast she walked it would be impossible to solve without both sets and therefore C?

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Re: Late for Dinner [#permalink]

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New post 13 Jul 2009, 11:41
According to me, the answer should be C, we need to know the rate at which atleast one person walked.

Guys please clarify the answer..

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Re: Late for Dinner [#permalink]

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New post 13 Jul 2009, 13:12
I go with C as well. Wendy's speed is required to find the time. Wendy can walk @ 2 or 5 miles/hr.

Wait time is dependent on speed, IMHO.

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Re: Late for Dinner [#permalink]

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New post 13 Jul 2009, 19:57
mbaMission wrote:
Bob and Wendy planned to walk from their home to a restaurant for dinner together. However, Bob was delayed at work, and Wendy left for the restaurant before Bob did. If the restaurant is 3 miles from their home and Bob left for the restaurant a half-hour after Wendy did, how long did Wendy have to wait for Bob at the restaurant?

(1) Wendy walked at a constant pace of 4 miles per hour

(2) Bob walked at a constant pace of 1 mile per hour faster than Wendy.

MGMAT/CAT1/DS

Answer has to be C.
only when we combine both the statements together we will come to know the time when each will reach the hotel....
hence C plz share OA/OE

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Re: Late for Dinner [#permalink]

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New post 15 Jul 2009, 05:34
For this problem we can set up an RTD (rate x time = distance) chart for Bob and Wendy. The difficulty is that Bob and Wendy will be traveling the same distance — it is the amount of time that each will be traveling is unknown.

One way that we can get around this problem is by assigning two different variables for time: one for the amount of time that they are both walking (let's say t), and a second for the amount of time that only Bob is walking (in other words, for the amount of time that Wendy is waiting.) Let's call that variable x. The question, rephrased, is "What is x?" In addition, we may want to "link" Bob and Wendy's rates to see whether the difference in the rates is sufficient. To do that, instead of assigning variables for Bob and Wendy's rates separately, we should assign a variable for Wendy's rate (rw) and a second variable for the difference between Bob and Wendy's rates (y). Bob's rate is then rw + y:

Bob Wendy
R rw + y rw
T t + x t + 1/2
D 3 3

Since rate x time = distance, we have:

(rw + y)(t + x) = 3
rw(t+1/2) = 3

Since they are traveling the same distance, we can set up the equations equal to each other and solve for x.

(rw + y)(t + x) = rw(t+1/2)
rwt + yt + xy + rwx = rwt + 1/2rw
yt + xy + rwx = 1/2rw
x(y + rw) = 1/2rw – yt
x = (1/2rw – t)/(rw + y)

Indeed the difference in rates is not sufficient. Therefore the question can be expressed/rephrased as follows:

What are Wendy's rate and the difference between Bob and Wendy's rate? (Or alternately, what are Bob and Wendy's individual rates?) Note that we can calculate t whenever we know Wendy's rate.

(1) INSUFFICIENT: This statement does not tell us anything about Bob's speed.

(2) INSUFFICIENT: As we saw earlier, knowing the difference between the rates is insufficient. We need to know both of Bob's and Wendy's rates.

(1) and (2) SUFFICIENT: Based on the formula above, x = [1/2(4) – 1·t]/[4 + 1] = (2 – t)/5

But since rw(t+1/2) = 3 and rw = 4, we know that t = 1/4. Therefore x = (2 – 1/4)/5 = 7/20 hours, or 21 minutes.

The correct answer is C.

===========
Just chk once ... i have just copied and pasted it here.
===========
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Do not answer without sharing the reasoning behind ur choice
-----------------------------------------------------------
Working on my weakness : GMAT Verbal
------------------------------------------------------------
Ask:
Why, What, How, When, Where, Who
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Kudos [?]: 156 [0], given: 2

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Re: Late for Dinner [#permalink]

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New post 15 Jul 2009, 07:14
Natia wrote:

(1) from this information we can get wendy's rate, but this is not enough to solve the problem, we don't know Bob's rate.
(2) let's X be Wendy's rate and X+1 Bob's. the distance is the same. We know that Wendy left for a half hour earlier, so 0.5 + 3/x = 3/x+1
By knowing rates of both Wendy and Bob we can determine the time Wendy had to wait for Bob.
So (2) alone is sufficient to solve the problem.
P.s. I am almost sleeping now and hope didn;t write something very stupid above


In the "red" colored portion, you are assuming that they both reached together. Only then that equation holds true.

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Re: Late for Dinner [#permalink]

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New post 15 Jul 2009, 07:22
The ans should be C

The waiting time will decrease as the numerical values for their speeds increase.

Say Wendy walks @2, and Bob @ 3, then they reach together.
But if wendy walks @3, and bob @4, then bob reaches 15 minutes after wendy.


What Im trying to say is, if both walk slower, waiting time will be lesser, and this time will go on increasing as their respective speeds are increased. At different speeds, we get different waiting times.

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Re: Late for Dinner   [#permalink] 15 Jul 2009, 07:22
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