Confusing DS question

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Confusing DS question

by rishianand7 » Tue Aug 13, 2013 1:19 pm
Allison always leaves her office at 5.00pm and reaches her hotel in an hour, while Brittany leaves the same hotel sometime after 5, meets Allison and reaches the same office at 6.00pm. On Wednesday,just when Brittany met Allison, she realized she had left her laptop. So she went back to the hotel, picked up her laptop and reached the office late. If Allison and Brittany always travel at constant rates, at what time did Brittany reach the office on Wednesday?

(1) Usually, when Allison and Brittany meet, Allison has already covered 2/3rd of the distance between the hotel and office.

(2) Brittany always leaves the hotel at exactly 5.30pm
Source: — Data Sufficiency |

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by GMATGuruNY » Tue Aug 13, 2013 4:29 pm
rishianand7 wrote:Allison always leaves her office at 5.00pm and reaches her hotel in an hour, while Brittany leaves the same hotel sometime after 5, meets Allison and reaches the same office at 6.00pm. On Wednesday,just when Brittany met Allison, she realized she had left her laptop. So she went back to the hotel, picked up her laptop and reached the office late. If Allison and Brittany always travel at constant rates, at what time did Brittany reach the office on Wednesday?

(1) Usually, when Allison and Brittany meet, Allison has already covered 2/3rd of the distance between the hotel and office.

(2) Brittany always leaves the hotel at exactly 5.30pm
Statement 1: Usually, when Allison and Brittany meet, Allison has already covered 2/3rd of the distance between the hotel and office.
What USUALLY happens tells us nothing about what happened on Wednesday.
INSUFFICIENT.

Statement 2: Brittany always leaves the hotel at exactly 5.30pm.
To see the situation more clearly, plug in a value for the distance between the hotel and the office.
Let the distance = 6 miles.

Since Allison takes 1 hour -- from 5 to 6pm -- to travel the entire distance, Alison's rate = d/t = 6/1 = 6 miles per hour.
Since Brittany takes 1/2 hour -- from 5:30 to 6pm -- to travel the entire distance, Brittany's rate = d/t = 6/(1/2) = 12 miles per hour.
When the two travel toward each other, Brittany's rate : Allison's rate = 12:6 = 2:1.
Implication:
Of every 2+1=3 miles that are traveled when Brittany and Allison travel toward each other, Brittany will travel 2 miles and Allison will travel 1 mile.

Thus:
From 5pm to 5:30pm, the distance traveled by Allison alone = r*t = 6*(1/2) = 3 miles.
At this point, 3 miles remain between Allison and Brittany.
When Brittany and Allison travel toward each other starting at 5:30pm, Brittany will travel 2 of the 3 miles between them, as noted above.
When Brittany returns for her laptop, she must travel 2 miles back to the hotel.
Then, to reach the office, she must travel the entire 6 miles.
Thus, the total distance traveled by Brittany = 2+2+6 = 10 miles.
Time for Brittany to travel 10 miles = d/r = 10/12 = 5/6 of an hour.
Since Brittany leaves at 5:30pm -- and she requires 5 minutes to retrieve the laptop -- the time that Brittany reaches the office = 5:30pm + 5/6 of an hour + 5 minutes = 6:25pm.
SUFFICIENT.

The correct answer is B.
Last edited by GMATGuruNY on Thu Nov 19, 2015 7:29 am, edited 2 times in total.
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by shankar245 » Sun Dec 15, 2013 10:49 am
Hi ,

Just a small correction , the 5 minutes required to get the laptop is missed.

thanks

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by shankar245 » Sun Dec 15, 2013 3:19 pm
Hi ,

Just a small correction , the 5 minutes required to get the laptop is missed.

thanks