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If car X followed car Y across a certain bridge that is 21m https://gmatclub.com/forum/ifcarxfollowedcaryacrossacertainbridgethatis21m102552.html 
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Author:  niheil [ 09 Oct 2010, 12:13 ] 
Post subject:  If car X followed car Y across a certain bridge that is 21m 
If car X followed car Y across a certain bridge that is 21mile long, how many seconds did it take car X to travel across the bridge? (1) Car X drove onto the bridge exactly 3 seconds after car Y drove onto the bridge and drove off the bridge exactly 2 seconds after car Y drove off the bridge. (2) Car Y traveled across the bridge at a constant speed of 30 miles per hour. Additional info on the problem Source: Paper Test Test Code: 42 Section: 2 (Data Sufficiency) Problem: 5 
Author:  Bunuel [ 09 Oct 2010, 12:43 ] 
Post subject:  Re: Plz Help with Data Sufficiency problem 
niheil wrote: Hi guys, Please explain how to answer the following Data Sufficiency question: If car X followed car Y across a certain bridge that is 21mile long, how many seconds did it take car X to travel across the bridge? (1) Car X drove onto the bridge exactly 3 seconds after car Y drove onto the bridge and drove off the bridge exactly 2 seconds after car Y drove off the bridge. (2) Car Y traveled across the bridge at a constant speed of 30 miles per hour. Let the time needed for car X to travel across the bridge be \(t_x\) seconds and the time for Y \(t_y\) seconds. Question: \(t_x=?\) (1) Car X drove onto the bridge exactly 3 seconds after car Y drove onto the bridge and drove off the bridge exactly 2 seconds after car Y drove off the bridge > car X needs 1 second less to travel across the bridge than car Y > \(t_y=t_x+1\). Not sufficient to calculate \(t_x\). (2) Car Y traveled across the bridge at a constant speed of 30 miles per hour = \(\frac{30}{3600}=\frac{1}{120}\) miles per second > car Y needs \(t_y=\frac{21}{\frac{1}{120}}=21*120\) seconds to travel across the bridge. Not sufficient to calculate \(t_x\). (1)+(2) \(t_y=t_x+1\) and \(t_y=21*120\) > \(t_x=21*1201\). Sufficient. Answer: C. 
Author:  niheil [ 09 Oct 2010, 14:00 ] 
Post subject:  Re: Plz Help with Data Sufficiency problem 
Awesome! Thanks again, Bunuel. I wish you could take the GMAT for me, lol. 
Author:  PennState08 [ 02 Dec 2010, 14:09 ] 
Post subject:  Rate Problem 
All, This was a data sufficiency problem, but want to double check my math in case I see it in a problem solver. If Car X followed Car Y across a certain bridge that is \(\frac{1}{2}\) mile long, how many seconds did it take Car X to travel across the bridge? (1) Car X drove onto the bridge exactly 3 seconds after Car Y drove onto the bridge and drove off the bridge exactly 2 seconds after Car Y drove off the bridge. (2) Car Y traveled across the bridge at a constant speed of 30 miles per hour. C is the correct answer for data sufficiency, but I want to go through the problem in various ways (problem solving) to make sure my math is correct. Provided all the data; What are the rates of each Car? How many seconds would it take for Car X to catch Car Y? At what distance would Car X catch Car Y? Rate Car X: ~30.5 mph OR \(\frac{1}{118}\) miles per second Rate Car Y: 30 mph (given) OR \(\frac{1}{120}\) miles per second Second for Car X to catch Car Y: 180 seconds OR 3 minutes OR \(\frac{1}{20}\) hour Distance: 1.5 miles Explanations: Rate x Time(t) = Distance Car Y (given at 30 mph, so find \(t\) to solve for rate of car X) \(y\) x \(t\) = \(\frac{1}{2}\) 30 x \(t\) = \(\frac{1}{2}\) (need miles per second, not hour) \(\frac{1}{120}\) x \(t\) = \(\frac{1}{2}\) \(t\) = 60 Car X \(x\) x (\(t\)  1) = \(\frac{1}{2}\) ; 1 second for the time difference (waited 3 seconds after Y, finished 2 seconds after Y: 3  2 = 1) \(x\) x (60  1) = \(\frac{1}{2}\) \(x\) x 59 = \(\frac{1}{2}\) \(x\) = \(\frac{1}{118}\) Time (in seconds) Car X catches Car Y: \(\frac{1}{118}\) (\(t\)  3) = \(\frac{1}{120}t\) ; 3 = amount of seconds after Y left. \(\frac{1}{118} t\)  \(\frac{3}{118}\) = \(\frac{1}{120}t\) \(\frac{60}{7080}t\)  \(\frac{180}{7080}\) = \(\frac{59}{7080}\) \(\frac{1}{7080} t\) = \(\frac{180}{7080}\) \(t\) = 180 seconds 180 seconds for Car X to catch Car Y Distance (in miles) for Car X to catch Car Y Taking either equation and substituting 180 for \(t\) X) \(\frac{1}{118}\) x (1803) = 1.5 miles Y) \(\frac{1}{120}\) x (180) = 1.5 miles I have all this set up correctly, right? 
Author:  Bunuel [ 02 Dec 2010, 14:20 ] 
Post subject:  Re: Rate Problem 
Merging similar topics. The only difference is in the length of the bridge (21 miles in first question and 1/2 miles in the second one). But the answer for both of them is C. Please ask if anything remains unclear. 
Author:  Basshead [ 13 Oct 2020, 09:37 ] 
Post subject:  Re: If car X followed car Y across a certain bridge that is 21m 
(1) This tells us Car X crosses the bridge 1 second faster than Car Y. We can't determine the time it took for Car X to cross the bridge. (2) Car Y traveled across the bridge at a constant speed of 30 miles per hour. We can determine the time it takes Car Y to cross the bridge; however, this does not tell us anything about Car X (1 & 2) From Statement 2 we can determine the time it takes Car Y to cross the bridge. We know Car X crosses the bridge 1 second faster than Car Y. Therefore, with both statements, we can determine the time it takes Car X to cross the bridge. 
Author:  Hoozan [ 25 May 2021, 00:52 ] 
Post subject:  Re: If car X followed car Y across a certain bridge that is 21m 
EducationAisle based on (1) doesn't car X take 5 seconds longer that car Y? Why is this not correct? 
Author:  EducationAisle [ 25 May 2021, 02:52 ] 
Post subject:  Re: If car X followed car Y across a certain bridge that is 21m 
Hoozan wrote: EducationAisle based on (1) doesn't car X take 5 seconds longer that car Y? Why is this not correct? Not really 5 seconds Hoozan. Car X a) drove onto the bridge exactly 3 seconds after car Y drove onto the bridge and b) drove off the bridge exactly 2 seconds after car Y drove off the bridge. Notice that if X and Y had same speeds, then X would have driven off the bridge exactly 3 seconds after car Y drove off the bridge (because X drove onto the bridge exactly 3 seconds after car Y drove onto the bridge). However, since X drove drove off the bridge exactly 2 seconds after car Y drove off the bridge, this would mean that X "caught up" with Y, 1 second, on the bridge. So, car X actually took 1 second less than car Y, to travel the bridge. 
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Post subject:  Re: If car X followed car Y across a certain bridge that is 21m 
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