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The safety distance \(D\) (meters) between cars moving on a motorway depends on their speed \(V\) (meters per second): \(D = \frac{V^2}{100} + 9\) If two cars want to maintain a time interval of 1 second between each other, which of the following speeds will breach the safety regulation? (A) 5 m/s (B) 10 m/s (C) 25 m/s (D) 50 m/s (E) 90 m/s Source: GMAT Club Tests  hardest GMAT questions Can someone please explain this question? I tried to do it plugging in the answer choices (I was running behind time and the numbers looked easy enough)... I figured out D should be less than V per second. a) V = 5m/s D = (V^2)/100 + 9 = (5^2)/100 + 9 = 25/100 + 9 = 9.25 And this is greater than 5. So why is this wrong??! What mistake am I making? b) D = 10 (fine) c) 15.25 (!!!) D = (25*25)/100 + 9 = 25/4 + 9 = 6.25 + 9 = 15.25 d) 34 (!!) D = (50*50)/100 + 9 = 50/2 + 9 = 25 + 9 = 34 e) 90 (fine) So by plugging in values of V you can eliminate A, B and D. Any help?



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12 Jun 2009, 11:43
can you post the question in full.



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12 Jun 2009, 18:00
The safety distance \(D\) (meters) between cars moving on a motorway depends on their speed \(V\) (meters per second): \(D = \frac{V^2}{100} + 9\)
If two cars want to maintain a time interval of 1 second between each other, which of the following speeds will breach the safety regulation?
(C) 2008 GMAT Club  m09#31
* 5 m/s * 10 m/s * 25 m/s * 50 m/s * 90 m/s



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Re: M09 #31 [#permalink]
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13 Jun 2009, 05:49
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thinkblue wrote: The safety distance \(D\) (meters) between cars moving on a motorway depends on their speed \(V\) (meters per second): \(D = \frac{V^2}{100} + 9\)
If two cars want to maintain a time interval of 1 second between each other, which of the following speeds will breach the safety regulation?
(C) 2008 GMAT Club  m09#31
* 5 m/s * 10 m/s * 25 m/s * 50 m/s * 90 m/s Actual Distance = Velocity x 1 second (1) \(D = \frac{5^2}{100} + 9 \hspace{10} \Rightarrow Distance_{safe} \hspace{5} = \hspace{5} 9.25 \hspace{5} > \hspace{5} Distance_{actual} \hspace{3}(5 m)\) BREACH(2) \(D = \frac{10^2}{100} + 9 \hspace{10} \Rightarrow Distance_{safe} \hspace{5} = \hspace{5} 10 \hspace{5} = \hspace{5} Distance_{actual} \hspace{3}(10 m)\) On Par, FINE(3) \(D = \frac{25^2}{100} + 9 \hspace{10} \Rightarrow Distance_{safe} \hspace{5} = \hspace{5} 15.25 \hspace{5} < \hspace{5} Distance_{actual} \hspace{3}(25 m)\) FINE(4) \(D = \frac{50^2}{100} + 9 \hspace{10} \Rightarrow Distance_{safe} \hspace{5} = \hspace{5} 34 \hspace{5} < \hspace{5} Distance_{actual} \hspace{3}(34 m)\) FINE(5) \(D = \frac{90^2}{100} + 9 \hspace{10} \Rightarrow Distance_{safe} \hspace{5} = \hspace{5} 90 \hspace{5} = \hspace{5} Distance_{actual} \hspace{3}(90 m)\) On Par, FINEClearly, first choice of 5 m/s speed violates the regulation because distance maintained (5 m) is less than what the rule suggests. Answer A. I think you got confused which distance should be greater than the other.



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Re: M09 #31 [#permalink]
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28 Jul 2009, 14:47
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I understand that plugging values is a better way but this is how i did it.... "If two cars want to maintain a time interval of 1 second between each other" think of it this way => distant is given interms of time (1 sec)..ie ...'D' meters / 1 sec (or D m/s) hence velocity to be maintained = distance => substitute D=V or V=D in above equation => we get V^2  100V + 900 < 0 ( < sign indicates required condition ..ie..Speed which will not breach the safety barrier ) => on solving this we get (V90) (V10) < 0 => thus it is clear that any speed between 10 m/s to 90 m/s is OK but nothing out of this range....Hence answer is 5 m/s If you like my post...consider it for Kudos...
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Re: M09 #31 [#permalink]
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10 Feb 2010, 01:09
goldeneagle94 wrote: thinkblue wrote: The safety distance \(D\) (meters) between cars moving on a motorway depends on their speed \(V\) (meters per second): \(D = \frac{V^2}{100} + 9\)
If two cars want to maintain a time interval of 1 second between each other, which of the following speeds will breach the safety regulation?
(C) 2008 GMAT Club  m09#31
* 5 m/s * 10 m/s * 25 m/s * 50 m/s * 90 m/s Actual Distance = Velocity x 1 second (1) \(D = \frac{5^2}{100} + 9 \hspace{10} \Rightarrow Distance_{safe} \hspace{5} = \hspace{5} 9.25 \hspace{5} > \hspace{5} Distance_{actual} \hspace{3}(5 m)\) BREACH(2) \(D = \frac{10^2}{100} + 9 \hspace{10} \Rightarrow Distance_{safe} \hspace{5} = \hspace{5} 10 \hspace{5} = \hspace{5} Distance_{actual} \hspace{3}(10 m)\) On Par, FINE(3) \(D = \frac{25^2}{100} + 9 \hspace{10} \Rightarrow Distance_{safe} \hspace{5} = \hspace{5} 15.25 \hspace{5} < \hspace{5} Distance_{actual} \hspace{3}(25 m)\) FINE(4) \(D = \frac{50^2}{100} + 9 \hspace{10} \Rightarrow Distance_{safe} \hspace{5} = \hspace{5} 34 \hspace{5} < \hspace{5} Distance_{actual} \hspace{3}(34 m)\) FINE(5) \(D = \frac{90^2}{100} + 9 \hspace{10} \Rightarrow Distance_{safe} \hspace{5} = \hspace{5} 90 \hspace{5} = \hspace{5} Distance_{actual} \hspace{3}(90 m)\) On Par, FINEClearly, first choice of 5 m/s speed violates the regulation because distance maintained (5 m) is less than what the rule suggests. Answer A. I think you got confused which distance should be greater than the other. Great explanation! Thank you!



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Re: M09 #31 [#permalink]
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11 Feb 2010, 09:59
Yeah punching the numbers in for the velocity .This explation makes it very clear for option A to be correct.



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Re: M09 #31 [#permalink]
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01 Sep 2010, 18:58
I am getting confused with the equation
on LHS we have a variable (D) whose unit is m whereas on RHS we have (v^2) which is (m/s)^2... so how is the equation valid?? Please explain



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Re: M09 #31 [#permalink]
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06 Sep 2010, 07:50
One more Quick way to do. Looking at the equation...I choose the most easy value for plugging in..i.e., 10 m/s . D = 100/100 + 9 = 10, which exactly meeting the norms (Option B). Now, the fun part is you have 3 options (C,D,E) > just Ok value (10) 1 option (B) < just ok value (10). I didn't use my brain much analyzing which is exceeding or less than norms... All I know is there should be ONE & ONLY answer...hence option " A". Gimme a kudo if you like this



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Re: M09 #31 [#permalink]
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06 Sep 2011, 02:51
I did using the bhushan252 way.......but got confused in the end.....niw i understand it.......A



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Re: M09 #31 [#permalink]
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06 Sep 2011, 06:56
A
Not sure if I did this right but here goes:
V= d/t therefore V = d/1 therefore V=D D = (D^2/100) + 9 100D = D^2 + 900 D^2 100D + 900 D = 10 or 90 Therefore, to meet the specifications of the formula the answer must be between 10 and 90 so the answer must be A.
I have no idea if I did it right, but I did get the right answer!



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Re: M09 #31 [#permalink]
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06 Sep 2011, 07:06
Here is how I solved it:
T = D/R; (v^2+900)/100r = 1. v = 90, v = 10. Since the distance should be less, "A" must be correct.
Hope somebody gets this strange approach.



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Re: M09 #31 [#permalink]
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06 Sep 2011, 10:49
Nice explanation. Got confused too. goldeneagle94 wrote: thinkblue wrote: The safety distance \(D\) (meters) between cars moving on a motorway depends on their speed \(V\) (meters per second): \(D = \frac{V^2}{100} + 9\)
If two cars want to maintain a time interval of 1 second between each other, which of the following speeds will breach the safety regulation?
(C) 2008 GMAT Club  m09#31
* 5 m/s * 10 m/s * 25 m/s * 50 m/s * 90 m/s Actual Distance = Velocity x 1 second (1) \(D = \frac{5^2}{100} + 9 \hspace{10} \Rightarrow Distance_{safe} \hspace{5} = \hspace{5} 9.25 \hspace{5} > \hspace{5} Distance_{actual} \hspace{3}(5 m)\) BREACH(2) \(D = \frac{10^2}{100} + 9 \hspace{10} \Rightarrow Distance_{safe} \hspace{5} = \hspace{5} 10 \hspace{5} = \hspace{5} Distance_{actual} \hspace{3}(10 m)\) On Par, FINE(3) \(D = \frac{25^2}{100} + 9 \hspace{10} \Rightarrow Distance_{safe} \hspace{5} = \hspace{5} 15.25 \hspace{5} < \hspace{5} Distance_{actual} \hspace{3}(25 m)\) FINE(4) \(D = \frac{50^2}{100} + 9 \hspace{10} \Rightarrow Distance_{safe} \hspace{5} = \hspace{5} 34 \hspace{5} < \hspace{5} Distance_{actual} \hspace{3}(34 m)\) FINE(5) \(D = \frac{90^2}{100} + 9 \hspace{10} \Rightarrow Distance_{safe} \hspace{5} = \hspace{5} 90 \hspace{5} = \hspace{5} Distance_{actual} \hspace{3}(90 m)\) On Par, FINEClearly, first choice of 5 m/s speed violates the regulation because distance maintained (5 m) is less than what the rule suggests. Answer A. I think you got confused which distance should be greater than the other.
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Re: M09 #31 [#permalink]
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06 Sep 2011, 16:59
tough one to understand...



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Re: M09 #31 [#permalink]
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11 Sep 2011, 21:33
for safe distance  D >= V^2/100 + 9 D = V*T = V (T=1, given) D >= D^2/100 + 9 D^2 100D +9 =< 0 (D10)(D90) =< 0 10 =< D =< 90 10 =< V =< 90 (V=D) anything, out of this range would violate the speed only option A violates the speed limit
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Re: M09 #31 [#permalink]
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02 Nov 2011, 20:17
The biggest issue is the awkward wording of the question. It would be better if it was clarified that breach meant go in excess of as opposed to break. Since you can in theory breach it by going to close or too far. Since Only B and E work out to be exact it should be reworded as such that going over the amount is clear that its incorrect.











