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On a certain transatlantic crossing, 20 percent of a ship s
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On a certain transatlantic crossing, 20 percent of a ship's passengers held roundtrip tickets and also took their cars aboard the ship. If 60 percent of the passengers with roundtrip tickets did not take their cars aboard the ship, what percent of the ship's passengers held roundtrip tickets? (A) 33 1/3% (B) 40% (C) 50% (D) 60% (E) 66 2/3% OPEN DISCUSSION OF THIS QUESTION IS HERE: onacertaintransatlanticcrossing20percentofaships168577.html
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Originally posted by ritula on 20 Jun 2008, 00:16.
Last edited by Bunuel on 21 May 2014, 01:34, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.




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23 Nov 2010, 13:04
azule45 wrote: On a certain transatlantic crossing, 20 percent of a ship’s passengers held roundtrip tickets and also took their cars abroad the ship. If 60 percent of the passengers with roundtrip tickets did not take their cars abroad the ship, what percent of the ship’s passengers held roundtrip tickets?
a 33 1/3 % b 40% c 50% d 60% e 66 2/3% Let the total # of passengers be 100. Now, 20 passengers held roundtrip tickets AND cars. As 60% of the passengers with roundtrip tickets did not take their cars then 40% of the passengers with roundtrip tickets did take their cars, so # of passengers with roundtrip tickets AND cars is 40% of the passengers with roundtrip tickets. If we take # of the passengers with round trip tickets to be \(x\) then we'll have \(0.4x=20\) > \(x=50\). Answer: C.
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On a certain transatlantic crossing, 20 percent of a ship’s
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23 Nov 2010, 12:38
On a certain transatlantic crossing, 20 percent of a ship’s passengers held roundtrip tickets and also took their cars abroad the ship. If 60 percent of the passengers with roundtrip tickets did not take their cars abroad the ship, what percent of the ship’s passengers held roundtrip tickets? A. 33 1/3 % B. 40% C. 50% D. 60% E. 66 2/3%
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Re: ship passengers
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20 Jun 2008, 00:28
On a certain transatlantic crossing, 20 percent of a ship’s passengers held roundtrip tickets and also took their cars abroad the ship. If 60 percent of the passengers with roundtrip tickets did not take their cars abroad the ship, what percent of the ship’s passengers held roundtrip tickets?
Solution: Let total number of passengers be 100 According to Q stem 40% of passengers who had roundtrip tics have taken cars  let number of passengers with round trip be X then
40% of X = 20 => X= 50.
Thus 50% of passengers have roundtrip tics with them!!



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Re: ship passengers
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20 Jun 2008, 08:58
ritula wrote: On a certain transatlantic crossing, 20 percent of a ship’s passengers held roundtrip tickets and also took their cars abroad the ship. If 60 percent of the passengers with roundtrip tickets did not take their cars abroad the ship, what percent of the ship’s passengers held roundtrip tickets? A. 33 1/3% B. 40% C. 50% D. 60% E. 66 2/3% I also got C but with a slightly different approach. If 60% of the passengers with round trip tickets did not take their cars that means 40% of the passengers with round trip tickets took their car which equates to 20% of the total of ship passengers. This translates to 30% of the total of ship passengers who had round trip tickets but did not take their car. (60/40 x 20%) = 30% Thus to find the total nnumber of round trip ticket holders, you just add the two. 20% + 30% = 50% T



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Re: ship passengers
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20 Jun 2008, 13:39
ritula wrote: On a certain transatlantic crossing, 20 percent of a ship’s passengers held roundtrip tickets and also took their cars abroad the ship. If 60 percent of the passengers with roundtrip tickets did not take their cars abroad the ship, what percent of the ship’s passengers held roundtrip tickets? A. 33 1/3% B. 40% C. 50% D. 60% E. 66 2/3% C
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Re: On a certain transatlantic crossing, 20 percent of a ship's passengers
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11 May 2011, 11:17
Yeah that's correct.
20+.6x = x => x=50%
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Re: On a certain transatlantic crossing, 20 percent of a ship's passengers
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11 May 2011, 21:09
C C! R 20 X20 X R! 100x 100 = total 0.4 x = 20 gives x = 50 Hence C.
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Re: On a certain transatlantic crossing, 20 percent of a ship's passengers
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12 May 2011, 00:44
0.20P = RT + C 0.6(RT) = no C => 0.40(RT) had C 0.20P = 0.40(RT) RT/P = 1/2 Answer  C
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Re: On a certain transatlantic crossing, 20 percent of a ship’s
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Re: On a certain transatlantic crossing, 20 percent of a ship’s
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23 Nov 2011, 21:18
Baten80 wrote: Let round ticket = x so, 0.60x+0.20 = x x = 50% Ans. C
Please help if i am not correct. If the thought process was that x is the fraction of total passengers who have the round ticket and total passengers = 1 (theoretically since we need percentages), then the logic is sound. Otherwise, note that 0.20 on its own doesn't mean anything. It needs to be 0.20p where p is the total number of passengers. 0.60x + 0.20p = x x/p = .50 = 50% or you can say that since 60% people with round trip ticket did not take their car, 40% people with round trip ticket did take their car. 40% people with round trip ticket = 20% total people So people with round trip ticket are 1/2 (i.e.50%) of total people.
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Re: On a certain transatlantic crossing, 20 percent of a ship’s
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09 Dec 2011, 12:51
Suppose the whole number of passengers is 100
\(%20*100=20\) have round trip and car.
Based on the question, this number is equal to the \(%40*(total passengers who have round trip)\). So
\(20= %40*X\) X=50
and this is %50 of the 100
Answer C



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Re: On a certain transatlantic crossing, 20 percent of a ship’s
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02 Jan 2012, 13:13
Fairly easy question this. Easy to get to 50%, the correct answer. 60% of the people who have not taken the car also gives us 40% of the people who have taken the car (people with round trip tickets btw). From there, it is easy to get to 50%.
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Re: On a certain transatlantic crossing, 20 percent of a ship’s
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09 Jan 2012, 15:15
You can use a table approach as recommended by MGMAT. RnRTotal C20 nC0.6x Total x100 We get, 0.4x = 20 > x = 50%
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Re: On a certain transatlantic crossing, 20 percent of a ship’s
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01 May 2013, 03:37
Another method?
Take the total number of passengers to be 100. therefore those with both round trip tickets and cars is 20, leaving 80 passengers with either round trip ticket only and cars only (the question is silent on the number of passengers who have neither round trip tickets nor cars, so assuming that to be zero). let number of passengers with round trip tickets be  x therefore number of passengers with cars is  0.6x hence 80= x + 0.6x x=50%



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Re: On a certain transatlantic crossing, 20 percent of a ship’s
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25 May 2013, 12:20
can anyone please clarify this through a Venn diagram. I understand that this is a percent of a percent problem. I just need to picture it. Thanks
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Re: On a certain transatlantic crossing, 20 percent of a ship’s
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27 May 2013, 07:24
i might sound a little silly with my query but if one looks at the question and tries to translate it seems some what like this:
1)20 percent of a ship’s passengers held roundtrip tickets and also took their cars abroad the ship which means that 20% held roundtrip tickets and took their cars aboard the ship 2) 60 percent of the passengers with roundtrip tickets did not take their cars abroad the ship which means that another 60% held round trip tickets but didn't take their cars abroad. so aren't they like 2 different groups i dunno but i am so confused.



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Re: On a certain transatlantic crossing, 20 percent of a ship’s
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27 May 2013, 09:20
mohnish104 wrote: i might sound a little silly with my query but if one looks at the question and tries to translate it seems some what like this:
1)20 percent of a ship’s passengers held roundtrip tickets and also took their cars abroad the ship which means that 20% held roundtrip tickets and took their cars aboard the ship 2) 60 percent of the passengers with roundtrip tickets did not take their cars abroad the ship which means that another 60% held round trip tickets but didn't take their cars abroad. so aren't they like 2 different groups i dunno but i am so confused. Read the stem carefully: "60 percent of the passengers with roundtrip tickets did not take their cars abroad the ship..." So, 60% of some particular group did not take their cars abroad the ship.
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Re: On a certain transatlantic crossing, 20 percent of a ship’s
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27 Sep 2013, 11:07
Let total no of passengers be T Let the no of passengers with round trip tickets be X Given: 20% of total passengers have round trip tickets and took their cars aboard = 0.2*T 60% of passengers with round trip tickets did not take their cars => 40% of passengers with round trip tickets took their cars => 0.4*X=0.2*T => X=1/2*T => X=50% of T



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Re: On a certain transatlantic crossing, 20 percent of a ship s
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20 May 2014, 09:38
I think this is a poorly worded problem: They should make the distinctions in the second sentence more clearly that the 60% is only of the passangers WITH the roundtrip tickets would have been more obvious. Upon reading the problem, I ended up with 80%.
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