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On a certain transatlantic crossing, 20 percent of a ship’s [#permalink]
 23 Nov 2010, 12:38
Question Stats:
80% (02:09) correct
19% (06:14) wrong based on 21 sessions
On a certain transatlantic crossing, 20 percent of a ship’s passengers held round-trip tickets and also took their cars abroad the ship. If 60 percent of the passengers with round-trip tickets did not take their cars abroad the ship, what percent of the ship’s passengers held round-trip tickets? A. 33 1/3 % B. 40% C. 50% D. 60% E. 66 2/3%
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azule45 wrote: On a certain transatlantic crossing, 20 percent of a ship’s passengers held round-trip tickets and also took their cars abroad the ship. If 60 percent of the passengers with round-trip tickets did not take their cars abroad the ship, what percent of the ship’s passengers held round-trip 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 round-trip tickets AND cars. As 60% of the passengers with round-trip tickets did not take their cars then 40% of the passengers with round-trip tickets did take their cars, so # of passengers with round-trip tickets AND cars is 40% of the passengers with round-trip 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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Re: On a certain transatlantic crossing, 20 percent of a ship’s [#permalink]
21 Nov 2011, 09:56
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Re: On a certain transatlantic crossing, 20 percent of a ship’s [#permalink]
23 Nov 2011, 21:18
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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 [#permalink]
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 [#permalink]
09 Dec 2011, 22:39
More discussion below: My method is same as yours. problem-solving-84782.html
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Re: On a certain transatlantic crossing, 20 percent of a ship’s [#permalink]
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 [#permalink]
09 Jan 2012, 15:15
You can use a table approach as recommended by MGMAT. -----R---nR---Total C---20------------- nC--0.6x----------- Total x---------100 We get, 0.4x = 20 -> x = 50%
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Re: On a certain transatlantic crossing, 20 percent of a ship’s [#permalink]
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 [#permalink]
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
[#permalink]
25 May 2013, 12:20
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