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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 [#permalink]

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20 Jun 2008, 00:16
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On a certain transatlantic crossing, 20 percent of a ship's passengers held round-trip tickets and also took their cars aboard the ship. If 60 percent of the passengers with round-trip tickets did not take their cars aboard 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%

OPEN DISCUSSION OF THIS QUESTION IS HERE: on-a-certain-transatlantic-crossing-20-percent-of-a-ship-s-168577.html
[Reveal] Spoiler: OA

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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20 Jun 2008, 00:28
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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?

Solution: Let total number of passengers be 100
According to Q stem 40% of passengers who had round-trip 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 round-trip tics with them!!
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20 Jun 2008, 08:58
ritula 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%

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%

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20 Jun 2008, 13:39
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ritula 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%

C
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Re: On a certain transatlantic crossing, 20 percent of a ship s [#permalink]

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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 round-trip tickets would have been more obvious. Upon reading the problem, I ended up with 80%.
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Re: On a certain transatlantic crossing, 20 percent of a ship s [#permalink]

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20 May 2014, 10:16
Let total passengers on the ship be p.
Let passengers with round-trip tickets be r.

A person can either take a car or not take a car. There are only two options.

As given,

60% of r did not take cars.
=> 40% of r took cars.

Also give,

20% of all passengers had round trip tickets and cars.

=> 20% of p = 40% of r
=> 0.2 p = 0.4 r
=> r= p/2

Thus, half of all passengers have round-trip tickets.
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Re: On a certain transatlantic crossing, 20 percent of a ship s [#permalink]

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21 May 2014, 01:37
Expert's post
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tedchou12 wrote:
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 round-trip tickets would have been more obvious. Upon reading the problem, I ended up with 80%.

The question clearly says "60 percent of the passengers with round-trip tickets..."

On a certain transatlantic crossing, 20 percent of a ship's passengers held round-trip tickets and also took their cars aboard the ship. If 60 percent of the passengers with round-trip tickets did not take their cars aboard 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 $$0.4x=20$$ --> $$x=50$$.

OPEN DISCUSSION OF THIS QUESTION IS HERE: on-a-certain-transatlantic-crossing-20-percent-of-a-ship-s-168577.html
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Re: On a certain transatlantic crossing, 20 percent of a ship s   [#permalink] 21 May 2014, 01:37
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