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# On a certain transatlantic crossing, 20 percent of a ship's

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Senior Manager
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On a certain transatlantic crossing, 20 percent of a ship's [#permalink]

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06 May 2006, 04:51
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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?

33.3%
40%
50%
60%
66.6%
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16 May 2006, 19:37
getzgetzu 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?

33.3%
40%
50%
60%
66.6%

= 20%/(1-0.6)=50%
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16 May 2006, 23:46
Director
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17 May 2006, 03:42
ok...umm..let there be 100x all passengers.

Then 20x have round trip tickets and taking cars.

Now if 60% of passengers with round-trip tickets did not take cars, then the remaining 40% must have.

That 40% with cars is 20x
Hence, the remaining % (60% - without cars) = 30x

%=(20x+30x)/100x * 100 = 50%
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17 May 2006, 05:23
Frankly speaking i dint understand the language of this question. Please help me in understanding the same.

It says
"20 percent of a ship's passengers held round-trip tickets and also took their cars abroad the ship"

which means "20% of the total has round-trip ticket + cars"

I think this sentence should have been phrased in a much more clear way. Is this the way questions are phrased in GMAT also????
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17 May 2006, 05:53
What you have to do is take the following 2 statements independently and not relate them to one another:

1. 20 percent of a ship's passengers held round-trip tickets and also took their cars abroad the ship.

2. 60 percent of the passengers with round-trip tickets, did not take their cars abroad the ship.

From statement 2, from all the passengers with round trip tickets, if 60% did not take cars, then the remaining 40% must have taken cars.

This 40% (of passengers with roundtrip tickets) = 20% of total passengers.

Hope I explained it better...
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17 May 2006, 12:15
very strange))
20 percent of a ship's passengers held round-trip tickets and the question is what percent of the ship's passengers held round-trip tickets?
Confused me a little))
gmatmba gr8 decoding 10x
_________________

IE IMBA 2010

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18 May 2006, 04:55
I understood the question

Passangers with round tickets:

with cars: 20%
without cars: 0.6X%
Total: X%

20+0.6x=x
0.4x=20
x=50%
18 May 2006, 04:55
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