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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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Question Stats:

76% (02:43) correct
24% (01:16) wrong based on 80 sessions

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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?

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!!

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%

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%. _________________

Aiming for a 750. Let me know if this is possible. Ted Chou

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 \(0.4x=20\) --> \(x=50\).

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