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

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

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31 Jul 2014, 00:41

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%

a = total passenger b = passenger owning R ticket ( including car (C) and no car (NC))

passenger owning R ticket and C = 20%*a passenger owning R ticket and NC = 60%*b

so, b = 20%a + 60%b -> 40%b = 20%a -> b/a = 1/2 -> b = 50%
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On a certain transatlantic crossing, 20 percent of a ship’s [#permalink]

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30 Aug 2016, 22:45

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%

Needs to be "aboard" not abroad I guess
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Re: On a certain transatlantic crossing, 20 percent of a ship’s [#permalink]

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20 Jul 2017, 12:36

suppose total = x ; and round trip pass = r 0.2x = roundtrip + car ---- eq 1 0.6r = rountrip + no car so 0.4 r = roundtrip + car ---- eq2 as per eq 1 and eq 2 0.2x = 0.4r x = 2r => which means x is 50% of r

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

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26 Nov 2017, 11:39

Since the number of passengers on the ship is immaterial, let the number of passengers on the ship be 100 for convenience. Let x be the number of passengers that held round-trip tickets.

Then, since 20 percent of the passengers held a round-trip ticket and took their cars aboard the ship, 0.20(100) = 20 passengers held round-trip tickets and took their cars aboard the ship.

The remaining passengers with round-trip tickets did not take their cars aboard, and they represent 0.6x (that is, 60 percent of the passengers with round-trip tickets). Thus 0.6x + 20 = x, from which it follows that 20 = 0.4x, and so x = 50.

The percent of passengers with round-trip tickets is, then, 50/100 * 100=50%