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

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23 Nov 2010, 11:38

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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?

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

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]

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02 Jan 2012, 12:13

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]

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09 Jan 2012, 14:15

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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]

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01 May 2013, 02:37

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%

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

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25 May 2013, 11:20

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]

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27 May 2013, 06:24

i might sound a little silly with my query but if one looks at the question and tries to translate it seems some what like this:

1)20 percent of a ship’s passengers held round-trip tickets and also took their cars abroad the ship which means that 20% held roundtrip tickets and took their cars aboard the ship 2) 60 percent of the passengers with round-trip tickets did not take their cars abroad the ship which means that another 60% held round trip tickets but didn't take their cars abroad. so aren't they like 2 different groups i dunno but i am so confused.

i might sound a little silly with my query but if one looks at the question and tries to translate it seems some what like this:

1)20 percent of a ship’s passengers held round-trip tickets and also took their cars abroad the ship which means that 20% held roundtrip tickets and took their cars aboard the ship 2) 60 percent of the passengers with round-trip tickets did not take their cars abroad the ship which means that another 60% held round trip tickets but didn't take their cars abroad. so aren't they like 2 different groups i dunno but i am so confused.

Read the stem carefully: "60 percent of the passengers with round-trip tickets did not take their cars abroad the ship..." So, 60% of some particular group did not take their cars abroad the ship.
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Re: On a certain transatlantic crossing, 20 percent of a ship’s [#permalink]

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27 Sep 2013, 10:07

Let total no of passengers be T Let the no of passengers with round trip tickets be X

Given: 20% of total passengers have round trip tickets and took their cars aboard = 0.2*T

60% of passengers with round trip tickets did not take their cars => 40% of passengers with round trip tickets took their cars => 0.4*X=0.2*T => X=1/2*T => X=50% of T

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%