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

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

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20 May 2014, 09:16
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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, 00:37
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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]

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

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31 Jul 2014, 01:35
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SaraLotfy wrote:
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

Refer diagram below:

Lets say total Round trip passengers = x

Round trip with car = 20

So, round trip without car = x - 20

Given that 60% of Total Round trip passengers are without cars

$$\frac{60x}{100} = x-20$$

x = 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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13 May 2017, 06:05
took me 5 mins because of the wording
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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

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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%

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Re: On a certain transatlantic crossing, 20 percent of a ship’s   [#permalink] 26 Nov 2017, 11:39

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