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

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Bunuel
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On a certain transatlantic crossing, 20 percent of a ship's passengers [#permalink] New post  Updated on: 11 Mar 2022, 23:59
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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?

(A) 33 1/3%
(B) 40%
(C) 50%
(D) 60%
(E) 66 2/3%
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Originally posted by Bunuel on 20 Jun 2008, 00:16.
Last edited by Bunuel on 11 Mar 2022, 23:59, edited 3 times in total.
Edited the question.
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Re: On a certain transatlantic crossing, 20 percent of a ship's passengers [#permalink] New post  09 Mar 2014, 15:17
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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?

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

Answer: C.
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Re: On a certain transatlantic crossing, 20 percent of a ship's passengers [#permalink] New post  20 Jun 2008, 13:39
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ritula 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%


C
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Re: On a certain transatlantic crossing, 20 percent of a ship's passengers [#permalink] New post  23 Nov 2010, 13:04
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Re: On a certain transatlantic crossing, 20 percent of a ship's passengers [#permalink] New post  20 Jun 2008, 00:28
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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?

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!!
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Re: On a certain transatlantic crossing, 20 percent of a ship's passengers [#permalink] New post  21 Nov 2011, 09:56
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Let round ticket = x
so,
0.60x+0.20 = x
x = 50%
Ans. C

Please help if i am not correct.
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Re: On a certain transatlantic crossing, 20 percent of a ship's passengers [#permalink] New post  23 Nov 2011, 21:18
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Baten80 wrote:
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 passengers [#permalink] New post  09 Jan 2012, 15:15
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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 passengers [#permalink] New post  27 May 2013, 07:24
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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.
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Re: On a certain transatlantic crossing, 20 percent of a ship's passengers [#permalink] New post  27 May 2013, 09:20
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mohnish104 wrote:
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 passengers [#permalink] New post  09 Mar 2014, 19:59
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Answer = (C) 50%

Lets say total passengers = 100

Passengers with round ticket & cars = 20

If 60 percent of the passengers with round-trip tickets did not take their cars aboard the ship which means 40% with round trip tickets took cars

40x/100 = 20

x = 50

Answer = 50%
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Re: On a certain transatlantic crossing, 20 percent of a ship's passengers [#permalink] New post  09 Mar 2014, 20:14
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20% of the passengers were RT passengers with cars aboard. 60% of RT passengers did not bring cars. Thus, 20% of all passengers corresponds to 40% of RT passengers.

r = RT
t = total

2r/5 = t/5

r = 5t/10

= t/2

= 50%

OR 40% = 2(20%). 10% of all passengers are 20% of RT, so 100% of RT are 50% of passengers.
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Re: On a certain transatlantic crossing, 20 percent of a ship's passengers [#permalink] New post  20 May 2014, 09:38
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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%.
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Re: On a certain transatlantic crossing, 20 percent of a ship's passengers [#permalink] New post  21 May 2014, 01:37
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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..."
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Re: On a certain transatlantic crossing, 20 percent of a ship's passengers [#permalink] New post  15 Jun 2015, 11:07
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We can solve this problem using Double matrix. Unknown NUmber is here the number of pasengers with round trip tickets - let it be X, then 60% of X = 0,6X
Let the Total be 100, then people with Round Trip Tickets and Cars = 20% of 100 = 20

See the attached screenshot and then solve.

20 + 0,6X = X
X = 50
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Re: On a certain transatlantic crossing, 20 percent of a ship's passengers [#permalink] New post  06 Apr 2018, 08:46
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Bunuel wrote:

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%


We can let the number of passengers = 100.

Thus, 20 passengers held round trip tickets and took their cars aboard the ship. Since we are given that 60 percent of the passengers with round-trip tickets did not take their cars aboard the ship, the 20 passengers who held round trip tickets and took their cars aboard the ship represent 40 percent of the passengers with round-trip tickets. If n = the number of passengers with round-trip tickets, we have:

20 = 0.4n

n = 20/0.4 = 200/4 = 50

Thus, 50/100 = 50% of the ship’s passengers held round-trip tickets.

Answer: C
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