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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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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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09 Mar 2014, 19:14

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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.

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

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09 Mar 2014, 23:28

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Option C. If 60% of passengers with round-trip do not have cars,40% of passengers with round-trip tickets will have their cars on board. 40% of total=20 =50

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

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15 Jun 2015, 10:07

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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

Attachments

Temp.JPG [ 12.22 KiB | Viewed 4518 times ]

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

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05 Jun 2016, 12:59

pepo wrote:

On a certain boat, 20 % of the passengers have round-trip tickets and are trasporting cars. if 60% of the passengers who have round-trip tickets are not transporting cars, what percent of all the passengers have round-trip ticketsOn a certain boat, 20 % of the passengers have round-trip tickets and are trasporting cars. if 60% of the passengers who have round-trip tickets are not transporting cars, what percent of all the passengers have round-trip tickets?

A 40% B 50% C 60% D 70% E 80%

Let's suppose total customers are 100 and customers who have round-trip tickets are r

if 60% of the passengers who have round-trip tickets are not transporting cars or 40% who have round trips are transporting cars.

i.e. .6r are not transporting cars and .4r are transporting cars.

Its given that 20 % of the passengers have round-trip tickets and are transporting cars (20 out of 100)

.4r=20 r=50%

C is the answer
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Re: On a certain transatlantic crossing, 20 percent of a ship's [#permalink]

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10 Aug 2016, 11:37

I have a slight difficulty in understanding the question: 1.On a certain transatlantic crossing, 20 percent of a ship's passengers held round-trip tickets and also took their cars aboard the ship. 2.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?

>> So 2nd statement means that 40% have Round trip ticket and also took car.

But in the first sentence it said that percentage of such people who have round trip ticket and brought their cars is 20% ?

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

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21 Aug 2017, 04:19

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