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

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

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Re: On a certain transatlantic crossing, 20 percent of a ship's passengers [#permalink]
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Baten80 wrote:
Let round ticket = x
so,
0.60x+0.20 = x
x = 50%
Ans. C

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]
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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]
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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]
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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]
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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

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tik.jpg [ 23.4 KiB | Viewed 72942 times ]

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Re: On a certain transatlantic crossing, 20 percent of a ship's passengers [#permalink]
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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]
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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]
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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]
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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 70622 times ]

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

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