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

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VP
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On a certain transatlantic crossing, 20 percent of a ship s [#permalink] New post 04 Oct 2007, 13:46
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%
Manager
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 [#permalink] New post 04 Oct 2007, 15:44
Is the answer C, 50% ?
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 [#permalink] New post 04 Oct 2007, 15:52
C is the answer...

As 60% of the passengers with round-trip tickets did not take their cars abroad the ship so 40% took their cars abroad the ship.
These 40% = 20% of total number of passengers (Given)
so these 100% which are having round-trip tickets = 50% of total number of passengers
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 [#permalink] New post 05 Oct 2007, 06:31
Another vote for C
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 [#permalink] New post 05 Oct 2007, 07:14
X held R. T. Tickets.

Assume 100 people on the ship.

20 have RTT and cars. .6X RTT and no cars.

20+ .6X = X

X=50 or 50%
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 [#permalink] New post 05 Oct 2007, 07:38
lets assume 120 passengers.

Round-trip tickets and car = 120*0.2 = 24

Total round-trip tickets = X

Round-trip tickets and no car =

X-24 = 0.6*X

0.4*X = 24

X = 24/0.4 = 60

what percent of the ship’s passengers held round-trip tickets

60/120 = 50%

the answer is (C)

:)
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 [#permalink] New post 05 Oct 2007, 10:04
another C

round trip = x
cars Y = 0.20
cars N = 0.60x

total round trip = cars Y + cars N
x = 0.20 + 0.60x
0.40x = 0.20
x = .50 or 50%

please confirm with the OA :)
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Re: ship passengers [#permalink] New post 05 Oct 2007, 10:18
C for me too

20 + 60X/100 = X
Re: ship passengers   [#permalink] 05 Oct 2007, 10:18
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