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

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VP
Joined: 22 Nov 2007
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On a certain transatlantic crossing, 20 percent of a ship s [#permalink]  06 Jan 2008, 08:06
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%
Director
Joined: 12 Jul 2007
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Re: counting & sets [#permalink]  06 Jan 2008, 08:28
.20 = 20% of ship's passengers who held RT tickets AND took their cars aboard
.60 = 60% of ship's passengers who held RT tickets AND did NOT take their cars aborad

.20 of total = (1-.6) of RT ticket holders
1/5 of total = 2/5 of RT
Total = 2 RT
RT = 1/2 of total

you can think of it this way

1/5 of total = 2/5 of RT this means that 1/10 of total = 1/5 of RT
multiply through by 5 and we get 5/10 of total = RT so 50%
CEO
Joined: 29 Mar 2007
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Re: counting & sets [#permalink]  06 Jan 2008, 09:08
marcodonzelli 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%

R NR T
C 20
NC .6x
T x 100

20+.6x=x --> 20=.4x --> 10/4*20=x x=50

Sorry my double set matrix probably won't look right on the post...

C
Re: counting & sets   [#permalink] 06 Jan 2008, 09:08
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