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# On a certain transatlantic crossing, 20 percent of a ships

Author Message
Manager
Joined: 19 Aug 2006
Posts: 204
On a certain transatlantic crossing, 20 percent of a ships  [#permalink]

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25 Dec 2006, 06:23
1
3
00:00

Difficulty:

(N/A)

Question Stats:

100% (00:55) correct 0% (00:00) wrong based on 100 sessions

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On a certain transatlantic crossing, 20 percent of a shipâ€™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 shipâ€™s passengers held round-trip tickets?

A. 33 1/3%
B. 40%
C. 50%
D. 60%
E. 66 2/3%

OPEN DISCUSSION OF THIS QUESTION IS HERE: on-a-certain-transatlantic-crossing-20-percent-of-a-ship-s-105277.html
Director
Joined: 28 Dec 2005
Posts: 891

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25 Dec 2006, 06:38
2
1
johnycute wrote:
On a certain transatlantic crossing, 20 percent of a shipâ€™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 shipâ€™s passengers held round-trip tickets?
A. 33 1/3%
B. 40%
C. 50%
D. 60%
E. 66 2/3%

I get C (50%)

20% held roundtrip tix and took cars.
Let x be total on ship then above can be represented as 0.2x.
Let r be number of people who took round trip.
60% of roundtrip passengers did not take cars => 40% of r took cars
=>0.4r = 0.2x

=> r/x (asked) = 0.2/0.4 = 0.5
thus in terms of percentage, 50%
SVP
Joined: 08 Nov 2006
Posts: 1529
Location: Ann Arbor
Schools: Ross '10

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25 Dec 2006, 07:53
I used venn diagram and got the same answer 50%
Director
Joined: 30 Nov 2006
Posts: 573
Location: Kuwait

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25 Dec 2006, 08:12
My reasoning:

T: total passengers
G1: those with round-trip tickets
G2: those with cars .... etc
B: both with cars and tickets [ G1 x G2 ]

By definition: those with tickets only are [ G1 - B ]
those with cars only are [ G2 - B ]

B = 0.2 T
G1 - B = 0.6 G1 [ G1 only ]
--> B = 0.4 G1

back to our initial equation
B = 0.4 G1 = 0.2 T
G1/T = 0.2 / 0.4 = 0.5 or 50 %

Intern
Joined: 24 Feb 2006
Posts: 48

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26 Dec 2006, 11:42
Got C

X*20/100 = Y*(100 -60)*/100
=> Y/X*100 = 50
VP
Joined: 21 Mar 2006
Posts: 1047
Location: Bangalore

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26 Dec 2006, 19:39
1
60% of people who had round trip tickets did not take their cars.
Thus, 40% of the people who had round trip tickets took their cars.
Number of people who had round trip tickets AND took their cars = 20

(40/100) * X = 20

<where X is the number of people who had round trip tickets>

X = 50.

Give me C!
Senior Manager
Joined: 05 Oct 2006
Posts: 258

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26 Dec 2006, 21:24
X = TOTAL no of passengers
y = total no of passengers with round trip tkt

then 20% of x = 40 % of y

hence reqd % = y/x * 100 = 50 %

choice C
Senior Manager
Status: Gonna rock this time!!!
Joined: 22 Jul 2012
Posts: 449
Location: India
GMAT 1: 640 Q43 V34
GMAT 2: 630 Q47 V29
WE: Information Technology (Computer Software)
Re: On a certain transatlantic crossing, 20 percent of a ships  [#permalink]

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05 Jan 2013, 00:02
how to solve this using a 2 set matrix?
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Math Expert
Joined: 02 Sep 2009
Posts: 51100
Re: On a certain transatlantic crossing, 20 percent of a ships  [#permalink]

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05 Jan 2013, 03:36
Sachin9 wrote:
how to solve this using a 2 set matrix?

Check here: on-a-certain-transatlantic-crossing-20-percent-of-a-ship-s-105277.html
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Joined: 09 Sep 2013
Posts: 9116
Re: On a certain transatlantic crossing, 20 percent of a ships  [#permalink]

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28 Nov 2018, 17:48
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: On a certain transatlantic crossing, 20 percent of a ships &nbs [#permalink] 28 Nov 2018, 17:48
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