It is currently 26 Jun 2017, 22:41

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
VP
Joined: 09 Jul 2007
Posts: 1100
Location: London
On a certain transatlantic crossing, 20 percent of a ship s [#permalink]

### Show Tags

04 Oct 2007, 14:46
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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
Joined: 16 May 2007
Posts: 142

### Show Tags

04 Oct 2007, 16:44
Is the answer C, 50% ?
Intern
Joined: 03 Oct 2007
Posts: 16

### Show Tags

04 Oct 2007, 16:52

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
Senior Manager
Joined: 27 Aug 2007
Posts: 253

### Show Tags

05 Oct 2007, 07:31
Another vote for C
Manager
Joined: 22 May 2007
Posts: 121

### Show Tags

05 Oct 2007, 08:14
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%
VP
Joined: 08 Jun 2005
Posts: 1145

### Show Tags

05 Oct 2007, 08:38
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%

Director
Joined: 11 Jun 2007
Posts: 914

### Show Tags

05 Oct 2007, 11:04
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%

Director
Joined: 08 Jun 2007
Posts: 573

### Show Tags

05 Oct 2007, 11:18
C for me too

20 + 60X/100 = X
Re: ship passengers   [#permalink] 05 Oct 2007, 11:18
Display posts from previous: Sort by