On a certain trasatlantic crossing, 20 percent of a ship's

Hi All,

We're told that on a certain transatlantic crossing, 20% of a SHIP'S PASSENGERS held round-trip tickets AND took their cars aboard the ship and 60% of the PASSENGERS WITH ROUND-TRIP TICKETS did NOT take their cars aboard the ship. We're asked what percent of the ship's passengers held round-trip tickets. While this question might seem a bit strange at first, the concept tested is "groups within groups" and can be solved by TESTing VALUES and just a bit of Algebra.

To start, there are 3 groups implied by the prompt:
Passengers with round-trip tickets ....
1) who ALSO took their cars.
2) who did NOT take their cars
and
3) Passengers WITHOUT round-trip tickets

The first group is 20% of the TOTAL passengers.
The second group is 60% of passengers with round-trip tickets (that sub-group is comprised of the first 2 groups, but not the 3rd group).

IF... there are 100 total passengers....

X = number of passengers with round-trip tickets
20 = those with round-trip tickets AND took their cars
.6X = those with round-trip tickets who did NOT take their cars

Thus, X = 20 + .6X
We can now solve for X...
X = 20 + .6X
.4X = 20
4X = 200
X = 50

Thus, 50 of the 100 passengers had round-trip tickets = 50%

Final Answer: C

GMAT assassins aren't born, they're made,
Rich