Problem Solving

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

AAPL wrote:Official Guide

On a certain transatlantic crossing, 20 percent of a ship's passenger held round-trip tickets and also took their cars aboard the ship. If 60 percent of the passengers with round-trip tickets did not take their cars aboard 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%

We can let the number of passengers = 100.

Thus, 20 passengers held round trip tickets and took their cars aboard the ship. Since we are given that 60 percent of the passengers with round-trip tickets did not take their cars aboard the ship, the 20 passengers who held round trip tickets and took their cars aboard the ship represent 40 percent of the passengers with round-trip tickets. If n = the number of passengers with round-trip tickets, we have:

20 = 0.4n

n = 20/0.4 = 200/4 = 50

Thus, 50/100 = 50% of the ship's passengers held round-trip tickets.

Answer: C