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%