azule45 wrote:

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%

a = total passenger

b = passenger owning R ticket ( including car (C) and no car (NC))

passenger owning R ticket and C = 20%*a

passenger owning R ticket and NC = 60%*b

so, b = 20%a + 60%b -> 40%b = 20%a -> b/a = 1/2 -> b = 50%

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