shanice wrote:On a certain transatlantic crossing, 20 percent of a ship's passengers 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%

Answer is C.


Can someone please explain to me what is the connection of sentence 1 to sentence 2?

Thank you in advance.

Let's say there were 100 passengers on the ship. Let the number of round trip ticket holders be x.
We need to find x.
Statement 1 tells us that 20 people had round-trip tickets and had cars aboard the ship.

Statement 2 tells us that 60% of x did not take cars aboard the ship. If 60% did not take cars aboard the ship, 40% of x must have taken the cars aboard the ship. This is where the connection occurs. In statement 2 we determine that 40%x have round-trip tickets and cars aboard the ship.
Statement 1 tells us that 20 people had round tickets and cars aboard the ship.
Since both 1 and 2 talk about the same quantity, we can equate one with the other to find x.

so:

40%x = 20
x = 20/(40%) = 20*100/40 = 50.

Hence 50% of the passengers had round trip tickets.

C is correct.

Eagleeye, that was a perfect explanation. I could not ask for more.

Thank you.