On a certain transatlantic crossing

ska7945 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. 33.3%
B. 40%
C. 50%
D. 60%
E. 66.3%

Percent question with an unknown total: let's pick 100 to get the ball rolling.

There are 100 total passengers.

20 of them have round-trip tickets AND cars.

Since 60 percent of the passengers with round-trip tickets did NOT take their cars abroad the ship, we know that 40% of the passengers with round-trip tickets DID take their cars.

Accordingly, if we let R = # of round trip ticket holders:

40%(R) = 20
(2/5)R = 20
R = 20(5/2)
R = 50

choose (C).

I felt its a good question and a great answer for that..

Stuart,

Jus one question,

20 of them have round-trip tickets AND cars.
So I assumed, 80 of them did have round trip ticket but no cars.

Ofcorse i got the answer as 75% which was wrong. My question is -

What was wrong with my assumption? what is difference between cond 1 and cond 2?

Thanks

schumi_gmat wrote:Stuart,

Jus one question,

20 of them have round-trip tickets AND cars.
So I assumed, 80 of them did have round trip ticket but no cars.

Ofcorse i got the answer as 75% which was wrong. My question is -

What was wrong with my assumption? what is difference between cond 1 and cond 2?

Thanks

You assumed that everyone on the ship has a round-trip ticket, which is never stated (and which is in fact disproven by the other information we're given). If that were the case, the answer would be 100%.

Stuart, Can we solve this by using the venn diagram concept.

Haaress wrote:Stuart, Can we solve this by using the venn diagram concept.

You can indeed treat this as an overlapping set problem and use venn diagrams, but you'd still want to pick 100 as the total number of passengers. Also, you'd have to be very careful about the numbers that you plugged in (i.e. you can't just use "60" for anything, you'd still have to solve as I did in my solution).