Pessengers with Roundtrip ticket

I arrive at choice C.

Reasoning: I always enjoy using actual numbers that have a common multiple, so in this case, I chose 120 (since 20% and 60% are in the text).

For 120 people, 20%, or 24 people, have both R&C (round-trip and car). Since 60% of those with round-trip did not take their cars, that means there are 40% passengers who have round-trip tickets that do have their cars. 24 people represent 40% of the total round-trip community, so 24/0.4 = 60 people.

There are 60 people who have round-trip tickets out of a total of 120, so 50%. Choice C.

Let the total % of passengers on ship = 100%

20% of the total ship passengers had round trip tickets and cars aboard.

total round trip passengers = x

0.6(x) did not have round trip tickets.

20 + 0.6(x) = x

x = 20/(0.4) = 50

total round trip passengers = x = 50% (C)

Thanks Guys.
OA is C. Thanks for the explanations

Hi,

For grouping problems I generally use a table of items (please refer the attachment)

The total number of passengers who have return trip tickets are X. 60%(or 3/5th) of these passengers do not have their cars aboard. Thus the cell corresponding to "No Car aboard" and "RT Ticket" has value (3/5)*X.

The rows and columns add up respectively, and then add up to the RHS lower corner cell which is the grand total (which i have taken 100 here)

From the table we get
20+(3/5)*X = X
=> X-(3/5)*X=20 => (2/5)*X = 20 => X=50

Thanks Vivek for showing such an methodology.

The method given by Vivek is the Best way to deal with overlapping sets.
MGMAT book for Word translations has explained this method beautifully and after readin the book I use only this method..