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%
OA will come later
Pessengers with Roundtrip ticket
This topic has expert replies
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.
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.
- ssmiles08
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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)
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)
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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
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
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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..
MGMAT book for Word translations has explained this method beautifully and after readin the book I use only this method..