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by jkwan » Thu Feb 05, 2009 10:09 pm
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%
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by DanaJ » Thu Feb 05, 2009 11:04 pm
You've got four groups here:
people with round trip tickets and cars = a
people with round trip tickets but no cars = b
people with cars but no round trip tickets = c
people with neither round trip tickets nor cars = d
a + b + c + d = t (total).

Now, you also have that a = 20%of t or that a = 0.2t
And again you have that 60% of a + b (people with round trip tickets) = b, meaning that 0.6(a + b) = b.
Now, what you are looking for is how much of t is (a+b).

Let's use what we have:
0.6 (a + b) = b
0.6a + 0.6b = b
0.6a = 0.4b
3a = 2b
b = 3a/2. Now replace a = 0.2t over here and you get that b = 0.3t.
Then add a and b and you get a + b = 0.2t + 0.3t = 0.5t or 50% of t.

IMHO answer is C[/spoiler]
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ShortCut

by gmat740 » Mon Feb 16, 2009 11:52 pm
Bingo!!
I have a trick for this prob:

20% of Ship Passenger(S) = C(cars) + Round-Ticket (R)
20% of S = C + R

60% R with no cars, so 40% R with Cars.

thus,

40% R = 20% S

1% R = 20/40 % S = 0.5% S


So, 100% R = 50% Of Ship Passenger
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by marouan » Tue Feb 17, 2009 5:35 am
Good job GMAT 740,
me too usualy I use the same starategy and it's easy if you draw it like that..you will reach :
20% S = 40% R
so R= 50% S.
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