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PDF800 SET1 question 18

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Senior | Next Rank: 100 Posts
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by zozo123 » Tue Jul 31, 2007 2:30 am
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 trandb » Tue Jul 31, 2007 5:59 am
is C (50 %) the answer ?
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by mindruna » Tue Jul 31, 2007 7:45 am
I would also say 50%.

The 20% of the (total) passengers hold a round trip ticket and have cars and 60% of round trip passengers have no cars. You can see that the 20% is equal to 40% of the total passengers and that means that those with round trip tickets without cars are equal to 30% of the total passengers.


30% +20%=50%
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by lalitgmat » Tue Jul 31, 2007 8:58 am
If you combine all AND clauses,
1> 20% of Ships's passengers (SP) have CAR aboard.
2> 60% of SP DO NOT have CAR aboard.
Now my argue is "how are you sure that ship does NOT have any other type of passenger". Basically how you deduce that "ALL passengers of ship have round-trip tickets".
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by mindruna » Tue Jul 31, 2007 9:02 am
I feel like you have to assume that there are no other types of passengers other than those with and without a car and those with and without round trip tickets. If not, there would be no way to solve the problem...
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by gabriel » Tue Jul 31, 2007 9:16 am
... guyz .. seriously ignore this question .. the construction of the question leaves a lot of room for ambiguity ..
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