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On a certain transatlantic crossing, 20 percent of a ship’s

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by xsmail321 » Fri May 01, 2009 11:47 am
What wrong with this approach?..
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%



0.4 R = 0.2 P, R/P * 100 = 50

Answer: C.

What wrong with this approach?
Make a table..

Whats wrong with this approach?
Round Not Round
With Car 0.2
No Car 0.6
0.8 1



Answer .8??
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Source: — Problem Solving |

by beatthegmat » Sat May 02, 2009 2:19 am
Moving this post to the PS forum.
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by DeepakR » Sat May 02, 2009 2:43 am
R with car=0.2P
It is also given that, 0.6 R did not take car with them which means 0.4 R took the car with them.

Hence 0.4 R=0.2 P so R/P=1/2=50%

-Deepak
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