ship

[ketkoag]

For passengers in a ship, 20% have round-tickets and put their cars on the ship board, 60% of who have round-tickets do not put their cars on the ship board. What's the percent of people who have round-tickers in the ship?

IMO:[spoiler] 80%[/spoiler] please confirm the answer..

[sreak1089]

Since those who put cars and those who donot is mutually exclusive, I also beleive answer is 80%.

[bluementor]

hmm... I'm getting a different answer...

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x = number of passangers with roundtrip tickets
y = number of passengers without roundtrip tickets

so total number of passangers = x + y

question: x/(x+y) ?

"60% of who have round-tickets do not put their cars on the ship board"

this means, 40% of those who have roundtrip tickets do put their cars on the ship.

40% of x = those with round tickets AND put their cars on the ship

and it is given that "For passengers in a ship, 20% have round-tickets and put their cars on the ship board":

20% of (x + y) = those who have round tickets AND put their cars on the ship.

so, we can equate both statements:

40% of x = 20% of (x+y)
0.4x = 0.2(x+y)
x/(x+y) = 0.2/0.4
x/(x+y) = 0.5

IMHO 50%.

-BM-

[tohellandback]

IMO 50%
check the picture
now 20 is 40% of the people who have round tickets and put the cars..so people who DO NOT put he cars on board and have round tickets is 30
30+20 is 50

[ketkoag]

got it thanks....

[truplayer256]

There are a total of x passengers aboard the ship. Of those x passengers, y passengers have round trip tickets.

.20(x)= Number of passengers who have round trip tickets and park their cars on board.

.60(y)= Number of passengers who have round trip tickets but do not park their cars on board. This means that .40(y) of the passengers have round trip tickets and park their cars on board, so:

.40(y)=.20(x)
2y=x

% of people who have round trip tickets on the ship: y/x

2y=x

2y/x=1

y/x=1/2 or 50%.