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According to a car dealer's sales report, 1/3 of the cars

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Intern
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According to a car dealer's sales report, 1/3 of the cars [#permalink] New post 19 Apr 2008, 16:51
According to a car dealer's sales report, 1/3 of the cars sold during a certain period were sedans and 1.5 of the other cars sold were station wagons. If N station wagons were sold during that period, how many sedans, in terms of N, were sold?
(A) 2/15 N
(B) 3/5 N
(C) 5/3 N
(D) 5/2 N
(E) 15/2 N

[I keep getting C... is there something I'm missing here?]
Director
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Re: OG 11th - Q#124 [#permalink] New post 19 Apr 2008, 19:40
According to a car dealer's sales report, 1/3 of the cars sold during a certain period were sedans and 1/5 of the other cars sold were station wagons. If N station wagons were sold during that period, how many sedans, in terms of N, were sold?


correction in the question -> it is 1/5 and not 1.5

now let us say there are X cars

and # of cars sold are Y

sedans sold -> Y/3

station wagons sold =1/5 of the other cars-> 1/5 *2/3* Y=2/15 Y

but 2/15 Y= N or Y =15/2 N

thus # of sold sedans = Y/3 => (15/2 N )/3 =>5/2 N

thus D
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Re: OG 11th - Q#124 [#permalink] New post 19 Apr 2008, 20:16
o.o wrote:
According to a car dealer's sales report, 1/3 of the cars sold during a certain period were sedans and 1.5 of the other cars sold were station wagons. If N station wagons were sold during that period, how many sedans, in terms of N, were sold?
(A) 2/15 N
(B) 3/5 N
(C) 5/3 N
(D) 5/2 N
(E) 15/2 N

[I keep getting C... is there something I'm missing here?]


D

"other cars" make this problem a bit unclear....

T = total cars
X = "other cars"
S = Sedan

(1) T/3 = S
(2) X/5 = N
(3) S + X = T
From (1) and (3), X = 2S
Plug in (2), 2S/5 = N
S = (5/2) N
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Re: OG 11th - Q#124 [#permalink] New post 19 Apr 2008, 21:45
Thanks for the quick reply guys, and I'm sorry about the typo in the question. ;o

I've analyzed this problem a little bit more and I've been able to pin-point the mistake I made (thanks to your explanations, of course); I didn't take into account the "other cars" variable. No wonder.

I tried to work it out a bit looking for a faster strategy that suits me and found out that this problem becomes really easy if you pick numbers.

T = Total cars.
X = Other cars.
S = Sedans.
N = Station Wagons.

With this information, we know from the question that:
T/3 = S ; X/5 = N

Now let us suppose there were 15 cars in total, thus T = 15.

Plugging into the equations above:
15/3 = 5 sedans sold.
15 - 5 = 10, or "Remaining cars", in our case "other cars".
Thus, 10/5 = 2 Station Wagons sold.

Knowing this we can determine the answer to our question:
Which would be 5/2 (2) = 10/2 = 5.

Or 5/2 N from the answer choices.


...any objections about this method?
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Re: OG 11th - Q#124 [#permalink] New post 20 Apr 2008, 03:49
Picking # is always a good way to go ahead.

Word of caution is - It is easy to solve numbers and get to the next level with out knowing that you are making a mistake. As long as one understands the question, either numbers or any other approach is purely personal and good!
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Re: OG 11th - Q#124   [#permalink] 20 Apr 2008, 03:49
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