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# How many weeks was the Carnival open last year? (1) The

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Manager
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How many weeks was the Carnival open last year? (1) The [#permalink]

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02 Jun 2009, 23:15
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How many weeks was the Carnival open last year?

(1) The ratio between the number of weeks it was open and closed last year is $$1:3$$.

(2) The multiple of the number of weeks the Carnival was open and the number of weeks the Carnival was closed is $$507$$.
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Senior Manager
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02 Jun 2009, 23:50

1, insuffucient X/y = 1/3 =>3X = Y but no idea of the x,y
2, insufficient.. x*y*k = 507 = 3*13*13

when k = 1, x=3 and y = 39 or x = 39 or y = 3

combining 1 and 2 ,

3x^2 K = 507
x^2 K = 169..

when k = 1, X = 13, y = 39/...

Whats the OA ?

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Director
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02 Jun 2009, 23:52
How many weeks was the Carnival open last year?

(1) The ratio between the number of weeks it was open and closed last year is $$1:3$$.

(2) The multiple of the number of weeks the Carnival was open and the number of weeks the Carnival was closed is $$507$$.

C

x=open
y=closed
x/y=1/3
xy=507
3x^2=507
x=13

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Manager
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02 Jun 2009, 23:55
Good attempts
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Director
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03 Jun 2009, 00:08
Good attempts

must be E..there are only 52 weeks in a year

52 factors to 13*2*2...no combo will give you 1/3 so cant be A
507 factors to 13*13*3 ...no sum adds to 52, cant be B

did you make this question up?

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Manager
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03 Jun 2009, 09:07
How many weeks was the Carnival open last year?

(1) The ratio between the number of weeks it was open and closed last year is .

(2) The multiple of the number of weeks the Carnival was open and the number of weeks the Carnival was closed is 507.

1 - one year -52 weeks
the ratio ==> x:3x

4x=52
x=52/4 =13

2.507=3x13x13
we can derive from B also. Number of weeks when open + number of weeks when closed should be 52 and change the multiples.

As 507 is a multiple 0f 13 and 52 is a multiple of 13, we can solve from stat 2.
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Manager
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03 Jun 2009, 09:30
1
KUDOS
$$X$$ - # of weeks open
$$Y$$ - # of weeks closed

$$X + Y = 52$$ (inferred)

Question:$$(X?)$$

(1) $$X/Y = 1/3$$

2 linear equations, 2 variables $$\longrightarrow SUFFICIENT$$ .

(2) $$XY = 507$$

Rearrange inferred equation to $$Y = 52-X$$ and substitute into above

$$X(52-X) = 507$$
$$52X - X^{2} = 507$$
$$X^{2} - 52X + 507 = 0$$
$$X^{2} - 13X - 39X + 507 = 0$$
$$X(X - 13) - 39(X - 13) = 0$$
$$(X - 13)(X - 39) = 0$$

$$\longrightarrow X=13$$ or $$X=39$$
$$\longrightarrow INSUFFICIENT$$.

Final Answer, $$A$$.

$$\surd$$
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Manager
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03 Jun 2009, 15:57
Thanks amolsk.

And yes bigtreezl, I actually did. My questions are sneaky MF'ers.
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03 Jun 2009, 23:04
i am sure i am gonna leave out such ones..

Hades, What is the source of such brilliant questions ..?

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04 Jun 2009, 00:00
Neochronic wrote:
i am sure i am gonna leave out such ones..

Hades, What is the source of such brilliant questions ..?

My brain my friend,
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Re: Tough DS 5   [#permalink] 04 Jun 2009, 00:00
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