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# A box contains blue and purple marbles only. What is the

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A box contains blue and purple marbles only. What is the [#permalink]

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23 Oct 2010, 22:29
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45% (medium)

Question Stats:

61% (01:04) correct 39% (01:09) wrong based on 133 sessions

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A box contains blue and purple marbles only. What is the total number of blue marbles in the box?

(1) The probability of selecting a blue marble at random is 2/5

(2) If three purple marbles were removed from the box, the probability of selecting a purple marble at random would be 2/3
[Reveal] Spoiler: OA

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24 Oct 2010, 03:49
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shrive555 wrote:
A box contains blue and purple marbles only. What is the total number of blue marbles in the box?

(1) The probability of selecting a blue marble at random is 2/5

(2) If three purple marbles were removed from the box, the probability of selecting a purple marble at random would be 2/3

(1) Says that 2 out of every 5 marbles is blue. Doesnt say how many marbles there are

(2) Let there be b blue and p purple marbles.
New prob = (p-3)/(b+p-3) = 2/3
3p-9=2b+2p-6
p=2b+3
Not enough to solve for p and b

(1+2) 1 also gives us the equation, b/p=2/5 or 5b=2p. The two equation combined can be solved to get the value of b

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24 Oct 2010, 04:24
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shrouded1 wrote:
shrive555 wrote:
A box contains blue and purple marbles only. What is the total number of blue marbles in the box?

(1) The probability of selecting a blue marble at random is 2/5

(2) If three purple marbles were removed from the box, the probability of selecting a purple marble at random would be 2/3

(1) Says that 2 out of every 5 marbles is blue. Doesnt say how many marbles there are

(2) Let there be b blue and p purple marbles.
New prob = (p-3)/(b+p-3) = 2/3
3p-9=2b+2p-6
p=2b+3
Not enough to solve for p and b

(1+2) 1 also gives us the equation, b/p=2/5 or 5b=2p. The two equation combined can be solved to get the value of b

I think the numbers must be wrong for this question.

Let the # of blue marbles be $$b$$ and the value of purple marbles be $$p$$

(1) $$\frac{b}{b+p}=\frac{2}{5}$$ (not b/p=2/5) --> $$5b=2b+2p$$ --> $$3b=2p$$, insufficient to find the value of $$b$$.
(2) $$\frac{p-3}{b+p-3}=\frac{2}{3}$$ --> $$3p-9=2b+2p-6$$ --> $$p=2b+3$$, insufficient to find the value of $$b$$.

(1)+(2) $$3b=2p$$ and $$p=2b+3$$ --> $$3b=2(2b+3)$$ --> $$b=-6$$, but # of blue marbles can not be negative.

Consider the following:
From (1) the probability of selecting a purple marble is $$\frac{p}{b+p}=\frac{3}{5}=\frac{9}{15}$$;
From (2) if three purple marbles were removed from the box, the probability of selecting a purple marble at random would be 2/3: $$\frac{p-3}{b+p-3}=\frac{2}{3}=\frac{10}{15}$$;

$$\frac{9}{15}<\frac{10}{15}$$ --> we removed 3 purple marbles and the probability of selecting a purple marble increased: this can not be true.

So there is something wrong with this question.
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24 Oct 2010, 04:49
actually to come to think of it, you don't even need to solve all this to know this question is wrong.

the probability of purples was 0.6
we remove some purples
instead of probability going down, it actually increases to 0.67

This is impossible

Bunuel wrote:
shrouded1 wrote:
shrive555 wrote:
A box contains blue and purple marbles only. What is the total number of blue marbles in the box?

(1) The probability of selecting a blue marble at random is 2/5

(2) If three purple marbles were removed from the box, the probability of selecting a purple marble at random would be 2/3

(1) Says that 2 out of every 5 marbles is blue. Doesnt say how many marbles there are

(2) Let there be b blue and p purple marbles.
New prob = (p-3)/(b+p-3) = 2/3
3p-9=2b+2p-6
p=2b+3
Not enough to solve for p and b

(1+2) 1 also gives us the equation, b/p=2/5 or 5b=2p. The two equation combined can be solved to get the value of b

I think the numbers must be wrong for this question.

Let the # of blue marbles be $$b$$ and the value of purple marbles be $$p$$

(1) $$\frac{b}{b+p}=\frac{2}{5}$$ (not b/p=2/5) --> $$5b=2b+2p$$ --> $$3b=2p$$, insufficient to find the value of $$b$$.
(2) $$\frac{p-3}{b+p-3}=\frac{2}{3}$$ --> $$3p-9=2b+2p-6$$ --> $$p=2b+3$$, insufficient to find the value of $$b$$.

(1)+(2) $$3b=2p$$ and $$p=2b+3$$ --> $$3b=2(2b+3)$$ --> $$b=-6$$, but # of blue marbles can not be negative.

Consider the following:
From (1) the probability of selecting a purple marble is $$\frac{p}{b+p}=\frac{3}{5}=\frac{9}{15}$$;
From (2) if three purple marbles were removed from the box, the probability of selecting a purple marble at random would be 2/3: $$\frac{p-3}{b+p-3}=\frac{2}{3}=\frac{10}{15}$$;

$$\frac{9}{15}<\frac{10}{15}$$ --> we removed 3 purple marbles and the probability of selecting a purple marble increased: this can not be true.

So there is something wrong with this question.

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24 Oct 2010, 05:10
yeah ... it gives negative number of purples! What do we select as the answer if at all a question like this comes up? I selected E

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24 Oct 2010, 05:14
hemanthp wrote:
yeah ... it gives negative number of purples! What do we select as the answer if at all a question like this comes up? I selected E

You won't see such question on GMAT.
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24 Oct 2010, 12:17
DS question asks if we have sufficient information or not. in this context can't we opt for C as it gives information regardless what sign comes at the end.
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24 Oct 2010, 12:27
shrive555 wrote:
DS question asks if we have sufficient information or not. in this context can't we opt for C as it gives information regardless what sign comes at the end.

Technically yes, we can find the value of b with (1)+(2) but again the question is flawed and there is no way you'll see such question on real test.
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25 Oct 2010, 10:02
tricky question though !

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Re: A box contains blue and purple marbles only. What is the [#permalink]

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03 Jan 2017, 18:44
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Re: A box contains blue and purple marbles only. What is the   [#permalink] 03 Jan 2017, 18:44
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