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23 Oct 2010, 22:29
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
62% (01:22) correct
38%
(01:30)
wrong
based on 159
sessions
History
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Not Attempted Yet
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) 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
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24 Oct 2010, 03:49
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shrive555
(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
Answer is (c)
24 Oct 2010, 04:24
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shrouded1
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.
