actually to come to think of it, you don't even need to solve all this to know this question is wrong.

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

Answer is (c) 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.