You have a bag which contains only red and green marbles. In this bag with x^2+5 marbles total, x+1 are red. Also, x-3 marbles have a scratch on them. The probability of drawing a red marble from the original bag is equal to that of drawing a marble with a scratch from the marbles left in the bag after twenty scratch-free marbles are taken out of the full bag. How many marbles were originally in the bag?

23 answers Show only top 10 answers

It's 30, not 50, liar!

its 30 liar

its 30

The answer's freaking 30

it's 30

30!!!!

the answer is 30. :/

It is 30

ok it's 30

Actually, it's 30. I checked on AoPS.

its 30 people who say 50

First of all don't cheat

It is 30 i can confirm 100% it is 30

THE FRIGGIN ANSWER IS 30 BELIEVE ME

first of all 50 is not even possible

Yeah it's 30

Here's the solution:
From the given equality of probabilities, we get $\frac{x+1}{x^{2}+5} = \frac{x-3}{x^{2}+5-20}$. Cross multiplying, we see that this is equivalent to $(x+1)(x^{2}-15) = (x^{2}+5)(x-3)$. Multiplying out each side with the distributive property, we obtain $x^{3}+x^{2}-15x-15 = x^{3}-3x^{2}+5x-15$. The $x^{3}$ terms cancel, and we end up with $4x^{2}=20x$. Dividing by 4 gives $x^2 = 5x$. Rearranging gives $x^2 - 5x = 0$, so $x(x-5)=0$, which gives us the solutions $x = 0$ and $x= 5$. However, since $x$ must be at least 3, we have $x=5$. Finally, we are asked for how many marbles were originally in the bag, which is $x^{2}+5 = \fbox{30}$.