Question

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?

Answer 1

(x + 1) / (x^2 + 5) = (x - 3) / (x^2 - 15)

x^3 + x^2 - 15x - 15 = x^3 - 3x^2 + 5x - 15

4 x^2 - 20 x = 0 ... x^2 - 5 x = 0

Answer 2

Thanks!

Answer 3

It's not 0.

Answer 4

ok it's 30

Answer 5

The answer is actually 50, stop cheating

Answer 6

its 30 liar

Answer 7

It's 30, not 50, liar!

Answer 8

The answer's freaking 30

Answer 9

it's 30

Answer 10

30!!!!

Answer 11

its 30

Answer 12

the answer is 30. :/

Answer 13

It is 30

Answer 14

no 50

Answer 15

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

Answer 16

its 30 people who say 50

Answer 17

First of all don't cheat

Answer 18

THE FRIGGIN ANSWER IS 30 BELIEVE ME

Answer 19

It is 30 i can confirm 100% it is 30

Answer 20

first of all 50 is not even possible

Answer 21

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 $x^{2}+5 = \fbox{30}$.

Answer 22

It's 30, lol.

Answer 23

its 30 yall stop cheating