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

Let the total be N = x^2 + 5, red count R = x+1, scratched S = x-3. After removing 20 scratch-free marbles, the remaining total is N - 20 = x^2 - 15 and scratched still S.

Equate the probabilities:
(x+1)/(x^2+5) = (x-3)/(x^2-15).

Cross-multiply and simplify:
(x+1)(x^2-15) = (x-3)(x^2+5)
=> x^3 + x^2 -15x -15 = x^3 -3x^2 +5x -15
=> 4x^2 -20x = 0
=> 4x(x-5) = 0.

x = 5 (x = 0 gives negative scratched marbles, so discard). Thus the original number of marbles is x^2 + 5 = 5^2 + 5 = 30.