Alejandra is designing a game of chance. For one part of the game, a

Question

Alejandra is designing a game of chance. For one part of the game, a player is to randomly choose 3 marbles, without replacement, from a box containing  B  blue marbles,  R  red marbles, and no other marbles. Alejandra correctly determined the positive integers B and R so that the number of possible selections in which 2 blue marble and 1 red marbles are chosen is twice the number of possible selections in which 1 blue marbles and 2 red marble are chosen.

The positive integers B and R that Alejandra determined must be such that  B  is the number that is __1__ the number that is __ 2 __  R .

Based on the information provided, select for 1 and for 2 the options that create the most accurate statement. Make only two selections, one in each column.

EugeneTheGuy · Accepted Answer

The formula is basically like this :
=> (B/B+R) x (B-1/B+R-1) x (R/B+R-2) = 2 x (B/B+R) x (R/B+R-1) x (R-1/B+R-2)
When we ignore the denominator, it becomes this
=> B^2R - BR = 2BR^2-2BR
=>BR(2R-1-B)=0
Since we only need the ratio between B and R, so we neglect BR.
=> B=2R-1 ##

This is how I solved it. Is there any faster method?

HarshavardhanR · Answer

GMAT-Club-Forum-q2x9o15m.png ___ Harsha

andrewjohn8 · Answer

similar question here - though the differences are strange. https://gmatclub.com/forum/alejandra-is ... 22148.html

Dbrunik · Answer

can you elaborate and break this down more please?

KarishmaB · Answer

B blue marbles, R red marbles P(2 Blue and 1 Red) = \frac{BC2 * RC1 }{ (B+R)C3} P(1 Blue and 2 Red) = \frac{BC1 * RC2 }{ (B+R)C3} Given: BC2 * RC1 = 2 * BC1 * RC2 \frac{B(B-1)}{2} * R = 2 * B * \frac{R(R - 1)}{2} B + 1 = 2R Let's say B + 1 = 2R = x So B is a number that is 1 less than the number (x) that is twice R. Try this same concept on another such question here: https://youtu.be/Mk9mabFnKHU

ravjaz · Answer

Hi, I thought this answer was so useful but I do not understand why we remove the denominators 2 in the second step, how is it possible that we can just remove those?

Bunuel · Answer

We don’t just drop the denominators, they cancel out.

Left-hand side:  \frac{B(B-1)R}{2} .

Right-hand side:  \frac{2BR(R-1)}{2} .

\frac{B(B-1)R}{2}=\frac{2BR(R-1)}{2}

BR(B-1)=2BR(R-1)

B-1=2(R-1)

B+1=2R

Stormik1 · Answer

@Bunel, the question should ask about the probability and not the number of selection. Please rectify.

Bunuel · Answer

The question is correct as written. It asks for the  number of possible selections , which is exactly what the combinations count.

If the question asked for probabilities instead, we would divide both counts by the same total number of possible 3-marble selections, so that denominator would cancel and give the same equation.

1	2
		2 more than
		1 more than
		1 less than
		2 less than
		half
		twice

Alejandra is designing a game of chance. For one part of the game, a

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