actleader wrote:

A box of balls originally contained 2 blue balls for every red ball. If after 12 red balls were added to the box, the ratio of red balls to blue balls became 5 to 2, how many balls were in the box before the additional 12 balls were added?

(A) 9

(B) 12

(C) 15

(D) 18

(E) 24

Another 2-variable solution...

Let R = the ORIGINAL number of red balls in the box

Let B = the ORIGINAL number of blue balls in the box

A box of balls originally contained 2 blue balls for every red ball.In other words, the ratio of red balls to blue balls is 1/2 (aka 1 : 2)

So, we can write: R/B = 1/2

Cross multiply to get:

B = 2RIf after 12 red balls were added to the box, the ratio of red balls to blue balls became 5 to 2After adding 12 red balls, the number of red balls = R + 12

The number of blue balls is still B

So, we can write: (R + 12)/B = 5/2

Cross multiply to get: 2(R + 12) = 5B

Expand to get:

2R + 24 = 5BWe now have:

B = 2R2R + 24 = 5BTake the bottom equation and replace B with 2R to get: 2R + 24 = 5(2R)

Simplify: 2R + 24 = 10R

Solve to get:

R = 3If

R = 3 and we know that

B = 2R, then

B = 6How many balls were in the box before the additional 12 balls were added?In other words, what is the value of R + B?

R + B =

3 +

6 = 9

Answer: A

Cheers,

Brent

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