BTGmoderatorDC 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

The ratio of red : blue = x : 2x. We can create the following equation:

(x + 12)/2x = 5/2

2(x + 12) = 10x

2x + 24 = 10x

24 = 8x

3 = x

Thus, there were originally 3 + 6 = 9 balls in the box.

Answer: A

BTGmoderatorDC 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

We can also solve by using 2 variables...

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 = 2R

If after 12 red balls were added to the box, the ratio of red balls to blue balls became 5 to 2
After 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 = 5B

We now have:
B = 2R
2R + 24 = 5B

Take the bottom equation and replace B with 2R to get: 2R + 24 = 5(2R)
Simplify: 2R + 24 = 10R
Solve to get: R = 3
If R = 3 and we know that B = 2R, then B = 6

How 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