BrentGMATPrepNow wrote:
A box contains red balls, blue balls, and green balls only. If 16 red balls were removed from the box, the ratio of the number of red balls to the number of blue balls would be 2 to 3. If 24 blue balls were added to the original collection of balls, the ratio of the number of blue balls to the number of green balls would be 5 to 2. What is the ratio of the number of red balls to the number of green balls in the original collection?
A) 5 to 8
B) 3 to 2
C) 2 to 1
D) 8 to 5
E) 5 to 3
Let R = the number of red balls in the (original) collection
Let B = the number of blue balls in the (original) collection
Let G = the number of green balls in the (original) collectionIf 16 red balls were removed from the box, the ratio of the number of red balls to the number of blue balls would be 2 to 3. We can write:
(R - 16)/B = 2/3Cross multiply:
3(R - 16) = 2BExpand:
3R - 48 = 2BIf 24 blue balls were added to the original collection of balls, the ratio of the number of blue balls to the number of green balls would be 5 to 2. We can write:
(B + 24)/G = 5/2Cross multiply:
2(B + 24) = 5GExpand:
2B + 48 = 5GWhat is the ratio of the number of red balls to the number of green balls in the original collection?In other words, what is the value of
R/G?
We have the following system of equations:
3R - 48 = 2B2B + 48 = 5GTake the top equation and add 48 both sides to get:
3R = 2B + 482B + 48 = 5GSince
3R and
5G are both set equal to
2B + 48, it must be the case that:
3R = 5GDivide both sides of the equation by G to get:
3R/G = 5Divide both sides of the equation by 3 to get:
R/G = 5/3Answer: E