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# A box contains red balls, blue balls, and green balls only. If 16 red

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Re: A box contains red balls, blue balls, and green balls only. If 16 red [#permalink]
BrentGMATPrepNow wrote:
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) collection

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.
We can write: (R - 16)/B = 2/3
Cross multiply: 3(R - 16) = 2B
Expand: 3R - 48 = 2B

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.
We can write: (B + 24)/G = 5/2
Cross multiply: 2(B + 24) = 5G
Expand: 2B + 48 = 5G

What 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 = 2B
2B + 48 = 5G

Take the top equation and add 48 both sides to get:
3R = 2B + 48
2B + 48 = 5G

Since 3R and 5G are both set equal to 2B + 48, it must be the case that: 3R = 5G
Divide both sides of the equation by G to get: 3R/G = 5
Divide both sides of the equation by 3 to get: R/G = 5/3

Great explanation BrentGMATPrepNow. Not sure is below calculation correct? Thanks Brent
R:B:G
2:3
5:2
10:15:6

R:G= 10:6 >> 5:3
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Re: A box contains red balls, blue balls, and green balls only. If 16 red [#permalink]
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Kimberly77 wrote:

Great explanation BrentGMATPrepNow. Not sure is below calculation correct? Thanks Brent
R:B:G
2:3
5:2
10:15:6

R:G= 10:6 >> 5:3

I believe your correct answer is simply a coincidence, because your solution does not address the removal of 16 red balls, and the addition of 24 blue balls.
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Re: A box contains red balls, blue balls, and green balls only. If 16 red [#permalink]
BrentGMATPrepNow wrote:
Kimberly77 wrote:

Great explanation BrentGMATPrepNow. Not sure is below calculation correct? Thanks Brent
R:B:G
2:3
5:2
10:15:6

R:G= 10:6 >> 5:3

I believe your correct answer is simply a coincidence, because your solution does not address the removal of 16 red balls, and the addition of 24 blue balls.

Get it. Thanks BrentGMATPrepNow for clarification.
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Re: A box contains red balls, blue balls, and green balls only. If 16 red [#permalink]
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Re: A box contains red balls, blue balls, and green balls only. If 16 red [#permalink]
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