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A box of balls originally contained 2 blue balls for every

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by BTGmoderatorDC » Sun Jul 29, 2018 4:29 pm

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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
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by GMATGuruNY » Mon Jul 30, 2018 1:50 am
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
A box of balls originally contained 2 blue balls for every red ball.
Let the number of red balls = x and the number blue balls = 2x, implying that the total number of balls = x+2x = 3x.

After 12 red balls were added to the box, the ratio of red balls to blue balls became 5 to 2.
Since 12 red balls are added, the new number of red balls = x + 12.
Since the ratio of red to blue becomes 5 to 2, we get:
(x+12)/2x = 5/2
2x + 24 = 10x
24 = 8x
x = 3.

Thus, the original number of balls = 3x = 3*3 = 9.

The correct answer is A.
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by Scott@TargetTestPrep » Wed Aug 01, 2018 3:35 pm
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

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by deloitte247 » Wed Aug 01, 2018 3:39 pm
Let the Red balls = x
$$Ratio\ of\ Red\ to\ Blue\ balls\ =\ x\ :\ 2x\ =\ \frac{x}{2x}$$
$$\frac{x\ +12}{2x}\ =\ \frac{5}{2}\ \left(cross\ multiply\right)$$
2(x +12) = 10x
2x +24 = 1ox
24 = 10x - 2x
$$\frac{24}{8}\ =\ \frac{8x}{8}$$
x = 3

Thus, originally there were 3 + 6 =9 balls in the box
Option A is the correct answer.
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by Brent@GMATPrepNow » Thu Aug 02, 2018 10:18 am
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
Brent Hanneson - Creator of GMATPrepNow.com
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