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A box of balls originally contained 2 blue balls for every r

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A box of balls originally contained 2 blue balls for every r

by M7MBA » Sat Feb 03, 2018 6:43 am
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 OA is the option A.

Experts, may you help me here? I don't know how to start to solve this PS question.

I'd appreciate your help.
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by [email protected] » Sat Feb 03, 2018 10:46 am
Hi MY7MBA,

We're told that a box of balls originally contained 2 blue balls for every red ball and that after 12 red balls were added to the box, the ratio of red balls to blue balls became 5 to 2. We're asked for the TOTAL number of balls in the box BEFORE the additional 12 balls were added. This question can be solved by TESTing THE ANSWERS.

Let's TEST Answer B: 12 total balls
Since we have 2 blue balls for every 1 red ball, that means that 2/3 of the balls are blue.
Starting Blue = (2/3)(12) = 8
Starting Red = 12 - 8 = 4
Adding 12 red balls gives us....
Blue = 12
Red = 4 + 12 + 16
Red:Blue = 16:12 = 4:3
This does NOT match what we were told (the ratio is supposed to be 5:2). To increase this ratio, we need the 'impact' of adding 12 red balls to be greater - which means that we need FEWER red balls to start. There's only one answer that does that, but here's the proof that it's correct:

Let's TEST Answer A: 9 total balls
Since we have 2 blue balls for every 1 red ball, that means that 2/3 of the balls are blue.
Starting Blue = (2/3)(9) = 6
Starting Red = 9 - 6 = 3
Adding 12 red balls gives us....
Blue = 6
Red = 3 + 12 + 15
Red:Blue = 15:6 = 5:2
This is an exact match for what we were told, so this MUST be the answer.

Final Answer: A

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Rich
Contact Rich at [email protected]
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by EconomistGMATTutor » Sat Feb 03, 2018 2:08 pm
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 OA is the option A.

Experts, may you help me here? I don't know how to start to solve this PS question.

I'd appreciate your help.
Hi M7MBA,
Let's take a look at your question.

Let b represents blue balls, r represents red balls and x represents the total balls originally in the box.
A box of balls originally contained 2 blue balls for every red ball can be represented as:
$$r+2b=x\ ...\ \left(i\right)$$

After 12 red balls were added to the box, the ratio of red balls to blue balls became 5 to 2
$$5r+2b=x+12\ ...\ \left(ii\right)$$

Subtracting eq(i) from (ii):
$$5r+2b-\left(r+2b\right)=x+12-x\ $$
$$5r+2b-r-2b=12$$
$$5r-r=12$$
$$4r=12$$
$$r=3$$

Therefore, there are 3 red ball in the box originally.
Since, the box of balls originally contained 2 blue balls for every red ball.
Therefore, there will be 6 blue balls.
Total balls in the box originally = 3 + 6 = 9

Therefore, Option A is correct.

Hope it helps.
I am available if you'd like any follow up.
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by Jeff@TargetTestPrep » Fri Feb 09, 2018 10:14 am
M7MBA 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

Jeffrey Miller
Head of GMAT Instruction
[email protected]

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