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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 [#permalink]

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New post 31 Oct 2012, 22:40
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Question Stats:

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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
[Reveal] Spoiler: OA
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Re: A box of balls originally contained 2 blue balls for every r [#permalink]

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New post 31 Oct 2012, 23:37
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actleader 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


No. of blue balls = B, No. of red balls = R,

So, \(B = 2R\)

\(\frac{R+12}{B} = \frac{5}{2}\)

\(\frac{R+12}{R} = 5\)

\(R = 3 ; B = 6 ; Total = 9\)

Answer is A.

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Re: A box of balls originally contained 2 blue balls for every r [#permalink]

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New post 01 Nov 2012, 00:52
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actleader 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

There is another quicker way to solve most of such problems. Note earlier the balls were in 2:1 ratio and after adding 12 balls the ratio is 5:2. What does it signify? the number of balls earlier were a multiple of 3 and now a multiple of 7. take LCM, Total number of balls now must be 21.

Hence Total number of balls before must be 21-12 =9.
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Re: A box of balls originally contained 2 blue balls for every r [#permalink]

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New post 02 Nov 2012, 12:40
Vips0000 wrote:
actleader wrote:
There is another quicker way to solve most of such problems. Note earlier the balls were in 2:1 ratio and after adding 12 balls the ratio is 5:2. What does it signify? the number of balls earlier were a multiple of 3 and now a multiple of 7. take LCM, Total number of balls now must be 21.

Hence Total number of balls before must be 21-12 =9.


Yep. I'd like such methods as you exampled, but where I can find smth about it?
I read smth about elimination in Princeton (Crack the GMAT) they call PI and PITA,
but as for me it is too little to become trained enough.
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Re: A box of balls originally contained 2 blue balls for every r [#permalink]

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New post 23 Aug 2013, 03:41
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No. of Red ball=x, Blue Ball=2x
After 12 red balls were added to the box, the ratio of red balls to blue balls=
(X+12)/2x=5/2.......x=3
Red=3
Blue=2.3=6
Total Ball=3+6=9
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Re: A box of balls originally contained 2 blue balls for every r [#permalink]

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New post 07 Sep 2014, 01:36
this ques took from which book ?
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Re: A box of balls originally contained 2 blue balls for every r [#permalink]

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New post 08 Sep 2014, 01:40
Red .................. Blue ............... Total
x ......................... 2x ................... 3x

Addition of 12 Red balls

x+12 ...................... 2x .................... 3x+12

New Ratio

\(\frac{x+12}{2x} = \frac{5}{2}\)

x = 3

3x = 9

Answer = A
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Re: A box of balls originally contained 2 blue balls for every r [#permalink]

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New post 09 Feb 2018, 11:07
actleader 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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Re: A box of balls originally contained 2 blue balls for every r   [#permalink] 09 Feb 2018, 11:07
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