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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, 21:40
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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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Re: A box of balls originally contained 2 blue balls for every r  [#permalink]

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New post 31 Oct 2012, 23:52
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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 31 Oct 2012, 22:37
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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 02 Nov 2012, 11:40
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, 02:41
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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, 00:36
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, 00:40
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, 10:07
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]

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New post 02 Aug 2018, 10:26
Top Contributor
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


Another 2-variable solution...

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
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Re: A box of balls originally contained 2 blue balls for every r &nbs [#permalink] 02 Aug 2018, 10:26
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