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A ball is randomly selected from a box containing white balls and

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A ball is randomly selected from a box containing white balls and [#permalink]

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14 Feb 2017, 07:06
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A ball is randomly selected from a box containing white balls and black balls only. If the probability of randomly selecting a white ball is 4/5, how many white balls must be added to the box so that the probability of randomly drawing a white ball is 7/8?

(1) The ratio of white balls to black balls is 4:1
(2) There are 27 more white balls than black balls

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[Reveal] Spoiler: OA

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Re: A ball is randomly selected from a box containing white balls and [#permalink]

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14 Feb 2017, 07:59
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GMATPrepNow wrote:
A ball is randomly selected from a box containing white balls and black balls only. If the probability of randomly selecting a white ball is 4/5, how many white balls must be added to the box so that the probability of randomly drawing a white ball is 7/8?

(1) The ratio of white balls to black balls is 4:1
(2) There are 27 more white balls than black balls

* kudos for all correct solutions

Hi

We are given (w - # of white blls, b - # of black balls, x - # of white balls to be added):

$$\frac{w}{w+b} = \frac{4}{5}$$

$$w = 4b$$ or $$w:b = 4:1$$

and

$$\frac{w + x}{w + b + x} = \frac{7}{8}$$

$$\frac{4b + x}{5b + x} = \frac{7}{8}$$

$$x = 3b$$

Now let's lok at our options:

(1) does not give us any additional information, we already know that w:b is 4:1. Insufficient.

(2) w = b + 27, combinig this with w=4b we get b=9 and x=27. Sufficient.

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A ball is randomly selected from a box containing white balls and [#permalink]

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14 Feb 2017, 08:15
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GMATPrepNow wrote:
A ball is randomly selected from a box containing white balls and black balls only. If the probability of randomly selecting a white ball is 4/5, how many white balls must be added to the box so that the probability of randomly drawing a white ball is 7/8?

(1) The ratio of white balls to black balls is 4:1
(2) There are 27 more white balls than black balls

* kudos for all correct solutions

Hi,

Let the ratio be $$\frac{4x}{5x}$$...
Add y white balls...
$$\frac{4x+y}{5x+y}=\frac{7}{8}$$..
$$32x+8y=35x+7y.......3x=y$$..
So y has to be multiple of 3.

Let's see the statements.
(1) The ratio of white balls to black balls is 4:1
the probability 4/5 itself means the ratio to be 4:1..
Nothing new..
Insufficient

(2) There are 27 more white balls than black balls

So 4x-x=27.... x=9..
Substitute in 3x=y...So y = 3*9=27
B

Edited..
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A ball is randomly selected from a box containing white balls and [#permalink]

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14 Feb 2017, 08:17
Ans. A tells me nothing new, so, just noodling with multiples of 4 & differences of 27 got me, eventually, to 36 & 9, whose sum is a multiple of 5, so there's the 4/5 probabilty . . . but Bunuel, et. al. will no doubt have a more elegant way. Also to get to the required 7/8, you will add . . . 27! This stuff is deep!

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Re: A ball is randomly selected from a box containing white balls and [#permalink]

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14 Feb 2017, 10:10
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GMATPrepNow wrote:
A ball is randomly selected from a box containing white balls and black balls only. If the probability of randomly selecting a white ball is 4/5, how many white balls must be added to the box so that the probability of randomly drawing a white ball is 7/8?

(1) The ratio of white balls to black balls is 4:1
(2) There are 27 more white balls than black balls

Target question: How many white balls must be added to the box so that the probability of randomly drawing a white ball is 7/8?

Given: The probability of randomly selecting a white ball is 4/5
Let W = number of white balls in the box
Let B = number of black balls in the box
So, the TOTAL number of balls = W+B
If the probability of selecting a white ball = 4/5, then we can write: W/(W+B) = 4/5
Cross multiply to get: 4(W+B) = 5W
Expand: 4W + 4B = 5W
Simplify: 4B = W

Statement 1: The ratio of white balls to black balls is 4:1
In other words: W/B = 4/1
Cross multiply to get 4B = W
This MATCHES the given information. In other words, statement 1 provides no new information.
As such, statement 1 is NOT SUFFICIENT

Statement 2: There are 27 more white balls than black balls
In other words, W = B + 27
Now take 4B = W and replace W with B+27, to get: 4B = B + 27
Solve to get: B = 9
So, there are presently 9 blacks in the box, which means there are 36 white balls in the box.
Now that we know exactly how many white and black balls are in the box, we can just keep add white balls to the box until P(selecting a white ball) = 7/8
So, we COULD answer the target question with certainty.
As such, statement 2 is SUFFICIENT

ASIDE: For "fun," let's determine how many white balls we need to add.
We presenty have 36 white balls and 9 black balls for a total of 45 balls.
Let's add x white balls to the box
So, 36+x = the new number of white balls in the box
And 45+x = TOTAL number of balls in the box
We want P(white ball) = 7/8
We can write: (36+x)/(45+x) = 7/8
Cross multiply to get: 8(36+x) = 7(45+x)
Expand: 288 + 8x = 315 + 7x
Solve: x = 27
So, we must add 27 white balls to the box so that P(white ball) = 7/8

Cheers,
Brent
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Re: A ball is randomly selected from a box containing white balls and [#permalink]

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14 Feb 2017, 10:21
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chetan2u wrote:
GMATPrepNow wrote:
A ball is randomly selected from a box containing white balls and black balls only. If the probability of randomly selecting a white ball is 4/5, how many white balls must be added to the box so that the probability of randomly drawing a white ball is 7/8?

(1) The ratio of white balls to black balls is 4:1
(2) There are 27 more white balls than black balls

* kudos for all correct solutions

Hi,

Ans is B but GMATPrepNow, statement II may not be correct completely that is the difference in white black balls will not be 27 but has to be some multiple of 8..
Below is the reason why!

Let the ratio be $$\frac{4x}{5x}$$...
Add y white balls...
$$\frac{4x+y}{5x}=\frac{7}{8}$$..
$$32x+8y=35x.......3x=8y$$..
So x has to be multiple of 8..

Let's see the statements..

(1) The ratio of white balls to black balls is 4:1
the probability 4/5 itself means the ratio to be 4:1..
Nothing new..
Insufficient

(2) There are 27 more white balls than black balls

So 4x-x=27.... x=9..
Substitute in 3x=8y...So y = 3*9/8..
B

Although B is sufficient but the balls cannot be in fraction..
So number 27 should be 24 or something of similar properties..

It took my quite a while to find the issue, but I believe there's a problem with your solution (highlighted above in red)

In your expression 4x/5x, the 4x represents the number of white balls and 5x represents the total number of balls
So, if we add y white balls, 4x+y represents the NEW number of white balls and 5x+y represents the NEW total number of balls
Your solution has just 5x as the NEW total number of balls

My solution (above this post) shows that the given numbers yield integer values for the numbers of balls.

ASIDE: Although it may take a little longer to do so, I prefer to use two variables when trying to solve these kinds of ratio/proportion questions. Otherwise, I end up forgetting what the variables stand for.

Cheers,
Brent
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Re: A ball is randomly selected from a box containing white balls and [#permalink]

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15 Feb 2017, 16:18
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GMATPrepNow wrote:
A ball is randomly selected from a box containing white balls and black balls only. If the probability of randomly selecting a white ball is 4/5, how many white balls must be added to the box so that the probability of randomly drawing a white ball is 7/8?

(1) The ratio of white balls to black balls is 4:1
(2) There are 27 more white balls than black balls

We are given that in a box containing only white and black balls, the probability of selecting a white ball is 4/5 and thus the probability of selecting a black ball is 1/5. We must determine how many white balls must be added to the box so the probability of drawing a white ball is 7/8.

Statement One Alone:

The ratio of white balls to black balls is 4:1.

This means that for some positive integer x, there are 4x white balls and x black balls in the box. Thus, there are a total of 5x balls in the box and the probability of selecting a white ball is (4x)/(5x) = 4/5. However, since we already know that the probability of selecting a white ball is 4/5, statement one does not provide any new information and thus is not sufficient to answer the question.

Statement Two Alone:

There are 27 more white balls than black balls.

We can let b = the number of black balls and w = the number of white balls, and thus:

w = b + 27

Furthermore we know:

w/(b+w) = 4/5

(b + 27)/(b + b + 27) = 4/5

(b + 27)/(2b + 27) = 4/5

5(b + 27) = 4(2b + 27)

5b + 135 = 8b + 108

27 = 3b

b = 9

Since b = 9 and w = 9 + 27 = 36, we can determine the number of white marbles that must be added to the box so the probability of selecting a white marble is 7/8.

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Re: A ball is randomly selected from a box containing white balls and [#permalink]

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15 Feb 2017, 16:20
Great post thank you !

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A ball is randomly selected from a box containing white balls and [#permalink]

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08 Apr 2017, 09:45
A ball is randomly selected from a box containing white balls and black balls only. If the probability of randomly selecting a white ball is 4/5, how many white balls must be added to the box so that the probability of randomly drawing a white ball is 7/8?

(1) The ratio of white balls to black balls is 4:1

ratio 4:1
the no of balls can be as below:
w : b
4 1
8 2 - in this case if we add 6 w balls ratio the probability of randomly drawing a white ball is 7/8.
12 3 - in this case if we add 9 w balls ratio the probability of randomly drawing a white ball is 7/8.
.......

.......

36 9

NS

(2) There are 27 more white balls than black balls

And, . If the probability of randomly selecting a white ball is 4/5..
definite no of w, b balls are known.

w : b

36 9- in this case if we add 27 w balls ratio the probability of randomly drawing a white ball is 7/8.

suff.

Ans. B

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Re: A ball is randomly selected from a box containing white balls and [#permalink]

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04 Jul 2017, 04:34
vitaliyGMAT wrote:
GMATPrepNow wrote:
A ball is randomly selected from a box containing white balls and black balls only. If the probability of randomly selecting a white ball is 4/5, how many white balls must be added to the box so that the probability of randomly drawing a white ball is 7/8?

(1) The ratio of white balls to black balls is 4:1
(2) There are 27 more white balls than black balls

* kudos for all correct solutions

Hi

We are given (w - # of white blls, b - # of black balls, x - # of white balls to be added):

$$\frac{w}{w+b} = \frac{4}{5}$$

$$w = 4b$$ or $$w:b = 4:1$$

and

$$\frac{w + x}{w + b + x} = \frac{7}{8}$$

$$\frac{4b + x}{5b + x} = \frac{7}{8}$$

$$x = 3b$$

Now let's lok at our options:

(1) does not give us any additional information, we already know that w:b is 4:1. Insufficient.

(2) w = b + 27, combinig this with w=4b we get b=9 and x=27. Sufficient.

Hi,

I have a question here. We already know that w:b=4:1 - agreed. But in statement 2, we use w=4b for the solution. As w=4b was also given in statement 1, I landed up opting choice C. I hope you understood my question. Please help

Thanks,
Uma

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Re: A ball is randomly selected from a box containing white balls and [#permalink]

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23 Sep 2017, 23:33
GMATPrepNow wrote:
A ball is randomly selected from a box containing white balls and black balls only. If the probability of randomly selecting a white ball is 4/5, how many white balls must be added to the box so that the probability of randomly drawing a white ball is 7/8?

(1) The ratio of white balls to black balls is 4:1
(2) There are 27 more white balls than black balls

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Tricky

You don't need statement 1 because it is already implied in the question stem

B

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A ball is randomly selected from a box containing white balls and [#permalink]

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03 Oct 2017, 23:24
Statement 1 says nothing new, i.e 4b= w.

Statement 2 says w-b = 27, combined with 4b=w (not exclusively from 1) b= 9, w= 36 and b+w = 45. Also makes sense, multiple of 5, just need to add 3b (27) worth of w to change the probability from 4/5 to 7/8.

Hence B.

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A ball is randomly selected from a box containing white balls and   [#permalink] 03 Oct 2017, 23:24
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