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GMAT Club Legend  V
Joined: 12 Sep 2015
Posts: 4121
A box contains black balls and white balls only. If a ball is randomly  [#permalink]

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4 00:00

Difficulty:   35% (medium)

Question Stats: 72% (01:43) correct 28% (01:34) wrong based on 285 sessions

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A box contains black balls and white balls only. If a ball is randomly selected from the box, what is the probability that the selected ball is black?

(1) There are 31 more black balls than white balls in the box
(2) If 19 black balls were removed from the box, the probability of selecting a black ball would be 0.6

*kudos for all correct solutions

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Math Expert V
Joined: 02 Aug 2009
Posts: 8282
Re: A box contains black balls and white balls only. If a ball is randomly  [#permalink]

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GMATPrepNow wrote:
A box contains black balls and white balls only. If a ball is randomly selected from the box, what is the probability that the selected ball is black?

(1) There are 31 more black balls than white balls in the box
(2) If 19 black balls were removed from the box, the probability of selecting a black ball would be 0.6

*kudos for all correct solutions

Hi everyone..

What are we looking for :-
either the number of both black and white balls..
OR just a ratio between the two
Or ratio of anyone with the TOTAL
..

Let's see the statements..

(1) There are 31 more black balls than white balls in the box
Nothing can be found as it does not give any information required on top..
Just one equation B=31+W
Insufficient.

(2) If 19 black balls were removed from the box, the probability of selecting a black ball would be 0.6
If numbers are B and W, then $$\frac{B-19}{B+W-19}=0.6=\frac{6}{10}=\frac{3}{5}$$...
5B-19*5=4B+3W-3*19.....B=3W+38
Again insufficient..

Combined..
We have two different equations and two variables, we can find value of number of each colour..
Suff

C
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Intern  B
Joined: 03 Jan 2017
Posts: 29
Re: A box contains black balls and white balls only. If a ball is randomly  [#permalink]

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1
Option A: insufficient since we don't know the number of white balls.

Option B:
B-19= 0.6(B-19+W)

We get the ratio of black to the total balls. As 0.6= 3/5
We get the ratio of black balls to the total balls.
Sufficient.
Ans B.

Sent from my MotoG3 using GMAT Club Forum mobile app
GMAT Club Legend  V
Joined: 12 Sep 2015
Posts: 4121
Re: A box contains black balls and white balls only. If a ball is randomly  [#permalink]

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Top Contributor
1
GMATPrepNow wrote:
A box contains black balls and white balls only. If a ball is randomly selected from the box, what is the probability that the selected ball is black?

(1) There are 31 more black balls than white balls in the box
(2) If 19 black balls were removed from the box, the probability of selecting a black ball would be 0.6

Target question: What is the probability that the selected ball is black?

Given: A box contains black balls and white balls only.
Let B = number of black balls in the box
Let W = number of white balls in the box
This means B+W = TOTAL number of balls in the box

So, P(selected ball is black) = B/(B+W)

Statement 1: There are 31 more black balls than white balls in the box
We can write: B = W + 31
Is this enough information to determine the value of B/(B+W)?
No.
Replace B with W + 31 to get: B/(B+W) = (W + 31)/(W + 31 + W)
= (W + 31)/(2W + 31)
Notice that, as we change the value of W, the probability changes.
For example, when W = 1, we get: (W + 31)/(2W + 31) = 32/33
When W = 2, we get: (W + 31)/(2W + 31) = 33/35
etc.
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: If 19 black balls were removed from the box, the probability of selecting a black ball would be 0.6
If 19 black balls were removed from the box, then B - 19 = new number of black balls
And the TOTAL number of balls would be B + W - 19
So, we can write: (B - 19)/(B + W - 19) = 0.6
Or we can write: we can write: (B - 19)/(B + W - 19) = 3/5

Is this enough information to determine the value of B/(B+W)?
No.
Cross multiply to get: 5(B - 19) = 3(B + W - 19)
Expand: 5B - 95 = 3B + 3W - 57
Rearrange to get: 2B - 3W = 38
As you might see, we cannot use this information to determine the value of B/(B+W)
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1 tells us that B = W + 31
Statement 2 tells us that 2B - 3W = 38
So, we have two different equations with 2 variables.
Since we COULD solve this system of equations for B and W, we COULD determine the value of B/(B+W)
Since we COULD answer the target question with certainty, the combined statements are SUFFICIENT

Cheers,
Brent
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Joined: 17 Nov 2016
Posts: 24
Re: A box contains black balls and white balls only. If a ball is randomly  [#permalink]

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GMATPrepNow wrote:
GMATPrepNow wrote:
A box contains black balls and white balls only. If a ball is randomly selected from the box, what is the probability that the selected ball is black?

(1) There are 31 more black balls than white balls in the box
(2) If 19 black balls were removed from the box, the probability of selecting a black ball would be 0.6

Target question: What is the probability that the selected ball is black?

Given: A box contains black balls and white balls only.
Let B = number of black balls in the box
Let W = number of white balls in the box
This means B+W = TOTAL number of balls in the box

So, P(selected ball is black) = B/(B+W)

Statement 1: There are 31 more black balls than white balls in the box
We can write: B = W + 31
Is this enough information to determine the value of B/(B+W)?
No.
Replace B with W + 31 to get: B/(B+W) = (W + 31)/(W + 31 + W)
= (W + 31)/(2W + 31)
Notice that, as we change the value of W, the probability changes.
For example, when W = 1, we get: (W + 31)/(2W + 31) = 32/33
When W = 2, we get: (W + 31)/(2W + 31) = 33/35
etc.
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: If 19 black balls were removed from the box, the probability of selecting a black ball would be 0.6
If 19 black balls were removed from the box, then B - 19 = new number of black balls
And the TOTAL number of balls would be B + W - 19
So, we can write: (B - 19)/(B + W - 19) = 0.6
Or we can write: we can write: (B - 19)/(B + W - 19) = 3/5

Is this enough information to determine the value of B/(B+W)?
No.
Cross multiply to get: 5(B - 19) = 3(B + W - 19)
Expand: 5B - 95 = 3B + 3W - 57
Rearrange to get: 2B - 3W = 38
As you might see, we cannot use this information to determine the value of B/(B+W)
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1 tells us that B = W + 31
Statement 2 tells us that 2B - 3W = 38
So, we have two different equations with 2 variables.
Since we COULD solve this system of equations for B and W, we COULD determine the value of B/(B+W)
Since we COULD answer the target question with certainty, the combined statements are SUFFICIENT

Cheers,
Brent

Hi Brent,

From Statement B, I believe I can infer the probability of w

If B-19/B+W-19= 0.6
then
W/B+W-19= 0.4

Then, I can calculate both W and B here.

What am I missing?
Math Expert V
Joined: 02 Aug 2009
Posts: 8282
Re: A box contains black balls and white balls only. If a ball is randomly  [#permalink]

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1
Zoser wrote:
GMATPrepNow wrote:
GMATPrepNow wrote:
A box contains black balls and white balls only. If a ball is randomly selected from the box, what is the probability that the selected ball is black?

(1) There are 31 more black balls than white balls in the box
(2) If 19 black balls were removed from the box, the probability of selecting a black ball would be 0.6

Target question: What is the probability that the selected ball is black?

Given: A box contains black balls and white balls only.
Let B = number of black balls in the box
Let W = number of white balls in the box
This means B+W = TOTAL number of balls in the box

So, P(selected ball is black) = B/(B+W)

Statement 1: There are 31 more black balls than white balls in the box
We can write: B = W + 31
Is this enough information to determine the value of B/(B+W)?
No.
Replace B with W + 31 to get: B/(B+W) = (W + 31)/(W + 31 + W)
= (W + 31)/(2W + 31)
Notice that, as we change the value of W, the probability changes.
For example, when W = 1, we get: (W + 31)/(2W + 31) = 32/33
When W = 2, we get: (W + 31)/(2W + 31) = 33/35
etc.
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: If 19 black balls were removed from the box, the probability of selecting a black ball would be 0.6
If 19 black balls were removed from the box, then B - 19 = new number of black balls
And the TOTAL number of balls would be B + W - 19
So, we can write: (B - 19)/(B + W - 19) = 0.6
Or we can write: we can write: (B - 19)/(B + W - 19) = 3/5

Is this enough information to determine the value of B/(B+W)?
No.
Cross multiply to get: 5(B - 19) = 3(B + W - 19)
Expand: 5B - 95 = 3B + 3W - 57
Rearrange to get: 2B - 3W = 38
As you might see, we cannot use this information to determine the value of B/(B+W)
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1 tells us that B = W + 31
Statement 2 tells us that 2B - 3W = 38
So, we have two different equations with 2 variables.
Since we COULD solve this system of equations for B and W, we COULD determine the value of B/(B+W)
Since we COULD answer the target question with certainty, the combined statements are SUFFICIENT

Cheers,
Brent

Hi Brent,

From Statement B, I believe I can infer the probability of w

If B-19/B+W-19= 0.6
then
W/B+W-19= 0.4

Then, I can calculate both W and B here.

What am I missing?

HI...

$$\frac{B-19}{B+W-19}= 0.6$$

subtract BOTH sides from 1..
$$1-\frac{B-19}{B+W-19}=1- 0.6$$
$$\frac{B+W-19-(B-19)}{B+W-19}= 0.4....................\frac{W}{B+W-19}=0.4$$, which is your second equation..
so finally you have just ONE equation and TWO variables...

it is same as a+b=1....... 1-(a+b)=0
these are not two equation but same..

Hope it helps
_________________
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Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 8606
Location: United States (CA)
Re: A box contains black balls and white balls only. If a ball is randomly  [#permalink]

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1
GMATPrepNow wrote:
A box contains black balls and white balls only. If a ball is randomly selected from the box, what is the probability that the selected ball is black?

(1) There are 31 more black balls than white balls in the box
(2) If 19 black balls were removed from the box, the probability of selecting a black ball would be 0.6

We need to determine the probability of selecting a black ball from a box containing only white and black balls. If we let b = the number of black balls in the box and w = the number of white balls in the box, then the probability of selecting a black ball is b/(b + w).

Statement One Alone:

There are 31 more black balls than white balls in the box.

This means b = w + 31 or w = b - 31. Substituting b - 31 for w in b/(b + w), we have:

b/(b + b - 31)

b/(2b - 31)

However, since we don’t know the value of b, we cannot determine the probability of selecting a black ball from the box. Statement one alone is not sufficient to answer the question.

Statement Two Alone:

If 19 black balls were removed from the box, the probability of selecting a black ball would be 0.6.

We can create the following equation:

(b - 19)/(b - 19 + w) = 6/10

(b - 19)/(b - 19 + w) = 3/5

Cross-multiplying, we have:

5(b - 19) = 3(b - 19 + w)

5b - 95 = 3b - 57 + 3w

2b = 38 + 3w

Since we can neither determine the value of b nor the value of w, statement two alone is not sufficient to answer the question.

Statements One and Two Together:

Using statements one and two, we have two equations:

b = w + 31

and

2b = 38 + 3w

We can substitute w + 31 for b in the equation 2b = 152 + 3w and we have:

2(w + 31) = 38 + 3w

2w + 62 = 38 + 3w

24 = w

Thus, b = 24 + 31 = 55, and therefore, the probability of selecting a black ball is 55/(55 + 24) = 55/79.

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