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
Source: GMAT Prep Now
Difficulty level: 650-700
Answer: C
A box contains black balls and white balls only.
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- Brent@GMATPrepNow
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Target question: What is the probability that the selected ball is black?Brent@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
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
Answer: C
Cheers,
Brent
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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).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
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.
Answer: C
Jeffrey Miller
Head of GMAT Instruction
[email protected]
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Let's say we have b black balls and w white balls in the box. We want to find b/(b + w).
S1:
b = w + 31. So b/(b+w) is (w + 31)/(2w + 31). We don't know w, so this is insufficient.
S2:
(b - 19)/(b + w - 19) = .6
w = (2b - 38)/3
Subbing that into b/(b + w), we get
b/(b + (2b - 19)/3), or
3b/(5b - 38)
That's closer, but not quite sufficient.
S1 + S2:
S1 told us w = b - 31, and S2 told us w = (2b - 38)/3. That gives us (b - 31) = (2b - 38)/3, which will give us a unique solution for b. Armed with that, we can also get w, so we're done!
S1:
b = w + 31. So b/(b+w) is (w + 31)/(2w + 31). We don't know w, so this is insufficient.
S2:
(b - 19)/(b + w - 19) = .6
w = (2b - 38)/3
Subbing that into b/(b + w), we get
b/(b + (2b - 19)/3), or
3b/(5b - 38)
That's closer, but not quite sufficient.
S1 + S2:
S1 told us w = b - 31, and S2 told us w = (2b - 38)/3. That gives us (b - 31) = (2b - 38)/3, which will give us a unique solution for b. Armed with that, we can also get w, so we're done!
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On test day, of course, don't fall into Brent's trap of bothering to write out those equations! Once you realize they're each linear and they're not identical, you're realize you're in the familiar "two equations, two variables" scenario, and that you'll be able to solve.
Once we have 2 equations with b and w can we just go straight to C ?Matt@VeritasPrep wrote:Let's say we have b black balls and w white balls in the box. We want to find b/(b + w).
S1:
b = w + 31. So b/(b+w) is (w + 31)/(2w + 31). We don't know w, so this is insufficient.
S2:
(b - 19)/(b + w - 19) = .6
w = (2b - 38)/3
Subbing that into b/(b + w), we get
b/(b + (2b - 19)/3), or
3b/(5b - 38)
That's closer, but not quite sufficient.
S1 + S2:
S1 told us w = b - 31, and S2 told us w = (2b - 38)/3. That gives us (b - 31) = (2b - 38)/3, which will give us a unique solution for b. Armed with that, we can also get w, so we're done!
I shouldve read to the end! Ignore my last question pleaseMatt@VeritasPrep wrote:On test day, of course, don't fall into Brent's trap of bothering to write out those equations! Once you realize they're each linear and they're not identical, you're realize you're in the familiar "two equations, two variables" scenario, and that you'll be able to solve.