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If 90 students auditioned for the school musical, how many were accepted?

(1) 2/3 of the boys and 1/3 of the girls who auditioned were accepted --> \(\frac{2}{3}*B+\frac{1}{3}(90-B)\) students were accepted: if there were 30 boys and 60 girls (90-30), then 40 students were accepted but if there were 60 boys and 30 girls, then 50 students were accepted. Not sufficient.

(2) 26 of the boys who auditioned were accepted. Clearly insufficient.

(1)+(2) From (2) we have that \(\frac{2}{3}*B=26\) --> \(B=39\), so \(\frac{2}{3}*39+\frac{1}{3}*51=43\) students were accepted. Sufficient.

Re: If 90 students auditioned for the school musical, how many [#permalink]

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31 Aug 2012, 03:53

The answer is C. The only way to solve this question is to use both information offered. 1) Insufficient. We don't know the ratio of boys to girls, so we cannot calculate the solution. 2)Insufficient. We don't know how many girls were accepted.

Combination of these information enables us to solve the problem.
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Re: If 90 students auditioned for the school musical, how many [#permalink]

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18 Nov 2013, 16:42

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Re: If 90 students auditioned for the school musical, how many [#permalink]

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02 May 2015, 04:40

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Re: If 90 students auditioned for the school musical, how many [#permalink]

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16 Jul 2016, 01:07

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If 90 students auditioned for the school musical, how many were accepted?

(1) 2/3 of the boys and 1/3 of the girls who auditioned were accepted. (2) 26 of the boys who auditioned were accepted.

We are given that 90 students auditioned for a musical, and we need to determine how many were accepted.

Statement One Alone:

2/3 of the boys and 1/3 of the girls who auditioned were accepted.

With statement one, we can set up two equations in which b = the number of the boys who auditioned and g = the number of girls who auditioned:

b + g = 90

(2/3)b + (1/3)g = total accepted

We see we do not have enough information to determine how many students were accepted. Statement one is not sufficient to answer the question. We can eliminate answer choices A and D.

Statement Two Alone:

26 of the boys who auditioned were accepted.

Without knowing how many girls were accepted, we do not have enough information to determine how many total students were accepted. Statement two is not sufficient to answer the question. We can eliminate answer choice B.

Statements One and Two Together:

From statements one and two we know the following:

b + g = 90

(2/3)b + (1/3)g = total accepted

We also know that 26 boys who auditioned were accepted. Thus we can say:

(2/3)b = 26

b = 26 x 3/2

b = 39

Since there are 39 boys and 90 total students, we know there are 90 – 39 = 51 total girls. We also know that 1/3 of the girls who auditioned were accepted. Thus, we know that 51 x 1/3 = 17, is the total number of girls who were accepted. So finally we can say:

A total of 39 boys + 17 girls = 56 students who auditioned were accepted.

The answer is C.
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Scott Woodbury-Stewart Founder and CEO

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Re: If 90 students auditioned for the school musical, how many [#permalink]

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05 Mar 2017, 11:55

I was able to solve this problem using only the given information from choice A as follows: Step 1: (2/3)x + (1/3)x = 90 - x Step 2: x = 90 - x Step 3: 2x = 90 Step 4: x = 45 ~ I used this value to plug into the fractions given from Step 1 Step 5: (2/3)(45) = 30 Step 6: (1/3)(45) = 13 Step 7: 30 + 13 = 43 total were accepted

I understand that my discreet values for male and female differ from the explanations given in the thread, but I am curious how this is incorrect given the question is asking for the total number (inclusive M + F), which strangely enough came out to be the same value as the solution. In short, why does it matter that number of female and male numbers are exact when I was able to come to the same total as the solution?

I was able to solve this problem using only the given information from choice A as follows: Step 1: (2/3)x + (1/3)x = 90 - x Step 2: x = 90 - x Step 3: 2x = 90 Step 4: x = 45 ~ I used this value to plug into the fractions given from Step 1 Step 5: (2/3)(45) = 30 Step 6: (1/3)(45) = 13 Step 7: 30 + 13 = 43 total were accepted

I understand that my discreet values for male and female differ from the explanations given in the thread, but I am curious how this is incorrect given the question is asking for the total number (inclusive M + F), which strangely enough came out to be the same value as the solution. In short, why does it matter that number of female and male numbers are exact when I was able to come to the same total as the solution?

Your initial equation isn't a good translation of the info in the question & statement. What is 'x', specifically, in your equation?

If x is 'the number of boys who auditioned', then you can't use the same variable in both 2/3 x and 1/3 x. That's because you don't know that the same number of girls and boys auditioned. You'd need to write something like '2/3 x + 1/3 y', instead.

Also, where is the 90-x coming from? The problem tells you that 2/3 x + 1/3 y sums to 'the number of students who were accepted'. We don't know that the number of students who were accepted is 90-x. (That would be like saying 'the number of students who were accepted is equal to the number of female students'. If x is the number of male students, then 90-x would be the number of female students.)
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Chelsey Cooley | Manhattan Prep Instructor | Seattle and Online

Re: If 90 students auditioned for the school musical, how many [#permalink]

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05 Mar 2017, 17:29

Hello Chelsey, You asked some great questions, so bear with me here. I did some further thinking on this question after posting my comment and I do completely agree with what you've pointed out. Yes, the population of boys and the population girls represented by x and y or b and g is not given explicitly in the stem so statement 1 most certainly is insufficient (or is it?).

First: 90-x = the number of kids who did not make the cut, so x equals the number of kids making it. Multiplying this x, regardless of sex (I will use gender neutrality as the basis of my argument - ok not really lol) by the fractions will give me a number of heads total, which at the end of the day, it's what the question is asking for. Also notice that the two fractions (2/3) and (1/3) add up to 1, maybe it's a coincidence maybe not. Now taking my value of x and plugging it in to test my values on both sides of the equation, which I didn't do at the time would show that it was indeed insufficient. This was kind of hard to wrap my head around after I had written this post and it's even stranger that I calculated the same total, but I believe it has something to do with the fact that both fractions add up to 1.

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