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At a certain state university last term, there were p [#permalink]

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19 Jun 2007, 22:40

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Question Stats:

64% (02:36) correct
36% (01:24) wrong based on 134 sessions

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At a certain state university last term, there were p students each of whom paid either the full tuition of x dollars or half the full tuition. What percent of the tuition paid by the p students last term was tuition from students who paid the full tuition?

(1) Of the p students, 20 percent paid the full tuition. (2) The p students paid a total of $91.2 million for tuition last term.

Re: At a certain state university last term, there were p [#permalink]

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20 Jun 2007, 18:07

Obtained from the question, total tuition paid = AX + BX/2
where A is the number of those who paid the full tuition and B is the number of those who paid half the tuition.
X is the total full tuition paid for each student
A+B = P

The question asks for: AX/[AX+0.5BX] = AX/X[A+0.5B]
= A/[A+0.5B]

(1) Of the p students, 20 percent paid the full tuition.
--------------------------------------------------------------
A/P = A/[A+B] = 0.2

A = 0.2A+0.2B
0.8A = 0.2 B --> B = 4A

A/[A+0.5B] = A/[A+0.5x4A] = A/3A = 1/3

statement 1 is sufficient

(2) The p students paid a total of $91.2 million for tuition last term.
------------------------------------------------------------------------------
AX + BX/2 = 91.2

Re: At a certain state university last term, there were p [#permalink]

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17 Apr 2011, 13:05

1

This post received KUDOS

Amit05 wrote:

At a certain state university last term, there were p students each of whom paid either the full tuition of x dollars or half the full tuition. What percent of the tuition paid by the p students last term was tuition from students who paid the full tuition?

(1) Of the p students, 20 percent paid the full tuition. (2) The p students paid a total of $91.2 million for tuition last term.

Let "k" students paid full tuition of "$x"

Tuition paid in full($x)

Tuition paid in half$(x/2)

Total

Number of students

k

p-k

p

Total Sum Received

kx

\(\frac{x}{2}(p-k)\)

\(kx+\frac{x}{2}(p-k)\)

Q: What percent of the tuition paid by the p students last term was tuition from students who paid the full tuition?

Tuition paid by "p" students \(=kx+\frac{x}{2}(p-k)\) Tuition paid by "k" students \(=kx\)

Q: What is \(\frac{kx}{kx+\frac{x}{2}(p-k)}\) OR, if you simplify, What is \(\frac{2k}{p+k}?\)

(1) Of the p students, 20 percent paid the full tuition.

\(\frac{k}{p}=\frac{1}{5}\) \(5k=p\) Substitute it in the main equation: \(\frac{2k}{p+k}=\frac{2k}{5k+k}=\frac{1}{3}=33\frac{1}{3}%\)

Sufficient.

(2) The p students paid a total of $91.2 million for tuition last term.

\(kx+\frac{x}{2}(p-k)=$91.2m\) Even though we have the denominator for \(\frac{kx}{kx+\frac{x}{2}(p-k)}\), we will not be able to find the value of the numerator "kx". Not Sufficient.

Re: At a certain state university last term, there were p [#permalink]

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31 Jul 2014, 10:48

Hello from the GMAT Club BumpBot!

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Re: At a certain state university last term, there were p [#permalink]

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26 Nov 2015, 21:56

What is the percentage of full tuition from total tuition:

Statement 1: 20% of students paid full tuition.

For every student that paid a full tuition of 1, 4 students paid a half tuition of 1/2. Total tuition would be 1 + 4*1/2 = 3. The percentage of full tuition from total tuition is 1 of 3 or 33.3' %.

Sufficient.

Statement 2: Total tuition 92.1 mil. No way to determine percentage of full tuition.

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