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

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At a certain state university last term, there were p [#permalink]  19 Jun 2007, 21:40
00:00

Difficulty:

55% (hard)

Question Stats:

63% (02:36) correct 37% (01:18) wrong based on 50 sessions
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. [Reveal] Spoiler: OA Math Forum Moderator Joined: 20 Dec 2010 Posts: 2027 Followers: 142 Kudos [?]: 1213 [1] , given: 376 Re: At a certain state university last term, there were p [#permalink] 17 Apr 2011, 12:05 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.

Ans: "A"
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Re: At a certain state university last term, there were p [#permalink]  20 Jun 2007, 00:00
St1:
Total tuition paid = 0.2px + 0.2*0.5px = 0.3px
full tuition = 0.2px
We can find the %. Sufficient.

St2:
Useless.

Ans A
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Re: At a certain state university last term, there were p [#permalink]  20 Jun 2007, 12:49
A is sufficient
B is not as you don't know what the original amount of tution is.
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Re: At a certain state university last term, there were p [#permalink]  20 Jun 2007, 17:07
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

So ?

statement 2 is insufficient

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Re: At a certain state university last term, there were p [#permalink]  17 Apr 2011, 20:46
Let n pay the full tuition

Fees paid = nx + (p-n)x/2

nx/{nx + (p-n)x/2} * 100 = ?

From (1) :

n = 0.2p, which is sufficient as the variables cancel froom Nominator and denominator

From(2)

nx + (p-n)x/2 = 91.2 * 10^6, which is not sufficient, as we don't know the break-up of nx and (p-n)x/2

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Re: At a certain state university last term, there were p [#permalink]  31 Jul 2014, 09:48
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Re: At a certain state university last term, there were p   [#permalink] 31 Jul 2014, 09:48
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