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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:

35% (medium)

Question Stats:

65% (02:38) correct 35% (01:02) wrong based on 17 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: 2048 Followers: 128 Kudos [?]: 899 [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" _________________ GMAT Club Legend Joined: 07 Jul 2004 Posts: 5097 Location: Singapore Followers: 17 Kudos [?]: 136 [0], given: 0 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 Senior Manager Joined: 21 Jun 2006 Posts: 289 Followers: 1 Kudos [?]: 14 [0], given: 0 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. Director Joined: 30 Nov 2006 Posts: 592 Location: Kuwait Followers: 11 Kudos [?]: 145 [0], given: 0 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

So the question asks :

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

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