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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~fluke

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