AMC12 2001 A

AMC12 2001 A · Q7

AMC12 2001 A · Q7. It mainly tests Fractions, Divisibility & factors.

A charity sells 140 benefit tickets for a total of \$2001. Some tickets sell for full price (a whole dollar amount), and the rest sell for half price. How much money is raised by the full-price tickets?

一个慈善机构出售了140张募捐票，总收入为\$2001。有些票按全价出售（票价为整数美元），其余的票按半价出售。全价票筹集了多少钱？

(A) $782$ $782$

(B) $986$ $986$

(D) $1219$ $1219$

(E) $1449$ $1449$

Answer

Correct choice: (A)

正确答案：（A）

Solution

(A) Let $n$ be the number of full-price tickets and $p$ be the price of each in dollars. Then $np+(140-n)\cdot\frac{p}{2}=2001$, so $p(n+140)=4002$. Thus $n+140$ must be a factor of $4002=2\cdot3\cdot23\cdot29$. Since $0\le n\le140$, we have $140\le n+140\le280$, and the only factor of $4002$ that is in the required range for $n+140$ is $174=2\cdot3\cdot29$. Therefore, $n+140=174$, so $n=34$ and $p=23$. The money raised by the full-price tickets is $34\cdot23=782$ dollars.

（A）设 $n$ 为全价票的张数，$p$ 为每张票的价格（美元）。则 $np+(140-n)\cdot\frac{p}{2}=2001$，所以 $p(n+140)=4002$。因此 $n+140$ 必须是 $4002=2\cdot3\cdot23\cdot29$ 的一个因子。由于 $0\le n\le140$，有 $140\le n+140\le280$，而在该范围内，$4002$ 唯一满足条件的因子是 $174=2\cdot3\cdot29$。所以 $n+140=174$，从而 $n=34$ 且 $p=23$。全价票筹得的金额为 $34\cdot23=782$ 美元。

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