AMC10 2011 B

AMC10 2011 B · Q6

AMC10 2011 B · Q6. It mainly tests Linear equations, Fractions.

On Halloween Casper ate $\frac{1}{3}$ of his candies and then gave 2 candies to his brother. The next day he ate $\frac{1}{3}$ of his remaining candies and then gave 4 candies to his sister. On the third day he ate his final 8 candies. How many candies did Casper have at the beginning?

在万圣节，Casper 吃了他的糖果的 $\frac{1}{3}$，然后给了兄弟 2 颗糖果。第二天，他吃了剩余糖果的 $\frac{1}{3}$，然后给了姐姐 4 颗糖果。第三天，他吃掉了最后的 8 颗糖果。Casper 一开始有多少糖果？

(A) 30 30

(B) 39 39

(D) 57 57

(E) 66 66

Answer

Correct choice: (A)

正确答案：（A）

Solution

Answer (A): Let $x$ be Casper’s original number of candies. After the first day he was left with $x-\left(\frac{1}{3}x+2\right)=\frac{2}{3}x-2$ candies. On the second day he ate $\frac{1}{3}\left(\frac{2}{3}x-2\right)$ candies, gave away 4 candies, and was left with 8 candies. Therefore $$ \frac{2}{3}x-2-\left(\frac{1}{3}\left(\frac{2}{3}x-2\right)+4\right)=8. $$ Solving for $x$ results in $x=30$.

答案（A）：设 $x$ 为 Casper 最初拥有的糖果数。第一天后他剩下 $x-\left(\frac{1}{3}x+2\right)=\frac{2}{3}x-2$ 颗糖。第二天他吃了 $\frac{1}{3}\left(\frac{2}{3}x-2\right)$ 颗糖，又送出 4 颗糖，最后剩下 8 颗糖。因此 $$ \frac{2}{3}x-2-\left(\frac{1}{3}\left(\frac{2}{3}x-2\right)+4\right)=8. $$ 解得 $x=30$。

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