Question

Ana 和 Bonita 的生日是同月同日，但出生年份相差 $n$ 年。去年 Ana 的年龄是 Bonita 的 $5$ 倍。今年 Ana 的年龄是 Bonita 年龄的平方。求 $n$ 。

Ana and Bonita were born on the same date in different years, $n$ years apart. Last year Ana was $5$ times as old as Bonita. This year Ana's age is the square of Bonita's age. What is $n?$

$3$

$5$

$9$

$12$

$15$

答案：D

知识点：年龄问题方程组二次方程

难度评级：960

解答：

设 Ana 今年 $a$ 岁，Bonita 今年 $b$ 岁。 $\begin{aligned} a-1&=5(b-1),\\ a&=b^2. \end{aligned}$

代入可得 $\begin{aligned} b^2-1&=5b-5,\\ b^2-5b+4&=0,\\ (b-4)(b-1)&=0. \end{aligned}$

一岁会使两人同龄，不符合题意。因为 $b \neq 1$ ，所以 $b = 4$ 。

于是 $a = 4^2 = 16$ ，年龄差为 $n = 16 - 4 = 12$ 。

所以正确答案是 D。

Let $a$ be Ana's current age and $b$ be Bonita's current age. Then $\begin{aligned} a-1&=5(b-1),\\ a&=b^2. \end{aligned}$

Substitution gives $\begin{aligned} b^2-1&=5b-5,\\ b^2-5b+4&=0,\\ (b-4)(b-1)&=0. \end{aligned}$

We can see that $b \neq 1$ since that would make Ana and Bonita the same age, so we know that $b = 4.$

This gives us that $a = 4^2 = 16$ and $n = 16 - 4 = 12.$

Thus, D is the correct answer.

2019 AMC 10A 第 3 题

答案：D

难度评级：960

解答：

其他年份的第 3 题