AMC10 2019 A

AMC10 2019 A · Q3

AMC10 2019 A · Q3. It mainly tests Linear equations, Word problems (algebra).

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 $n$?

Ana 和 Bonita 在不同年份的同一天出生，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。$n$ 是多少？

(A) 3 3

(B) 5 5

(D) 12 12

(E) 15 15

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): If $B$ denotes Bonita’s age last year, then Ana’s age last year was $5B$. This year Bonita’s age is $B + 1$, and Ana’s age is $5B + 1$. So the given condition is $5B + 1 = (B + 1)^2$. This quadratic equation simplifies to $B^2 = 3B$. Because the two girls were born in different years, $B \ne 0$, so $B = 3$. Last year Bonita was $3$ and Ana was $5 \cdot 3 = 15$, so they were born $15 - 3 = 12$ years apart. As a check, note that Ana’s age this year, $16$, is indeed the square of Bonita’s age this year, $4$.

答案（D）：如果用$B$表示Bonita去年年龄，那么Ana去年年龄是$5B$。今年Bonita的年龄是$B+1$，Ana的年龄是$5B+1$。因此题目给出的条件为$5B+1=(B+1)^2$。这个二次方程化简为$B^2=3B$。因为两个女孩出生年份不同，所以$B\ne0$，从而$B=3$。去年Bonita是$3$岁，Ana是$5\cdot3=15$岁，所以她们相差$15-3=12$岁出生。检验：Ana今年$16$岁，确实是Bonita今年$4$岁的平方。

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