AMC10 2003 A

AMC10 2003 A · Q13

AMC10 2003 A · Q13. It mainly tests Systems of equations, Word problems (algebra).

The sum of three numbers is 20. The first is 4 times the sum of the other two. The second is seven times the third. What is the product of all three?

三个数的和是 20。第一数是另外两个数之和的 4 倍。第二数是第三数的 7 倍。三者乘积是多少？

(A) 28 28

(B) 40 40

(D) 400 400

(E) 800 800

Answer

Correct choice: (A)

正确答案：（A）

Solution

(A) Let $a$, $b$, and $c$ be the three numbers. Replace $a$ by four times the sum of the other two to get $4(b+c)+b+c=20$, so $b+c=4$. Then replace $b$ with $7c$ to get $7c+c=4$, so $c=\frac{1}{2}$. The other two numbers are $b=7/2$ and $a=16$, and the product of the three is $16\cdot 7/2\cdot 1/2=28$.

（A）设 $a$、$b$、$c$ 为这三个数。把 $a$ 替换为另外两个数之和的 4 倍，得到 $4(b+c)+b+c=20$，所以 $b+c=4$。再把 $b$ 替换为 $7c$，得到 $7c+c=4$，所以 $c=\frac{1}{2}$。另外两个数是 $b=7/2$ 和 $a=16$，三个数的乘积为 $16\cdot 7/2\cdot 1/2=28$。

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