2003 AMC 10A 第 13 题

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13.

三个数的和为 2020。第一个数是另外两个数之和的 44 倍。第二个数是第三个数的七倍。三个数的乘积是多少?

The sum of three numbers is 20.20. The first is 44 times the sum of the other two. The second is seven times the third. What is the product of all three?

2828

4040

100100

400400

800800

答案:A
知识点:方程组换元法
难度评级:1310
解答:

设三个数为 aabbcc。由于 a=4(b+c)a = 4(b + c),可得 4(b+c)+(b+c)=204(b + c) + (b + c) = 20,所以 b+c=4b + c = 4,且 a=16a = 16

因为 b=7cb = 7c,所以 7c+c=47c + c = 4,得 c=12c = \dfrac{1}{2}b=72b = \dfrac{7}{2}

乘积为 167212=2816 \cdot \dfrac{7}{2} \cdot \dfrac{1}{2} = 28

所以正确答案是 A

Let the numbers be a,a, b,b, c.c. Since a=4(b+c),a = 4(b + c), we get 4(b+c)+(b+c)=20,4(b + c) + (b + c) = 20, so b+c=4b + c = 4 and a=16.a = 16.

With b=7c,b = 7c, we have 7c+c=4,7c + c = 4, so c=12c = \dfrac{1}{2} and b=72.b = \dfrac{7}{2}.

The product is 167212=28.16 \cdot \dfrac{7}{2} \cdot \dfrac{1}{2} = 28.

Thus, the correct answer is A.

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