1950 AMC 12 第 30 题

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所有题目均经美国数学协会(MAA)官方合法授权使用。

30.

一群男孩和女孩中先有 1515 名女孩离开,此时每名女孩对应两名男孩。随后又有 4545 名男孩离开,此时每名男孩对应 55 名女孩。最初的女孩人数为:

From a group of boys and girls, 1515 girls leave. There are then left two boys for each girl. After this 4545 boys leave. There are then 55 girls for each boy. The number of girls in the beginning was:

4040

4343

2929

5050

以上答案均不正确

None of these

答案:A
知识点:方程组比与比例
难度评级:1600
小提示:

设最初的女孩和男孩人数分别为 GGBB

Let GG and BB be the original numbers of girls and boys

大提示:

将两个比例写成 B=2(G15)B=2(G-15)G15=5(B45)G-15=5(B-45)

Translate the two ratios as B=2(G15)B=2(G-15) and G15=5(B45)G-15=5(B-45)

解答:

设最初有 GG 名女孩和 BB 名男孩。两个条件给出 B=2(G15),G15=5(B45) \begin{aligned} B&=2(G-15),\\ G-15&=5(B-45) \end{aligned}\text{。}将第一个方程代入第二个方程,得到 G15=5(2(G15)45)=10G375 \begin{aligned} G-15 &=5\bigl(2(G-15)-45\bigr)\\ &=10G-375 \end{aligned}\text{。}因此 9G=3609G=360,所以 G=40G=40

因此,正确答案是 A

Let the original counts be GG girls and BB boys. The two conditions give B=2(G15),G15=5(B45). \begin{aligned} B&=2(G-15),\\ G-15&=5(B-45). \end{aligned} Substituting the first into the second yields G15=5(2(G15)45)=10G375. \begin{aligned} G-15 &=5\bigl(2(G-15)-45\bigr)\\ &=10G-375. \end{aligned} Hence 9G=360,9G=360, so G=40.G=40.

Thus, the correct answer is A.

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