1954 AMC 12 第 28 题

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28.

mn=43\dfrac mn=\dfrac43rt=914\dfrac rt=\dfrac9{14},则 3mrnt4nt7mr\dfrac{3mr-nt}{4nt-7mr} 的值为:

If mn=43\dfrac mn=\dfrac43 and rt=914,\dfrac rt=\dfrac9{14}, the value of 3mrnt4nt7mr\dfrac{3mr-nt}{4nt-7mr} is:

512-5\dfrac12

1114-\dfrac{11}{14}

114-1\dfrac14

1114\dfrac{11}{14}

23-\dfrac23

答案:B
知识点:比与比例分数换元法
难度评级:1590
小提示:

由已知比值可得 mr:nt=(mn)(rt)mr:nt=(\frac{m}{n})(\frac{r}{t})

The given ratios imply mr:nt=(mn)(rt)mr:nt=(\frac{m}{n})(\frac{r}{t})

大提示:

写成 mr=6kmr=6knt=7knt=7k

Write mr=6kmr=6k and nt=7knt=7k

解答:

将两个已知比值相乘, mrnt=43914=67 \frac{mr}{nt}=\frac43\cdot\frac9{14}=\frac67\text{。}mr=6kmr=6knt=7knt=7k。则 3mrnt4nt7mr=18k7k28k42k=1114 \begin{aligned} \frac{3mr-nt}{4nt-7mr} &=\frac{18k-7k}{28k-42k}\\ &=-\frac{11}{14} \end{aligned}\text{。}

因此,正确答案是 B

Multiplying the two given ratios, mrnt=43914=67. \frac{mr}{nt}=\frac43\cdot\frac9{14}=\frac67. Put mr=6kmr=6k and nt=7k.nt=7k. Then 3mrnt4nt7mr=18k7k28k42k=1114. \begin{aligned} \frac{3mr-nt}{4nt-7mr} &=\frac{18k-7k}{28k-42k}\\ &=-\frac{11}{14}. \end{aligned}

Thus, the correct answer is B.

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