AMC12 2009 B

AMC12 2009 B · Q10

AMC12 2009 B · Q10. It mainly tests Fractions, Basic counting (rules of product/sum).

A particular $12$-hour digital clock displays the hour and minute of a day. Unfortunately, whenever it is supposed to display a $1$, it mistakenly displays a $9$. For example, when it is 1:16 PM the clock incorrectly shows 9:96 PM. What fraction of the day will the clock show the correct time?

某种 $12$ 小时制的数字时钟显示一天中的小时和分钟。不幸的是，每当它应该显示一个 $1$ 时，它会错误地显示成一个 $9$。例如，当时间是下午 1:16 时，时钟会错误显示为下午 9:96。一天中有多少比例的时间该时钟会显示正确时间？

(A) \frac{1}{2} \frac{1}{2}

(B) \frac{5}{8} \frac{5}{8}

(D) \frac{5}{6} \frac{5}{6}

(E) \frac{9}{10} \frac{9}{10}

Answer

Correct choice: (A)

正确答案：（A）

Solution

The clock will display the incorrect time for the entire hours of $1, 10, 11$ and $12$. So the correct hour is displayed $\frac 23$ of the time. The minutes will not display correctly whenever either the tens digit or the ones digit is a $1$, so the minutes that will not display correctly are $10, 11, 12, \dots, 19$ and $01, 21, 31, 41,$ and $51$. This amounts to fifteen of the sixty possible minutes for any given hour. Hence the fraction of the day that the clock shows the correct time is $\frac 23 \cdot \left(1 - \frac {15}{60}\right) = \frac 23 \cdot \frac 34 = \boxed{\frac 12}$. The answer is $\mathrm{(A)}$. The required fraction is the number of correct times divided by the total times. There are 60 minutes in an hour and 12 hours on a clock, so there are 720 total times. We count the correct times directly; let a correct time be $x:yz$, where $x$ is a number from 1 to 12 and $y$ and $z$ are digits, where $y<6$. There are 8 values of $x$ that will display the correct time: 2, 3, 4, 5, 6, 7, 8, and 9. There are five values of $y$ that will display the correct time: 0, 2, 3, 4, and 5. There are nine values of $z$ that will display the correct time: 0, 2, 3, 4, 5, 6, 7, 8, and 9. Therefore there are $8\cdot 5\cdot 9=40\cdot 9=360$ correct times. Therefore the required fraction is $\frac{360}{720}=\frac{1}{2}\Rightarrow \boxed{\mathrm{(A)}}$.

该时钟在整点为 $1, 10, 11$ 和 $12$ 的整个小时都会显示错误。因此小时显示正确的时间占 $\frac 23$。分钟在十位或个位为 $1$ 时不会显示正确，因此不正确的分钟为 $10, 11, 12, \dots, 19$ 以及 $01, 21, 31, 41,$ 和 $51$。这相当于每小时 60 个可能分钟中的 15 个。于是一天中时钟显示正确时间的比例为 $\frac 23 \cdot \left(1 - \frac {15}{60}\right) = \frac 23 \cdot \frac 34 = \boxed{\frac 12}$。答案是 $\mathrm{(A)}$。所求比例等于正确显示的时刻数除以总时刻数。每小时 60 分钟，时钟上有 12 小时，所以总共有 720 个时刻。直接计数正确的时刻；设一个正确时刻为 $x:yz$，其中 $x$ 是 1 到 12 的数，$y$ 和 $z$ 是数字，且 $y<6$。有 8 个 $x$ 的取值会正确显示：2, 3, 4, 5, 6, 7, 8, 和 9。有 5 个 $y$ 的取值会正确显示：0, 2, 3, 4, 和 5。有 9 个 $z$ 的取值会正确显示：0, 2, 3, 4, 5, 6, 7, 8, 和 9。因此正确时刻共有 $8\cdot 5\cdot 9=40\cdot 9=360$ 个。因此所求比例为 $\frac{360}{720}=\frac{1}{2}\Rightarrow \boxed{\mathrm{(A)}}$。

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