专家免费评估留学案例雅思保分免费备考资料留学热线 400-880-6200

AMC10每日一题（2001年真题#07）

2018-05-22
750 人浏览
分享
收藏

2001 AMC 10 竞赛试题/第07题

Problem

When the decimal point of a certain positive decimal number is moved four places to the right, the new number is four times the reciprocal of the original number. What is the original number?

$\textbf{(A) }0.0002\qquad\textbf{(B) }0.002\qquad\textbf{(C) }0.02\qquad\textbf{(D) }0.2\qquad\textbf{(E) }2$

Solution

If $x$ is the number, then moving the decimal point four places to the right is the same as multiplying $x$ by $10000$ . This gives us the equation $10000x=4\cdot\frac{1}{x}$ This is equivalent to $x^2=\frac{4}{10000}$ Since $x$ is positive, it follows that $x=\frac{2}{100}=\boxed{\textbf{(C)}\ 0.02}$