2001 AMC 10 Problems/Problem 7

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 1

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: $10000x=4\cdot\frac{1}{x} \implies x^2=\frac{4}{10000}$ Since $x>0\implies x=\frac{2}{100}=\boxed{\textbf{(C)}\ 0.02}$

Solution 2

Alternatively, we could try each solution and see if it fits the problems given statements.

After testing, we find that $\boxed{\textbf{(C)}\ 0.02}$ is the correct answer.

Video Solution by Daily Dose of Math

https://youtu.be/M-mjlmRQByI?si=8jC0mYpCej1cRTIQ

~Thesmartgreekmathdude

2001 AMC 10 (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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2001 AMC 10 Problems/Problem 7

Contents

Problem

Solution 1

Solution 2

Video Solution by Daily Dose of Math

See Also