M08-01

Official Solution:

The price of a certain commodity increased at a constant rate of \(x\%\) per year between 2000 and 2004, where \(x > 0\). If the commodity was priced at \(m\) dollars in 2001 and \(n\) dollars in 2003, where \(mn \neq 0\), then, in terms of \(m\) and \(n\), what was the price of the commodity in 2002?

A. \(\sqrt{mn}\)
B. \(n\sqrt{\frac{n}{m} }\)
C. \(n\sqrt{m}\)
D. \(\frac{nm}{\sqrt{n} }\)
E. \(nm^{\frac{3}{2} }\)


The price in 2001 = \(m\);

The price in 2002 = \(m*(1+\frac{x}{100})\). Our goal is to express this in terms of \(m\) and \(n\), so we need to express \(1+\frac{x}{100}\) using \(m\) and \(n\).

The price in 2003 = \(m*(1+\frac{x}{100})*(1+\frac{x}{100})=n\), which leads to \(m*(1+\frac{x}{100})^2=n\).

From the above equation: \((1+\frac{x}{100}) = \sqrt{\frac{n}{m} }\)

Therefore, the price in 2002 = \(m*(1+\frac{x}{100}) = m*\sqrt{\frac{n}{m} }=\sqrt{mn}\)



Alternatively, we can use the plug-in method.

Suppose the price in 2001 is 100 and the annual rate is 10%. Then:

\(2001 = 100 = m\)

\(2002 = 110\)

\(2003 = 121 = n\)

Now, plug 100 and 121 into the answer choices to see which one yields 110:

A. \(\sqrt{mn}=\sqrt{100*121}=10*11=110\), which is the correct answer.

P.S. For the plug-in method, it might happen that for some specific numbers, more than one option may give the "correct" answer. In this case, simply pick other numbers and check those "correct" options again.


Answer: A