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 = \(m*(1+\frac{x}{100}) = m*\sqrt{\frac{n}{m} }=\sqrt{mn}\)
Hi
Bunuel,
I understood the solution algebraically. I have one doubt as it is stated in the question time from 2000 to 2004 then in the compounding will be as per below in my understanding
2000 - a
2001 - a * (1+x/100) = m
2002 - a * (1+x/100)^2
2003 - a * (1+x/100)^3 = n
Then why we are considering directly from 2001 instead of 2000 while finding the value of 2002.