Arithmetic Mean
The arithmetic mean (or average) of a set of $n$ numbers is calculated by dividing the sum of those numbers by $n$. The arithmetic mean of the numbers \( x_1, x_2, x_3, \cdots, x_n \) is defined as:
\[ \frac{x_1 + x_2 + x_3 + \cdots + x_n}{n} \]
Specifically, the arithmetic mean of two numbers \( x_1 \) and \( x_2 \) is:
\[ \frac{x_1 + x_2 }{2} \]
Question 1
The arithmetic mean of two numbers, $x$ and $y$, is \(2 \sqrt{ 3} \), and their geometric mean is $3$. Based on this information, what is the value of the expression:
\[ \frac{1}{x^2} + \frac{1}{y^2} \]
\[
\text{A)} 1 \quad
\text{B) }\frac{1}{2} \quad
\text{C) } \frac{3}{5} \quad
\text{D) } \frac{10}{27} \quad
\text{E) } \frac{15}{32}
\]
Solution:
From the definition of the Arithmetic Mean:
\[ \frac{x+y}{2} = 2 \sqrt{3} \Rightarrow x + y = 4 \sqrt{3} \]
From the definition of the Geometric Mean:
\[ \sqrt{xy} = 3 \Rightarrow xy = 9 \]
Now, we rewrite the target expression by finding a common denominator and using algebraic identities:
\[ \frac{1}{x^2} + \frac{1}{y^2} = \frac{x^2 + y^2}{x^2 \cdot y^2} = \frac{(x+y)^2 – 2xy}{(xy)^2} \]
Substitute the values of \((x+y)\) and \((xy)\) into the equation:
\[ = \frac{(4 \sqrt{3})^2 – 2 \cdot 9}{9^2} = \frac{48 – 18}{81} = \frac{30}{81} = \frac{10}{27} \]
\(\textbf{Answer: D} \)
Question 2
\[
\text{A)} 2 \quad
\text{B) } 4 \quad
\text{C) } 6 \quad
\text{D) } 8 \quad
\text{E) } 10
\]
Solution:
Let the sum of the original 5 numbers be $x$, and the sum of the 2 removed numbers be $y$. Consequently, the sum of the remaining 3 numbers will be \(x – y \). We can set up the following equations:
\[ \frac{x}{5} = a \Rightarrow x = 5a \]
\[ \frac{x-y}{3} = \frac{5a}{3} – 4 \Rightarrow \frac{5a-y}{3} = \frac{5a-12}{3} \Rightarrow 5a – y = 5a – 12 \Rightarrow y = 12 \]
The arithmetic mean of the two removed numbers is calculated by dividing their sum ($y$) by 2:
\[ \frac{y}{2} = \frac{12}{2} = 6 \]
\(\textbf{Answer: C} \)
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