Math Problem Statement

The AM of two observations is 25, and GM is 15. Find the two observations and their HM.

Solution

Given:

  • Arithmetic Mean (AM) of two observations = 25
  • Geometric Mean (GM) of two observations = 15

Let the two observations be xx and yy.

Step 1: Use the AM to form the first equation

The AM of xx and yy is given by:

x+y2=25\frac{x + y}{2} = 25

Multiplying both sides by 2:

x+y=50(Equation 1)x + y = 50 \quad \text{(Equation 1)}

Step 2: Use the GM to form the second equation

The GM of xx and yy is given by:

xy=15\sqrt{xy} = 15

Squaring both sides:

xy=225(Equation 2)xy = 225 \quad \text{(Equation 2)}

Step 3: Solve the system of equations

We now have two equations:

x+y=50x + y = 50 xy=225xy = 225

These equations can be solved using the quadratic formula. Consider the quadratic equation:

t2(x+y)t+xy=0t^2 - (x+y)t + xy = 0

Substituting the values from Equations 1 and 2:

t250t+225=0t^2 - 50t + 225 = 0

Step 4: Solve the quadratic equation

To solve this, use the quadratic formula:

t=b±b24ac2at = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

For our equation t250t+225=0t^2 - 50t + 225 = 0:

t=50±5024×1×2252×1t = \frac{50 \pm \sqrt{50^2 - 4 \times 1 \times 225}}{2 \times 1} t=50±25009002t = \frac{50 \pm \sqrt{2500 - 900}}{2} t=50±16002t = \frac{50 \pm \sqrt{1600}}{2} t=50±402t = \frac{50 \pm 40}{2}

So, the two roots (observations) are:

t1=50+402=45t_1 = \frac{50 + 40}{2} = 45 t2=50402=5t_2 = \frac{50 - 40}{2} = 5

Therefore, the two observations are x=45x = 45 and y=5y = 5.

Step 5: Calculate the Harmonic Mean (HM)

The Harmonic Mean (HM) of xx and yy is given by:

HM=2xyx+y\text{HM} = \frac{2xy}{x + y}

Substituting the values:

HM=2×45×545+5=45050=9\text{HM} = \frac{2 \times 45 \times 5}{45 + 5} = \frac{450}{50} = 9

Final Answer

  • The two observations are 45 and 5.
  • The Harmonic Mean (HM) is 9.

Would you like any more details on this solution, or do you have any further questions?

Here are 5 related questions you might find interesting:

  1. How does the AM-GM-HM inequality relate to the values found here?
  2. What are the conditions under which the AM equals the GM or HM?
  3. How would the solution change if the GM were different?
  4. How can you use these means to estimate the range of two observations?
  5. What is the relationship between the AM, GM, and HM in a set of more than two observations?

Tip: The Harmonic Mean is most useful when averaging rates or ratios, particularly when the quantities have the same units but are distributed unevenly.

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Math Problem Analysis

Mathematical Concepts

Arithmetic Mean
Geometric Mean
Harmonic Mean
Quadratic Equations

Formulas

Arithmetic Mean formula
Geometric Mean formula
Harmonic Mean formula
Quadratic formula

Theorems

AM-GM inequality

Suitable Grade Level

Grades 9-12