Math Problem Statement

The AM of two observations is 25, and GM is 15. Find the two observations and their HM.Without using quadratic equation

Solution

Given:

  • Arithmetic Mean (AM) of two observations aa and bb is 25.
  • Geometric Mean (GM) of aa and bb is 15.

We are to find the two observations aa and bb, and their Harmonic Mean (HM).

Step 1: Use the Arithmetic Mean (AM) formula

The formula for the AM of two numbers aa and bb is:

AM=a+b2\text{AM} = \frac{a + b}{2}

Given AM=25\text{AM} = 25:

a+b2=25\frac{a + b}{2} = 25

Multiplying both sides by 2:

a+b=50a + b = 50

Step 2: Use the Geometric Mean (GM) formula

The formula for the GM of two numbers aa and bb is:

GM=ab\text{GM} = \sqrt{ab}

Given GM=15\text{GM} = 15:

ab=15\sqrt{ab} = 15

Squaring both sides:

ab=225ab = 225

Step 3: Find the two observations aa and bb

We now have the system of equations:

a+b=50a + b = 50 ab=225ab = 225

To find aa and bb, observe that these are the roots of the quadratic equation x2(a+b)x+ab=0x^2 - (a+b)x + ab = 0. However, since we are not using the quadratic formula, we proceed by solving it directly.

Assume aa and bb are the roots of the equation:

x250x+225=0x^2 - 50x + 225 = 0

We can factor this by finding two numbers that multiply to 225 and add to 50. These numbers are 25 and 9.

Thus, the two observations are: a=25andb=9a = 25 \quad \text{and} \quad b = 9

Step 4: Calculate the Harmonic Mean (HM)

The formula for the Harmonic Mean (HM) of two numbers aa and bb is:

HM=2aba+b\text{HM} = \frac{2ab}{a + b}

Substituting the values of a=25a = 25 and b=9b = 9:

HM=2×25×925+9=4503413.24\text{HM} = \frac{2 \times 25 \times 9}{25 + 9} = \frac{450}{34} \approx 13.24

Final Answer

The two observations are 25 and 9, and their Harmonic Mean is approximately 13.2413.24.

Would you like more details or have any questions?

Related Questions

  1. How do you derive the formula for the Harmonic Mean?
  2. Can you find the observations if the GM was 20 instead of 15?
  3. What is the relationship between AM, GM, and HM for any two positive numbers?
  4. How would the solution change if the AM was 30 and the GM was 18?
  5. Can you demonstrate the same problem using a graphical method?

Tip: When working with means (AM, GM, HM), remember that AM ≥ GM ≥ HM for any set of positive numbers.

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Math Problem Analysis

Mathematical Concepts

Arithmetic Mean
Geometric Mean
Harmonic Mean
Quadratic Equations
Factorization

Formulas

Arithmetic Mean formula
Geometric Mean formula
Harmonic Mean formula
Quadratic equation factorization

Theorems

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Suitable Grade Level

Advanced High School