Question

The arithmetic mean and geometric mean between two numbers are 75 and 21 respectively, then the numbers are

• A. 63,87
• B. 73,77
• C. 17,133
• D. 3,147

Answer 1

The Correct Option is D

Step 1: Recall the formulas for arithmetic mean (AM) and geometric mean (GM).

The arithmetic mean of two numbers \( a \) and \( b \) is given by:

\[ \text{AM} = \frac{a + b}{2}. \]

The geometric mean of two numbers \( a \) and \( b \) is given by:

\[ \text{GM} = \sqrt{a \cdot b}. \]

Step 2: Use the given values of AM and GM to set up equations.

From the problem, the arithmetic mean is 75:

\[ \frac{a + b}{2} = 75 \implies a + b = 150. \tag{1} \]

From the problem, the geometric mean is 21:

\[ \sqrt{a \cdot b} = 21 \implies a \cdot b = 21^2 = 441. \tag{2} \]

Step 3: Solve the system of equations.

We now have the following system of equations:

\[ a + b = 150, \quad a \cdot b = 441. \]

Let \( a \) and \( b \) be the roots of a quadratic equation. The quadratic equation with roots \( a \) and \( b \) is:

\[ x^2 - (a + b)x + a \cdot b = 0. \]

Substitute \( a + b = 150 \) and \( a \cdot b = 441 \):

\[ x^2 - 150x + 441 = 0. \]

Step 4: Solve the quadratic equation.

Use the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \]

where \( a = 1 \), \( b = -150 \), and \( c = 441 \). Substitute these values:

\[ x = \frac{-(-150) \pm \sqrt{(-150)^2 - 4(1)(441)}}{2(1)} = \frac{150 \pm \sqrt{22500 - 1764}}{2}. \]

Simplify:

\[ x = \frac{150 \pm \sqrt{20736}}{2} = \frac{150 \pm 144}{2}. \]

Compute the two roots:

\[ x = \frac{150 + 144}{2} = \frac{294}{2} = 147, \quad x = \frac{150 - 144}{2} = \frac{6}{2} = 3. \]

Step 5: Identify the numbers.

The two numbers are \( 147 \) and \( 3 \).

Final Answer: The numbers are \( \mathbf{3 \text{ and } 147} \), which corresponds to option \( \mathbf{(4)} \).