Question

The sum and difference of the arithmetic mean and the geometric mean of two positive integers are respectively, \(18\) and \(8\). Then the values of the two numbers are

• A. \(12\) and \(24\)
• B. \(2\) and \(24\)
• C. \(6\) and \(20\)
• D. \(8\) and \(18\)
• E. \(1\) and \(25\)

Hint:

For two numbers, once you know the arithmetic mean and geometric mean, first find their sum and product. Then use a quadratic equation to recover the original numbers.

Answer 1

Answer: (E)

Step 1: Let the arithmetic mean and geometric mean be \(A\) and \(G\).
According to the question, the sum of the arithmetic mean and geometric mean is \(18\), and their difference is \(8\). Therefore:
\[ A+G=18 \] and \[ A-G=8 \]

Step 2: Solve these two equations.
Add the two equations:
\[ (A+G)+(A-G)=18+8 \] \[ 2A=26 \] \[ A=13 \] Now substitute into \(A+G=18\):
\[ 13+G=18 \] \[ G=5 \]

Step 3: Express arithmetic mean and geometric mean in terms of the numbers.
Let the two positive integers be \(x\) and \(y\). Then:
\[ \frac{x+y}{2}=13 \] and \[ \sqrt{xy}=5 \]

Step 4: Convert these into equations in \(x\) and \(y\).
From \[ \frac{x+y}{2}=13, \] we get:
\[ x+y=26 \] Also, from \[ \sqrt{xy}=5, \] we get:
\[ xy=25 \]

Step 5: Form the quadratic equation.
If two numbers have sum \(26\) and product \(25\), then they are roots of the equation:
\[ t^2-26t+25=0 \]

Step 6: Solve the quadratic equation.
Factorizing:
\[ t^2-26t+25=(t-1)(t-25)=0 \] Hence,
\[ t=1 \quad \text{or} \quad t=25 \] So the two numbers are \(1\) and \(25\).

Step 7: Match with the options.
The pair \(1\) and \(25\) appears in option \((5)\). Therefore, the correct answer is:
\[ \boxed{1 \text{ and } 25} \]