Question

Two positive integers differ by 5. The sum of their reciprocals is 9/14. Then one of the numbers is:

• A. 3
• B. 2
• C. 1
• D. 4
• E. 6

Answer 1

The Correct Option is B

To solve the problem, we are given two positive integers that differ by 5, and the sum of their reciprocals is \(\frac{9}{14}\).

Let's denote these two integers by \(x\) and \(x + 5\). According to the problem statement, their reciprocals sum to \(\frac{9}{14}\). Therefore, we can set up the equation: 

\(\frac{1}{x} + \frac{1}{x+5} = \frac{9}{14}\)

To solve for \(x\), we need to combine the left-hand side fractions:

\(\frac{x + 5 + x}{x(x+5)} = \frac{9}{14}\)

This simplifies to:

\(\frac{2x + 5}{x(x+5)} = \frac{9}{14}\)

Cross-multiply to eliminate the fractions:

\(14(2x + 5) = 9x(x + 5)\)

Expand both sides:

\(28x + 70 = 9x^2 + 45x\)

Rearranging all terms to one side gives:

\(9x^2 + 45x - 28x - 70 = 0\)

Which simplifies to:

\(9x^2 + 17x - 70 = 0\)

We can solve this quadratic equation using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 9\), \(b = 17\), and \(c = -70\).

Calculate the discriminant:

\(b^2 - 4ac = 17^2 - 4 \cdot 9 \cdot (-70)\)

\(= 289 + 2520 = 2809\)

The discriminant is a perfect square, so we proceed:

\(x = \frac{-17 \pm \sqrt{2809}}{18}\)

\(\sqrt{2809} = 53\), so:

\(x = \frac{-17 \pm 53}{18}\)

Calculating the two potential solutions:

1. \(x = \frac{36}{18} = 2\)

2. \(x = \frac{-70}{18}\) (which is a negative number, and since we are looking for positive integers, we discard this solution)

Therefore, one of the numbers is \(2\).