\(If a^2 + b^2 = 111 ,\) a × b = 27, and a > b, find the value of \(\frac{a -b}{a+b}\)

Question

Question

This question was previously asked in

RRB NTPC Graduate Level CBT-I Official Paper (Held On: 06 Jun, 2025 Shift 2)

Answer 1

Given:

a² + b² = 111

a × b = 27

a > b

Formula used:

(a - b)² = a² + b² - 2ab

(a + b)² = a² + b² + 2ab

Calculations:

First, find the value of (a - b)²:

(a - b)² = a² + b² - 2ab

⇒ (a - b)² = 111 - 2 × 27

⇒ (a - b)² = 111 - 54

⇒ (a - b)² = 57

Since a > b, (a - b) must be positive:

⇒ a - b = \(\sqrt{57}\)

Next, find the value of (a + b)²:

(a + b)² = a² + b² + 2ab

⇒ (a + b)² = 111 + 2 × 27

⇒ (a + b)² = 111 + 54

⇒ (a + b)² = 165

Assuming a + b is positive (as implied by the options):

⇒ a + b = \(\sqrt{165}\)

Now, find the value of \(\frac{a - b}{a + b}\):

\(\frac{a - b}{a + b} = \frac{\sqrt{57}}{\sqrt{165}}\)

⇒ \(\frac{a - b}{a + b} = \sqrt{\frac{57}{165}}\)

∴ The value of \(\frac{a - b}{a + b}\) is \(\sqrt{\frac{57}{165}}\).