Simplify \(\frac{1}{2+2p} + \frac{1}{2+2q}+\frac{1}{2+2r}\), where p = \(\frac{x}{y+z}\), if q = \(\frac{y}{z+x}\) and r = \(\frac{z}{x+y}\).

Question

Question

This question was previously asked in

SSC CGL 2023 Tier-I Official Paper (Held On: 24 Jul 2023 Shift 4)

Answer 1

Shortcut Trick

let x = y = z = 1

So, p = 1/2, q = 1/2, z = 1/2

Now,

\(\frac{1}{2+2p} + \frac{1}{2+2q}+\frac{1}{2+2r}\)

⇒ \(\frac{1}{2+1} + \frac{1}{2+1}+\frac{1}{2+1}\)

⇒ \(\frac{1}{3} + \frac{1}{3}+\frac{1}{3}\) = 1

∴ The correct answer is option (1).

Alternate Method

Calculation

\(\frac{1}{2+2p} + \frac{1}{2+2q}+\frac{1}{2+2r}\)

⇒ 1/(2 + 2 × (x/y + z)) + 1/(2 + 2 × (y/x+ z)) + 1/(2 + 2 × (z/x+ y))

⇒ [(y + z)/2x + 2y + 2z] + [(x + z)/2x + 2y + 2z] + [(y + x)/2x + 2y + 2z]

⇒ (2x + 2y + 2z)/(2x + 2y + 2z)

⇒ 1

The value is 1.

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