If abc = 5, what is the value of (\({1 \over 1 \ + \ a \ + \ b^{-1} }\) + \({1 \over 1 \ + \ b \ + \ 5 c^{-1}}\) +\(\frac{1}{1+\frac{c}{5}+a^{-1}}\) \(\))?

Question

Question

This question was previously asked in

SSC CGL 2022 Tier-I Official Paper (Held On : 08 Dec 2022 Shift 3)

Answer 1

Given:

If abc = 5

Concept used:

abc = 5

⇒ ac = \(5\over b \)

⇒ (b)^-1

⇒ b = \(5\over ac\)

Calculation:

\({1 \over 1 \ + \ a \ + \ b^{-1} }\) + \({1 \over 1 \ + \ b \ + \ 5 c^{-1}}\) + \(\frac{1}{1+\frac{c}{5}+a^{-1}}\)

⇒\(\frac{1}{1+\frac{ac}{5}+a}\) + \(\frac{1}{1+\frac{5}{ac}+5c^{-1}}\) + \(\frac{1}{1+\frac{1}{ab}+a^{-1}}\)

⇒\(5\over5+5a+ac \)\(+\) \(ac\over5+5a+ac \) \(+\) \(a\over5+a+ac \)

⇒\(a+5+ac\over5+a+ac \)

⇒\(1\)

Hence, the value of ( \({1 \over 1 \ + \ a \ + \ b^{-1} }\)+\({1 \over 1 \ + \ b \ + \ 5 c^{-1}}\)+\(\frac{1}{1+\frac{c}{5}+a^{-1}}\)) is 1.

Shortcut Trick let a = b = 1 and c = 5

now, as per the question,

⇒ \({1 \over 1 \ + \ 1 \ + \ 1}\) + \({1 \over 1 \ + \ 1 \ + {5 \over 5 }}\) + \(\frac{1}{1+\frac{5}{5}+ {1}}\)

⇒ 1/3 + 1/3 + 1/3

⇒ 3/3 = 1