যদি abc = 5 হয়, তাহলে (\({1 \over 1 \ + \ a \ + \ b^{-1} }\) + \({1 \over 1 \ + \ b \ + \ 5 c^{-1}}\) +\(\frac{1}{1+\frac{c}{5}+a^{-1}}\) \(\)) এর মান কত?

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Question

This question was previously asked in

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

Answer 1

প্রদত্ত:

যদি abc = 5

অনুসৃত ধারণা:

abc = 5

⇒ ac = \(5\over b \)

⇒ (b)^-1

⇒ b = \(5\over ac\)

গণনা:

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

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

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

⇒\(a+5+ac\over5+a+ac \)

⇒\(1\)

সুতরাং, ( \({1 \over 1 \ + \ a \ + \ b^{-1} }\)+\({1 \over 1 \ + \ b \ + \ 5 c^{-1}}\)+\(\frac{1}{1+\frac{c}{5}+a^{-1}}\)) এর মান হল 1

Shortcut Trick ধরি, a = b = 1 এবং c = 5

প্রশ্নানুযায়ী,

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

⇒ 1/3 + 1/3 + 1/3

⇒ 3/3 = 1