If ( {p^{-1} q^{2}}/{p^{3} q^{-2}} )^{ \frac{1}{3}}\div ( {p^{6} q^{-3}}/{p^{-2} q^{3}} )^{ \frac{1}{3}}=p^{a}q^b then the value of a + b, where p and q are different positive primes, is

SSC CGL Tier II 25 Oct 2015 Paper 1

Question : 58 of 100

Marks: +1, -0

(\ {p^{-1} q^{2}}/{p^{3} q^{-2}} )^{\ {1}/{3}}÷ ( {p^{6} q^{-3}}/{p^{-2} q^{3}} )^{ {1}/{3}}=p^{a}q^b

then the value of a + b, where p and q are different positive primes, is

Solution:

(2)

(\ {p^{-1} q^{2}}/{p^{3} q^{-2}} )^{\ {1}/{3}}÷ ( {p^{6} q^{-3}}/{p^{-2} q^{3}} )^{ {1}/{3}}=p^{a}q^b

⇒(p^{-1-3} q^{2+2} )^{ {1}/{3}} ÷(p^{6+2} q^{-3-3} )^{ {1}/{3}}=p^a q^b

⇒(p^{-4} q^{4} )^{ {1}/{3}}÷(p^{8} q^{-6} )^{ {1}/{3}}=p^{a} q^{b}

⇒{p^{- {4}/{3}} q^{ {4}/{3}}}/{p^{ {8}/{3}} q^{ {-6}/{3}}}=p^{a} q^{b}

⇒p^{ {-4}/{3}- {8}/{3}} q^{ {4}/{3}+ {6}/{3}}=p^{a} q^{b}

⇒ p^{-4} q^{\ {10}/{3}}=p^{a} q^{b}

⇒ a=-4, b= {10}/{3}

∴ a+b=-4+ {10}/{3}= {-2}/{3}

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