Answer

Correct Answer : b ) 2

According to the question

Given,

\(x={6ab \over a+b}\)

\({x \over 3a}={2b \over a+b}\)

By using componendo and dividendo,

\({x+3a \over x-3a}={2b+a+b \over 2b-a-b} ⇒{a+3b \over b-a} …. (1)\)

Similarly,

\(x={6ab \over a+b}\)

\({x \over 3b}={2a \over a+b}\)

Taking componendo and dividendo

\({x+3b \over x-3b} ={2a+a+b \over 2a-a-b}⇒{3a+b \over a-b}…….(2)\)

Adding equation (1) and (2)

\({x+3a \over x-3a}+ {x+3b \over x-3b} = {a+3b \over b-a} + {3a+b \over a-b}\)

⇒\({x+3a \over x-3a}+ {x+3b \over x-3b} = {a+3b \over b-a} -{(3a+b) \over b-a}\) 

⇒\({x+3a \over x-3a}+ {x+3b \over x-3b} = {a+3b-3a-b \over b-a}\)

⇒\({2b-2a \over b-a}\)

Taking 2 common,

⇒\({2(b-a) \over b-a}\)

= 2

Hence, (b) is the correct answer

Such type of question is asked in various government exams like SSC CGL, SSC MTS, SSC CPO, SSC CHSL, RRB JE, RRB NTPC, RRB GROUP D, RRB OFFICER SCALE-I, IBPS PO, IBPS SO, RRB Office Assistant, IBPS Clerk, RBI Assistant, IBPS RRB OFFICER SCALE 2&3, UPSC CDS etc.

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