a hydrocarbon A ( molecular formula C5H10) yields 2- methylbutane on catalytic hydrogenation. a adds hbr compound B which on reaction with silver hydroxide forms an alcohol C (C5H12). alcohol C on oxidation gives a ketone D. deduce the structure of A, B, C and D.

The molecular formula of A is C₅H₁₀. As it fits into general formula C_nH_2n therefore it could be an alkene or cycloalkane.
As it undergoes hydrogenation therefore it is an alkene. 
Since on hydrogenation we get 2-methylbutane therefore a double bond may be present on any of the carbon but the basic skeletal of the compound would be same as that of 2-methylbutane.
As C i.e. Alcohol on oxidation give ketone therefore carbon bearing -OH group should be secondary. 
Hence, the compound which satisfies all the above conditions is 3-methylbut-1-ene. 
The scheme for above reactions is as follows:

Hence, A is 3-methylbut-1-ene, B is 2-Bromo-3-methylbutane, C is 3-methylbutan-2-ol (please note the molecular formula that you have provided is incorrect, it should be C₅H₁₂O) and D is 3-methylbut-2-one. 
During the conversion of alkyl bromide to alcohol the reagent mentioned in you question is not clear, you may use aqueous KOH or NaOH for this reaction.