NIMCET 2015 Question Paper with solutions

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Question : 116 of 120

Marks: +1, -0

Which of the following is equivalent to the expression

({{X+Y}↖{–––––––}+Z}↖{––––––––––––})

$(X↖{-}+Y↖{-})$ Z
(X + Y) $Z↖{-}$
$(X↖{-}+Y↖{-})$ $Z↖{-}$
(X + Y) Z

Solution:

Given, expression

({{x+y}↖{–––––––}+z↖{–}}↖{––––––––––––})

As, we know De-morgan's law says

(\ov{a+b}=\ov{a}⋅\ov{b})

So,let

x+y=a

z=b

⇒{{{x+y}↖{––––––}+z↖{–}}↖{–––––––––––––}}={x+y}↖{––––––}↖{––––––}=z↖{-}↖{-}

↓\;\;\;\;\;\;↓\;\;\;\;\;\;\;↓\;↓

–––––––– – –

\a\;\;+\;\;\b

\a\;\;⋅\;\b

⇒{x+y}↖{–––––}↖{–––––}⋅z↖{-}↖{-}=(x+y)⋅z

Hence

{{x+y}↖{–––––}+z↖{-}}↖{–––––––––––}=(x+y)⋅z

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