GATE Civil Engineering (CE) 2018 Shift 2 Solved Paper

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Question : 1 of 65

Marks: +1, -0

{a+a+a+. . .+a}↙{╰───────🡫──────╯}↙{\text "n times"}=a^{2} b

and

{b+b+b+. . .+b}↙{╰───────🡫──────╯}↙{\text "n times"}=a b^{3}

,the value of

(\;{m+m+m+. . .+m}↙{╰───────🡫──────╯}↙{\text "n times"})(\;{n+n+n+. . .+n}↙{╰───────🡫──────╯}↙{\text "n times"}) ?

$2 a^{2} b^{2}$
$a^{4} b^{4}$
$a b(a+b)$
$a^{2}+b^{2}$

Solution:

So,

{m+m+. . . . . .+m}↙{╰───────🡫──────╯}↙{\text "n times"} × {n+n+. . . . . .+n}↙{╰───────🡫──────╯}↙{\text "n times"}

\;{a+a+a+. . .+a}↙{╰───────🡫──────╯}↙{\text "n times"} =n a=a^{2} b

⇒ \;\; n=a b

⏟{b+b+b+. . .+b}↙{╰───────🡫──────╯}↙{\text "n times"} =m b=b^{2} a

So,

\;{m+m+. . . . . .+m}↙{╰───────🡫──────╯}↙{\text "n times"} × {n+n+. . . . . .+n}↙{╰───────🡫──────╯}↙{\text "n times"}

i.e.

m n × m n=(m n)^{2}

from (i) and (ii)

m n=a^{2} b^{2}

So, result,

(m n)^{2}=(a^{2} b^{2})=a^{4} b^{4}

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