# GATE2016-EC_2 [closed]

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closed as a duplicate of: GATE2016 EC-2: GA-5

closed
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Ans is given B).
0
yes , B) will be ans

Consider efficiency of M as X , no. of Days E worked as D. $\therefore$ no. of days M worked= $\frac{D}{2}$

$\therefore$ Efficiency of E=$\frac{X}{2}$

$\therefore$ Contribution(M)=X $\times$ $\frac{D}{2}$ $\times$ 6 = 3XD [ $\because$ he has 6 hour shifts ]

$\therefore$ Contribution(E)=$\frac{X}{2}$ $\times$ D $\times$ 12=6XD  [ $\because$ he has 12 hour shifts ]

$\frac{ Contribution(M)}{Contribution(E)}$=$\frac{3XD}{6XD}$=$\frac{1}{2}$

selected by

contribution of a worker =number of days he/she worked$*$number of hours worked per day$*$work done per hour.

Now coming to problem,

Let S,E,F does work of x units/hour.Since M is twice as efficient as others,M does work of 2x units/hour.

Let E works for y days.Then M works for y$/$2 days.

M works for 6 hrs in a day while E works for 12 hrs in a day.

$\therefore$ Contribution of M = (y$/$2)$*$2x$*$6 = 6xy

and Contribution of E = y$*$x$*$12$=$12xy

Contribution of M$/$Contribution of E $=$1$/$2

The work done can be any thing.It can be eating 2 apples per hour or typing 3 pages per hour...depending on the project.

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