GATE ECE 2016 Set 2 | GA | Question: 5

Question

$S$, $M$, $E$ and $F$ are working in the shifts in a team to finish a project. $M$ works with twice the efficiency of others but for half as many days as $E$ worked. $S$ amd $M$ have $6$ hour shifts in a day, whereas $E$ and $F$ have $12$ hours shifts. What is the ratio of contribution of $M$ to contribute of $E$ in the project?

• A. $1:1$
• B. $1:2$
• C. $1:4$
• D. $2:1$

Comments

Ans is given B).

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yes , B) will be ans

Answer 1

Let the efficiency of $S, E$ and $F$ be $r$ units/hour.
So the efficiency of $M = 2r$ units/hour.

Let the number of days $E$ works be $d$, so that of $M$ is $d/2$.

The amount of work done by $M,$ $W_{m} = 2r \times (d/2) \times 6.$
(As $M$ is working $6$ hours per day)

The amount of work done by $E,$ $W_{e} = r \times d \times 12.$

$W_{m}$ : $W_{e}$ = $\frac{2r \times (d/2) \times 6}{r \times d \times 12} = 1 : 2.$

Correct Answer: $B$

Comments

I don't understand where they have mentioned they are telling the efficiency of a day or efficiency of an hour. 

$M$ works with twice the efficiency of others but for half as many days as $E$ worked.

as I can understand from this sentence, in the later part of the sentence, they were talking about days, so I think they were talking about the efficiency of a day.

let, the efficiency of $S, E$ and $R = e$ units / day
and number of days $E$ worked $= 2d$ days

so, the efficiency of $M = 2e$ units / day
and number of days $M$ worked $= d$ days

$E$ worked $12$ hours a day, so $E's$ hourly efficiency $= \dfrac{e}{12}$ units / hour 

and $M$ worked $6$ hours a day, so $M's$ hourly efficiency $= \dfrac{2e}{6}$ units / hour 

work done by $E$: $W_{E} = 2d \times 12 \times  \dfrac{e}{12} = 2de$

and work done by $M$: $W_{M} = d \times 6 \times  \dfrac{2e}{6} = 2de$

hence, $W_{M} : W_{E} = 2de : 2de = \boxed{1:1}$

Answer 2

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 ]

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

Comments

Nice approach 😀

Answer 3

For sol. Please check this link

https://books.google.co.in/books?id=J4rCCwAAQBAJ&pg=PR98&lpg=PR98&dq=log+tan+1+)%2B+log+tan+2+)+gate+2015&source=bl&ots=y2xZJmm-cm&sig=VSKvumx-kIU_NAzCgCvCa0y6nMQ&hl=en&sa=X&ved=0ahUKEwjN6qCrrJfVAhWEv7wKHTmGCugQ6AEIKDAD#v=onepage&q=S%2C%20M%2C%20E%20and%20F%20are%20working%20in%20shifts%20in%20a%20team%20to%20finish%20a%20proje&f=false

Answer 4

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

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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.

Answer 5

Option B

Comments

Now explain:)

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Dmte/w is always constant.d=days m=men/women t=time e=efficiency w=work
So 6*2*Ee*De/2*12*Ee*De=1:2 here Ee is efficiency of e,De is no of days of e.

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Can you tell what is the constant thing you are mentioning here.I know work done is constant but here they are doing different work?

Answer 6

Ans. B = 1:2 efficiency.