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by Rahul@gurome » Thu Oct 21, 2010 9:50 pm
At a loading dock, each worker on the night crew loaded 3/4 as many boxes as each worker on the day crew. If the night crew has 4/5 as many workers as the day crew, what fraction of all the boxes loaded by the two crews did the day crew load?
(A) 1/2
(B) 2/5
(C) 3/5
(D) 4/5
(E) 5/8
Say, number of workers in the day crew = w
=> Number of workers in the night crew = 4w/5.

Say, total number of boxes loaded by the day crew = x
=> Number of boxes loaded by each of the workers in the day crew = x/w.
=> Number of boxes loaded by each of the workers in the night crew = (3/4)*(x/w).
=> Total number of boxes loaded by the night crew = Number of workers in the night crew * Boxes loaded by each of the workers in the night crew = (4w/5)*[(3/4)*(x/w)] = 3x/5.

Therefore, total number of boxes loaded by the two crews = Boxes loaded by the day crew + Boxes loaded by the night crew = x + 3x/5 = 8x/5.

Therefore, required fraction = Boxes loaded by the day crew/Total number of boxes loaded by the two crews = x/(8x/5) = 5/8.

The correct answer is E.[/quote]
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by pradeepkaushal9518 » Fri Oct 22, 2010 4:07 am
thanks rahul now its clear to me
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by pesfunk » Sun Oct 31, 2010 2:48 am
A quicker approach:

Just convert all into quick numbers

Day shift: 4 Boxes
Night Shift: 3 Boxes

Day Shift: 5 Employees
Night Shift: 4 Employees

Total Day shift = 20 boxes
Total Night shift = 12 boxes

Day / Total = 20 / 32 = 5/8
pradeepkaushal9518 wrote:thanks rahul now its clear to me