Here is a problem solving question

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by Night reader » Wed Dec 08, 2010 6:12 pm
hittingthegmat wrote: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 ?
Night crew load (NC) 3/7
Day crew load (DC) 4/7
3/4 => 3/7+4/7

Night crew workers (N) 4/9
Day crew workers (D) 5/9
4/5 => 4/9+5/9

find (DC*D)/(DC*D+NC*N)

(4/7 * 5/9) / ([4/7 *5/9] + [3/7 *4/9]) = 5/8

answer 5/8

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by hittingthegmat » Wed Dec 08, 2010 7:06 pm
Nifty solution in deed what was the reasoning behind your solution though.

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by Night reader » Wed Dec 08, 2010 7:26 pm
hittingthegmat wrote:Nifty solution in deed what was the reasoning behind your solution though.
we are given the weights for the Crews and Boxes => we need to find the product of Crews add Boxes in the right proportion => ratio Crew*Boxes for Day time workers to Crew*Boxes for Day time and Night time workers

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by GMATGuruNY » Thu Dec 09, 2010 7:19 am
hittingthegmat wrote: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 ?
When the answer choices are fractions and there are unknowns in the problem, we can plug in our own values for the unknowns.

...the night crew has 4/5 as many workers as the day crew..
Plug in 5 workers for the day crew, (4/5)*5 = 4 workers for the night crew.

...each worker on the night crew loaded 3/4 as many boxes as each worker on the day crew.

Plug in 4 boxes per worker for the day crew, (3/4)*4 = 3 boxes per worker for the night crew.

Boxes loaded by day crew = 5*4 = 20.
Boxes loaded by night crew = 4*3 = 12.
Total boxes loaded = 20+12 = 32.

Day crew/Total boxes = 20/32 = 10/16 = 5/8.
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