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At a loading dock, each worker on the night crew loaded 3/4 [#permalink]
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18 May 2006, 14:50
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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 OPEN DISCUSSION OF THIS QUESTION IS HERE: ataloadingdockeachworkeronthenightcrewloaded128407.html
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Re: PSLoading dock (From OG11) [#permalink]
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18 May 2006, 15:10
buckkitty 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?
A. 1/2 B. 2/5 C. 3/5 D. 4/5 E. 5/8
My strategy of answering this question was not efficient. I am curious to know if you all approach the same as OG. OA to follow.
Haven't checked OG but my method is below and i suppose it would be the standard way.
Ans is E.
Method: X = No. of boxes loaded by day crew.
Boxes by night crew = 3/4 * 4/5 X = 3/5 X
% loaded by day crew = X/(X + 3/5X) = 5/8



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Re: PSLoading dock (From OG11) [#permalink]
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18 May 2006, 16:01
num ppl at daytime x wrk per worker/ num ppl at night x wrk per worker
= 1/(4/5x3/4) = 5/3.
Therefore ratio is 5/8.
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the night crew can load 4/5 * 3/4 = 3/5 the amount of boxes the day crew can (d)
3d/5 + d = 1
solve for d
d=5/8



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Picking numbers:
Assume Day Crew has 5 workers > Nigh Crew has 4 workers
Assume Day crew loads 4 boxes/worker > Night crew loads 3 boxes/worker
Total boxes loaded by Day crew = 5*4 = 20
Total boxes loaded by Night crew = 4*3 = 12
Requried Ratio = 20/32 = 5/8.



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Day crew:
# of workers = w
# of boxes loaded per worker = d
# of boxes loaded (total) = wd
Night crew:
# of workers = 4w/5
# of boxes loaded per worker= 3d/4
# of boxes loaded (total) = 3wd/5
Total loaded in a day = wd + 3wd/5 = 8wd/5
Fraction loade by day crew = wd/(8wd/5) = 5/8



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I do such problems by picking numbers.
No. of boxes by each day worker = 4. Hence each night worker = 3
No of day workers = 5. Hence no of night workers = 4
Total output daytime = 5*4 = 20
Total output nighttime = 3*4 = 12
Fraction by daytime = 20/(20+12) = 20/32 = 5/8



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When I see problems such as these I generally pick #'s. I guess my problem was that I picked bad #'s and that is the key to the method working. Must note that in error log Anyway, OG solved as Saha (and most of you) did. OA is E. Thanks



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PS  Loading Dock [#permalink]
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02 Sep 2006, 21:24
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



Senior Manager
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E 5/8
Take sample numbers..
Day crew = 4 boxes/person * 10 people = 40
then Night Crew = 3 boxes/persion * 8 people = 24
Hence fraction of day crew = 40/64 = 5/8



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PS  Loading Dock [#permalink]
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02 Sep 2006, 23:36
gmatornot wrote: E 5/8
Take sample numbers..
Day crew = 4 boxes/person * 10 people = 40 then Night Crew = 3 boxes/persion * 8 people = 24
Hence fraction of day crew = 40/64 = 5/8
Thanks gmatnort, your explaination was much simpler than the OE



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E.
I took 60 workers who move 1 box a day for the Day crew, and then just worked through the problem...



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Re: PS  Loading Dock [#permalink]
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03 Sep 2006, 08:06
uvs_mba wrote: gmatornot wrote: E 5/8
Take sample numbers..
Day crew = 4 boxes/person * 10 people = 40 then Night Crew = 3 boxes/persion * 8 people = 24
Hence fraction of day crew = 40/64 = 5/8 Thanks gmatnort, your explaination was much simpler than the OE
You are welcome !



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PS  1000Q [#permalink]
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25 Oct 2006, 11:50
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



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Re: PS  1000Q [#permalink]
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25 Oct 2006, 13:26
SimaQ 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?
(A) 1/2 (B) 2/5 (C) 3/5 (D) 4/5 (E) 5/8
Argh  something is wrong with me as I keep on getting 4/7??? What am I doing wrong?
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Answer E [#permalink]
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25 Oct 2006, 13:56
NC/DC=4/5
Assue 9 workers in total crew 4 in night, 5 in Day
N/D=3/4
Asume 7 total work Night 3 Day 4
Work in night = 4*3=12
Work in day = 5*4=20
Total work = 12+20=32
Ratio Dayto total =20/32=5/8



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Re: PS  1000Q [#permalink]
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25 Oct 2006, 14:39
Matrix02 wrote: SimaQ 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?
(A) 1/2 (B) 2/5 (C) 3/5 (D) 4/5 (E) 5/8 Argh  something is wrong with me as I keep on getting 4/7??? What am I doing wrong?
matrix,
this should be baby stuff for you, reread the question and you should get it. i did it wrong, reread the question and presto, it's quite simple.



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Re: PS  1000Q [#permalink]
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25 Oct 2006, 14:41
kidderek wrote: Matrix02 wrote: SimaQ 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?
(A) 1/2 (B) 2/5 (C) 3/5 (D) 4/5 (E) 5/8 Argh  something is wrong with me as I keep on getting 4/7??? What am I doing wrong? matrix, this should be baby stuff for you, reread the question and you should get it. i did it wrong, reread the question and presto, it's quite simple.
You are right Kidderek
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Joined: 11 Jul 2006
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agreed, 5/8
again with smaller numbers and a different assumption:
assume, again, that there were 4 night shifters and 5 day shifters.
to simplify, assume every day shifter only loaded 1 box.
then evey night shifter loaded 3/4 of a box, or 4*(3/4)=3 between the two of them.
day shifters loaded 5/(5+3)=5/8 of total.



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Re: PS  1000Q [#permalink]
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26 Oct 2006, 05:47
SimaQ 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?
(A) 1/2 (B) 2/5 (C) 3/5 (D) 4/5 (E) 5/8
E
Pick numbers.
Let there be 5 day workers and so, 4 night workers
Let each day worker load 4 boxes and so, each night worker loads 3 boxes
Total boxes loaded = 5*4 + 4*3 = 32.
20/32







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