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At a loading dock, each worker on the night crew loaded 3/4

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At a loading dock, each worker on the night crew loaded 3/4 [#permalink] New post 18 May 2006, 13:50
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A
B
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D
E

Difficulty:

  25% (medium)

Question Stats:

80% (02:43) correct 20% (02:12) wrong based on 41 sessions
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: at-a-loading-dock-each-worker-on-the-night-crew-loaded-128407.html
[Reveal] Spoiler: OA
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Re: PS-Loading dock (From OG11) [#permalink] New post 18 May 2006, 14:10
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: PS-Loading dock (From OG11) [#permalink] New post 18 May 2006, 15:01
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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 [#permalink] New post 18 May 2006, 15:53
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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 [#permalink] New post 18 May 2006, 19:26
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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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 [#permalink] New post 18 May 2006, 19:33
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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 [#permalink] New post 19 May 2006, 02:09
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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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 [#permalink] New post 19 May 2006, 12:29
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] New post 02 Sep 2006, 20:24
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
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 [#permalink] New post 02 Sep 2006, 21:43
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] New post 02 Sep 2006, 22:36
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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 [#permalink] New post 02 Sep 2006, 23:31
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] New post 03 Sep 2006, 07:06
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] New post 25 Oct 2006, 10:50
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] New post 25 Oct 2006, 12:26
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] New post 25 Oct 2006, 12:56
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] New post 25 Oct 2006, 13:39
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, re-read the question and you should get it. i did it wrong, re-read the question and presto, it's quite simple.
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Re: PS - 1000Q [#permalink] New post 25 Oct 2006, 13:41
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, re-read the question and you should get it. i did it wrong, re-read the question and presto, it's quite simple.



You are right Kidderek :oops:
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 [#permalink] New post 25 Oct 2006, 13:43
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] New post 26 Oct 2006, 04:47
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
Re: PS - 1000Q   [#permalink] 26 Oct 2006, 04:47
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