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

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Re: At a loading dock, each worker on the night crew loaded [#permalink]

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New post 28 May 2016, 12:22
I plugged in numbers to make this one easier.

Each night worker loads 3 boxes while each day worker loads 4 boxes. There is a total of 4 night works and 5 days workers.

Boxes Loaded Day = 5*4 = 20
Boxes Loaded Night = 3*4 = 12
Total Boxes Loaded = 32
Fraction of Boxes Loaded by Day Crew = 20/32 = 5/8

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Re: At a loading dock, each worker on the night crew loaded [#permalink]

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New post 21 Jun 2016, 10:31
jeepsdaddy 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


The easiest way to solve this question is to choose convenient values for the number of day workers and for the number of boxes loaded by each day worker.

We are given that the night crew loaded 3/4 as many boxes as each worker on the day crew and that the night crew has 4/5 as many workers as the day crew. We can set up an equation relating the number of day workers and the number of night workers:

4/5(number of day workers) = number of night workers

Let's choose the convenient value of 20 for the number of day workers and substitute it into the equation above:

4/5 x 20 = number of night workers

16 = number of night workers

Next, we set up the equation for the number of boxes loaded by each worker.

(¾) x (number of boxes loaded by each day worker) = number of boxes loaded by each night worker

Now let's choose the convenient value of 8 for the number of boxes loaded by each day worker and substitute it into the above equation:

¾ x 8 = number of boxes loaded by each night worker

6 = number of boxes loaded by each night worker

Thus, the total number of boxes loaded by each type of worker is as follows:

Day workers = 20 x 8 = 160

Night workers = 16 x 6 = 96

Total boxes loaded = 160 + 96 = 256

The question asks us to determine the fraction of all the boxes loaded by the day crew, and that is:

(Total boxes loaded by day crew)/(Total boxes loaded)

160/256

40/64

5/8

Answer is E.
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Re: At a loading dock, each worker... [#permalink]

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New post 04 Sep 2016, 21:11
Hi Manonamission,

This question can be solved by TESTing VALUES.

We're told that the night crew consists of 4/5 as many workers as the day crew:

Night crew = 4 workers
Day crew = 5 workers

We're also told that each member of the night crew loaded 3/4 as many crates as each member of the day crew:

Each Night crew member = 3 crates/each
Each Day crew member = 4 crates/each

We're asked to find the fraction of the TOTAL CRATES loaded by the Day crew...

Day crew total = (5)(4) = 20 crates
Night crew total = (4)(3) = 12 crates

(Day crew total)/(All crates) = 20/(20+12) = 20/32 = 5/8

Final Answer:
[Reveal] Spoiler:
E


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Re: At a loading dock, each worker... [#permalink]

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New post 04 Sep 2016, 22:16
Manonamission 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
(0 3/5
(D) 4/5
(E) 5/8

Note : Apologies for the typo error in the initial Q. Modified numbers in red.


Hi,
Let the number of workers in day be x and each loads y boxes, so total is x*y....

While in night, the workers are 4/5*x and each loads 3/4*y, so total = 4x/5 * 3y/4= 3xy/5....

TOTAL= xy + 3/5* xy =8/5*xy..

Fraction day workers loads xy/(8/5)xy=5/8..
E..

The other way is of course take some numeric value for each, but these values should be multiples of the denominator to avoid confusion....
So day worker should be 4 or its multiple and # of day workers be 5...
Day workers load 4*5=20..
Ni workers load 3/4*4*4/5*5=3*4=12..
So TOTAL=12+20=32..
Fraction of day work=20/32=5/8
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Re: At a loading dock, each worker on the night crew loaded [#permalink]

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New post 27 Apr 2017, 07:16
jeepsdaddy 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



I think the best approach would be to use numbers. Let the day workers be 10. Then night workers would be 8.

Let the number of boxes loaded by day worker be 16. then boxes for night workers would be 12.

Now, total number of boxes from day crew is 160. Total number of boxes from both the crews 96+160.

thus, 160/256 =5/8

Answer: 5/8
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At a loading dock, each worker on the night crew loaded [#permalink]

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Set up the ratios given in the prompt:
1. the ratio for # of boxes: Let n stand for the # of boxes of the night crew and d for the # of boxes of the day crew:
n=3/4d --> 4n=3d
2. Now set up the ratio for the # of workers of the two crews: Same as above
n=4/5d --> 5n=4d
Now look for the minimum values that satisfy the equation.
For ratio 1. it would be 4*(3)=3(4) and for ratio 2. 5(4)=4(5).
The question asks for the fraction of all the boxes loaded by the two crews, the day crew loaded:
Choose the minimum values and you obtain: total # boxes loaded by day crew/ total # of boxes loaded by both crews
= 4*5/(4*5)+(3*4)
= 20/(20 + 12)
= 20/32
= 5/8
Answer E!

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At a loading dock, each worker on the night crew loaded [#permalink]

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New post 07 Sep 2017, 10:40
Day:
X - number of workers
Y - number of boxes per worker
Night:
4/5X - number of workers
3/4Y-number of boxes per worker

XY/(XY+4/5X*3/4Y) -->5/8

BUT what I do not understand (since I am not a native speaker) is what would be wording of the question stem if it meant to say that the number of night workers was more than the number of day workers for 4/5? I spent a lot of time trying to figure out what does it mean "as many as"

Could you please shed some light?

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At a loading dock, each worker on the night crew loaded [#permalink]

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New post 12 Sep 2017, 06:20
jeepsdaddy 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


n = 3/4 x (d)
n = 4/5 x (d)

So,
n= 3 box
d = 4 box
again

n = 4 worker
d = 5 worker

Now

d / (d + n)

(4 x 5) / (4 x 5) + (3 x 4)
=5/8

So, cheers ...

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At a loading dock, each worker on the night crew loaded   [#permalink] 12 Sep 2017, 06:20

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