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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, 11:22
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, 09:31
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, 20:11
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

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

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New post 04 Sep 2016, 21:16
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, 06:16
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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New post 04 Jun 2017, 01:49
1
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, 09:40
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, 05:20
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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Re: At a loading dock, each worker on the night crew loaded  [#permalink]

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New post 23 Aug 2018, 23:51
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


This is how I solved

Lets put in some values as ans is required in Fractions

LEts assume the Day crew to be 5 people . then night crew = 4/5* 5 = 4 people
LEts assume the boxes loaded by DC to be 12 per person then no of boxes for NC = 3/4* 12= 9 box per person
Total Boxes = DC = 5*12 = 60 + NC = 4* 9 = 36 = 60 +36 = 96

Target Ans= No of boxes by DC/ Total Boxes
= 60/96 = 5/8 = E
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Re: At a loading dock, each worker on the night crew loaded  [#permalink]

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New post 18 Oct 2018, 20:54
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



Looks easy:

DAY(assume W workers and B boxes which each loads)

So, NIGHT(3/4)B*(4/5)W .

Now fraction of boxes loaded during day(DAY/(Day +Night)
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Re: At a loading dock, each worker on the night crew loaded  [#permalink]

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New post 27 Jun 2019, 23:27
I think the problem is"....what fraction of all the boxes loaded by the two crews did the day crew load?"
it is misunderstanding because it may be 2 crews as individual ! but it means two =day + night !
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Re: At a loading dock, each worker on the night crew loaded  [#permalink]

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New post 27 Jun 2019, 23:53
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



let each person is the day crew load x boxes and let there be y workers in the day crew

no of boxes each night crewloads= 3/4x
no of workers in night crew= 4/5y
total no of boxes loaded by both crews= 8/5xy
no of boxes loaded by day crew = xy
hence option E 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 02 Jul 2019, 10:19
Night Crew strength is 4/5 of Day crew. So suppose:
Day crew = 5 (to make calculation simpler)
which implies Night crew = 4/5 * 5 = 4

Now suppose Boxes packed by each Day crew member = x
==> Total boxes by day crew = 5x

Also implies that Boxes packed by each night crew member = 3/4*x
==> Total boxes packed by night crew = 3/4*x* 4= 3x (4= total number of night crew)

So total boxes packed = 3x+5x =8x
Question asked = Total Day crew boxes / Total boxes = 5x/8x =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 08 May 2020, 01:38
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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

Hello Experts,
EMPOWERgmatRichC, VeritasKarishma, IanStewart
To solve this question, we're just going to divide the "total box loaded by day crew" by "all the boxes loaded by the two crews". That means: \(\frac{4×5}{(12+20)}=5/8\). But the question did not specify the "total" job loaded by day crew; it simply said "did the day crew load".
Am I missing anything to comprehend whole the scenario?
Thanks__
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Re: At a loading dock, each worker on the night crew loaded  [#permalink]

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New post 08 May 2020, 10:43
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Asad wrote:
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

Hello Experts,
EMPOWERgmatRichC, VeritasKarishma, IanStewart
To solve this question, we're just going to divide the "total box loaded by day crew" by "all the boxes loaded by the two crews". That means: \(\frac{4×5}{(12+20)}=5/8\). But the question did not specify the "total" job loaded by day crew; it simply said "did the day crew load".
Am I missing anything to comprehend whole the scenario?
Thanks__


Hi Asad,

The phrase "all the boxes loaded by the two crews" can be reasonably broken down into two "pieces":

"ALL of the boxes loaded by the DAY CREW" + "ALL of the boxes loaded by the NIGHT CREW"

There's nothing in the prompt that's asking us to consider anything other than the TOTAL number of boxes loaded by the Day Crew vs. the TOTAL of ALL BOXES loaded by both crews, so in the absence of that extra information (which again, is NOT there), we have to assume that the question is referring to ALL of the boxes loaded by the Day Crew.

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

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New post 11 May 2020, 01:40
2
Asad wrote:
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

Hello Experts,
EMPOWERgmatRichC, VeritasKarishma, IanStewart
To solve this question, we're just going to divide the "total box loaded by day crew" by "all the boxes loaded by the two crews". That means: \(\frac{4×5}{(12+20)}=5/8\). But the question did not specify the "total" job loaded by day crew; it simply said "did the day crew load".
Am I missing anything to comprehend whole the scenario?
Thanks__


The question doesn't need to say "total boxes loaded by day crew". It is obvious.

Think about it - you have two types of toys - red toys and green toys.
Say, there are 4 red toys and 6 green toys.

what fraction of all the toys you have are the red toys?
You will say 4/10 right?

Do you need to say "total red toys"? No. When you want the red toys out of total, you are automatically considering all red toys.
Same is the case here:
what fraction of all the boxes loaded by the two crews did the day crew load?
Since you are considering ALL boxes, some were loaded by day crew (whenever and by whichever day crew member) and others by night crew. You need the fraction loaded by day crew. You will need all boxes loaded by day crew.
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Re: At a loading dock, each worker on the night crew loaded  [#permalink]

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New post 30 Jun 2020, 12:57
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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


Since the two pieces of information regarding the night shift are related to the information regarding the day shift, let's assign some nice values to the day shift.

Number of workers

Day shift: 5 workers (this is an easy number to find 4/5 of)
Night shift: 4 workers (4/5 of 5 = 4)

Boxes loaded per worker
Day shift: 4 boxes per worker
Night shift: 3 boxes per worker (3/4 of 4 = 3)

Total boxes loaded
Day shift: 5 workers times 4 boxes per worker = 20 boxes
Night shift: 4 workers times 3 boxes per worker = 12 boxes

Combined total boxes for both shifts = 20 + 12 = 32

Of the 32 boxes, the day shift loaded 20 of them.
20/32 = 5/8

Answer: E

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Re: At a loading dock, each worker on the night crew loaded   [#permalink] 30 Jun 2020, 12:57

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