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At a loading dock, each worker on the night crew loaded [#permalink]
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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
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Originally posted by jeepsdaddy on 07 Oct 2010, 08:12.
Last edited by Bunuel on 26 Sep 2012, 11:56, edited 3 times in total.
Edited the question and added the OA



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Re: Help me with the ANSWER .. Rates + Ratio Question. [#permalink]
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Re: Help me with the ANSWER .. Rates + Ratio Question. [#permalink]
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07 Oct 2010, 10:05
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Just take the Box Loaded by each Night crew is Bn and Box Loaded by each Day Crew is Bd.Then According to question (Bn)=(3/4)(Bd) i.e (Bn)/(Bd)=3/4
Now again let Night Crew Has Nn number of crew members and Day crew has Nd members.Then according to question (Nn)=(4/5)(Nd) i.e (Nn)/(Nd)=4/5
As number of box carried by the group will be the product of Number of crew multiplied by the number of boxes carried by each member, According to question we need to find out { (Nd)(Bd) } / { (Nn)(Bn)+(Nd)(Bd) }
Dividing the Nominator and denominator by (Nd)(Bd),
We get, 1 / { (Nn)(Bn)/(Nd)(Bd) + 1 } i.e 1 / { (Nn)/(Nd)*(Bn)/(Bd) + 1 }
i.e 1 / { (3/4)*(4/5) + 1} =5/8
So answer is E. Thanks for a valued Question.
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Re: Help me with the ANSWER .. Rates + Ratio Question. [#permalink]
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07 Oct 2010, 23:13
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its 5/8, Work done by day crew will be 4x (let x=1), the count of day crew = 5x Work done by day crew will be 3x (let x=1), the count of day crew = 4x Total amount work done by both is 5*4+4*3 = 20+12 = 32 Ratio of amount of work done by day crew to total amount of work done is = 20/32 =5/8.



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Re: PS: loading dock [#permalink]
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14 Feb 2011, 12:13
el1981 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 boxes loaded by two crews did the day crew load? A 1/2 B 2/5 C 3/5 D 4/5 E 5/8 night crew loads 3/4 of what day crew loads. Also, night crew is 4/5th day crew. Therefore if one day crew member loads x boxes Night crew is 4/5th of (1 ) * 3/4 = 3/5x Therefore total number of boxes per crew member (1 night + one day crew) 3/5x +x = 8/5x. The x/8/5x = 5/8



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Re: Please help me understand this question in terms of solving [#permalink]
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26 Mar 2011, 05:33
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Let the number of boxes loaded by each worker in the day crew be x Then the number of boxes loaded by each worker in the night crew is 3x/4 The number of people in the day crew = y The number of people in the night crew = 4y/5 Total number of boxes loded: Day crew = xy Night crew = 3x/4 * 4y/5 = 3xy/5 Total = xy + 3xy/5 = 8xy/5 Fraction loaded by the day crew = xy/(8xy/5) = 5/8 Ans  'E'
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Re: Please help me understand this question in terms of solving [#permalink]
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26 Mar 2011, 05:40
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scbguy 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 Please explain the step by step working of this question. Thanks Say, Day crew = 5 people Night crew = (4/5)*5 = 4 people Each person in day crew loaded = 4 boxes. Each person in night crew loaded = (3/4)*4= 3 boxes. Total boxes loaded by day crew = 5*4 = 20 Total boxes loaded by night crew = 4*3 = 12 Fraction of boxes loaded by day crew = 20/(20+12) = 20/32 = 5/8 Ans: "E"
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Re: Please help me understand this question in terms of solving [#permalink]
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27 Mar 2011, 19:42
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scbguy 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 Please explain the step by step working of this question. Thanks Workers at night crew loaded 3/4 as many boxes as did day crew so they loaded fewer boxes. Night crew has 4/5 as many workers as the day crew so night crew has fewer workers too. Overall, night crew loaded 3/4 * 4/5 = 3/5 as many boxes as did day crew. So out of a total of 8 boxes, Night Crew loaded 3 and Day Crew loaded 5 so Day Crew loaded 5/8 of total boxes. Or Boxes loaded by Night Crew: Boxes loaded by Day Crew = 3/5 : 1 = 3:5 So day crew loaded 5/8 of total boxes. If you are uncomfortable with ratios, check out the following posts. They explain ratios and their applications: http://www.veritasprep.com/blog/2011/03 ... ofratios/http://www.veritasprep.com/blog/2011/03 ... osintsd/http://www.veritasprep.com/blog/2011/03 ... problems/*Didn't see fluke's response there! The concept used is pretty similar.
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Re: OG12 PS 138 Loading dock and workers [#permalink]
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22 Sep 2011, 13:59
Hi, I'm trying to plug in for Question # 138 in OG 12. Could anyone point out where i'm going wrong in my reasoning? I understand the solutions posted in the earlier posts but would like to know where i went wrong. No. of Boxes loaded by Night workers = 3/4 of Day workers No. of Night workers = 4/5 of Day workers. Let the number of Day workers = 100 and they load a total of 100 boxes. No. of boxes loaded by Night workers = 3/4 x 100 = 75 No. of Night workers = 4/5 x 100 = 80 total no. of boxes loaded by both workers = 175 total no. of boxes loaded by day workers = 100 fraction of boxes loaded by day workers out of the total = 100/175 = 4/7 The closest answer is 4/5 which is D. But the correct answer is 5/8 which is E.



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Re: OG12 PS 138 Loading dock and workers [#permalink]
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24 Sep 2011, 09:53
gmatyro wrote: Hi, I'm trying to plug in for Question # 138 in OG 12. Could anyone point out where i'm going wrong in my reasoning? I understand the solutions posted in the earlier posts but would like to know where i went wrong. No. of Boxes loaded by Night workers = 3/4 of Day workers No. of Night workers = 4/5 of Day workers. Let the number of Day workers = 100 and they load a total of 100 boxes. No. of boxes loaded by Night workers = 3/4 x 100 = 75 No. of Night workers = 4/5 x 100 = 80 total no. of boxes loaded by both workers = 175 total no. of boxes loaded by day workers = 100 fraction of boxes loaded by day workers out of the total = 100/175 = 4/7 The closest answer is 4/5 which is D. But the correct answer is 5/8 which is E. Here is the problem in your solution: It is given that EACH night worker loads 3/4 of the boxes loaded by EACH day worker. (not that total number of boxes loaded by night workers is 3/4 of the boxes loaded by day workers) If I want to take values and I say, each day worker loads 100 boxes, then each night worker loads 75 boxes. Total no of boxes loaded by day workers = 100*100 = 10,000 Total no of boxes loaded by night workers = 80*75 = 6000 Fraction of total boxes loaded by day workers = 10,000/16,000 = 5/8 Though, if I want to take numbers and plug in, I would take easier numbers. Say, there are 5 day workers. Then there are 4 night workers. Say, each day worker loads 4 boxes in a day. Each night worker then loads 3 boxes in a day. Total no of boxes loaded by day workers = 5*4 = 20 Total no of boxes loaded by night workers = 4*3 = 12 Fraction of boxes loaded by day workers = 20/32 = 5/8
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Re: OG12 PS 138 Loading dock and workers [#permalink]
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24 Sep 2011, 21:50
Night Crew rate = (3/4)* Day Crew rate. 1
Number of night Crew workers = (4/5)*Day crew workers.2
Total Work done by Day crew workers = Number of day crew workers * day crew worker rate.
Total Work done by Night crew workers = Number of night crew workers * night crew worker rate
= (4/5) Day crew workers * (3/4) * Day crew rate
= (3/5)* Total work done by Day crew workers
fraction of all the boxes loaded by the two crews did the day crew load = 1/(1+(3/5)) = 5/8
Answer is E.



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Re: OG12 PS 138 Loading dock and workers [#permalink]
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01 Mar 2012, 04:30
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I think that the biggest problem for solving this kind of questions is the language understanding (i.e. misunderstanding). I've had a lot of trouble with this one until I figured out that the night crew did 3/4 and the day crew did 4/4 of the work. And that the night crew had 4/5 workers and the day crew 5/5 workers. It can become quite messy for a nonnative person.
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#138 in OG 12 edition problem solving word problem [#permalink]
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12 Apr 2012, 19:21
If you don't have the book in front of you the question is:
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 didn't know how to answer this question so I looked at the answer explanations and I still do not understand how to get the answer. The explanation says:
From this, the workers on the night crew will load 3/4(4/5)= 3/5 as many boxes as the day crew. The total loaded by both the day and night crews is thus 1 + 3/4= 5/5 + 3/5= 8/5 of the day crew's work. Therefore, the fraction of all the boxes loaded by the two crews that was done by the day crew was 1/8/5 = 1(5/8) = 5/8.
I understand the first sentence saying that the night crew will load 3/5 as many boxes as the day crew. Where I get mixed up is in the next sentence, The total loaded by both the day and night crews is thus 1 + 3/5 = 5/5 + 3/5 = 8/5 of the day crew's work. I do not understand where the 1 comes from. I am figuring that the number 1 represents the day crew. Is it saying the night crew loads 3/5 as many as the day crew so the day crew loads 5/5? Can someone please explain this to me??
Thanks,
Josh



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Re: #138 in OG 12 edition problem solving word problem [#permalink]
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12 Apr 2012, 20:03
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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 SEE IT THIS WAY: day crew loads=x boxes/ member night crew loads = 3x/4 boxes/ member day crew workers= y night crew workers= 4y/5 total boxes loaded/ day= xy+ (3x/4)*(4y/5)= xy+3xy/5= 8xy/5 required fraction= xy/(8xy/5) = 5/8 Hence E Hope this clarifies..!!
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Re: At a loading dock, each worker on the night crew loaded [#permalink]
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09 Oct 2013, 06:38
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I picked some smart numbers and used a table to organize the given information. Just another way to visualize the problem for me... So, let's say we have 5 day workers. That would give us 4 night workers (4/5 of day). Say the day workers load 12 boxes (I picked a smartish number), then the night workers would load 9 boxes (3/4 of what the Day workers load). Put it all in a table and see what you have.
 Day  Night  Total People 5 4 Boxes  12 9  Total 603696
I arrived at the Totals on the bottom by multiplying the people and boxes (5x12, 4x9).
So if you want to figure out what fraction of the total the day workers loaded, use:
\(\frac{60}{96}\)
This is (number of boxes the day crew loaded) divided by (total number of boxes)
This simplifies to \(\frac{5}{8}\). Thus, E is the answer.
Hope this method helps.



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At a loading dock, each worker on the night crew loaded [#permalink]
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07 Jan 2015, 14:36
Let b  boxes and w  workers
Night crew loaded 0.75b per worker with 0.8w workers => 0.6bw Day crew loaded b per worker with w workers => bw Total loaded by both => 1.6bw Finally, divide bw / 1.6bw <=> 1/1.6 = 5/8



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At a loading dock, each worker on the night crew loaded [#permalink]
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07 Jan 2015, 22:53
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Refer diagram below Attachment:
frac.png [ 3.33 KiB  Viewed 27228 times ]
Let boxes in day per worker = 1 Let workers in day = 1 Total boxes loaded in day = 1*1 = 1 Similarly, total boxes loaded in night \(= \frac{3}{4} * \frac{4}{5} = \frac{3}{5}\) Total boxes loaded \(= 1+\frac{3}{5} = \frac{8}{5}\) Fraction \(= \frac{1}{\frac{8}{5}} = \frac{5}{8}\) Answer = E
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At a loading dock, each worker on the night crew loaded [#permalink]
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15 Jun 2015, 01:01
Question: Okay so I got that far that the night crew loads 3/5 as many boxes as the day crew. However I then calculated 3/5 divided by 5/5 (total in my opinion). I looked at the OG's solution and I don't get why we first add 5/5 to 3/5. It does not make sense in my head as the total number of boxes would then be greater than 100%? Maybe its a language problem as the problem is kind of wordy...



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At a loading dock, each worker on the night crew loaded [#permalink]
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18 Jun 2015, 01:17
I just calculated the number of boxes (for ONE worker) for the number or workers (based on the information in the stem), and found #boxes*#workers for each shift:
..............#Boxes........#Workers......Work done Night..........3.................4...................12....... Day............4.................5...................20....... Total.................................................32.......
In the end it is: 20/32 = 5/8. ANS E



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Re: At a loading dock, each worker on the night crew loaded [#permalink]
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01 Nov 2015, 14:44
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What I did was construct a matrix with top columns as Boxes l Workers l Total the rows were then Day and Night. Next I filled in the day row under boxes as X and under workers as W For the Night row I did 3/4X under boxes column and 4/5 W under workers column. I assigned X to be 100 and W to be 100 to have some even numbers to work with. I then reasoned that for Day Crew Total boxes would be WX= 10,000 and night crew would be 75 X 80 = 6,000 = 16,000 total for both 10,000/16,000= 5/8Any suggestions or if this helps kudos please




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