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There are 5 packers A, B, C, D, and E in a company. The five

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There are 5 packers A, B, C, D, and E in a company. The five  [#permalink]

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New post 07 Apr 2011, 09:50
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

62% (02:26) correct 38% (03:04) wrong based on 46 sessions

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There are 5 packers A, B, C, D, and E in a company. The five packers A, B, C, D and E charge $66, $52, $46, $32, and $28, respectively, to pack each item. The time taken by the packers to pack one item is 20 minutes, 24 minutes, 30 minutes, 40 minutes and 48 minutes, respectively. All the items are sold at the end of the day. Each item earns a profit of 100 dollars, and the packers are paid from this profit. If each packer works 8 hours a day, which packer contributes the most to the net profit of the company?

A. Packer A
B. Packer B
C. Packer C
D. Packer D
E. Packer E

Spoiler: :: "OA/OE"
Ans: "B"

Sol:
Each item earns a profit of 100 dollars for the company. If a packer charges x dollars for each item, the profit that the company would effectively get on each item packed would be 100 – x
Each worker works for 8 hours= 480 minutes. Hence, if a worker takes "t" minutes to pack an item, then he would pack "480/t" items each day. So, the net profit on the \(\frac{480}{t}\) items is \(\frac{480}{t}(100-x)\). The expression \(\frac{480}{t}(100-x)\) is a maximum when \(\frac{1}{t}(100-x)\) is maximum.

We just need to check for what value of "x" and "t", this expression \(\frac{100-x}{t}\) will be maximum.

A: Packer A; t=20 minutes, x=$66 : \(\frac{100-x}{t}=\frac{100-66}{20}=\frac{34}{20}=\frac{17}{10}\)

B: Packer B; t=24 minutes, x=$52 : \(\frac{100-x}{t}=\frac{100-52}{24}=\frac{48}{24}=2\)

C: Packer C; t=30 minutes, x=$46 : \(\frac{100-x}{t}=\frac{100-46}{30}=\frac{54}{30}=\frac{9}{5}\)

D: Packer D; t=40 minutes, x=$32 : \(\frac{100-x}{t}=\frac{100-32}{40}=\frac{68}{40}=\frac{17}{10}\)

E: Packer E; t=48 minutes, x=$28 : \(\frac{100-x}{t}=\frac{100-28}{48}=\frac{72}{48}=\frac{3}{2}\)

B is maximum at 2 and is the answer.
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Re: There are 5 packers A, B, C, D, and E in a company. The five  [#permalink]

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New post 07 Apr 2011, 12:44
1
the answer is B
A B C D E
66 54 46 32 28
20 24 30 40 48

A: (100-66)*60*8/20=816
B (100-54)*8*60/24=960
did thesame with C D and E and got 864 and 816 and 720
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Re: There are 5 packers A, B, C, D, and E in a company. The five  [#permalink]

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New post 06 Jun 2012, 01:34
1
fluke wrote:
There are 5 packers A, B, C, D, and E in a company. The five packers A, B, C, D and E charge $66, $52, $46, $32, and $28, respectively, to pack each item. The time taken by the packers to pack one item is 20 minutes, 24 minutes, 30 minutes, 40 minutes and 48 minutes, respectively. All the items are sold at the end of the day. Each item earns a profit of 100 dollars, and the packers are paid from this profit. If each packer works 8 hours a day, which packer contributes the most to the net profit of the company?

A. Packer A
B. Packer B
C. Packer C
D. Packer D
E. Packer E



Actually, the question is fair and could be a GMAT question. It takes less than 2 minutes to figure out but I would expect the values to be better in actual GMAT. Let me explain.

Ratio of time taken by the 5 packers is 20:24:30:40:48 = 10:12:15:20:24
Ratio of their speeds must be 1/10:1/12:1/15:1/20:1/24 = 12:10:8:6:5 (by multiplying by LCM = 120)

Obviously, there are packers who work faster than others so they will charge more. The rates would be fair if they charge in the ratio of their speeds i.e. in the ratio 12:10:8:6:5.

But they charge in the ratio 66:52:46:32:28 = 33:26:23:16:14 (this is where I say the numbers could have been better e.g. this ratio could have been 36:28:24:18:15 and the answer would have been a clear B)
Anyway, we work with the numbers we have.

Ratio of their speeds is 12:10:8:6:5 = 36:30:24:18:15 (multiplying by 3 to make it close to the ratio of their charges)

Their speeds ratio is 36:30:24:18:15
Their charges ratio is 33:26:23:16:14
A charges 3/36 = 1/12 less than what he could.
B charges 4/30 = 1/7.5 less than what he could.
C charges 1/24 less than what he could.
D charges 2/18 = 1/9 less than what he could.
E charges 1/15 less than what he could.

Since B gives the maximum discount to the company, he adds most to the company's profit.
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Re: There are 5 packers A, B, C, D, and E in a company. The five  [#permalink]

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New post 07 Apr 2011, 21:10
So if a packer is given from $100, then we have to see for whom the margin of (profit - payout) is greatest

A -> (8 * 60)/20 * (100-66) = (8 * 60)/20 * 34

B -> (8 * 60)/24 * (100-52) = (8 * 60)/24 * 48

C -> (8 * 60)/30 * (100-46) = (8 * 60)/30 * (54)

D -> (8 * 60)/40 * (100-32) = (8 * 60)/40 * (78)

E -> (8 * 60)/48 * (100-28) = (8 * 60)/48 * (72)


Clearly B has max value, so answer - B
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Re: There are 5 packers A, B, C, D, and E in a company. The five  [#permalink]

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New post 07 Apr 2011, 23:39
fluke wrote:
There are 5 packers A, B, C, D, and E in a company. The five packers A, B, C, D and E charge $66, $52, $46, $32, and $28, respectively, to pack each item. The time taken by the packers to pack one item is 20 minutes, 24 minutes, 30 minutes, 40 minutes and 48 minutes, respectively. All the items are sold at the end of the day. Each item earns a profit of 100 dollars, and the packers are paid from this profit. If each packer works 8 hours a day, which packer contributes the most to the net profit of the company?

(A) Packer A
(B) Packer B
(C) Packer C
(D) Packer D
(E) Packer E

Spoiler: :: "OA/OE"
Ans: "B"

Sol:
Each item earns a profit of 100 dollars for the company. If a packer charges x dollars for each item, the profit that the company would effectively get on each item packed would be 100 – x
Each worker works for 8 hours= 480 minutes. Hence, if a worker takes "t" minutes to pack an item, then he would pack "480/t" items each day. So, the net profit on the \(\frac{480}{t}\) items is \(\frac{480}{t}(100-x)\). The expression \(\frac{480}{t}(100-x)\) is a maximum when \(\frac{1}{t}(100-x)\) is maximum.

We just need to check for what value of "x" and "t", this expression \(\frac{100-x}{t}\) will be maximum.

A: Packer A; t=20 minutes, x=$66 : \(\frac{100-x}{t}=\frac{100-66}{20}=\frac{34}{20}=\frac{17}{10}\)

B: Packer B; t=24 minutes, x=$52 : \(\frac{100-x}{t}=\frac{100-52}{24}=\frac{48}{24}=2\)

C: Packer C; t=30 minutes, x=$46 : \(\frac{100-x}{t}=\frac{100-46}{30}=\frac{54}{30}=\frac{9}{5}\)

D: Packer D; t=40 minutes, x=$32 : \(\frac{100-x}{t}=\frac{100-32}{40}=\frac{68}{40}=\frac{17}{10}\)

E: Packer E; t=48 minutes, x=$28 : \(\frac{100-x}{t}=\frac{100-28}{48}=\frac{72}{48}=\frac{3}{2}\)

B is maximum at 2 and is the answer.


One way is to calculate hourly profit of each worker and compare.

Profit per item for each worker is 100 - cost per item of each worker.

Similarly we know the number of items per hour for each worker as the rate is given.

Hourly profits for each is as below:

A = 34*3 = 102
B = 48*2.5 = 120
C= 54*2 = 108
D=68*1.5 = 102
E=72*1.25 = 90

So, answer is B
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Re: There are 5 packers A, B, C, D, and E in a company. The five  [#permalink]

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New post 16 Jun 2011, 01:03
A 24(100-66)

B 20(100-52) B gives max
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Re: There are 5 packers A, B, C, D, and E in a company. The five  [#permalink]

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New post 26 Jun 2011, 10:35
B is the answer..

but isnt there any other easy way?
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Re: There are 5 packers A, B, C, D, and E in a company. The five  [#permalink]

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New post 29 Jun 2011, 11:07
excuse me, i thought profit=selling price -cost price...
however, you guys all say that profit 100 here is selling price and i really dont understand
can you guy please explain more details
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Re: There are 5 packers A, B, C, D, and E in a company. The five  [#permalink]

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New post 03 Jun 2012, 10:23
The problem has a lots of information but you can handle it in this simple way.
So

A ---> 66$ each pack work for 8 hours so 480 minutes and each pack require 20 minutes ------> so 480/20 = 24 packs total. Now the profit is 100* 24 packs = 2400 minus the costs that are 66$ * 24 packs = 1584 ----> NET PROFIT 816 $

B ----> the same reasoning lead you to NET PROFIT of 960

At this poit if you look at C D and E lead you to a NET PROFIT less than A or B (if you calculate as above you have less values).

B is the answer. Simple. ;)

PS: kudos if you want :)
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Re: There are 5 packers A, B, C, D, and E in a company. The five  [#permalink]

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New post 04 Jun 2012, 06:22
Hi,

Data is represented in the tabular format:

Answer (B)

Regards,
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Re: There are 5 packers A, B, C, D, and E in a company. The five  [#permalink]

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New post 05 Jun 2012, 05:14
B it is. This took me more than 5 mins. A guess would have been a more efficient approach...
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Re: There are 5 packers A, B, C, D, and E in a company. The five  [#permalink]

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New post 05 Jun 2012, 05:36
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Re: There are 5 packers A, B, C, D, and E in a company. The five  [#permalink]

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New post 06 Jun 2012, 05:24
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Re: There are 5 packers A, B, C, D, and E in a company. The five   [#permalink] 06 Jun 2012, 05:24
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