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PS-Work (m08q10)

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PS-Work (m08q10) [#permalink] New post 02 Dec 2008, 05:05
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Machine A can produce 50 components a day while machine B only 40. The monthly maintenance cost for machine A is $1500 while that for machine B is $550. If each component generates an income of $10 what is the least number of days per month that the plant has to work to justify the usage of machine A instead of machine B?

(A) 6
(B) 7
(C) 9
(D) 10
(E) 11

[Reveal] Spoiler: OA
D

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Re: PS-Work [#permalink] New post 02 Dec 2008, 11:26
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vivektripathi wrote:
Machine A can produce 50 components a day while
machine B only 40. The monthly maintenance cost for
machine A is $1500 while that for machine B is $550. If
each component generates an income of $10 what is the
least number of days per month that the plant has to work
to justify the usage of machine A instead of machine B?

(A) 6
(B) 7
(C) 9
(D) 10
(E) 11


Let d = days worked

Machine A monthly profit = 50*10*d - 1500
Machine B monthly profit = 40*10*d - 550

50*10*d - 1500 = 40*10*d - 550
500d - 400d = 950
100 d = 950
d = 9.5

Machine A must work 9.5 days to make the amount of money as Machine B.

D
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Re: PS-Work (m08q10) [#permalink] New post 05 Apr 2010, 07:32
To Proabir !

your conclusion about 29/3 post is right
It was a special case of power 2 (Nr²)
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Re: PS-Work (m08q10) [#permalink] New post 06 Apr 2010, 21:55
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To justify use of machine A, its maintenance cost must be lower than Machine B's (Assuming machine maintenance costs are constant)

Diff. in maintenance costs = 1500 - 550 = 950

Diff. in Revenue per day = 50(10) - 40(10) = 100

Min. number of days required to justify the cost diff. = 950/100 = 9.5

Correct choice = 10 = D
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Re: PS-Work (m08q10) [#permalink] New post 09 Apr 2010, 19:55
Let x = days worked

Machine A monthly profit = 50*10*x - 1500
Machine B monthly profit = 40*10*x - 550

50*10*x - 1500 = 40*10*x - 550
500d - 400x = 950
100 x = 950
x = 9.5

Machine A must work 9.5 days to make the amount of money as Machine B.

D
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Re: PS-Work (m08q10) [#permalink] New post 20 Jun 2010, 15:37
Hi All,

How do we know that the monthly cost is fixed, and not variable by the number of days? eg. why don't we create the following equation: [15000-1500]*m=[12000-550]*m

Do we always assume that monthly cost is fixed?
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Re: PS-Work (m08q10) [#permalink] New post 07 Apr 2011, 04:23
Revenue from Machine A should be >= Revenue from Machine B

=> x*50*10 - 1500 >= x*40*10 - 550

=> 10x * 10 >= 950

=> x >= 9.5

so x = 10

Answer - D
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Re: PS-Work (m08q10) [#permalink] New post 07 Apr 2011, 07:13
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Another way of doing it (might be complex) is the difference in amount earned by each machine is $100. and the difference in the maintanece amount is 950.
so 950/100 takes 9.5 days to make up for the cost.
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Re: PS-Work (m08q10) [#permalink] New post 07 Apr 2011, 10:22
Solve by Reverse PLUG-IN

9 days : A's profit = 3000
B's profit = 3050
10 days : A's profit = 3500
B's profit = 3450

hence, Ans= D
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Re: PS-Work (m08q10) [#permalink] New post 23 Apr 2011, 13:56
revenue from A should be greater than B.

lets assume is the x least number of days plant needs to run to justify A's cost

=> 50*x*10 -1500 = 40*x*10 - 550

=> x>9.5
=> x = 10

Answer is D.
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Re: PS-Work (m08q10) [#permalink] New post 05 Oct 2011, 21:05
honkergirl wrote:
Hi All,

How do we know that the monthly cost is fixed, and not variable by the number of days? eg. why don't we create the following equation: [15000-1500]*m=[12000-550]*m

Do we always assume that monthly cost is fixed?


this is a solid point and the question stem MUST mention that, irrespective of the number of days worked, the monthly maintenance cost would be fixed and not be pro rated !
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Re: PS-Work (m08q10) [#permalink] New post 08 Nov 2011, 05:34
I chose D, although I used a different logic. I tried to backsolve.

I first picked D) 10. If the plant works for 10 days, then the following is true:

10 days x 50 components per day x $10/component = $5,000 - $1,500 = $3,500 in revenue for machine A
10 days x 40 components per day x $10/component = $4,000 - $550 = $3,450 in revenue for machine B

Hence, the plant needs to work for at least 10 days to justify the use of machine A.

Am I right in my logic?
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Re: PS-Work (m08q10) [#permalink] New post 26 Feb 2012, 06:33
Hi,

why we do not divide the monthly cost by 30???? the question asks us how many DAYS, therefore we should divide divide monthly cost by 30......

Thanks,
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Re: PS-Work (m08q10) [#permalink] New post 26 Feb 2012, 06:52
Expert's post
Saurajm wrote:
Hi,

why we do not divide the monthly cost by 30???? the question asks us how many DAYS, therefore we should divide divide monthly cost by 30......

Thanks,


Monthly costs for machine A and for machine B are fixed. Meaning that even if plant doesn't work at all it'll still have these maintenance costs. The questions basically asks about minimum # of days (d) that plant should operate so that the profit from A is more than or equal to the profit from B.

Machine A can produce 50 components a day while machine B only 40. The monthly maintenance cost for machine A is $1500 while that for machine B is $550. If each component generates an income of $10 what is the least number of days per month that the plant has to work to justify the usage of machine A instead of machine B?
(A) 6
(B) 7
(C) 9
(D) 10
(E) 11

Profit from machine A in d days: 50*10*d-1,500;
Profit from machine B in d days: 40*10*d-550;

50*10*d-1,500\geq{40*10*d-550} --> d\geq{9.5}. Hence minimum # of days is 10.

Answer: D.

Hope it's clear.
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Re: PS-Work (m08q10) [#permalink] New post 12 Apr 2012, 06:49
D

Value generated by machine A in x days = 50*10*x - 1500
Value generated by machine B in x days = 40*10*x -550
Equate these and solve for x
x=9.5
so x is 10
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Re: PS-Work (m08q10) [#permalink] New post 12 Apr 2012, 09:29
assume d= least days
for machine A= 50 x 10 x d -1500
for machine B= 40 x 10 x d - 550

now work of machine A is require to check whether it is equal profitable (in at least for d days) with compare to machine B...

than it must be,

Revenue Generated my Machine A in d days = Revenue Generated my Machine B in d days
50 x 10 x d -1500 = 40 x 10 x d - 550

By solving above equations we will get d = 9.5

so answer is D = 10
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Re: PS-Work (m08q10) [#permalink] New post 12 Apr 2013, 05:43
Assuming x is the least number of days by which net income from A becomes more than net income from B.
So 50 * x * 10 - 1500 > 40 * x * 10 - 550
--> 100 x > 950
--> x > 9.5
--> Least number of days by which usage of A can be justified over B is 10 days.

Answer is D.
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Re: PS-Work (m08q10) [#permalink] New post 12 Apr 2013, 07:29
Machine A can produce 50 components a day while machine B only 40. The monthly maintenance cost for machine A is $1500 while that for machine B is $550. If each component generates an income of $10 what is the least number of days per month that the plant has to work to justify the usage of machine A instead of machine B?

(A) 6
(B) 7
(C) 9
(D) 10
(E) 11

Machine A:
Income for x days = 50 * 10 * x
Cost = 1500

Machine B:
Income for x days = 40 * 10 * x
Cost = 550

Usage of Machine A instead of B:
Income form A- Income from B >0
=>500x - 1500-(400x - 550) >0
=>x>9.5

Hence 10 days is the least, Option D
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Re: PS-Work (m08q10) [#permalink] New post 12 Apr 2013, 07:50
Income from A = x*10*50 - 1500
Income from B = x*10*40 - 550

use of A is justified iff
income from A > income from B
=> 500x -1500 > 400x -550
=> 100x > 950
=>x > 9.5
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Re: PS-Work (m08q10) [#permalink] New post 13 Apr 2013, 02:18
Machine A:
Cap: 50 Rev: 10$ total= 50x10xdays worked
Machine B:
Cap: 40 Rev: 10$ total = 40x10xdays worked

now A has to compensate B so => Revenue should be equal in total = > RA= RB

40x10xT.D -550= 50x10xTD -1500
TD = 9.5 :) :)

kudos please, if this helps !!!
Re: PS-Work (m08q10)   [#permalink] 13 Apr 2013, 02:18
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