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M08-10

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M08-10  [#permalink]

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New post 15 Sep 2014, 23:36
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A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

59% (01:33) correct 41% (02:04) wrong based on 80 sessions

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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 the cost 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 operate to justify the usage of Machine A instead of Machine B?

A. 6
B. 7
C. 9
D. 10
E. 11

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Re M08-10  [#permalink]

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New post 15 Sep 2014, 23:37
1
Official Solution:

Machine A can produce 50 components a day while Machine B only 40. The monthly maintenance cost for Machine A is $1500 while the cost 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 operate to justify the usage of Machine A instead of Machine B?

A. 6
B. 7
C. 9
D. 10
E. 11


Monthly maintenance costs for machine A and for machine B are fixed. Meaning that even if the plant doesn't operate 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.

Profit from machine A in d days: \(50*10*d-1,500\);

Profit from machine B in d days: \(40*10*d-550\);

\(50*10*d-1,500 \ge 40*10*d-550\), which leads to \(d \ge 9.5\). Hence the minimum # of days is 10.


Answer: D
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Re: M08-10  [#permalink]

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New post 19 Jan 2016, 05:54
Cover up the higher fixed cost, machine A needs to operate for at least a number days . But what is the number of days?
Profit=revenue -cost=price*quantity-cost
here only fixed cost counts.
profit for A=10*quantity-1500=10*50*d-1500
profit for B=10*40*d-550
10*50*d-1500≥10*40*d-550
or,d≥9.5
so, 9.5 or higher gives us 10 days.
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Re: M08-10  [#permalink]

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New post 19 Jan 2016, 10:07
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attached
>> !!!

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Re: M08-10  [#permalink]

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New post 19 Jan 2016, 18:05
Monthly maintenance cost for A is 950$ more than for B.
But every day A generates 100$ more income than B.
Then A needs to run at least 950/100 = 9.5 (rounded to 10) days to cover the difference in maintenance cost.



Bunuel wrote:
Machine A can produce 50 components a day while Machine B only 40. The monthly maintenance cost for Machine A is $1500 while the cost 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 operate to justify the usage of Machine A instead of Machine B?

A. 6
B. 7
C. 9
D. 10
E. 11
Intern
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Re: M08-10  [#permalink]

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New post 13 Aug 2018, 10:41
"If each component generates an income of $10, what is the least number of days per month that the plant has to operate to justify the usage of Machine A instead of Machine B?"

This is a poor quality question, as the number of days requested is unclear. The question would be MUCH clearer if read:

"If each component generates an income of $10, what is the least number of additionaldays per month that the plant would have to operate to justify the usage of Machine A instead of Machine B?"
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Re: M08-10  [#permalink]

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New post 01 Oct 2018, 22:16
With this post I want to explain exactly what is happening in this question:-

Attachment:
Mac A vs B.jpg


The blue line (profit for machine A) is above the orange line (profit for machine B) after 10 days (approx)

Attachment:
Mac A vs B (2).jpg

>> !!!

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Re: M08-10 &nbs [#permalink] 01 Oct 2018, 22:16
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