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# M08-10

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Math Expert
Joined: 02 Sep 2009
Posts: 64314

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15 Sep 2014, 23:36
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Difficulty:

55% (hard)

Question Stats:

66% (02:13) correct 34% (02:40) wrong based on 77 sessions

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 _________________ Math Expert Joined: 02 Sep 2009 Posts: 64314 Re M08-10 [#permalink] ### Show Tags 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.

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Joined: 22 Aug 2014
Posts: 38

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19 Jan 2016, 05:54
1
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.
Intern
Joined: 03 Jul 2015
Posts: 27

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19 Jan 2016, 10:07
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attached
>> !!!

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Intern
Joined: 04 Jan 2015
Posts: 9

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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 Joined: 03 Aug 2018 Posts: 2 Re: M08-10 [#permalink] ### Show Tags 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?" VP Joined: 11 Feb 2015 Posts: 1145 Re: M08-10 [#permalink] ### Show Tags 01 Oct 2018, 22:16 1 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 >> !!! You do not have the required permissions to view the files attached to this post. _________________ Manager Joined: 09 Nov 2018 Posts: 93 Schools: ISB '21 (A) Re M08-10 [#permalink] ### Show Tags 10 Aug 2019, 11:29 I think this is a high-quality question and I agree with explanation. Senior Manager Status: PhD-trained. Education, Research, Teaching, Training, Consulting and Advisory Services Joined: 23 Apr 2019 Posts: 439 Re: M08-10 [#permalink] ### Show Tags 14 Oct 2019, 14:58 jgoldman9 wrote: "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?"

Agreed. The original question lacks the key words of "additional days" that A has to operate to recover costs.
Re: M08-10   [#permalink] 14 Oct 2019, 14:58

# M08-10

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