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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?

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.

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.

Re: PS-Work (m08q10) [#permalink]
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

Re: PS-Work (m08q10) [#permalink]
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 !

Re: PS-Work (m08q10) [#permalink]
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.

Re: PS-Work (m08q10) [#permalink]
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.

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

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.

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