Let no. of days worked be x

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

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

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

D

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.

M08-10
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Just Trying TSD instead of T&M

the relative difference in earning per day for Machine A and B = 50*10-40*10=100
The relative cost that Machine A has to cover= 1500-550=950
no of days required=950/100=9.5 ~10

We are given that Machine A can produce 50 components per day, and that each component brings revenue of $10 per component. Thus, we know that Machine A’s daily revenue is (50)(10) = 500. We are also given that Machine A’s maintenance fee is $1500 per month.

Machine B produces 40 components per day and each component brings revenue of $10 per component. Thus, we know that Machine B’s daily revenue is (40)(10) = 400. Machine B’s maintenance fee is $550 dollars per month.

Since we need to justify the usage of Machine A, we need to determine the minimum number of days it will take until Machine A’s profit is greater than Machine B’s profit. We can let t = the number of days until this happens and create the following inequality:

50(10)(t) - 1500 > 40(10)(t) - 550

500t - 1500 > 400t - 550

100t > 950

t > 9.5

Thus, the minimum number of days needed is 10.

Answer: D
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Solution:
Given: Machine A can produce 50 components a day and each component generates income of $10.Similarly, Machine B can produce 40 components a day and each component generates income of $10.
Also given that; monthly maintenance cost for machine A and machine B is $1500 and $550 respectively.
Approach:
Machine A’s daily revenue \(= 50 ×10=500$\)
Machine B’s daily revenue \(= 40 ×10=400$\)
We need to determine the minimum number of days it will take until Machine A’s profit is greater than Machine B’s profit. Let “x” be the number of days until this happens and let’s create the inequality as follows:
\(500(x)- 1500>400(x)- 550\)
\(100x>950\)
\(x>9.5\)

Therefore the minimum number of days required is 10.

The correct answer option is “D”.

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