ashiima 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

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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