Difficult Math Problem #104 - Work

[800guy]

from diff math doc. oa coming when some people answer with explanations:

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

[jayhawk2001]

Let x be the number of days per month required for A to generate more
revenue than B

(50 * 10 * x - 1500) - (40 * 10 * x - 550) > 0

Solving for x
500x - 1500 - 400x + 550 > 0
100x - 950 > 0
or
x > 9.5

So answer should be D (10 days)

[slimsohn]

I think that the answer would be 10, (D).

Machine A will generate 50 * $10 = $500/day
Machine B will generate 40* $10 = $400/day

So every day that they both run, machine A makes $100 more revenue. The difference in maintenance cost is $950, so in order to justify using Machine A, it will have to run for at least 10 days which will result in $1000 extra revenue.

That $1000 will offset the difference in maintenance costs plus give added revenue.

[800guy]

oa:

Let x be the number of days that need to be worked
500x - 1500 > 400x - 550
100x > 950
x > 9.5 (D)

[BTGmoderatorRO]

The problem can be solved as a word problem

Machine A produces 50 components on a daily basis each at $10
therefore, Machine A will generate 50 * $10 = $500 on a daily basis

Machine B on the other hand produce 40components in a day each at $10
machine B will generate 40 * $10 = $400

monthly maintenance fee of A is $1500
monthly maintenance fee of B is $ 550

Based on their daily productivity, A earn $100 more than B
difference in the maintenance cost is $1500 - $550 = $950

for us to justify the usage of Machine A over B, let us consider the difference in the income earned by A on a daily basis. i.e $10
machine A spend more on maintenance and will only to operate for at least of 10 days which will gives $100 turnover and will cover for the difference in maintenance cost between A and B.