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Question 1100047: Ginelle is operating a tutorial center. She is operating at a loss of 15,000 pesos. Her tutorial fees are at 450 pesos per hour, and she has to pay a maintenance fee of 250 pesos per hour. What is the function that illustrates the flow of money of the tutorial center? Let x be the number of hours. Combine like terms. How many hours of tutorial service does her center have to operate in order to gain a profit of 15,000 pesos?
Found 2 solutions by richwmiller, Theo:
Answer by richwmiller(17219)   (Show Source):
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! profit = revenue minus cost.
her revenue is 450 pesos per hour.
her cost is equal to a fixed cost plus a variable cost of 250 pesos per hour.
if we allow:
p = profit
r = revenue
f = fixed cost
v = variable cost
x = number of hours
then the formula becomes p = r*x - f - v*x
her revenue is 450 pesos per hour and her variable cost is 250 pesos per hour.
the formula becomes p = 450*x - f - 250*x
if we combine like terms, then the formula becomes p = 200 * x - f
if p is positive, then she has a profit.
if p is negative, then she has a loss.
you say she's operating at a loss of 15000 pesos.
the formula becomes -15000 = 200*x - f
we now need to know how many hours she was teaching when she was operating at a loss of 15000 pesos.
if it was 0 hours, then her fixed cost had to be 15000 pesos.
the formula would become -15000 = 200*0 - f
we would solve for f to get f = 15000.
the formula becomes -15000 = 200*x - 15000.
we would solve for x to get x = 0/200 = 0.
you then want to know how many hours she would have to teach in order to get a profit of 15000 pesos (positive 15000).
the formula would become 15000 = 200 * x - 15000
add 15000 to both sides of this equation to get 30000 = 200 * x
divide both sides of this equation by 200 and solve for x to get x = 30000 / 200 = 150.
she would have to teach for 150 hours to make a profit of 15000 pesos, if her fixed cost was 15000 pesos.
you did not specify what the fixed cost was, so i assumed 15000.
it could have been anything.
for example:
assume the fixed cost was 100,000 pesos.
the formula would then become p = r*s - f - v*x which would becomes p = 450*x - 100,000 - 250 * x which would be simplified to p = 200*x - 100,000.
if she was operating at a loss of 15000 pesos, then the formula would become -15000 = 200*x - 100,000.
you would add 100,000 to both sides of this equation to get 85,000 = 200*x and you would solve for x to get x = 85,000 / 200 = 425 hours.
when the fixed cost was 100,000 pesos, she would be operating at a loss of 15000 pesos if she was tutoring for 425 hours.
p = 200*425 - 100,000 = -15000.
in order to be operating at a profit of 15000, the same formula would be used, except p = 15000 and you would get 15000 = 200*x - 100,000
you would then add 100,000 to both sides to get 115,000 = 200*x and you would solve for x to get x = 115,000 / 200 = 575 hours.
she would have to tutor for 575 hours to make a profit of 15000 pesos.
p = 200*575 - 100,000 = 15000.
bottom line is the problem wasn't defined very well since i wasn't sure what the fixed cost was.
i assumed 15000 and got one answer.
i assumed 100,000 and got another answer.
the formula, however, is the same, so if you know what the fixed cost is supposed to be, you can solve for the he rest.
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