SOLUTION: The Calculator Club bought a $54 calculator for club use. If there had been three more students, each would have had to donate 20 cents less. How many students were in the club? HE

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Question 143385This question is from textbook algebra: structure and method: book1
: The Calculator Club bought a $54 calculator for club use. If there had been three more students, each would have had to donate 20 cents less. How many students were in the club? HELP MUCH APPRECIATED!!!! This question is from textbook algebra: structure and method: book1

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
The Calculator Club bought a $54 calculator for club use. If there had been three more students, each would have had to donate 20 cents less. How many students were in the club?
:
Let x = actual no. of students
:
Actual cost per student = 54%2Fx
:
Cost with 3 more students = 54%2F%28%28x%2B3%29%29
:
Actual cost - 20 cents = cost with 3 more students
54%2Fx - .20 = 54%2F%28%28x%2B3%29%29
Multiply equation by x(x+3), results:
54(x+3) - .2x(x+3) = 54x
:
54x + 162 - .2x^2 - .6x = 54x
:
-.2x^2 +54x - 54x - .6x + 162 = 0
:
-.2x^2 - .6x + 162 = 0
:
Multiply equation by -5; makes coefficient of x^2 = +1
x^2 + 3x - 810 = 0
:
Factors to:
(x+30)(x-27) = 0
:
Positive solution:
x = 27 students
:
:
Check solution by finding the costs:
54/27 = $2.00
54/30 = $1.80; 20 cents less