Bunuel
A small company is planning to rent either computer A or computer B to print customer mailing lists. Both computer A and computer B must be rented on an hourly basis. The rental fee is based only on the amount of time the computer is turned on. It will cost 40 percent more per hour to rent computer A than to rent computer B. Computer B would, however, require 20 hours more than computer A to do the job. If either computer A, or computer B were rented the total cost to rent the computer would be $660.00. What would be the approximate hourly charge to rent computer B?
(A) $9.40
(B) $11.30
(C) $13.20
(D) $17.80
(E) $22.10
Let the cost per hour to rent A and B be a and b respectively => a=1.4b, and
the time required by A and B be x and y respectively =>y=x+20
Cost for A = \(ax=1.4bx=660\)
Cost for B = \(by=b(x+20)=bx+20b=660\)
Firstly, since amount is same =>\(1.4bx=bx+20b........0.4bx=20b........x=50\)
Cost for A=\(1.4bx=1.4b*50=660.........7b=66<70\), so b<10..
Only $9.40 fits in.
We could also check by calculating b=66/7 = 9.43