mattocks wrote:Can someone explain the best way to approach this problem?

It takes machine A x hours to manufacture a deck of cards that machine B can manufacture in 1/x hours. If machine A operates alone for y hours and is then joined by machine B until 100 decks are finished, for how long will the two machines operate simultaneously?

a) (100y-x)/x^2 +1
b) (100x-y)/x^2 + 1
c) (100y-x^3-x)/x^2 + 1
d) (100y-x^2y-y)/x^2 +1
e) 100x/x^2+1

IMO B

solution attached

Here is my approach:

Given : Machine A takes x hrs to manufacture a deck of cards
so in 1 hr how much work is done ? 1/x

same way we can get for Machine B's 1 hour work which is 1/1/x = x

now machine A operates alone for y hours so the work done by machine A alone is
= work done in 1 hour * y (hours) = y*1/x = y/x

then machine A is joined by Machine B but machine A has already done some work alone. We subtract the work done by A from 100 (which is the total work i.e the decks) = 100-y/x = (100x-y)/x

now to find out how much time the two machines operate together :

work left to do / combine work done by machines per hour = [(100x-y)/x]/x+1/x
= (100x-y)x/x(x^2+1)
= (100x-y)/x^2+1
= (B)

Please do feel free to correct me if I am doing anything wrong