until 100 decks are finished

sanju09 wrote: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) (100 y – x)/ (x^2 + 1)
(B) (100 x – y)/ (x^2 + 1)
(C) (100 y – x^3 – x)/ (x^2 + 1)
(D) (100 y – x^2 y – y)/ (x^2 + 1)
(E) 100 x/ (x^2 + 1)

In 1 hour A manufactures 1/x deck of cards
In 1 hour B manufactures x deck of cards
In 1 hour A and B together manufacture (1/x + x) deck of cards

Hence, to manufacture a single deck of card, A and B together will take 1/(1/x + x) hours = x/(x² + 1) hours

Now, A alone works for y hours.
In y hours, A will manufacture y/x deck of cards.

The remaining (100 - (y/x)) deck of cards A and B manufactured together.
Hence, time required to do so = (100 - (y/x))*(x/(x² + 1)) hours
= (100x - y)/(x² + 1) hours

The correct answer is B.