Speed of Machine A = 1/x decks/hour
Speed of Machine B = 1/y decks/hour

Combined Speed of both machines i/x + 1/y = decks/hour

Machine A worked for z hours, so the number of decks produced in z hours =z/x decks

Decks remaining to be produced = 100-z/x

time taken to finish the work = Number of decks left/Combined Speed = (100-z/x)/i/x + 1/y i.e

b) y(100x - z)/(x+y) .

Thanks Guerrero, that is the best way to solve these type of problems.

But here in this case options allow me not to solve the question algebraically and we can answer in less than 30 seconds.

Options:-

a) (100xy - z)/(x+y) -- Not Possible as numerator is (hour^2 - hour), which is against the law of adding/subtracting expressions with similar units.
b) y(100x - z)/(x+y) -- Correct!!
c) (x+y)/(100xy - z) -- Not Possible as denominator is (hour^2 - hour), which is against the law of adding/subtracting expressions with similar units.
d) xz(100y - z)/(x+y+z) -- Not Possible as numerator is (hour^3 - hour^3) and denominator is hour , making overall expression as hour^2 but we need the answer in hour.
e) (x+y-z)/(100xy) -- Not Possible as numerator is hour and denominator is hour^2, making overall expression as hour^-1 but we need the answer in hour.

Hope this makes sense!!!

puneetkhurana2000 wrote:Question: - It takes machine A x hours to manufacture a deck of cards that machine B can manufacture in y hours. If machine A operates alone for z hours and is then joined by machine B until 100 decks are finished, for how long will the two machines operate simultaneously?

Options:-

a) (100xy - z)/(x+y)
b) y(100x - z)/(x+y)
c) (x+y)/(100xy - z)
d) xz(100y - z)/(x+y+z)
e) (x+y-z)/(100xy)

This is from MGMAT.

Let one deck = 2 units.

Let x = 1 hour.
Then A's rate = w/t = 2/1 = 2 units per hour.
Let y = 2 hours.

Since each deck = 2 units, 100 decks = 100*2 = 200 units.
Let A complete the ENTIRE JOB.
Time for A to produce 200 units at a rate of 2 units per hour = w/r = 200/2 = 100 hours.
Thus, z=100.

The question stem asks for the time that the machines work together.
Since A completes the entire job, the number of hours that the machines work together = 0. This is our target.

Now we plug x=1, y=2 and z=100 into the answers to see which yields our target of 0.
A quick scan reveals that only B works:
y(100x - z)/(x+y) = 2(100*1 - 100)/(1+2) = 0.

The correct answer is B.

Thanks Mitch, can you please have a look at my method of solving the question.

Hope it makes sense!!!

Puneet