Work problem

[umaa]

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?

(100xy - z)/ (x + y)



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




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



x + y /(100xy - z)



x + y - z / (100xy)

OA is B

[papgust]

A takes z hours to manufacture alone. So, its z/x decks

Remaining work to be done by both A and B is (100 - z/x) decks

Together, A and B takes (x+y)/xy hours to manufacture 1 deck. Hence, to manufacture (100 - z/x) decks,

(100 - z/x) * xy/(x+y)

which is nothing but,

y*(100x - z)/(x + y)

[mehravikas]

Awesome explanation.

A takes z hours to manufacture alone. So, its z/x decks

Remaining work to be done by both A and B is (100 - z/x) decks

Together, A and B takes (x+y)/xy hours to manufacture 1 deck. Hence, to manufacture (100 - z/x) decks,

(100 - z/x) * xy/(x+y)

which is nothing but,

y*(100x - z)/(x + y)

[mcdesty]

Here is how I would organize this on my scratch paper. Stay organized and don't fear the algebra.

[[email protected]]

Hi mcdesty et al.,

Since this post was last "active" over 4 years ago, my guess is that the original users are no longer on the site.

However, for anyone new who might be interested in a strategic approach to this question, it's actually perfect for TESTing VALUES. Here's how:

Machine A takes X hours to make a deck of cards.
Machine B takes Y hours to make a deck of cards.

Since the answers are suitably complex-looking, let's choose really small, easy numbers to work with:

X = 1
Y = 2

So...
Machine A takes 1 hour to make a deck of cards
Machine B takes 2 hours to make a deck of cards
When both machines work together for 2 total hours, 3 decks of cards are made.

The question goes on to state that Machine A will work alone for Z hours, then be joined by Machine B until 100 decks are made.

Z = 1

In that first hour, Machine A will produce 1 deck of cards, leaving 99 decks to go. Since the two machines together can produce 3 decks every 2 hours, the remaining 99 decks will take...
2 hours x 33 sets = 66 hours.

We're looking for the answer that equals 66 when we plug in X=1, Y=2 and Z=1 into the answer choices.

While it "looks like" there's a lot of math to be done, most of the answers are way too small to be 66 (and it shouldn't take too long to figure that out). Only one answer equals 66

Final Answer: B

GMAT assassins aren't born, they're made,
Rich

[GMATGuruNY]

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?

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)

Let x = 1 hour.
Since A takes 1 hour to produce a deck, A's rate = 1 deck per hour.
Let y = 2 hours.

Let A complete the ENTIRE JOB.
Since A produces 1 deck per hour, the time for A to produce all 100 decks = 100 hours.
Thus, z=100.

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

Now 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.

[GMATinsight]

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?

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)

A's 1 hour work = 1/X
B's 1 hour work = 1/Y
1 Hour work of both A and B together = (1/X)+(1/Y) = (X+Y)/XY

A's Z hour work = Z/X
B's T hour work = T/Y [T is the assumed time that B worked with A i.e. both A and B worked Simultaneously]

The total work done by A and B together in T hours = T*(X+Y)/XY

Total work = (Z/X)+{T*(X+Y)/XY} = {ZY+T(X+Y)}/(XY) = 100 Decks

T = (100XY-ZY)/(X+Y) = Y(100X-Z)/(X+Y)

Answer: Option B

[Scott@TargetTestPrep]

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?

A. (100xy - z)/(x + y)
B. y(100x - z)/(x + y)
C. 100y(x - z)/(x + y)
D. (x + y)/(100xy - z)
E. (x + y - z)/100xy

We can let the rate of machine A = 1/x and the rate of machine B = 1/y. We are given that machine A operates for z hours, so machine A completes (1/x)(z) = z/x decks of cards when operating alone. Thus, there will be 100 - z/x decks left to complete when machines A and B work together.

We can let n = the number of hours that machines A and B work together to complete (100 - z/x) decks and we can create the following equation:

(1/x + 1/y)(n) = 100 - z/x

Multiplying the entire equation by xy, we have:

(y + x)(n) = 100xy - zy

n = y(100x - z)/(y + x)

Answer: B