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It takes machine A 'x' hours to manufacture a deck of cards that

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New post 15 Apr 2018, 18:26
There is an easier method I guess.

Its usually used in Physics, but can be used here also I guess.

Parallelism of units.

See numerator of Option A,
100xy - z

x, y and z have hours units. We cannot add or subtract 100 (hour) (hour) with only (hour)

That is an unit inconsistency. The numerator has to contain two units of hours for each term, and the entire numerator should be divided by a term which has only one hour unit.

By seeing this, we can eliminate A. --> (100xy-z)
We can also eliminate D. --> (100xy-z) in the denominator.
We can also eliminate E. --> The resulting units will be 1/hour and not hour, as we desire.

We are left with B and C.

Analyzing the differences in the terms, we have either y(100x - z) or 100y(x - z)

As soon as we get a "100x - z" term in our working out of the answer, we can eliminate the other.

Let's start working.

A takes x hours to do what B takes y hours.

A operates for z hours. So, decks produced --> z/x

Remaining decks --> 100 - z/x

--> (100x - z)/x

Stop here. We got our term. The final answer, whatever it is, will have a variation of this term.

Hence, from B and C. B has the term 100x - z.

So, B.

No need to solve further.

Bunuel wrote:
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. \(\frac{100xy – z}{x + y}\)

B. \(\frac{y(100x – z)}{x + y}\)

C. \(\frac{100y(x – z)}{x + y}\)

D. \(\frac{x + y}{100xy – z}\)

E. \(\frac{x + y – z}{100xy}\)


Note that we are asked: "for how long will the two machines operate simultaneously?".

In first \(z\) hours machine A alone will manufacture \(\frac{z}{x}\) decks. So there are \(100-\frac{z}{x}=\frac{100x-z}{x}\) decks left to manufacture. Combined rate of machines A and B would be \(\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}\) decks/hour, (remember we can easily sum the rates).

As \(time=\frac{job}{rate}\), then \(time=\frac{100x-z}{x}*\frac{xy}{x+y}=\frac{y(100x-z)}{x+y}\).

Answer: B.

Hope it's clear.
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New post 19 Nov 2018, 17:30
sjayasa wrote:
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. \(\frac{100xy – z}{x + y}\)

B. \(\frac{y(100x – z)}{x + y}\)

C. \(\frac{100y(x – z)}{x + y}\)

D. \(\frac{x + y}{100xy – z}\)

E. \(\frac{x + y – z}{100xy}\)


I think by plugging in numbers I found that the questions is really asking, how long does it take for both machines to finish the job, as opposed to, "how long will the two machines operate simultaneously?"

For example I chose the numbers:
x=2
y=10
z=100

This lead me to find that the entire job would be completed in approximately 84 hours, but they would only be working simultaneously for 34 hours because the first 50 are completed by machine A alone.

When plugging in my numbers B was the closest with the answer of 83.333 so I chose B, but that was the answer that machines A and B took to complete the entire job. They only worked together for 34 hours.

Please let me know if anyone else found the wording misleading. Thank you!
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Re: It takes machine A 'x' hours to manufacture a deck of cards that  [#permalink]

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New post 22 Nov 2019, 01:12
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Re: It takes machine A 'x' hours to manufacture a deck of cards that   [#permalink] 22 Nov 2019, 01:12

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