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Working at their respective constant rates, machine A makes

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Working at their respective constant rates, machine A makes [#permalink]

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New post 09 Mar 2011, 15:03
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Working at their respective constant rates, machine A makes 100 copies in 12 minutes and machine B makes 100 copies in 15 minutes. If a number of x machine A and a number of y machine B work simultaneously at their respective rates for 2 hours, what is the total number of copies that they will produce?

(1) x = y.
(2) 5x + 4y = 90.
[Reveal] Spoiler: OA
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Re: Working at their respective constant rates, machine A makes [#permalink]

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New post 09 Mar 2011, 17:00
banksy wrote:
268. Working at their respective constant rates, machine A makes 100 copies in 12 minutes and machine B makes 100 copies in 15 minutes. If a number of x machine A and a number of y machine B work simultaneously at their respective rates for 2 hours, what is the total number of copies that they will produce?
(1) x = y.
(2) 5x + 4y = 90.

can't find the the total copies that they will produce without knowing the value of x and y. So with (1) and (2), it is sufficient to find values.
Answer C.
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banksy wrote:
268. Working at their respective constant rates, machine A makes 100 copies in 12 minutes and machine B makes 100 copies in 15 minutes. If a number of x machine A and a number of y machine B work simultaneously at their respective rates for 2 hours, what is the total number of copies that they will produce?
(1) x = y.
(2) 5x + 4y = 90.



Rate of Machine A is 100/12 copies per minute and that of Machine Y is 100/15 copies per minute.

When x number of A machines and y number of B machines work simultaneously at their respective rates for 2 hours (120 minutes), they will produce \(120*((100/12)*x+(100/15)*y)\)copies

\(120*((100/12)*x+(100/15)*y)\) on simplification becomes \(200*(5x+4y)\)

Statement 1 says \(x=y\), which doesn't help us get the exact numbers, so insufficient

Statement 2 says \(5x+4y = 90\), we cant solve for x and y from this, but all we need to answer the question is value of 5x+4y which we get from this statement, so sufficient.

Answer B.
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Re: Working at their respective constant rates, machine A makes [#permalink]

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New post 10 Mar 2011, 00:03
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(1) is clearly insufficient.

For (2), several answers are possible:

5 * 14 + 4 * 5 = 90

5 * 10 + 4 * 8 = 90

But (1) and (2) together give x = y = 10, so answer is C
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Re: Working at their respective constant rates, machine A makes [#permalink]

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New post 10 Mar 2011, 00:11
Mongolia2HBS wrote:
banksy wrote:
268. Working at their respective constant rates, machine A makes 100 copies in 12 minutes and machine B makes 100 copies in 15 minutes. If a number of x machine A and a number of y machine B work simultaneously at their respective rates for 2 hours, what is the total number of copies that they will produce?
(1) x = y.
(2) 5x + 4y = 90.

can't find the the total copies that they will produce without knowing the value of x and y. So with (1) and (2), it is sufficient to find values.
Answer C.




subhashghosh wrote:
(1) is clearly insufficient.

For (2), several answers are possible:

5 * 14 + 4 * 5 = 90

5 * 10 + 4 * 8 = 90

But (1) and (2) together give x = y = 10, so answer is C


Guys, we are not concerned here with exact values of x and y. All we need to find is the total number of copies the machines working together would produce. So, anything that gives us their combined relationship is sufficient. Statement 2 effectively is telling us about that relationship.

So, statement B is sufficient to answer the given question.

Irrespective of whatever value x and y individually may have, as long as 5x+4y is 90, the number of pages they will produce would remain 200*90

We can easily check the same by putting different values for x and y.
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Re: Working at their respective constant rates, machine A makes [#permalink]

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New post 10 Mar 2011, 08:30
As per question:

A makes 1000 pages in 2 hours
B makes 800 pages in 2 hours

x machines of A and y machines of B are used.

Total output = 1000x + 800y......(1)

Given by statement 2: 5x + 4y = 90...........(2)

From 1 and 2,

The answer will be 90 * 200 = 1800 pages.

Hence B alone is sufficient.
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Re: Working at their respective constant rates, machine A makes [#permalink]

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New post 10 Mar 2011, 20:45
A "lone wolf" trap. :-D
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New post 11 Mar 2011, 23:24
I was wrong, B alone is sufficient.
Thanks to beyondgmatscore and pesfunk.
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Re: Working at their respective constant rates, machine A makes [#permalink]

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Re: Working at their respective constant rates, machine A makes [#permalink]

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New post 29 Dec 2015, 10:50
Hello from the GMAT Club BumpBot!

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Re: Working at their respective constant rates, machine A makes [#permalink]

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New post 04 Feb 2016, 14:35
Rate of A \(\frac{100}{12}\)
Rate of B \(\frac{100}{15}]\)

\(\frac{100x}{12}*120 + \frac{100y}{15}*120\)
\(1000x + 800y\)

1. x=y Tell us nothing about the actual values.
2. 5x + 4y = 90

\(1000x + 800y\)
\((5x+4y=90)*200\)
\(1000x+800y=18000\)
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Re: Working at their respective constant rates, machine A makes [#permalink]

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New post 13 Feb 2016, 22:05
banksy wrote:
Working at their respective constant rates, machine A makes 100 copies in 12 minutes and machine B makes 100 copies in 15 minutes. If a number of x machine A and a number of y machine B work simultaneously at their respective rates for 2 hours, what is the total number of copies that they will produce?

(1) x = y.
(2) 5x + 4y = 90.


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Re: Working at their respective constant rates, machine A makes [#permalink]

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New post 29 Feb 2016, 20:17
banksy wrote:
Working at their respective constant rates, machine A makes 100 copies in 12 minutes and machine B makes 100 copies in 15 minutes. If a number of x machine A and a number of y machine B work simultaneously at their respective rates for 2 hours, what is the total number of copies that they will produce?

(1) x = y.
(2) 5x + 4y = 90.


X*2*(100/12) + Y*2*(100/15) - No.of copies.

Multiply by 60.

200*(5X+4Y)=No. of copies. B Sufficient.
Re: Working at their respective constant rates, machine A makes   [#permalink] 29 Feb 2016, 20:17
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