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Working at their respective constant rates, machine A makes [#permalink]
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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.
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Re: Working at their respective constant rates, machine A makes [#permalink]
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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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Re: Working at their respective constant rates, machine A makes [#permalink]
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09 Mar 2011, 23:56
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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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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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10 Mar 2011, 00:11
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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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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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10 Mar 2011, 20:45
A "lone wolf" trap.



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Re: Working at their respective constant rates, machine A makes [#permalink]
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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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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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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. One big trap .A great leveller of sort.
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Re: Working at their respective constant rates, machine A makes [#permalink]
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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.



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