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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, 14:03
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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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New post 09 Mar 2011, 22:56
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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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New post 09 Mar 2011, 16:00
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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New post 09 Mar 2011, 23:03
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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 09 Mar 2011, 23:11
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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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New post 10 Mar 2011, 07:30
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, 19:45
A "lone wolf" trap. :-D
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Re: Working at their respective constant rates, machine A makes  [#permalink]

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New post 11 Mar 2011, 22:24
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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New post 04 Feb 2016, 13:35
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, 21:05
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, 19:17
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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Re: Working at their respective constant rates, machine A makes  [#permalink]

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New post 08 Jul 2018, 19:39
A : 100 copies in 12 minute or 500 copies in 1hr
B: 100 copies in 15 minute or 400 copies in 1hr
The question basically asks
500*2*x+400*2*y = ?
or
the value of 1000x+800y
Statement 1
X=Y
Clearly not sufficient.
Statement 2:
5x + 4y = 90.
multiply 200 on both sides
5x*200+4y*200=90*200
=>1000x+800y=18000
Hence sufficient
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Re: Working at their respective constant rates, machine A makes  [#permalink]

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New post 09 Jul 2018, 08:12
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.




Rate of M/c A = 100 copies/12 min = 100*60/12 = 500 copies/hr

Rate of M/c B = 100 copies/15 min = 100*60/15 = 400 copies/hr

Hence in 2 hrs, one M/c A can make 1000 copies, So x M/c's will make 1000x copies

Similarly, in 2 hrs, one M/c B can make 800 copies, So y M/c's will make 800y copies

Total # of copies = 1000x + 800y


Statement 1: x = y

Total # of copies = 1000x + 800x = 1800x

Statement 1 is not Sufficient.


Statement 2: 5x + 4y = 90

Multiplying both Sides by 200

we get, 1000x + 800y = 1800

Statement 2 is Sufficient.


Answer B.



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GyM
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Re: Working at their respective constant rates, machine A makes &nbs [#permalink] 09 Jul 2018, 08:12
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