The constant rate at which machines A works is 3/4 of the constant rate at which machine B works. Under normal conditions, machine A would complete x lots in 2 days. However, when a technician attempts to use machines A and B simultaneously, there is not enough electrical power to run both machines at full power. As a result, running the machines simultaneously reduces the work rate of each machine by 20%.
How long will it take the machines to complete 3x lots, working simultaneously?
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talaangoshtari
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MartyMurray
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One great, pretty much fail safe way to handle work problems is to start by finding the rates of each machine or person doing the work.
A Normal Rate: x/2 per day
B Normal Rate: (x/2)/(3/4) = 2x/3 per day
Rate Together: A Rate + B Rate = 3x/6 + 4x/6 = 7x/6 Per Day
Together Running At Reduced Rate: 7x/6 x 4/5 = 28x/30 = 14x/15
3x/(14x/15) = 3/(14/15) = 45/14 Days Working Simultaneously
A Normal Rate: x/2 per day
B Normal Rate: (x/2)/(3/4) = 2x/3 per day
Rate Together: A Rate + B Rate = 3x/6 + 4x/6 = 7x/6 Per Day
Together Running At Reduced Rate: 7x/6 x 4/5 = 28x/30 = 14x/15
3x/(14x/15) = 3/(14/15) = 45/14 Days Working Simultaneously
Last edited by MartyMurray on Tue Apr 05, 2016 11:35 am, edited 1 time in total.
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Hi talaangoshtari,
When posting GMAT questions, you should make sure to post the 5 answer choices as well. In many cases, the answers provide 'clues' as to how you can approach the question. Without having that option, we're essentially forced to 'do math' (as Marty has shown), although you could also TEST VALUES (choose a value for X and then calculate from there).
GMAT assassins aren't born, they're made,
Rich
When posting GMAT questions, you should make sure to post the 5 answer choices as well. In many cases, the answers provide 'clues' as to how you can approach the question. Without having that option, we're essentially forced to 'do math' (as Marty has shown), although you could also TEST VALUES (choose a value for X and then calculate from there).
GMAT assassins aren't born, they're made,
Rich
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talaangoshtari
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Hi.
I know that. But, this is a GRE question. In GRE we have some questions without the answer choices.
I know that. But, this is a GRE question. In GRE we have some questions without the answer choices.
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regor60
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Marty Murray wrote:One great, pretty much fail safe way to handle work problems is to start by finding the rates of each machine or person doing the work.
A Normal Rate: x/2 per day
B Normal Rate: (x/2)/(3/4) = 2x/3 per day
Rate Together: A Rate + B Rate = 3x/6 + 4x/6 = 7x/6 Per Day
Together Running At Reduced Rate: 7x/6 x 3x/4 = 21x/24 = 7x/8
3x/(7x/8) = 3/(7/8) = 24/7 Days Working Simultaneously
Question indicates 20% reduction not 25%. Would think T is 45/14
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MartyMurray
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Yup. Fixed.regor60 wrote:Question indicates 20% reduction not 25%. Would think T is 45/14
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Matt@VeritasPrep
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Another popular approach is picking a value. Suppose x = 60. Then
A does 30 lots a day
B does 40 lots a day
after the power reduction,
A does 24 lots a day
B does 32 lots a day
So we've got 56 lots a day with the two together, and 180 lots for them to do.
180/56 => 4*45 / 4*15 => 45/14
A does 30 lots a day
B does 40 lots a day
after the power reduction,
A does 24 lots a day
B does 32 lots a day
So we've got 56 lots a day with the two together, and 180 lots for them to do.
180/56 => 4*45 / 4*15 => 45/14
