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Machines A and B, working simultaneously at their respective constant [#permalink]
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28 May 2015, 06:02
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Re: Machines A and B, working simultaneously at their respective constant [#permalink]
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28 May 2015, 06:38
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Hello all My attempt:Let the rate of \(Machine A\) be \(a\) units per hour and that of \(Machine B\) be \(b\) units per hour. From first statement we have \(a*1 + b*1 = m\)\(........equation I\) From the second statement we have \(2m/a  2m/b = 3\)\(..........equation II\) We need to find out the value of \(\frac{m}{a}\). Solving \(I\)and \(II\) and substituting \(x\) for \(\frac{m}{a}\) we get \(2x^2  7x + 3 = 0\) we get \(x = 0.5\) or \(3\) and none of these exist in the options so perhaps I am on the wrong path. Any help appreciated. My answers are \(0.5 hours\) or \(3 hours\)
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Re: Machines A and B, working simultaneously at their respective constant [#permalink]
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Re: Machines A and B, working simultaneously at their respective constant [#permalink]
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28 May 2015, 07:45
Bunuel wrote: Jackal wrote: Hello all My attempt:Let the rate of \(Machine A\) be \(a\) units per hour and that of \(Machine B\) be \(b\) units per hour. From first statement we have \(a*1 + b*1 = m\)\(........equation I\) From the second statement we have \(2m/a  2m/b = 3\)\(..........equation II\) We need to find out the value of \(\frac{m}{a}\). Solving \(I\)and \(II\) and substituting \(x\) for \(\frac{m}{a}\) we get \(2x^2  7x + 3 = 0\) we get \(x = 0.5\) or \(3\) and none of these exist in the options so perhaps I am on the wrong path. Any help appreciated. My answers are \(0.5 hours\) or \(3 hours\) The question is: how long does it take machine A to produce 5m units, not m units. Thanks. Then the answer is \(E\) \(15 hours\). \(x=0.5 or 3\) But we need \(5x\) therefore \(2.5 hours\) or \(15 hours\). We reject the solution \(x=0.5 hours\) because it invalidates the statement 1.
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Re: Machines A and B, working simultaneously at their respective constant [#permalink]
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28 May 2015, 07:47
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Bunuel wrote: Machines A and B, working simultaneously at their respective constant rates produce m units in 1 hour. If working independently it takes machine B 3 hours less than machine A to produce 2m units, how long does it take machine A to produce 5m units?
A. 1 hour B. 2.4 hours C. 2.5 hours D. 6 hours E. 15 hours hi, i think the answer should be 3 for A and 1.5 for B... Since A and B together produce m units together in 1 hr so individually it has to be more than 1.. secondly, the difference in producing 2m is 3 , so in producing m the difference will be 1.5... this means ans for A> 1+1.5 ie 2.5... only D and E are there which dont fit in... let A take x hr , B will take x1.5 hr.....so 1/x + 1/(x1.5)=1... 3 fits in... answer should be 3... since answer is for 5m ans is 5*3=15
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Re: Machines A and B, working simultaneously at their respective constant [#permalink]
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28 May 2015, 07:55
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Rate of machine A is 2m/t Rate of machine B is 2m / t3
2m/t + 2m / t3 = m t here must equal 6 for this to be true
a quick ratio expression solves it 6/2m = x/5m x= 15
Answer:E



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Re: Machines A and B, working simultaneously at their respective constant [#permalink]
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28 May 2015, 08:14
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together A and B in 1 Hour can produce M units
A's 1 Hr + B's 1 Hr = M  Equation 1
Let A takes x Hours to Produce 2M => A in one hour can produce 2M/X As per the statement B takes X3 hours to produce 2M => B in one hour can produce 2M/(X3)
Putting A's 1 hour and B's 1 Hour in equation 1
2M/X + 2M/(X3) = M
on simplifying we will get X = 1 or X = 6
X cannot be equal to 1, since as per the condition B takes (X3) hours i.e 2 Hours to produce 2 M (Not possible)
Hence X = 6
A takes 6 Hours to Produce 2M and hence 15 Hours to produce 5M
Option E



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Re: Machines A and B, working simultaneously at their respective constant [#permalink]
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28 May 2015, 08:20
Jackal wrote: Bunuel wrote: Jackal wrote: Hello all My attempt:Let the rate of \(Machine A\) be \(a\) units per hour and that of \(Machine B\) be \(b\) units per hour. From first statement we have \(a*1 + b*1 = m\)\(........equation I\) From the second statement we have \(2m/a  2m/b = 3\)\(..........equation II\) We need to find out the value of \(\frac{m}{a}\). Solving \(I\)and \(II\) and substituting \(x\) for \(\frac{m}{a}\) we get \(2x^2  7x + 3 = 0\) we get \(x = 0.5\) or \(3\) and none of these exist in the options so perhaps I am on the wrong path. Any help appreciated. My answers are \(0.5 hours\) or \(3 hours\) The question is: how long does it take machine A to produce 5m units, not m units. Thanks. Then the answer is \(E\) \(15 hours\). \(x=0.5 or 3\) But we need \(5x\) therefore \(2.5 hours\) or \(15 hours\). We reject the solution \(x=0.5 hours\) because it invalidates the statement 1. BTW the question has 1minute shortcut solution. Without any equations.
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Re: Machines A and B, working simultaneously at their respective constant [#permalink]
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28 May 2015, 08:24
Bunuel wrote: BTW the question has 1minute shortcut solution. Without any equations.
Would love to know some tricks! I am poor at these work type problems. I usually take the clean approach because I know this is not a strength area for me and thus my mind should not get muddled up after spending few seconds on the question. Thanks!
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Re: Machines A and B, working simultaneously at their respective constant [#permalink]
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28 May 2015, 08:29



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Re: Machines A and B, working simultaneously at their respective constant [#permalink]
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28 May 2015, 08:35
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Bunuel wrote: Machines A and B, working simultaneously at their respective constant rates produce m units in 1 hour. If working independently it takes machine B 3 hours less than machine A to produce 2m units, how long does it take machine A to produce 5m units?
A. 1 hour B. 2.4 hours C. 2.5 hours D. 6 hours E. 15 hours This is how I approached the question. combined rate is m units in 1 hr and B takes 3 hrs less then A to produce 2m units. Combined rate will take 2 hr to produce 2m units ( just to make same number of units) . Say A takes x hrs to produce 2m units, so B will take x3 hrs to produce 2m units. so we have equation (combined rate) 1/2 = 1/x3 + 1/x ( individual rates of A and B) solving this we have => x^2 7x +6 = 0, so we have two values of x = 1 & 6 ( 1 is not possible as this will make B's rate ve) so we have x=6, this is the rate of A. Now A takes 6 hrs to produce 2m units and for 5m units he will take 15 hrs. Soln E



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Re: Machines A and B, working simultaneously at their respective constant [#permalink]
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28 May 2015, 09:30
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Bunuel wrote: Jackal wrote: Bunuel wrote: BTW the question has 1minute shortcut solution. Without any equations.
Would love to know some tricks! I am poor at these work type problems. I usually take the clean approach because I know this is not a strength area for me and thus my mind should not get muddled up after spending few seconds on the question. Thanks! Consider this: if together A and B produce 5m units in 5 hours (m in 1 hour). Can only A produce 5m in LESS than 5 hours? This eliminates A, B and C right away. Now, can you tell me how can we eliminate the remaining incorrect answer? 6 hours is also not possible, if A takes 6 hrs to produce 5m => to produce 2m, A needs 12/5 hrs = 2.4 hrs.. Now B needs 3 hours lesser than A => 2.4  3 = negative value, which is not possible To check E, if A takes 15 hrs to produce 5m => to produce 2m, A needs 6 hrs, and B needs 3 hours. (to cross check, A's rate = 1/3 and B's rate = 2/3 => m/hour rate) Hence the answer is E



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Re: Machines A and B, working simultaneously at their respective constant [#permalink]
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28 May 2015, 09:33
Bunuel wrote: BTW the question has 1minute shortcut solution. Without any equations.
hi, Since A and B together produce m units together in 1 hr so individually it has to be more than 1.. secondly, the difference in producing 2m is 3 , so in producing 5m the difference will be 7.5... this means ans for A> 1+7.5 ie 8.5... only E LEFT... ans 15 E
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Re: Machines A and B, working simultaneously at their respective constant [#permalink]
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29 May 2015, 11:54
I got caught up in the algebra behind the problem without even glancing at the answer choices.
Once you understand the relationships between A and B, you can quickly eliminate incorrect ans. choices and get on to the correct answer.
Nice question.



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Re: Machines A and B, working simultaneously at their respective constant [#permalink]
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29 May 2015, 12:29
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Quick 1 minute answer.
Reading through the question you know that it takes machine A greater than 3 hours to product 2m. This can be inferred from the statement that "working alone, it takes machine B 3 hours less than machine A to produce 2m units".
Therefore to produce 5m units the time machine A takes must be greater than 7.5 hours.
The only answer choice that fits is E. 15



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Re: Machines A and B, working simultaneously at their respective constant [#permalink]
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01 Jun 2015, 03:02
Bunuel wrote: Machines A and B, working simultaneously at their respective constant rates produce m units in 1 hour. If working independently it takes machine B 3 hours less than machine A to produce 2m units, how long does it take machine A to produce 5m units?
A. 1 hour B. 2.4 hours C. 2.5 hours D. 6 hours E. 15 hours OFFICIAL SOLUTION:On the PS section always look at the answer choices before you start to solve a problem. They might often give you a clue on how to approach the question.If it takes machine B 3 hours less than machine A to produce 2m units, then to produce 2m*2.5 = 5m units machine B will take 3*2.5 = 7.5 hours less than machine A. Thus time of A to produce 5m units must be more than 7.5 hours. Only E fits. Answer: E.
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Re: Machines A and B, working simultaneously at their respective constant [#permalink]
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15 Dec 2016, 23:28
M units > 1 hour > 1/A + 1/B............(1) 1/A  1/B = 1/3 > 2m units....................(2) Therefore, 1/A = 1/3 + 1/B, Considering (1), 1/A +1/B = 2/3  1/B = 1/A Adding both, 1/a=3/3 therefore, m = 3 hours, 5 m = 15 hours



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Re: Machines A and B, working simultaneously at their respective constant [#permalink]
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23 Dec 2016, 05:32
x/A = 2m multiplied by 2,5 2,5x/A=5m , thus A must be divisible by 2,5h , option C & E are possible, however, since it is known that B worked 3 less more to produce 5m qty, it should be possible to subtract 3 from A, so only E option is possible. Answer is E.



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