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Machines A and B, working simultaneously at their respective constant

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Machines A and B, working simultaneously at their respective constant  [#permalink]

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New post 28 May 2015, 06:02
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A
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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

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Re: Machines A and B, working simultaneously at their respective constant  [#permalink]

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New post 01 Jun 2015, 03:02
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5
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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New post 28 May 2015, 06:38
1
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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New post 28 May 2015, 07:38
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.
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Re: Machines A and B, working simultaneously at their respective constant  [#permalink]

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New post 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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New post 28 May 2015, 07:47
3
2
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 x-1.5 hr.....so 1/x + 1/(x-1.5)=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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New post 28 May 2015, 07:55
1
1
Rate of machine A is 2m/t
Rate of machine B is 2m / t-3

2m/t + 2m / t-3 = 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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New post 28 May 2015, 08:14
1
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 X-3 hours to produce 2M => B in one hour can produce 2M/(X-3)

Putting A's 1 hour and B's 1 Hour in equation 1

2M/X + 2M/(X-3) = M

on simplifying we will get
X = 1 or X = 6

X cannot be equal to 1, since as per the condition B takes (X-3) 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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New post 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 1-minute 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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New post 28 May 2015, 08:24
Bunuel wrote:

BTW the question has 1-minute 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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New post 28 May 2015, 08:29
1
Jackal wrote:
Bunuel wrote:

BTW the question has 1-minute 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?
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Re: Machines A and B, working simultaneously at their respective constant  [#permalink]

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New post 28 May 2015, 08:35
2
1
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 x-3 hrs to produce 2m units.

so we have equation

(combined rate) 1/2 = 1/x-3 + 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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New post 28 May 2015, 09:30
2
1
Bunuel wrote:
Jackal wrote:
Bunuel wrote:

BTW the question has 1-minute 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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New post 28 May 2015, 09:33
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1
Bunuel wrote:

BTW the question has 1-minute 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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New post 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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New post 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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New post 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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New post 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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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] 15 Jul 2019, 18:52
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