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Working simultaneously and independently at an identical constant rate

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Working simultaneously and independently at an identical constant rate, 4 machines of a certain type can produce a total of x units of product P in 6 days. How many of these machines, working simultaneously and independently at this constant rate, can produce a total of 3x units of product P in 4 days?

(A) 24
(B) 18
(C) 16
(D) 12
(E) 8

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Re: Working simultaneously and independently at an identical constant rate [#permalink]

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Working simultaneously and independently at an identical constant rate, 4 machines of a certain type can produce a total of x units of product P in 6 days. How many of these machines, working simultaneously and independently at this constant rate, can produce a total of 3x units of product P in 4 days?

tough one, let's see...
4 machines do x units in 6 days

we have x/6 => rate of the 4 machines

we know that we need to have 3x units in 4 days
therefore, we need to get to 3x/4 rate of the machines.

rate of one machine is x/6*1/4 = x/24.

now, we need to know how many machines need to work simultaneously, to get 3x done in 4 days.
4x/4 work needs to be done by machines that work at x/24 rate.

let's assign a constant Y for the number of machines:
(x/24)*y = 3x/4

y = 3x/4 * 24/x
cancel 4 with 24, and x with x and get -> 18. Answer choice B

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Re: Working simultaneously and independently at an identical constant rate [#permalink]

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New post 18 Oct 2015, 14:52
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Bunuel wrote:
Working simultaneously and independently at an identical constant rate, 4 machines of a certain type can produce a total of x units of product P in 6 days. How many of these machines, working simultaneously and independently at this constant rate, can produce a total of 3x units of product P in 4 days?

(A) 24
(B) 18
(C) 16
(D) 12
(E) 8

Kudos for a correct solution.


If it takes 4 machines 6 days to produce x units
then it takes 4 machines 18 days to produce 3x units
then it takes \(M\) machines 4 days to produce 3x units

There is an inverse relationship between the amount of time and the number of machines required to do the task
So if the task is 3x units, then (4 machines)*(18 days)=(M machines)*(4 days) --> M = 18

Correct answer choice is B.

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Re: Working simultaneously and independently at an identical constant rate [#permalink]

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New post 18 Oct 2015, 22:41
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Bunuel wrote:
Working simultaneously and independently at an identical constant rate, 4 machines of a certain type can produce a total of x units of product P in 6 days. How many of these machines, working simultaneously and independently at this constant rate, can produce a total of 3x units of product P in 4 days?

(A) 24
(B) 18
(C) 16
(D) 12
(E) 8

Kudos for a correct solution.



we have x units by 4 machine in 6 days

=> x units in 24 machine days
=> 3x units in 72 machine days

therefore machine required = 72 / 4 = 18

Answer choice B

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Re: Working simultaneously and independently at an identical constant rate [#permalink]

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New post 18 Oct 2015, 23:39
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Bunuel wrote:
Working simultaneously and independently at an identical constant rate, 4 machines of a certain type can produce a total of x units of product P in 6 days. How many of these machines, working simultaneously and independently at this constant rate, can produce a total of 3x units of product P in 4 days?

(A) 24
(B) 18
(C) 16
(D) 12
(E) 8

Kudos for a correct solution.



Using the method discussed in this post: http://www.veritasprep.com/blog/2013/02 ... variation/

4 machines --- x units --- 6 days
? machines --- 3x units -- 4 days

Machines required = 4 * (3x/x) * (6/4) = 18
Answer (B)
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Re: Working simultaneously and independently at an identical constant rate [#permalink]

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New post 19 Oct 2015, 00:09
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Let work done per machine per day = m

Work done by 4 machines in a day, 4m = x/6
Work done by 1 machine in a day , m = x/24

Since work needs to completed in 4 days , work done by 1 machine in 4 days = x/6
Number of machines required = Total work to be completed in 3 days/ Work done by a single machine in 4 days
= 3x/(x/6) = 18

Answer B
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New post 11 Dec 2015, 13:44
Since we need to do 3 times as much work (3x vs x) we will need 3 times as many m/cs. so multiply by a factor of 3
Now we need to do this work faster i.e. in 4 days vs 6days, so we will need 6/4 as many more m/cs. so multiply by a factor of 6/4
No of mc/ required= 4*3*(6/4)=18
ans B
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one machine can do x/24 units in one day
3x/(x/24)(4)=18 machines

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Re: Working simultaneously and independently at an identical constant rate [#permalink]

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New post 18 Dec 2015, 21:31
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VeritasPrepKarishma wrote:
Bunuel wrote:
Working simultaneously and independently at an identical constant rate, 4 machines of a certain type can produce a total of x units of product P in 6 days. How many of these machines, working simultaneously and independently at this constant rate, can produce a total of 3x units of product P in 4 days?

(A) 24
(B) 18
(C) 16
(D) 12
(E) 8

Kudos for a correct solution.



Using the method discussed in this post: http://www.veritasprep.com/blog/2013/02 ... variation/

4 machines --- x units --- 6 days
? machines --- 3x units -- 4 days

Machines required = 4 * (3x/x) * (6/4) = 18
Answer (B)


Quote:
In the end, you took 6/4 because the machines have fewer days to do more (3x) work, so more machines will be needed and therefore one has to multiply with the value greater than 1, i.e. 6/4 and not 4/6. Is that a correct reasoning?


Forget "work" here - whether it is more or less. Focus on the relation between the unknown and the variable in question only. So focus on machines and days only.

You have fewer days so you will need more machines to complete the work. So you multiply by 6/4, a quantity greater than 1. This will increase the number of machines.
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4 machines take 6 days to complete x amount of work, so 1 machine will take \(6*4=24\) days to complete same work. 3x, i.e. three times of work will require that single machine to work for \(24*3=72\) days. Now in order to do 72 days worth of work in 4 days, we'd need \(72/4=18\) machines.
Answer: B

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Re: Working simultaneously and independently at an identical constant rate [#permalink]

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New post 02 May 2016, 14:55
given : 4 machines rate = x stuff / 6 days or 4m = x/6

need to find y machines = 3x stuff / 4 days = ym = 3x/4

two equations, two unknowns, solvable now.

4 = x/6 & y = 3x/4
4 = x/6
24 = x (now plug into second equation)

y = 3(24) / 4
y = 3(6)
y = 18

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New post 03 May 2016, 05:52
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Bunuel wrote:
Working simultaneously and independently at an identical constant rate, 4 machines of a certain type can produce a total of x units of product P in 6 days. How many of these machines, working simultaneously and independently at this constant rate, can produce a total of 3x units of product P in 4 days?

(A) 24
(B) 18
(C) 16
(D) 12
(E) 8

Kudos for a correct solution.


We are given that 4 machines can produce a total of x units in 6 days. Since rate = work/time, the rate of the 4 machines is x/6. We need to determine how many machines, working at the same rate, can produce 3x units of product P in 4 days. Thus, we need to determine how many machines are needed for a rate of 3x/4.

Since we know that all the machines are working at an identical constant rate, we can create a proportion to determine the number of machines necessary for a rate of 3x/4. The proportion is as follows:

“4 machines are to a rate of x/6 as n machines are to a rate of 3x/4."

4/(x/6) = n/(3x/4)

24/x = 4n/(3x)

When we cross multiply, we obtain:

72x = 4xn

18 = n

Answer: B
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Re: Working simultaneously and independently at an identical constant rate [#permalink]

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New post 17 May 2016, 19:58
Attached is a visual that should help.
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Screen Shot 2016-05-17 at 7.55.31 PM.png
Screen Shot 2016-05-17 at 7.55.31 PM.png [ 126.58 KiB | Viewed 4842 times ]


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Re: Working simultaneously and independently at an identical constant rate [#permalink]

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New post 24 Aug 2016, 00:29
Bunuel wrote:
Working simultaneously and independently at an identical constant rate, 4 machines of a certain type can produce a total of x units of product P in 6 days. How many of these machines, working simultaneously and independently at this constant rate, can produce a total of 3x units of product P in 4 days?

(A) 24
(B) 18
(C) 16
(D) 12
(E) 8

Kudos for a correct solution.



Let individual rate be 'a'
therefore, for 4 machines we have:
\(\frac{1}{a} +\frac{1}{a} + \frac{1}{a} +\frac{1}{a} = \frac{x}{6}\)
-->\(\frac{4}{a} = \frac{x}{6}\)
--> ax = 6 x 4 = 24 ......(eq1)
now If Z machines of same rate were used to produce 3x work in 4 days, we would have:
Z(\(\frac{1}{a}\)) = \(\frac{3x}{4}\)
--> Z = \(\frac{3x(a)}{4}\)
--> Z = \(\frac{3 (24)}{4}\) ......(using eq1)
--> Z = 18

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Re: Working simultaneously and independently at an identical constant rate [#permalink]

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New post 28 Aug 2016, 21:52
The problem is not dependent on the rate of each machine. So let us say each machine takes 1 day to produce each product. So 4 machines produce 4 per day. In 6 days they will produce 24 machines. Take 72 machines = 24*3.
now the combined rate has to be 72/4 = 18. Since each machine still produces 1 per day we need 18 machines to have a rate of 18 per day.

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Re: Working simultaneously and independently at an identical constant rate [#permalink]

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New post 29 Aug 2016, 00:16
Bunuel wrote:
Working simultaneously and independently at an identical constant rate, 4 machines of a certain type can produce a total of x units of product P in 6 days. How many of these machines, working simultaneously and independently at this constant rate, can produce a total of 3x units of product P in 4 days?

(A) 24
(B) 18
(C) 16
(D) 12
(E) 8

Kudos for a correct solution.


4 Machines produce x units in 6 days
N Machines produce 3x units in 4 days
4/x/6 = N/3x/4
N = 4 * 3x/x * 6/4
N= 72/ 4 = 18
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Re: Working simultaneously and independently at an identical constant rate [#permalink]

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New post 03 Oct 2016, 08:24
M1D1T1/W1 =M2D2T2/W2
M-MEN
D-DAYS TAKEN
T-TIME TAKEN PER DAY
W-TOTAL WORK

(4*6)/X=(?*4)/3X
SO ?=18 MEN

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Re: Working simultaneously and independently at an identical constant rate [#permalink]

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New post 08 Oct 2016, 21:06
=4*(3x/x)*(6/4) = 18
(3x/x) because no of units went up, so more no of machines are required
(6/4) no of days to get work done went down, so more no of machines required
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Re: Working simultaneously and independently at an identical constant rate [#permalink]

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New post 23 Aug 2017, 07:03
No. of men- 4M
Work- X
Time- 6
Rate of 1 men = x/6
Rate of 4 men= x/24

Apply this rate in the second equation
No. of men needed - M
Work- 3x
Time- 4

So, Work =Rate x time
Apply rate that was calculated above as x/24
3x=x/24 * 4 * (M) no of men

After solving we get

M=18 and that’s the answer.

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Re: Working simultaneously and independently at an identical constant rate   [#permalink] 23 Aug 2017, 07:03
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