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Working simultaneously and independently at an identical constant rate

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Working simultaneously and independently at an identical  [#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?

A. 24
B. 18
C. 16
D. 12
E. 8

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

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New post 18 Oct 2015, 12:12
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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

The rate of 4 machines is rate=job/time=x/6 units per day --> the rate of 1 machine 1/6*(x/6)=x/24 units per day;

Now, again as {time}*{combined rate}={job done} then 4*(m*x/24)=3x --> m=18.

Or as 3 times more job should be done in 1.5 times less days than 3*1.5=4.5 times more machines will be needed 4*4.5=18.

Answer: 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, 13:52
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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: Rate problem  [#permalink]

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New post 10 Dec 2010, 05:33
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ajit257 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


4 machines and 6 days so the total work done is 24

4x6=24
now 3 times the work is 72

So in 4 days if the work is to be completed its 72/4 = 18
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Re: Rate problem  [#permalink]

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New post 21 Sep 2012, 10:16
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Rate*Time = Work
(4R)*(6)=x ( Total 4 machines)
therefore R=x/24

now again (n)*(x/24)*(4) = 3x

solving for n, we get n=18 nos.
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Re: Rate problem  [#permalink]

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New post 21 Sep 2012, 11:42
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ajit257 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



4 machines----------x units-----------6days
4 machines---------3x units-----------3*6 = 18 days
M machines---------3x unites----------4 days
The number of days and the number of machines which produce a certain number of units (in this case 3x) are inversely proportional.
This is because all the machines have the same constant rate.
Necessarily 4*18=M*4, therefore M = 18.

Answer B
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Re: Rate problem  [#permalink]

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New post 21 Sep 2012, 14:19
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This is a variation of the Distance = Rate * Time

Except we modify and say Output = # Machines * (Rate * Time)

X = # * r*t

X = 4 * r * 6
x = 24r
r = x/24

Solve for r because we want to find out the rate...then apply that rate to the new situation, which is output of 3x in 4 days

3x = N * r * 4

3x = N * (x/24) * 4
3x = N * (x/6)

18x = N *x
18 = N
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Re: Working simultaneously and independently at an identical  [#permalink]

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New post 14 Sep 2015, 05:54
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ajit257 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


solving in shortcut

m1 d1 h1 / w1 = m2 d2 h2/w2
4x6/x = mx4/3x

solving we get m=18

answer b

press kudos if you love my shortcut :-D
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Re: Working simultaneously and independently at an identical constant rate  [#permalink]

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New post 18 Oct 2015, 12:52
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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, 21:41
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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, 22:39
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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 18 Oct 2015, 23:09
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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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Re: Working simultaneously and independently at an identical constant rate  [#permalink]

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New post 18 Dec 2015, 20:31
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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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Working simultaneously and independently at an identical constant rate  [#permalink]

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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  [#permalink]

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New post 09 Apr 2016, 14:05
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4 machines can do x work in 6 days
4 machines can do x/6 work in 1 day
1 machine can do x/6*1/4 work in 1 day

so,

n machines in 4 days= 3x work
n machines in 1 day = n * (x/6*1/4)
n machines in 4 days = n * (x/6*1/4)*4

Hence, n machines can do n*(x/6*1/4)*4 work in 4 days i.e. n*(x/6*1/4)*4 =3x ====>>>> n=18
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Re: Working simultaneously and independently at an identical constant rate  [#permalink]

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New post 02 May 2016, 13:55
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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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Re: Working simultaneously and independently at an identical constant rate  [#permalink]

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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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Attached is a visual that should help.
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Screen Shot 2016-05-17 at 7.55.31 PM.png [ 126.58 KiB | Viewed 33222 times ]

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

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New post 23 Aug 2016, 23:29
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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.



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  [#permalink]

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New post 08 Jan 2017, 16:41
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Here's a way to solve this question quickly:



Here's a more in-depth look at a reliable approach for combined work questions with multiple identical machines:

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Re: Working simultaneously and independently at an identical   [#permalink] 08 Jan 2017, 16:41

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