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Working simultaneously and independently at an identical

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Senior Manager
Joined: 28 Aug 2010
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Working simultaneously and independently at an identical [#permalink]

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10 Dec 2010, 06:21
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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
[Reveal] Spoiler: OA

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Ajit

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10 Dec 2010, 06:37
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Expert's post
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

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.

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

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14 Sep 2015, 06:54
3
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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

press kudos if you love my shortcut
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10 Dec 2010, 06:33
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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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21 Sep 2012, 15:19
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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
Manager
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21 Sep 2012, 11:16
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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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21 Sep 2012, 12:42
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1
This post was
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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.

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

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09 Apr 2016, 15:05
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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 [#permalink]

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30 Jan 2017, 04:45
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1) First let's set up an equation for the 4 machines and find a rate of each machine: $$4r*6=x; 4r=\frac{x}{6}, r=\frac{x}{6}*\frac{1}{4}, r=\frac{x}{24}$$
2) Now let's set up an equation for 3x units: let N be the number of machines, then $$N*\frac{x}{24}*4=3x; N*\frac{x}{6}=3x, N=\frac{3x}{x/6}=18$$

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

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14 Nov 2012, 06:50
$$\frac{4}{m}=\frac{x}{6}==>m=\frac{24}{x}$$

Calculate number of machines to produce 3x in 4 days:

$$N(\frac{x}{24})(4)=3x==> N=18$$
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Re: Working simultaneously and independently at an identical [#permalink]

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09 Sep 2015, 19:47
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Re: Working simultaneously and independently at an identical [#permalink]

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08 Jan 2017, 17:41
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:

Director
Affiliations: Target Test Prep
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Re: Working simultaneously and independently at an identical [#permalink]

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02 Feb 2017, 11:58
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

We are given that 4 machines can complete x units in 6 days. Thus, the rate of the 4 machines is x/6.

Next we need to determine the number of machines needed to produce a rate of 3x/4. To calculate that number of machines, we can use the following proportion in which the value in each numerator is the number of machines and the value in each denominator is the corresponding rate of those machines. We can let n = the number of machines needed:

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

24/x = 4n/3x

72x = 4nx

18 = n

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Re: Working simultaneously and independently at an identical   [#permalink] 02 Feb 2017, 11:58
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