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Working simultaneously and independently at an identical

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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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Re: Rate problem [#permalink]

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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


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: Rate problem [#permalink]

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New post 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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Re: Rate problem [#permalink]

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

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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: Rate problem [#permalink]

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New post 10 Dec 2010, 06:33
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, 15:19
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 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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Re: Working simultaneously and independently at an identical [#permalink]

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New post 09 Apr 2016, 15:05
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
Re: Working simultaneously and independently at an identical   [#permalink] 09 Apr 2016, 15:05
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