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# Working simultaneously and independently at an identical constant rate

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Math Expert
Joined: 02 Sep 2009
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Working simultaneously and independently at an identical constant rate [#permalink]

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18 Oct 2015, 13:12
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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

Kudos for a correct solution.

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

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18 Oct 2015, 13:52
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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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Joined: 01 Jan 2015
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Re: Working simultaneously and independently at an identical constant rate [#permalink]

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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

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Joined: 01 Mar 2015
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Re: Working simultaneously and independently at an identical constant rate [#permalink]

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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

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

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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
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Verbal Forum Moderator Status: Greatness begins beyond your comfort zone Joined: 08 Dec 2013 Posts: 1979 Location: India Concentration: General Management, Strategy Schools: Kelley '20, ISB '19 GPA: 3.2 WE: Information Technology (Consulting) Re: Working simultaneously and independently at an identical constant rate [#permalink] ### Show Tags 19 Oct 2015, 00:09 2 This post received KUDOS 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 _________________ When everything seems to be going against you, remember that the airplane takes off against the wind, not with it. - Henry Ford The Moment You Think About Giving Up, Think Of The Reason Why You Held On So Long +1 Kudos if you find this post helpful Retired Moderator Joined: 29 Oct 2013 Posts: 273 Concentration: Finance GPA: 3.7 WE: Corporate Finance (Retail Banking) Working simultaneously and independently at an identical constant rate [#permalink] ### Show Tags 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 _________________ Please contact me for super inexpensive quality private tutoring My journey V46 and 750 -> http://gmatclub.com/forum/my-journey-to-46-on-verbal-750overall-171722.html#p1367876 Director Joined: 07 Dec 2014 Posts: 998 Re: Working simultaneously and independently at an identical constant rate [#permalink] ### Show Tags 13 Dec 2015, 10:50 1 This post was BOOKMARKED one machine can do x/24 units in one day 3x/(x/24)(4)=18 machines Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8064 Location: Pune, India Re: Working simultaneously and independently at an identical constant rate [#permalink] ### Show Tags 18 Dec 2015, 21:31 1 This post received KUDOS Expert's post 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. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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

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17 Feb 2016, 19:09
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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.
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Re: Working simultaneously and independently at an identical constant rate [#permalink]

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

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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

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

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17 May 2016, 19:58
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Attached is a visual that should help.
Attachments

Screen Shot 2016-05-17 at 7.55.31 PM.png [ 126.58 KiB | Viewed 7990 times ]

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

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24 Aug 2016, 00:29
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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 constant rate [#permalink]

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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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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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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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23 Aug 2017, 07:03
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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

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

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17 Dec 2017, 13:39
Let m = number of machine, d = number of days, x = number of units

If $$m = 4$$ and $$d = 6$$ then $$x = 24$$

Now if $$d = 4$$ and $$x = 3x$$ or $$72$$ then...

$$4m = 72$$

$$m = \frac{72}{4} = 18$$

Thus, the number of machines needed is 18
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Re: Working simultaneously and independently at an identical constant rate [#permalink]

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17 Dec 2017, 16:15
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Let’s say x=600 units of P. Then, there are 100 units produced per day (600/6).

And since there are 4 machines, each machine produces 25 units per day (100/4).

We want to know how many machines it takes to produce 3x. Since we assumed x is 600, 3x is 1800 units of P. There must be 450 units produced per day (1800/4). And because one machine produces 25 units per day we will need 18 machines (450/25).

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Re: Working simultaneously and independently at an identical constant rate   [#permalink] 17 Dec 2017, 16:15

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