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Six machines, each working at the same constant rate

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Re: Six machines, each working at the same constant rate  [#permalink]

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New post 01 Dec 2018, 11:33
Another option to solve this question is using Backsolving.

W= R x T
W= 6 machines x 12 days = 72 days for 1 machine to complete the job.

Now, we need to find "extra" machines that match "72 days per machine".

If we select:

A) 2 we will have 8 machines x 8 days = 64 days for 1 machine to complete the job. Not a match! Out!!

B) 3 we will have 9 machines x 8 days = 72 days for 1 machine to complete the same job in the same rate!! That's a match!

Answer: B


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Six machines, each working at the same constant rate  [#permalink]

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New post 08 Sep 2019, 06:13
Hi VeritasKarishma

I wish I could send you a gratitude note for your awesome time efficient approach mentioned
here :)

Please confirm your two cents:
Logic: more the nos of machines -> less no of days

Start with no of machines in initial stage:
6

Now we need to decide to multiply by 12/8 or 8/12
Since we need more no of machines to do the work in less time from 12 to 8 days , we need an improper fraction
hence
6 * (12/8) = 9 no of machines.

Additional machines = 9 - 6 = 3

Thanks for detailed post here too
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Re: Six machines, each working at the same constant rate  [#permalink]

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New post 09 Sep 2019, 06:17
1
adkikani wrote:
Hi VeritasKarishma

I wish I could send you a gratitude note for your awesome time efficient approach mentioned
here :)

Please confirm your two cents:
Logic: more the nos of machines -> less no of days

Start with no of machines in initial stage:
6

Now we need to decide to multiply by 12/8 or 8/12
Since we need more no of machines to do the work in less time from 12 to 8 days , we need an improper fraction
hence
6 * (12/8) = 9 no of machines.

Additional machines = 9 - 6 = 3

Thanks for detailed post here too


Yes, that's correct!
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Re: Six machines, each working at the same constant rate  [#permalink]

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New post 30 Sep 2019, 15:47
zz0vlb wrote:
Six machines, each working at the same constant rate, together can complete a certain job in 12 days. How many additional machines, each working at the same constant rate, will be needed to complete the Job in 8 days?

A. 2
B. 3
C. 4
D. 6
E. 8


6 machines --> 6r (rate) * 12 (time) = 1 Work
1 machine --> 1r * 72 = 1, we can say 1m = 72r
We need the n, number of machines to do the 1 Work in t=8
nr*8 = 72r, divide by r
n*8 = 72
n = 9
9-6 = 3
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Re: Six machines, each working at the same constant rate  [#permalink]

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New post 14 Oct 2019, 04:30
LalaB wrote:
6 machines in 12 hours can make 6*12=72 units.
x machines in 8 hours can make x*8=72 units x=9

12-9=3 more machines are needed
hope it helps


9-6=3
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Re: Six machines, each working at the same constant rate  [#permalink]

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New post 04 Feb 2020, 03:25
How about the following for simplicity and speed.
Job = Rate * Time ( J = R * T ).

6 machines working at constant rate R for the same Job J in T = 12 days
J = 6 * R * 12

(x) additional machines to the 6, (6+x), needed to complete the same Job in t = 8 days. In other words
J = (6+x) * R * 8

J = J
(6+x) * R * 8 = 6 * R * 12
(6+x) * 8 = 6 * 12
48 + 8x = 72
x = 24/8
x = 3

Answer choice B
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Re: Six machines, each working at the same constant rate  [#permalink]

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New post 15 May 2020, 22:56
I kept on getting this wrong until I started to make Work, Rate Time problems as a sentence

What we know.
> 6 Machines (m) work x (unknown) rate for 12 days to complete (=) 1 job
>> hence 6 machines will do 1/12 of the job in 1 day a rate of X
>> hence 1 machine will do 1/(12*6) of the job a rate of X
>>> hence x = 1(12*6)
>>> hence X = 1/(72)

> (6 original machines + new machine[s]) work x (same rate) for 8 days to complete (=) 1 job
>> hence (6 original machines + new machine[s]) will do 1/8 of the job in 1 days
>> hence 1 machine will do 1/8* (6 original machines + new machine[s]) of the in 1 day
>>> hence X = 1/(8* (6+NM)
>>> hence X = 1/(48 + 8NM)

Now we solve

Finally, since X is the same in both we can do the simple equation

Part 1 X = Part 2 X
> 1/(72) = 1/(48 + 8NM)
> 1*(48 + 8NM) =1*(72)
> 8NM = 72-48
> 8NM = 24
> NM = 24/8
> NM = 3
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Re: Six machines, each working at the same constant rate  [#permalink]

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New post 17 May 2020, 17:59
I kept on getting this wrong until I started to make Work, Rate Time problems as a sentence

What we know.
> 6 Machines (m) work x (unknown) rate for 12 days to complete (=) 1 job
>> hence 6 machines will do 1/12 of the job in 1 day a rate of X
>> hence 1 machine will do 1/(12*6) of the job a rate of X
>>> hence x = 1(12*6)
>>> hence X = 1/(72)

> (6 original machines + new machine[s]) work x (same rate) for 8 days to complete (=) 1 job
>> hence (6 original machines + new machine[s]) will do 1/8 of the job in 1 days
>> hence 1 machine will do 1/8* (6 original machines + new machine[s]) of the in 1 day
>>> hence X = 1/(8* (6+NM)
>>> hence X = 1/(48 + 8NM)

Finally, since X is the same we can do the simple equation

Part 1 X = Part 2 X
> 1/(72) = 1/(48 + 8NM)
> 1*(48 + 8NM) =1*(72)
> 8NM = 72-48
> 8NM = 24
> NM = 24/8
> NM = 3
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Re: Six machines, each working at the same constant rate   [#permalink] 17 May 2020, 17:59

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