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Six machines each working at same constant rate together can

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Six machines each working at same constant rate together can

by imane81 » Wed Feb 17, 2010 12:20 pm
Thanks for helping on this one

Six machines each working at 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
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by ajith » Wed Feb 17, 2010 12:23 pm
imane81 wrote:Thanks for helping on this one

Six machines each working at 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
Total work is 12*6 = 72 machine days
9 machines will do it in 8 days (9*8 = 72)

no of additional machines needed = 9-6 =3
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by harsh.champ » Thu Feb 18, 2010 2:28 pm
imane81 wrote:Thanks for helping on this one

Six machines each working at 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
The formal approach would be:-
Speed of a single machine = 12/6 = 2 days when six of them are working together.[Alone the speed will be 2 x 6 =12days]
Let x machines be added,then speed of a single machine is 8/(x+6) days.
Speed of machine working alone will be [8/(x+6)]*(x+6)

I am getting confused in the last step.Can anyone help in the formal approach??
It takes time and effort to explain, so if my comment helped you please press Thanks button :)



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by ironsferri » Tue Jun 29, 2010 2:20 pm
Hi All!

got the process, but didn't get the meaning - can please someone shed some light on this thought process?

1)Use RTD
2) Then find Rate each
3)then multiply the rate by NEW days and find additional machines

In the last step I find 1/9 - I don't understand how I can "read" 3 more machines looking at 1/6 and 1/9.

Thanks guys!

Chris
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by GMATGuruNY » Tue Jun 29, 2010 5:33 pm
A helpful technique for work problems:

When the job is undefined, try plugging in your own number for the job.

Let's say each machine produces one unit each day. Then the 6 machines together would produce 6 units each day. Over 12 days, 6 * 12 = 72 units would be produced.

If we want to produce the 72 units in only 8 days, we'll need 72/8 = 9 units to be produced each day.

Since we'll need 3 more units to be produced each day, and each machine produces 1 unit per day, we'll need 3 more machines.

The correct answer is B.
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by ironsferri » Wed Jun 30, 2010 10:54 am
Thanks Mitch! I see it now.
Just for reference, working with fractions as I did the first time, what should make me do the connection of saying "yes, 3 machines!" looking at 1/6 to 1/9. I just want to cover that pan w/ a lid of understanding... ;)
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by anaid1708 » Wed Jun 30, 2010 11:21 am
I am not sure if i approached this right, please tell me if this is a correct method?

6R X12days =1job
6R= 1/12
R= 1/72

so rate of one machine is 1/72
now
1/72M x 8 days=1 job
1/72M=1/8
M=1/8X72
M=9
If 9 machines will do the job in 8 days, then 9 - 6 =3 will number of additional machines
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by ironsferri » Wed Jun 30, 2010 11:34 am
Hi Anaid,
i thing you did it right - i followed the same process, 'til the point where I forgot to set the M as unknown.

We know that with RTD, 99% time (or so) questions at the end ask to find either R, t or D/W. In this case instead was asking something different, and because of the routine it's easy to get side-tracked.
Thanks for posting your process, helped claryfing my doubts... ;)
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by GMATGuruNY » Wed Jun 30, 2010 3:27 pm
anaid1708 wrote:I am not sure if i approached this right, please tell me if this is a correct method?

6R X12days =1job
6R= 1/12
R= 1/72

so rate of one machine is 1/72
now
1/72M x 8 days=1 job
1/72M=1/8
M=1/8X72
M=9
If 9 machines will do the job in 8 days, then 9 - 6 =3 will number of additional machines
Looks good.

Here's another way:

(number of machines) x (number of days) always has to yield the same amount of work.

So we could set up this equation:

(number of machines) x (number of days) = (number of machines) x (number of days)

6 * 12 = x * 8
72 = 8x
x = 9

Since we'll need 9 machines altogether, and we currently have 6, we'll need 9-6=3 more machines.

For all you math lovers: In math terms, the number of machines is inversely proportional to the number of days. When two values are inversely proportional, as one value goes up, the other must go down, so that the product of the two values is always the same. So in the problem above, as the number of machines goes up, the number of days must go down, so that we're always getting the same amount of work done.
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by Thouraya » Sun Jul 18, 2010 11:20 pm
Hi Mitch,

I like your second approach if it always works. I have a question on your first approach, please:

If I want to assume that every machine produces two units (instead of 1), then 6 machines produce 12 units a day, and 12 x 12= 144 units in 12 days.

If I want to produce the 144 units in 8 days instead, then 144/8 ?

Also, I cant cross multiple in work problems cuz I have three variables: days, machines, and units?

Thank you!
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by gmatmachoman » Sun Jul 18, 2010 11:57 pm
harsh.champ wrote:
imane81 wrote:Thanks for helping on this one

Six machines each working at 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


six machines one day's work : ( 1/12 ) part of work done

One m/c one day work : 1/ (12*6) : 1/72 part of work done

So it means one M/c will take 72 days to complete the whole work

To complete the work in 8 days we need to increase the M/c.

1m/c ----- 72 days

?? ---- 8 days

Inverse proportionality

72/8 = 9
So 9 m/c are required to complete the work in 8 days.

Extra m'c : 9-6 = 3

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by GMATGuruNY » Mon Jul 19, 2010 5:24 am
Thouraya wrote:Hi Mitch,

I like your second approach if it always works. I have a question on your first approach, please:

If I want to assume that every machine produces two units (instead of 1), then 6 machines produce 12 units a day, and 12 x 12= 144 units in 12 days.

If I want to produce the 144 units in 8 days instead, then 144/8 ?

Also, I cant cross multiple in work problems cuz I have three variables: days, machines, and units?

Thank you!
The second approach that I used is just a variation of the rate formula: rate * time = work.

In the problem above, rate = number of machines, so (number of machines) * (number of days) = work.

Since the work must remain the same no matter how many machines are used, we get the following inverse proportion:

(number of machines) * (number of days) = (number of machines) * (number of days)

If the number of machines goes up, the number of days must go down, so that the amount of work stays the same. If the number of machines goes down, the number of days must go up, so that the amount of work stays the same. This is the definition of an inverse proportion. As one value goes up, the other goes down, so that the product never changes.

1*30 = 30
2*15 = 30
3*10 = 30
etc.

Regarding the first approach (plugging in for the work being done by each machine), your numbers are correct. If each machine produces 2 units/day, then to complete the job in 9 days, 144/8 = 18 units must be produced each day. Since each machine produces 2 units/day, 9 total machines will be needed, so we'll need to add 3 machines -- the same answer that I got when I plugged in that each machine produces 1 unit/day.

Hope this helps!

(What exactly did you want to cross multiply?)
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by rbansal » Sun May 15, 2011 2:25 pm
anaid1708 wrote:I am not sure if i approached this right, please tell me if this is a correct method?

6R X12days =1job
6R= 1/12
R= 1/72

so rate of one machine is 1/72
now
1/72M x 8 days=1 job
1/72M=1/8
M=1/8X72
M=9
If 9 machines will do the job in 8 days, then 9 - 6 =3 will number of additional machines

I am unsure how you were able to take the 6R and multiply it with the 12 with fractions dont you have to multiply with numerator not denominator? Please explain?
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by rbansal » Sun May 15, 2011 2:28 pm
I am getting very confused with this question. I rationalized it as, it takes 2 days for each machine to complete the job in 12 days. If they are to complete the job in 8 days they will need two more machines both working at a rate of 2days a job. Where is my logic incorrect?



GMATGuruNY wrote:A helpful technique for work problems:

When the job is undefined, try plugging in your own number for the job.

Let's say each machine produces one unit each day. Then the 6 machines together would produce 6 units each day. Over 12 days, 6 * 12 = 72 units would be produced.

If we want to produce the 72 units in only 8 days, we'll need 72/8 = 9 units to be produced each day.

Since we'll need 3 more units to be produced each day, and each machine produces 1 unit per day, we'll need 3 more machines.

The correct answer is B.
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by Thouraya » Thu Sep 15, 2011 10:30 pm
Hi Guys,

I am still stuck on this problem. Is it 700-Level? I am not quite sure how it differs from the regular work/rate problems (in that they don't mention how much each machine can perform on its own? so I can consider that they are approaching the problem the other way around?). Would appreciate any guiding tips, thank you so much!
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