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Twelve identical machines, running continuously at the same

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Twelve identical machines, running continuously at the same [#permalink]

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Twelve identical machines, running continuously at the same constant rate, take 8 days to complete a shipment. How many additional machines, each running at the same constant rate, would be needed to reduce the time required to complete a shipment by two days?

A. 2
B. 3
C. 4
D. 6
E. 9
[Reveal] Spoiler: OA

Last edited by Bunuel on 29 Sep 2013, 08:57, edited 2 times in total.
Renamed the topic, edited the question and added the OA.

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Re: Rate & Work [#permalink]

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New post 31 May 2010, 15:55
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One way is to think "faster by what ratio?" or "the same job in what fraction of the time?"

We want to reduce the time by 2 days from the original 8 days: 6 days to complete.

6 days is 3/4 of the original 8 days. Since R = W/T, if R is constant, T of 3/4 as much implies W of 4/3 as much. In other words, if each machine doesn't speed up individually, you have to have 4/3 as many machines doing the work.

4/3 of the original number of machines (12) is 16 machines, or 4 additional machines.

Another way is to pick some smart numbers to make the problem more "real."

We have 12 machines working 8 days to make complete one "shipment." Let's say the shipment is 96 widgets (that's just 12*8). So the 12 machines make 12 widgets a day, which means each machine makes 1 widget a day (easiest number to work with!).

To complete the 96 widgets in 6 days, then, you'd need 16 widgets each day, or 16 machines working. 4 additional machines.
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Re: Rate & Work [#permalink]

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New post 31 May 2010, 18:36
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Another approach:

consider 12 machines to be 1 man (suppose).

Now the question is if 1 man can complete a task in 8 days how many men are needed to complete the same task in 6 days.

Simple unitary method:

To complete the work in 8 days - 1 man is needed
To complete the work in 1 day - (1X8) men needed
To complete the work in 6 days - (1x8)/6

Which means that if I want to complete the same work in 6 days I would need 4/3 times initial effort i.e. if I had 12 machines initially I would now need (4/3)x12=16

Thus 4 more machines needed

OA C

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Re: Rate & Work [#permalink]

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12-machines--8 days --1 shipment
1machine --96 days --1 shipment
x machines --6 days --1 shipment

work --rate*time

96=6x

x=16
4 more machines needed from original number of machines

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Re: Rate & Work [#permalink]

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New post 05 Jun 2010, 08:57
we know that w = R*T
work is constant here in both the cases. so R = 1/T
12 -----> 1/8
x -----> 1/6
12/x = 6/8
x= 16 (so 16-12 = 4 more machines will be required.)

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Re: Rate & Work [#permalink]

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New post 05 Jun 2010, 14:11
Thanks. Your help is much appreciated.

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Re: Rate & Work [#permalink]

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New post 28 Sep 2013, 10:21
Rate(12) = 1/8
Rate(1) = 1/96

Rate * Time = Work

1/96 * Time = 1

1/96 * 6 = 1

What number of machines must be multiplied by 1/96 to make it 1/6

16, and knowing that we used 12 machined already the additional machine number is 4.
Hence (C) :-D
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Re: Twelve identical machines, running continuously at the same [#permalink]

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Rate = job / time

12 machines do 1 job in 8 days: 12x = 1/8 ==> x = 1/96
Rate of one machine = 1/96

Job reduced by 2 days means the job needs to be completed in 6 days. How many machines can do the job in 6 days?
==> 1/96*x = 1/6 ==> x = 96/6 ==> x = 16

Additional machines: 16-12 = 4 machines

Answer: C

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Re: Twelve identical machines, running continuously at the same [#permalink]

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Re: Twelve identical machines, running continuously at the same [#permalink]

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Another way to look at it...

Assume each machine completes 1 unit of work a day. 12 machines complete 12 units of work a day. in 8 days number of units completed is 12*8 = 96 units.

Now number of machines required to complete 96 units in 6 days is 96/6 = 16. Additional number of machines required will be 16-12 = 4.
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Re: Twelve identical machines, running continuously at the same [#permalink]

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New post 11 Jan 2015, 06:35
12 machines * 8 days = 96 work days
N machines * 6 days = 96 work days
N = 96/6 = 16
i.e. 4 more machines
Answer: C

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Twelve identical machines, running continuously at the same [#permalink]

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New post 01 Mar 2015, 00:29
Since rate is constant, therefore
12 machines * 8 days = (x + 12)machines * 6days

x= additional machine required
6days(8days -2days)=Number of days left after reducing by 2 days

96 = 6x + 72
x = 4 , OPTION C
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Re: Twelve identical machines, running continuously at the same [#permalink]

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New post 13 Mar 2016, 01:42
12*8=6*x
96=6x
x=16
16-12=4 Additional Machines required :)

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Re: Twelve identical machines, running continuously at the same [#permalink]

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New post 13 Mar 2016, 03:51
gsaxena26 wrote:
Twelve identical machines, running continuously at the same constant rate, take 8 days to complete a shipment. How many additional machines, each running at the same constant rate, would be needed to reduce the time required to complete a shipment by two days?

A. 2
B. 3
C. 4
D. 6
E. 9


12 machines take 8 days to complete the task.
We need to find the number of machines to complete the same task in 6 days.
Machines and time are in inverse proportion.
To increase the number of machines multiply with increasing ratio'
12*8/6 = 16 machines
Therefore, 4 additional machines are required.

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Re: Twelve identical machines, running continuously at the same [#permalink]

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New post 03 Apr 2016, 00:27
Use formula
M1xD1/w1=M2xD2/W2

W1=W2=1
M1=12
D1=8
M2=?
D2=8-2=6
put above values in the formula ans is 16. Machine already in use 12 difference =4 additional machines are required.

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Re: Twelve identical machines, running continuously at the same [#permalink]

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New post 03 Apr 2016, 04:36
gsaxena26 wrote:
Twelve identical machines, running continuously at the same constant rate, take 8 days to complete a shipment.


12 *r*8 = 96r

gsaxena26 wrote:
How many additional machines, each running at the same constant rate, would be needed to reduce the time required to complete a shipment by two days?



(12+m)*r*6 = 96r
12+m = 16
m = 4

So, We require 4 additional workers.
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Re: Twelve identical machines, running continuously at the same [#permalink]

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New post 29 Aug 2016, 12:31
I think I have an easier way...

From the question you know that 12R = 1/8. The question asks you (partially) to make the rate from 1/8 to 1/6 (drop from 8 day to 6). So the only thing that you need to do is to find the magic number than can convert 1/8 to 1/6.

So 1/8 * x = 1/6 (1 equation with one unknown). So by solving this you get x = 8/6 or 4/3. Thats it then! Take the magic number 4/3 and multiply BOTH sides of the original equation and you have:

12*(4/3)*R = (4/3) * 1/8

4 * 4 * R = 1/6, Hence 16R = 1/6, therefore 4 more machines! :-D

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Re: Twelve identical machines, running continuously at the same [#permalink]

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New post 29 Aug 2016, 14:02
let m=total machines needed
rate of 1 machine=1/96
m*1/96*6=1
m=16
16-12=4 additional machines needed

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Re: Twelve identical machines, running continuously at the same [#permalink]

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New post 29 Aug 2016, 19:08
gsaxena26 wrote:
Twelve identical machines, running continuously at the same constant rate, take 8 days to complete a shipment. How many additional machines, each running at the same constant rate, would be needed to reduce the time required to complete a shipment by two days?

A. 2
B. 3
C. 4
D. 6
E. 9


12 M----- 8 D---- 96 MD
96/6 = 16 Machines
ie 4 additional MAchines are required
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Re: Twelve identical machines, running continuously at the same [#permalink]

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New post 17 Sep 2016, 22:28
Since 12 machines take 8 days to complete the task, then in terms of rate 12 (1/X) = 1/8 [1/X is the rate of machine & 1/8 is the work completed by the 12 machines at rate of 1/X].

Solving for X, X=96. If 12 machines take 8 days to complete 96 units of work. Then I would need 4 additional machines to complete the work before 2 days.

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Re: Twelve identical machines, running continuously at the same   [#permalink] 17 Sep 2016, 22:28

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