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In a workshop 10 machines working simultaneously can finish a work of

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In a workshop 10 machines working simultaneously can finish a work of  [#permalink]

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New post Updated on: 15 May 2017, 04:55
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In a workshop, 10 machines working simultaneously can finish a work of manufacturing 3000 components in 10 days while working 9 hours a day. What is the minimum number of additional machines required at the same workshop to finish the work of manufacturing 5000 similar components if work needs to be delivered in 6 days and maximum number of hours that machines can operate in a day is 10?

A) 4
B) 5
C) 10
D) 15
E) 25

SOURCE: http://www.GMATinsight.com

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Originally posted by GMATinsight on 15 May 2017, 03:23.
Last edited by GMATinsight on 15 May 2017, 04:55, edited 2 times in total.
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In a workshop 10 machines working simultaneously can finish a work of  [#permalink]

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New post 15 May 2017, 04:38
GMATinsight wrote:
If 10 machines working simultaneously can finish a work of manufacturing 3000 components in 10 days while working 9 hours a day. Then what is the minimum number of additional machines required to finish the work of manufacturing 5000 similar components if work needs to be delivered in 6 days and maximum number of hours that machines can operate in a day is 6.

A) 4
B) 5
C) 10
D) 15
E) 25

SOURCE: http://www.GMATinsight.com


Hi
When you say addl machines, does it mean these 10 are working for 9 hrs as earlier or these also wk for 6hrs.

Irrespective of either 6 or 9 ans cannot be D..
Let's take all for 9 hrs, so these 10 machine will cater for days 10 to 6 by 10*10/6..
Now to cater for increase in components from 3000 to 5000.... 10*10/6*5000/3000=500/18~28..
So ATLEAST 18 even if we take time as 9 hrs for all...
1) If we take all to be in 6 hrs, 500/18 * 9/6=1500/36~42 so addl 42-10=32..
2) if these 10 continue for 9 hrs, we can see how many these 10 make and rest can be seen with 6h.

After the necessary amendments...
To cater for change from 3000 to 5000 components
10 machines will become \(\frac{10*5000}{3000}\)
This will further cater for change of 10days to 6days..... Machines required \(\frac{10*5000*10}{3000*6}\)
Finally to cater for extra 1hr from 9 to 10hr...... Machines becomes \(\frac{10*5000*10*9}{3000*6*10}\)
=25..
So addl required=25-10=15

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Re: In a workshop 10 machines working simultaneously can finish a work of  [#permalink]

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New post 15 May 2017, 04:45
chetan2u wrote:
GMATinsight wrote:
If 10 machines working simultaneously can finish a work of manufacturing 3000 components in 10 days while working 9 hours a day. Then what is the minimum number of additional machines required to finish the work of manufacturing 5000 similar components if work needs to be delivered in 6 days and maximum number of hours that machines can operate in a day is 6.

A) 4
B) 5
C) 10
D) 15
E) 25

SOURCE: http://www.GMATinsight.com


Hi
When you say addl machines, does it mean these 10 are working for 9 hrs as earlier or these also wk for 6hrs.

Irrespective of either 6 or 9 ans cannot be D..
Let's take all for 9 hrs, so these 10 machine will cater for days 10 to 6 by 10*10/6..
Now to cater for increase in components from 3000 to 5000.... 10*10/6*5000/3000=500/18~28..
So ATLEAST 18 even if we take time as 9 hrs for all...
1) If we take all to be in 6 hrs, 500/18 * 9/6=1500/36~42 so addl 42-10=32..
2) if these 10 continue for 9 hrs, we can see how many these 10 make and rest can be seen with 6h.


Last number is 10 hours. Sorry for a typo

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Re: In a workshop 10 machines working simultaneously can finish a work of  [#permalink]

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New post 15 May 2017, 04:47
GMATinsight wrote:
If 10 machines working simultaneously can finish a work of manufacturing 3000 components in 10 days while working 9 hours a day. Then what is the minimum number of additional machines required to finish the work of manufacturing 5000 similar components if work needs to be delivered in 6 days and maximum number of hours that machines can operate in a day is 10

A) 4
B) 5
C) 10
D) 15
E) 25

SOURCE: http://www.GMATinsight.com


As per above information, i am getting 32 as the answer.

Kindly share your explanation.
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Re: In a workshop 10 machines working simultaneously can finish a work of  [#permalink]

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New post 15 May 2017, 04:51
GMATinsight wrote:
chetan2u wrote:
GMATinsight wrote:
If 10 machines working simultaneously can finish a work of manufacturing 3000 components in 10 days while working 9 hours a day. Then what is the minimum number of additional machines required to finish the work of manufacturing 5000 similar components if work needs to be delivered in 6 days and maximum number of hours that machines can operate in a day is 6.

A) 4
B) 5
C) 10
D) 15
E) 25

SOURCE: http://www.GMATinsight.com


Hi
When you say addl machines, does it mean these 10 are working for 9 hrs as earlier or these also wk for 6hrs.

Irrespective of either 6 or 9 ans cannot be D..
Let's take all for 9 hrs, so these 10 machine will cater for days 10 to 6 by 10*10/6..
Now to cater for increase in components from 3000 to 5000.... 10*10/6*5000/3000=500/18~28..
So ATLEAST 18 even if we take time as 9 hrs for all...
1) If we take all to be in 6 hrs, 500/18 * 9/6=1500/36~42 so addl 42-10=32..
2) if these 10 continue for 9 hrs, we can see how many these 10 make and rest can be seen with 6h.


Last number is 10 hours. Sorry for a typo

Posted from my mobile device

Posted from my mobile device


Taking this information into account, I am getting 15 as the answer. Is it correct?
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Re: In a workshop 10 machines working simultaneously can finish a work of  [#permalink]

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New post 15 May 2017, 05:03
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GMATinsight wrote:
In a workshop, 10 machines working simultaneously can finish a work of manufacturing 3000 components in 10 days while working 9 hours a day. What is the minimum number of additional machines required at the same workshop to finish the work of manufacturing 5000 similar components if work needs to be delivered in 6 days and maximum number of hours that machines can operate in a day is 10?

A) 4
B) 5
C) 10
D) 15
E) 25

SOURCE: http://www.GMATinsight.com


OE

CONCEPT: \(\frac{(Machine_Power * Time)}{Work} = Constant\)
i.e. \(\frac{(M_1 * T_1)}{W_1} = \frac{(M_2 * T_2)}{W_2}\)

i.e. \(\frac{(10 * (10*9)}{3000} = \frac{(M_2 * (6*10))}{5000}\)

i.e. \(T_2 = 25\)

i.e. Additional Machines required = 25-10 = 15

Answer: option D
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Re: In a workshop 10 machines working simultaneously can finish a work of  [#permalink]

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New post 17 May 2017, 13:56
GMATinsight wrote:
In a workshop, 10 machines working simultaneously can finish a work of manufacturing 3000 components in 10 days while working 9 hours a day. What is the minimum number of additional machines required at the same workshop to finish the work of manufacturing 5000 similar components if work needs to be delivered in 6 days and maximum number of hours that machines can operate in a day is 10?

A) 4
B) 5
C) 10
D) 15
E) 25

SOURCE: http://www.GMATinsight.com

Nice question! For these questions, I almost always find the individual machine or worker rate, using a slightly changed version of the RTW table. Just add one column for "Number of workers/machines," and it's easy:
Attachment:
Revised Work Formula Table.jpg
Revised Work Formula Table.jpg [ 40.42 KiB | Viewed 1202 times ]

Revised formula: (# of workers)*(rate)*(time) = Work

In this case, you have to use HOURS for units of time.
Calculate first row: 10 days * 9 hrs/day = 90 hours total
Calculate second row: 6 days * 10 hrs/day = 60 hours total

1. Find individual machine rate ==>

(10)*(R)*(90) = 3000

R = \(\frac{3000}{900}\) =\(\frac{10}{3}\)

2. Use that rate in second row to find TOTAL number of machines needed for new task ==>

(# of machines TOTAL)*(\(\frac{10}{3}\))*(60) = 5000

# of machines = \(\frac{5000}{200}\) = 25 TOTAL

Question asks, how many more machines needed for new task. 25 (need) - 10 (have) = 15.
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Re: In a workshop 10 machines working simultaneously can finish a work of   [#permalink] 13 Jan 2019, 00:15
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