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

Question

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: https://www.GMATinsight.com

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

SOURCE:  https://www.GMATinsight.com

GMATinsight · Accepted Answer

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

chetan2u · Answer

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

D

GMATinsight · Answer

Last number is 10 hours. Sorry for a typo

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gmatexam439 · Answer

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

Kindly share your explanation.

gmatexam439 · Answer

Taking this information into account, I am getting  15  as the answer.  Is it correct?

generis · Answer

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: Revised Work Formula Table.jpg 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. Answer D

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