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. \(\frac{[6 * (30*9)]}{W} = \frac{[M_2 * (25*8)]}{(10*W)}\)

i.e. \(M_2 = 81\)

Answer: option D
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Or you can do it using joint variation:

Number of men required = 6 * (30/25) * (9/8) * 10 = 81

Check this link to see how to solve such questions: http://www.veritasprep.com/blog/2013/02 ... variation/
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In this type of problems, converting for unit values will make it easy to solve.

Take a look at the first sentence. We can say, to do the work in 1 day of 9 hours each will take 6*30 men.
To do the work in 25 days will take (6*30) /25 men
Similarly adding the info, that to do the work in a day of 8 hours instead of 9 hours each will take (6 *30*9)/ (25*8) men
To do 10 times the amount of work will take 10 * (6*30*9) / (25*8) = 81 men
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Holistic and Systematic Approach

One version of the rate formula approach . . .
Use standard W = RT formula, but add (# of workers or machines) to R*T, thus: W = #*R*T

Work = (# of workers) * individual rate * time

Scenario 1

Get individual rate from first scenario, where t = (30 * 9hrs = 270 hrs)

Work = # * r * t

1 = 6 * ?? * 270

1 = r * 1620

r = \(\frac{1}{1620}\)

Scenario 2

At that rate, how many men will it take to do 10 times the amount of work if they work for 25 days of 8 hours?
t = (25 * 8 hrs = 200 hrs)

Work = # * r * t

10 = ?? * \(\frac{1}{1620}\) * 200

10 = # * \(\frac{200}{1620}\)

# of workers = \(\frac{10}{(\frac{200}{1620})} =\)

# of workers =\((10 * \frac{1620}{200})=(\frac{1620}{20})=\frac{162}{2}= 81\)

Answer D
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