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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Amount of work done by 6 men working 30 days working 9 hrs each = 30 * 9 * 6
Let the number of people required for the 2nd case = X

Then X * 25 * 8 = 30 * 9 * 6 * 10
= > X = 81 - 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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Can someone please explain any other method, preferably the w=rt method. Work rate problems always seem so simple but I always go wrong or am stuck. I was not able to grasp it even after going through the two approaches. Thank you.

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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If 6 men can do a piece of work in 30 days of 9 hours each, how many men will it take to do 10 times the amount of work if they work for 25 days of 8 hours?

A. 61
B. 64
C. 84
D. 81
E. 94

6*30*9=1620 total man hours to do work
let m=number of men needed
m*25*8*1/1620=10
m=81 men
D.

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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The hourly rate of the 6 men is 1/(30 x 9) = 1/270. If we let x = the number of men needed to do 10 times the work when they work for 25 days of 8 hours, then the hourly rate of these x men must be 10/(25 x 8) = 10/200 = 1/20. We can create the following proportion and solve for x:

6/(1/270) = x/(1/20)

1620 = 20x

x = 1620/20 = 81

Answer: D
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The best way to tackle any of these problems is to follow this sequence (be it any question)

No. of workers ----- Time ----- work
6 ----- 30x9 hrs ----- 1

we need to find:

? ----- 25x8 ----- 10

so, by unitary method:

6 ----- 1 ----- 1/ (30x9)

1 ----- 1 ----- 1/ (30x9)*(6)

1 ----- 25x8 ----- 25x8/ (30x9)*(6)

so,

{(10)* (30x9)*(6)/ (25x8)} ----- 25x8 ----- 10
=81

Used a different method here not sure if its correct please validate

6m x 9hr x30 = 1620 units of work

Works get increased by 10 times

new work = 16200, hrs given 25days x 8 hrs = 200 hrs

16200/200 = 81 machines

Yes, it is correct. You can put the question in terms of units of work.
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