Let the required number of units of work = k
According to the formula,
W1= m and W2 = k
M1T1D1W2 = M2T2D2W1
? m x m x m x k = n x n x n x m
? x= (m x n3) / m3 = n3/m2
4th observation i.e., 8 is the median.
47/10000 = .0047
.02 =(2/100) x 100% =2%
Let N / 11 = 233
Then, N = 233 x 11 = 2563
? Missing digit is 5.
121012 = 12 x 10084 + 4
? remainder = 4
We have the important relation, More work, More time (days)
? A piece of work can be done in 6 days.
? Three times of work of same type can be done in 6 x 3
= 18 days
? = 750.0003 ÷ 19.999
? ? ? 750 ÷ 20
? ? ? 375 ? 38
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