Let M, N and O worked together for x days.
From the given data,
M alone worked for 8 days
M,N,O worked for x days
N, O worked for 1 day
But given that
M alone can complete the work in 18 days
N alone can complete the work in 36 days
O alone can complete the work in 54 days
The total work can be the LCM of 18, 6, 54 = 108 units
M's 1 day work = 108/18 = 6 units
N's 1 day work = 108/36 = 3 units
O's 1 day work = 108/54 = 2 units
Now, the equation is
8 x 6 + 11x + 5 x 1 = 108
48 + 11x + 5 = 108
11x = 103 - 48
11x = 55
x = 5 days.
Hence, all M,N and O together worked for 5 days.
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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