Suppose A, B and C take x, x/2 and x/3 respectively to finish the work.
Then, (1/x + 2/x + 3/x) = 1/2
=> 6/x = 1/2 => x = 12
So, B takes 6 hours to finish the work.
We have M1 D1 H1 / W1 = M2 D2 H2 / W2 (Variation rule)
(60 x 30 x 8)/ 150 = (35 x D2 x 6) / 200
D2 = (60 x 30 x 8 x 200) / (150 x 35 x 6) => D2 = 91.42 days =~ 91.5 days.
For 30 days he receives
30 x 25 = 750
Actually he gets 425
Therefore, Fine is 750 - 425 = 325
=> No of days absent = 325/(25+7.50)= 10
As the work is same
M1D1 = M2D2
Here Let the number of men employed was M
=> 11M = 7(M+4)
=> 11M - 7M = 28
=> M = 7
Suppose A, B and C take x, x/2 and x/3 respectively to finish the work.
Then, (1/x + 2/x + 3/x) = 1/2
6/x = 1/2 => x = 12
So, B takes 6 hours to finish the work.
A can complete the work in 12 days working 8 hours a day
=> Number of hours A can complete the work = 12×8 = 96 hours
=> Work done by A in 1 hour = 1/96
B can complete the work in 8 days working 10 hours a day
=> Number of hours B can complete the work = 8×10 = 80 hours
=> Work done by B in 1 hour = 1/80
Work done by A and B in 1 hour = 1/96 + 1/80 = 11/480
=> A and B can complete the work in 480/11 hours
A and B works 8 hours a day.
Hence total days to complete the work with A and B working together
= (480/11)/ (8) = 60/11 days
K's one day's work = 1/30
L's one day's work = 1/45
(K + L)'s one day's work = 1/30 + 1/45 = 1/18
The part of the work completed in 3 days = 3 (1/18) = 1/6.
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