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.
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
Given K=4L
-->K+L = 4L+L = 5L
These 5L people can complete the work in 24 days, which means L alone can do the work in (24 x 5)=120 days.
Hence, K alone can do the work in 120/4= 30 days.
Comments
There are no comments.Copyright ©CuriousTab. All rights reserved.