1 man's 1 day work = 1/108
12 men's 6 day's work = 1/9 x 6 = 2/3
Remaining work = 1 - 2/3 = 1/3
16 men's 1 day work = 1/108 x 16 = 4/27
4/27 work is done by them in 1 day.
1/3 work is done by them in 27/4 x 1/3 = 9/4 days.
Let 1 man's 1 day work = x and 1 woman's 1 day work = y.
Then, 4x + 6y = 1/8 and 3x + 7y = 1/10
Solving these two equations, we get:
x = 11/400 and y = 1/400
10 woman's 1 day work = (1/400 x 10) = 1/40.
Hence, 10 women will complete the work in 40 days.
Given 4men, 12 women and 20 children work for 2 days.
Workdone for 2 days by 4men, 12 women and 20 children =
Therefore, remaining work = 1 - =
To complete the same work by only men in 1 day,
We know that M1 x D1 = M2 x D2
Here M1 = 6 , D1 = 12 and M2 = M , D2 = 1
12 x 6 = M x 1
=> M = 12 x 6 = 72
=> But the remaining work = 1/2
Men required => 1/2 x 72 = 36
Only men required to Complete the remaining work in 1 day = 36.
Given 10 men take 15 days to complete a work
=> Total mandays = 15 x 10 = 150
Let the work be 150 mandays.
=> Now 37 men can do 150 mandays in 150/37 =~ 4 days
Arun has completed rd of the work in 8 days
Then he can complete the total work in
---- 8
1 ---- ?
= 24 days
But given Akhil is only 60% as efficient as Arun
Akhil =
Akhil can complete the total work in 40 days
Now, remaining 2/3rd of work can be completed in
1 ------ 40
------ ?
= 26.66 days.
(P+Q)'s 1 day work = 1/24
P's 1 day work = 1/32
=> Q's 1 day work = 1/24 - 1/32 = 1/96
Work done by (P+Q) in 8 days = 8/24 = 1/3
Remainining work = 1 - 1/3 = 2/3
Time taken by Q to complete the remaining work = 2/3 x 96 = 64 days.
M = 10 days
The ratio of efficiency of M & N are 3 : 2
Hence, the time rquired for N alone = 15 days
=> Required time taken t=by both to complete the work = M x N / M + N
= 10 x 15/ 10 + 15
= 150/25
= 6 days.
Efficiency of P= 100/20= 5% per hour
Efficiency of Q= 100/25= 4% per hour
Efficiency of R= 100/40= 2.5% per hour
Efficiency of S=100/50= 2% per hour
Cistern filled till 10 am by P, Q and R
Thus, at 10 am pipe P,Q and R filled 30.5% of the cistern.
Rest of cistern to be filled = 100 - 30.5 = 69.5%
Now, the time taken by P,Q,R and S together to fill the remaining capacity of the cistern
= 69.5 / (5+4+2.5+2) = 5 Hours and 9 minutes(approx).
Therefore, total time =4 hrs + 5hrs 9 mins = 9 hrs and 9 mins
It means cistern will be filled up at 3:09 pm
Ratio of times taken by P & Q = 100 : 130 = 10:13
Let Q takes x days to do the work
Then, 10:13 :: 23:x
=> x = 23x13/10
=> x = 299/10
P's 1 day's work = 1/23
Q's 1 day's work = 10/299
(P+Q)'s 1 day's work = (1/23 + 10/299) = 23/299 = 1/13
Hence, P & Q together can complete the work in 13 days.
9M + 12B ----- 12 days ...........(1)
12M + 12B ------- 10 days........(2)
10M + 10B -------?
108M + 144B = 120M +120B
24B = 12M => 1M = 2B............(3)
From (1) & (3)
18B + 12B = 30B ---- 12 days
20B + 10B = 30B -----? => 12 days.
Days remaining 124 ? 64 = 60 days
Remaining work = 1 - 2/3 = 1/3
Let men required men for working remaining days be 'm'
So men required = (120 x 64)/2 = (m x 60)/1 => m = 64
Men discharge = 120 ? 64 = 56 men.
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