We know that, when a pipe fills a tank in m h, then the part of tank filled in 1 h = 1/m
Here, m = 6
? Required part of the tank to be filled in 1h = 1/6 part
Part filled A in 1 hr= (1/45)
Part filled B in 1 hr= (1/36)
Part filled by (A+B) together in 1 hr=(1/45)+(1/36)=1/20
So, The tank will be full in 20 hr.
Let the work lasted for N days,
Then, Rashmi's 4 day's work + Ravina (N - 3) day's + Gitika's N day's work = 1
? (4/16) + (N - 3) / (124/5) + N/32 = 1
? 5(N - 3)/64 + N/32 = 1 - 1/4
? [5(N - 3) + 2N] / 64 = 3/4
? 7N - 15 = 48
? N= (48 + 15)/7 = 63/7 = 9 days
Let the work be finished in N days.
Then, A's N day's work + B's (N - 1) days + work + C's (N - 2) day's work = 1
? N/16 + (N - 1/32) + (N - 2/48) = 1
? N/1 + N - 1/2 + N - 2/3 = 16
? (6N + 3N - 3 + 2N - 4)/6 = 16
? 11N - 7 = 96
? 11N = 103
? N = 103/11 = 94/11 days
Remaining work = 1 - (4/7) = 3/7
Remaining period = (92 - 66) = 26 days
Let the number of addition men = N
Given, M1 = 234, D1 = 66, T1 = 16, W1= 4/7,
M2 = (234 + x), D2 = 26, T2 = 18, W2 = 3/7
According to the question,
M1W2T1D1= M2W1T2D2
? 234 x (3/7) x 16 x 66 = (234 x N) x (4/7) x 18 x 26
? 234 + N = (3 x 66 x 16 x 234) / (4 x 26 x 18)
? 234 + N = 36 x 11 = 396
? N = 396 - 234 = 162
Additional men to be employed = 162
Given, M1 = 120, M2 = 120 - n, D1 = 64, D2 = 60
W1= 2/3 and W2 = 1/3
According to the formula,
(M1D1) / W1 = (M2D2) / W2
? [120 x (64/2)] / 3 = [(120 - n) x 60] / (1/3)
? (120 x 64) / (2 x 60) = (120 - n)
? (120 - n) = 64
? n = 120 - 64 = 56
Now, 56 men can be discharged to finish the work in time.
Work done by leak in 1 hour = (1/8 - 1/10) = 1/40
? The leak will empty the cistern in 40 hours.
Required answer = (12 x 16) /(12 + 16) = 48 / 7 = 66/7 minutes.
Here x = 8 hrs. and y = 8 + 2 = 10 hrs.
Now, applying the given rule, we have the
Required answer = (8 x 10) /(10 - 8) = 40 hrs.
We know that, when a pipe fills 1/m part of a tank in 1 h, then the pipe takes m h to fill the tank.
Here, 1/m = 1/8 ? m = 8
? Required time to fill the tank = 8 h
Time taken by the pipe to empty the cistern = 3 h
Then, time taken by the pipe to empty the 2/3 part = 3 x 2/3 = 2 h
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