A's 1 days 's work = 1/8
B ' s 1 days ' s = 1/10
C ' s day 's work = 1/20
(A + B + C)'s 1 days 's work = 1/8 + 1/10 + 1/20 = (5 + 4 + 2)/40 = 11/40
? (A + B + C ) can finish the work in 40/11 days or 3 7/11 days.
Let the original number of men = N
Time taken by N = 80 days
Now, (N - 10) men can finish the work in the (80 + 20) = 100 days
Here, M1 = N, M2 = (N - 10), D1 = 80 and D2 = 100
According to the formula.
M1D1 = M2D2
? N x 80 = 100 x (N - 10)
? 8N = 10N - 100
? 10N - 8N = 100
? N = 50
M1 = 250, D 1 = 33 days
per day meal W1 = W2 = 125 g
M2 = (250 + 80) =330 and D2=?
According to the formula
M1D1W2 = M2D2W1
250 x 33 x 125 = 330 x D2x 125
? D2 = (250 x 33)/330
? D2 = 25 days
Efficiency of Kavita = 5%
Efficiency of Babita = 1.66%
Efficiency of Samita = 3.33%
Work done in 5 days by K + B + S = 5 x 10 = 50%
Work done in 3 days by K + B = 3 x 6.66 = 20%
Remaning work (30%) done by Kavita alone = 30/5 = 6 days
Efficiency of Kareena and Krishna = 11.11 + 5.55 = 16.66%
Work done in 3 days = 3 x 16.66 = 50%
Rest work done by Kareena, Krishna and Shahid = 50/50 = 1 day
(Since efficiency of Shahid = 33%)
Thus in ( 3 + 1) days they have completed the work = 4 days.
Krishna's efficiency = 10%
Mohan's efficiency = 5%
Work done by Krishana and mohan together in 3 days = 15 x 3 = 45%
Now, number of days in which B completed rest (55%) work alone = 55/5 = 11
Total No. of days in which B worked = 3 + 11 = 14
Now No. of days required by B, when A and B both worked together = 100/15 = 62/3
? Required difference in No. of days = (11) - (62/3)
= 13/3 = 4 1/3 days
10 day's work by 15 men = 10/210 = 1/21
At the end of every 10 days 15 additional men are employed i.e., for the next 10 days we have 15 + 15 = 30 men
? Next 10 day's work by 30 men = 2/21
Hence in 20 days only (1/21 + 2/21) = 3/21 work is completed
To complete the whole work we have to reach the value of (21/21) work.
Now, (1/21) + (2/21) + (3/21) + ......... (6/21) = 21/21 = 1
Hence total time to complete the whole wok 10 + 10 + 10 + 10 + 10 + 10 = 60 days
Ramesh alone finish 1/2of the work in 10 days.
Remaining 1/2 of the job was finished by Ramesh and Dinesh together in 2 days.
Therefore, they both together can finish the complete job in 4 days.
(A + B)'s 20 day's work = (20 x 1/30) = 2/3
Remaining work = (1 - 2/3) = 1/3
1/3 work is done by A in 20 days
Whole work can be done by A in (3 x 20) days = 60 days.
Efficiency is proportional to work done per day. Work done per day N number of days
= Amount of work done. Considering efficience of A and B initially as 1.
Let A alone can do the work in M days and B alone can do the same work in N days.
Then, 5/M + 5/N = Total work done = 1
Since efficiency of A and B are 2 and 1/3 respectively
? 6/M + 1/N = 1 ....(i)
and 1/M + 1/N = 5 .....(ii)
Now, subtracting equation (ii) from equation (i), we have M = 25/4 = 61/4 days.
? In 20 days the work is completed by = 16 men + 12 women
? In 1 day the work is completed by = 20 x (16 men + 12 women )
= 320 men + 240 women
In 40 days the work is complete by 18 women
? 1 day the work is complete by = 18 x 40 = 720 women
? 720 women = 320 men + 240 women
? (720 - 240)women = 320 men
? 480 women = 320 men
? 1 men = 480/320 = 3/2 women
? 12 men + 27 women = 12 x (3/2) + 27 = 45 women
? 18 women complete work in 40 days
? 45 women complete 1 work = 40 x 18/45 = 16 days
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