A. 42/4 days
B. 41/4 days
C. 55 days
D. None of these
A. 15 days
B. 10 days
C. 8 days
D. 20 days
(4/m)+(2/w) = 1/2 2/m = 1/5 1/m = 1/10 (4/10)+(2/w) = (1/2) 2/w = 1/10 2 women alone take 10 days.
A. 32 hours
B. 48 hours
C. 3 days
D. 4 days
Part of task completed by A in a day = 1/x Part of task completed by B in a day = 3/x (B is thrice as efficient as A) (1/x)+(3/x) = 1 4/x = 1 x = 4 B completes 3/4th of the task in a day. He takes 4/3 days to complete the entire task. 4/3 days = (4/3*24) hours = 32 hours
A. 4(2/3) days
B. 4 days
C. 5(1/3) days
D. 1(1/3) days
(1/A)+(1/B) = (1/6) (1/B)+(1/C) = (1/12) (1/A)+(1/C) = (1/8) Solving the three equations,
(1/A)+(1/B)+(1/C) = (3/16) Together they take 16/3 = 5(1/3) days to complete the task.
A. 12 days
B. 15 days
C. 16 days
D. 18 days
A. 10 days
B. 11 days
C. 15 days
D. 20 days
A. 6 days
B. 10 days
C. 15 days
D. 20 days
Work done by X in 4 days = ( 1/20X 4 ) = 1/5 Remaining work = (1 – 1/5 ) = 4/5 (X + Y)’s 1 day’s work = 1/20 + 1/12 = 8/60 = 2/15 Now, 2/15 work is done by X and Y in 1 day. So, 4/5 work will be done by X and Y in 15/2 X 4/5 = 6 days Hence, total time taken = (6 + 4) days = 10 days.
A. 3 : 4
B. 4 : 3
C. 5 : 3
D. Data inadequate
A. 80 days
B. 100 days
C. 120 days
D. 150 days
A. 8 days
B. 10 days
C. 12 days
D. 15 days
(A.+ B)’s 1 day’s work = ( 1/15 + 1/10 ) = 1/6 Work done by A and B in 2 days = ( 1/6 X 2 ) = 1/3 Remaining work = ( 1 – 1/6 ) = 2/3
Now, 1/15 work is done by A in 1 day. so 2/3 work will be done by A in ( 15 X 2/3 ) = 10 days Hence, total time taken = (10 + 2) = 12 days.
