- Suppose A and B together can do a job in 12 days, while B alone can finish a job in 24 days. In how many days can A alone finish the work?
Solution:
Number of days in which A and B together can finish the work = 12 days
Number of days in which B alone can finish the work = 30
1/T = 1/m + 1/n
1/12 = 1/m + 1/30
1/m = 1/30 – 1/12 = 5-2/60 = 3/60 = 1/20
A can finish the work in 20 days.
Suppose A is twice as good a workman as B and together they can finish a job in 24 days. How many days A alone takes to finish the job?
Solution:
A is twice as good a workman as B
i.e if B can finish a work in t days A can finish it in 1/2 days
1/T = 1/m + 1/n
1/24 = 1/t/2 + 1/t = 2/t + 1/t = 3/t
1/24 = 3/t
t = 24 x 3 = 72
i.e, B takes 72days to finish the job
A takes 72/2 = 36 days to finish it
- Suppose B is 60% more efficient them A. if A can finish a job in 15 days how many days B needs to finish the same job?
Solution:
A can finish a work in 15 days.
Work done A in 1 day = 1/15
B is 60% more efficient
Work done by B in 1 day
1/15 + 1/15 x 60/100
= 1/15 (1 + 60/100)
= 1/15 ( 8/5)
= 8/75
Number of days in which B alone can finish the work = 1/(8/75) = 75/8 = 93/8 days
- Suppose A can do a piece of work in 14 days while B can do it in 21 days. They begin together and worked at it for 6 days. Then A fell ill B had to complete the work alone. In how many days was the work completed?
Solution:
M = 14 days
N = 21 days
Part of work done in 6 days
= (1/14 + 1/21)6
= 6(3+2/42) = 5×6/42 = 5/7
Remaining part of the work = 1-5/7 = 2/7
Days taken by B to finish
2/7 part of the work = (2/7)/(1/21) = 2/7 x 21/7 = 6 days
Total number of days in which the work is completed = 6+6 = 12 days
- Suppose A takes twice as much time as B and thrice as much time as C to complete a work. If all of them work together they can finish the work in 2 days. How much time B and C working together will take to finish it?
Solution:
If A alone takes to t1 days to do the work , B finishes it in t1/2 and C is t1/3 days
1/T = 1/t1 +1/t2 + 1/t3
= 1/t1 +1/(t1/2) + 1/(t1/3)
= 1/t1 +2/t1 + 3/t1
= 6/t1
1/T = 1/2
1/2 = 6/T1
i.e. t1 = 12 dyas
B takes 12/2 = 6days
C takes 12/3 = 4 days
Part of work done by B
In one day = 1/6
Part of work done by C in one day = 1/4
If B and C together takes t days to finish the work 1/T = 1/6 + 1/4 = 2+3/12 = 5/12
T = 12/5 = 2.4 days
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