A can finish 2/3 of work done by B in 1/4 of time taken by B. B can finish 4/3 of work done by C in 3/5 of time taken by C. Working together they can finish the work in 60 days. In how many days they will finish the work, working alone?
Answers
Answer:
Let x be the number of days that would take C to complete the job. In one day he does 1x of the job.
B can do 43 of any work done by C in 35 of the time that takes C to complete that much work (i.e. B is faster than C), therefore, in 3x5 days B does 43 jobs. Thus, B does the job in
3x5⋅34=9x20
days. In one day he does 209x of the job.
A can do 23 of any work done by B in 14 of the time that takes B to finish that much work (i.e. A is faster than B and the fastest of the three), hence, in
9x20⋅14=9x80
days A does 23 of the job. Consequently, A does the job in
9x80⋅32=27x160
days. In one day he does 16027x of the job.
We are now ready to calculate how much time it takes C to complete one job and thenceforth to compute the time needed by B and A to finish that much work by solving the following equation:
60⋅(16027x+209x+1x)=1
32009x+4003x+60x=1
3200+1200+540=9x
x=49409=20⋅2479
So C does the job in
49409
days (approximately 548.889 days).
B does the job in
9x20=920⋅49409=247
days.
A does the job in
27x160=27160⋅49409=3⋅2478
days (92.625 days).
We can verify this solution by adding their daily work outputs and checking whether that sum equals 160 :
83⋅247+1247+920⋅247=8⋅20+60+9⋅360⋅247=24760⋅247=160
hope this helps you