Question

A and B can do a piece of work in 12 days, B and C can do it in 15 days while C and A can finish it in 20 days. If A, B and C works together, in how many days will they finish the work? In how many days will each one of them finish it, working alone?​

Answer 1

Answer:

See below.

Step-by-step explanation:

Time taken by (A+B) to finish the work = 12 days

$\sf (A+B's) \: 1 \: day's \: work= \frac{1}{12}$

Time Taken by (B+C) to finish the work = 15 days

$\sf \: (B + C) 's\: 1 \: day's \: work = \frac{1}{15}$

Time taken by (C+A) to finish the work = 20 days

$\sf \:(C+A)'s \: 1 \: day's \: work = \frac{1}{20}$

Therefore,2 ( A + B + C)'s 1 day's work-

$\sf = ( \frac{1}{12 } + \frac{1}{15} + \frac{1}{20} )$

$\sf \: = \frac{5 + 4 + 3}{60}$

$\sf \: = \frac{12}{60}$

$\sf \: = \frac{1}{5}$

$\sf \: = > (A+B + C)'s \: work = \: (\frac{1}{2} \times \frac{1}{5} ) \:= \frac{1}{10}$

Therefore, A, B and C together can finish the work in 10 days.

Now,

A's 1 day's work = (A+B+C)'s 1 day's work - (B+C)'s 1 day's work

$\sf \: = ( \frac{1}{10} - \frac{1}{15} )$

$\sf \: = \frac{3 - 2}{30}$

$\sf \: = \frac{1}{30}$

• Hence,A alone can finish the work in 30 days.

B's 1 day's work = (A+B+C)'s 1 day's work - (A+C')s 1 day's work

$\sf \: = (\frac{1}{10} - \frac{1}{20} )$

$\sf \: = \frac{2 - 1}{20}$

$\sf \: = \frac{1}{20}$

• Hence,B alone can finish the work in 20 days.

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Answer 2

Answer:

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