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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. In how many days will A, B and C finish it, working together? In how many days will each of them finish it working alone?

User Null
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Answer:

  • It takes A 30 days to finish the work alone
  • It takes B 20 days to finish the work alone
  • It takes C 60 days to finish the work alone

  • Working together, it takes them 10 days to finish the work

Explanation:

Given;


(1)/(A) + (1)/(B) =(1)/(12) \\\\(1)/(B) + (1)/(C) = (1)/(15)\\\\(1)/(A) + (1)/(C) = (1)/(20)\\\\

From given equations above;


(1)/(A) +(1)/(B) = (1)/(12)\\\\But, (1)/(B) =(1)/(15) -(1)/(C)\\\\ (1)/(A) +(1)/(15) -(1)/(C)= (1)/(12)\\\\Also, (1)/(C) = (1)/(20) -(1)/(A) \\\\ (1)/(A) +(1)/(15) -((1)/(20) -(1)/(A))= (1)/(12)\\\\ (1)/(A) +(1)/(15) -(1)/(20) +(1)/(A)= (1)/(12)\\\\ (1)/(A) +(1)/(A) = (1)/(12) +(1)/(20)-(1)/(15)\\\\(2)/(A) = (300+ 180-240)/(3600) \\\\(2)/(A) =(240)/(3600)\\\\


(2)/(A) = (1)/(15) \\\\A = 30 \ days

solve for B;


(1)/(A) + (1)/(B) = (1)/(12)\\\\ (1)/(B) = (1)/(12) - (1)/(A) \\\\ (1)/(B) = (1)/(12) - (1)/(30)\\\\ (1)/(B) =(5-2)/(60) \\\\ (1)/(B) =(3)/(60)\\\\B = 20 \ days

Solve for C;


(1)/(C) =(1)/(20) -(1)/(A)\\\\ (1)/(C) = (1)/(20) -(1)/(30)\\\\ (1)/(C) =(3-2)/(60) \\\\ (1)/(C) =(1)/(60)\\\\C = 60 \ days

When they work together;


(1)/(A) +(1)/(B) +(1)/(C) =(1)/(y) \\\\(1)/(30) +(1)/(20) +(1)/(60) = (1)/(y)\\\\(2+ 3+ 1)/(60) =(1)/(y)\\\\(6)/(60) =(1)/(y)\\\\y = 10\ days

User Darryl Peterson
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