a and b together can do a work in 12 days.b and c together do it in 15 days.if a's efficiency is twice that of c,then the days required for b alone to finish the work is how many days?

• 20 days
  
  1/a + 1/b = 1/12 ---(i)
  1/b + 1/c = 1/15 ---(ii)
  As a's efficiency is twice that of c, so days required by a is half that of c => a=c/2 or c=2a
  Substituting c=2a in (ii), 1/b + 1/2a = 1/15 ---(iii)
  On subtracting (iii) from (ii), 1/a=1/30 , so 1/b=1/20 and 1/c=1/60
  Hence b alone requires 20 days to finish the work.
• 7 years agoHelpfull: Yes(14) No(4)
• A+B= 1/ 12------1
  A=2C--------------2
  FROM 1&2
  2C+B=1/12
  B=1/12-2C---------3
  NOW,B+C= 1/15
  (1/12-2C) + C= 1/15
  -C= 1/15 -1/12
  - C= -1/60
  C=1/60
  PUTTING THIS VALUE IN B+C= 1/15
  B+1/60=1/15
  B=1/20
  Therefore B finishes the work in 20 days
  Ans ->20 days
• 7 years agoHelpfull: Yes(1) No(0)
• ans : 60 days
  
  one day work of a & b .......................1/(a+b)= 1/12
  one day work of b & c .........................1/(b+c) = 1/15
  therefore .............one day work of a & c ....................1/(a+c)= [1/(a+b)] - [1/(b+c)] =1/60
  given a=2c
  [1/(a+c)] = 1/3c=1/60
  so 1/c= 1/20 ( c takes 20 days to finish the work)
  and [1/(b+c)] - 1/c = 1/60 ( b takes 60 days to finish the same work)
• 7 years agoHelpfull: Yes(0) No(10)