SPEED and VELOCITY can complete a piece of work respectively in FEW days and LITTLE days while working separately. The amount of work done by SPEED in 21 days exceeds the amount of work done by VELOCITY in 8 days by one-sixth of the amount of work done by SPEED in FEW days.

SPEED and VELOCITY began the work together but SPEED left after LITTLE/2 days forcing VELOCITY to complete the remaining work all alone in FEW/9 days.

Determine FEW and LITTLE.

    F*S = L*V                           (1)

   21*S = 8*V + F*S/6                   (2)

    L*V = L*(S+V)/2 + F*V/9             (3)

Substituting S = L*V/F in (2) and (3) gives
 
  126*L = 48*F + L*F                    (4)

  9*L*F = 9*L^2 + 2*F^2                 (5)

Substituting F = 126*L/(48+L) in (5) gives

  (48+L)^2 - 126*(48+L) + 3528 = 0      (6)

Solving for 48+L,

  48 + L = 63 +- 21

Therefore,

  LITTLE = 36 days  and  FEW = 54 days

Also,

  SPEED = (2/3)*VELOCITY  

Posted by Bractals on 2006-06-25 12:26:35