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 A Time And Work Problem (Posted on 2006-06-25)
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.

 See The Solution Submitted by K Sengupta Rating: 3.0000 (2 votes)

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 Solution | Comment 1 of 2
`    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

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