$D_1$ and $D_2$ can do a piece of work together in 16 days, $D_2$ and $D_3$ can do the same work together in 24 days, while $D_3$ and $D_1$ can do it together in 48 days. In how many days can all 3 , working together, do $\frac{5}{6}$ of the work?

$\frac{40}{3}$ days

D1 + D2 = 16 days,

D2 + D3 = 24 days,

D3 + D4 = 48 days,

⇒ Adding all the efficiencies, we get,

⇒ 2(D1 + D2 + D3) = 6,

⇒ (D1 + D2 + D3) = 3,

⇒ \(\frac{5}{6}\) of total work = \(\frac{5}{6}\) x 48 = 40 units,

⇒ Time required by (D1 + D2 + D3) to complete the 40 units of work = \(\frac{40}{3}\) days,   ..(\(\frac{Work}{Efficiency}\) = Time)