Question : D1 and D2 can do a piece of work together in 16 days. D2 and D3 can do the same work together in 24 days, while D3 and D1 can do it together in 48 days. In how many days can all 3, working together, do $\frac{5}{6}$ of the work?
Option 1: $\frac{31}{3}$ days
Option 2: $\frac{40}{3}$ days
Option 3: $\frac{35}{3}$ days
Option 4: $\frac{38}{3}$ days
Correct Answer: $\frac{40}{3}$ days
Solution :
Let $W$ be the total work to be done.
The work done by D
1
+ D
2
in one day = $\frac{W}{16}$
The work done by D
2
+ D
3
in one day = $\frac{W}{24}$
The work done by D
3
+ D
1
in one day = $\frac{W}{48}$
⇒ 2x [(D
1
+ D
2
+ D
3
)'s 1 day work] = $\frac{W}{16}$ +$\frac{W}{24}$+$\frac{W}{24}$
= $\frac{6W}{48}$
⇒ (D
1
+ D
2
+ D
3
)'s 1 day work = $\frac{W}{16}$
⇒ Time = $\frac{\text{Total Work}}{\text{Combined Rate}}$ = $\frac{\text{$\frac{5W}{6}$}}{\text{$\frac{W}{16}$}}$ = $\frac{5}{6}$ × $\frac{16}{1}$ = $\frac{40}{3}$ days
Hence, the correct answer is $\frac{40}{3}$days.
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