Question : P, Q, and R alone can do a piece of work in 5, 6, and 4 days, respectively. All three of them began the work together but Q left 1 day before completion of the work. In how many days was the work completed?
Option 1: $\frac{70}{37}$ days
Option 2: $\frac{100}{27}$ days
Option 3: $3$ days
Option 4: $\frac{80}{23}$ days
Correct Answer: $\frac{70}{37}$ days
Solution :
Let the total number of days be t.
Rate of P = $\frac{1}{5}$ per day
Rate of Q = $\frac{1}{6}$ per day
Rate of R = $\frac{1}{4}$ per day
⇒ Combined rate = $\frac{1}{5}$ + $\frac{1}{6}$ + $\frac{1}{4}$ = $\frac{12}{60}$ + $\frac{10}{60}$ + $\frac{15}{60}$ = $\frac{37}{60}$
Q leaves 1 day before the completion of the work.
So, P, Q, and R worked together for (t – 1) days, and P and R worked on the last day.
Total work = 1 = $\frac{37}{60}$ × (t – 1) + $\frac{9}{20}$
⇒ 1 – $\frac{9}{20}$ = $\frac{37}{60}$ × (t – 1)
⇒ $\frac{11}{20}$ = $\frac{37}{60}$ × (t – 1)
⇒ $\frac{11×60}{20×37}$ = (t – 1)
⇒ $\frac{33}{37}$ + 1 = t
⇒ t = $\frac{70}{37}$
Hence, the correct answer is $\frac{70}{37}$ days.
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