S can do a piece of work in 10 days and T in 30 days. They begin together but S leaves 3 days before the completion of the work. The remaining work will be done in how many days?6. 75 days11. 25 days8. 5 days15.5 day
Question
S can do a piece of work in 10 days and T in 30 days. They begin together but S leaves 3 days before the completion of the work. The remaining work will be done in how many days?6. 75 days11. 25 days8. 5 days15.5 day
Solution
First, let's find out how much work S and T can do in a day.
S can complete the work in 10 days, so S can do 1/10 of the work in a day. T can complete the work in 30 days, so T can do 1/30 of the work in a day.
When they work together, they can do (1/10 + 1/30) = 4/30 = 2/15 of the work in a day.
Let's say the total work takes D days. We know that S leaves 3 days before the completion of the work, so S works for (D-3) days.
So, the work done by S is (D-3) * (1/10).
Since T works for all D days, the work done by T is D * (1/30).
The total work done (which is 1, since they complete the work) is the sum of the work done by S and T, so we have the equation:
(D-3) * (1/10) + D * (1/30) = 1
Solving this equation for D gives us D = 11.25 days.
So, the work will be completed in 11.25 days.
However, since S leaves 3 days before the completion of the work, the remaining work after S leaves will be done in 3 days.
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