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?
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?
Solution
Let's solve this step by step:
Step 1: Find the one day's work of S and T. S's one day's work = 1/10 T's one day's work = 1/30
Step 2: Find their combined one day's work. S and T's one day's work together = 1/10 + 1/30 = 4/30 = 2/15
Step 3: Find out how much work is done by S and T together in the days before S leaves. S leaves 3 days before the work is completed, so they work together for (10-3) = 7 days. Work done in 7 days = 7 * 2/15 = 14/15
Step 4: Find out how much work is left. Remaining work = 1 (total work) - 14/15 = 1/15
Step 5: Find out how many days T will take to complete the remaining work. Time taken by T to complete the remaining work = Remaining work / T's one day's work = 1/15 / 1/30 = 2 days
So, the remaining work will be done in 2 days.
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