A worker can complete a piece of work in 50 d | vedclass

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(C) Let the total work be $50$ units.Since one worker completes the work in $50$ days,the efficiency of one worker is $1$ unit per day.On day $1$,the

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$10$

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(C) Let the total work be $50$ units.Since one worker completes the work in $50$ days,the efficiency of one worker is $1$ unit per day.On day $1$,the number of workers is $1$,so work done = $1$ unit.On day $2$,the number of workers is $2$,so work done = $2$ units.On day $3$,the number of workers is $3$,so work done = $3$ units.This follows an arithmetic progression where the work done on day $n$ is $n$ units.The total work done in $n$ days is the sum of the first $n$ natural numbers: $S_n = \frac{n(n+1)}{2}$.We need to find $n$ such that $S_n \ge 50$.$\frac{n(n+1)}{2} \ge 50 \implies n(n+1) \ge 100$.For $n = 9$,$9 	imes 10 = 90 For $n = 10$,$10 	imes 11 = 110 \ge 100$.Thus,the work will be completed on the $10^{th}$ day.

$A$ worker can complete a piece of work in $50$ days. On the first day,he works alone. On the second day,one more person joins him,on the third day,one more person joins them,and this process continues until the work is completed. How many days are needed to complete the work?

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