X complete a piece of work in $20$ days and Y complete it in $30$ days and Z complete it in $60$ days. At first day to every third day Y and every fourth day Z helped X. At how many days will complete the work?

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Solution

X's one day work = $\frac{1}{20}$
Y's one day work = $\frac{1}{30}$
Z's one day work = $\frac{1}{60}$
Work done by X in first two days = $\frac{2}{20} = \frac{1}{10} t h$ part
Work done by X and Y on third day = $\frac{1}{20} + \frac{1}{30} = \frac{1}{12} t h$ part
Work done by Z and X on fourth day = $\frac{1}{20} + \frac{1}{60} = \frac{1}{15} t h$ part.
Work Done in one time unit (one time unit = 4 days)
$= \frac{1}{10} + \frac{1}{12} + \frac{1}{15} = \frac{1}{4}$
As $\frac{1}{4} t h$ part of work has been finished in $1$ unit of time then to complete work they need $4$ unit of time.
$∴ 4$ unit of time = $4 \times 4 = 16$ days

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X complete a piece of work in 20 days and Y complete it in 30 days and Z complete it in 60 days. At first day to every third day Y and every fourth day Z helped X. At how many days will complete the work?

X complete a piece of work in $20$ days and Y complete it in $30$ days and Z complete it in $60$ days. At first day to every third day Y and every fourth day Z helped X. At how many days will complete the work?