anurag356 wrote:

A person X working alone can complete a work in 10 days. A person Y completes the same amount of work in 20 days and a person Z when working alone can complete the same amount of work in 30 days. How many more days are required if all the people started working together but X and Y leave the work after 2 days ?

a) 19

b) 20

c) 21

d) 22

e) 25

Alternate Nomenclature (but similar in way to Veritas' explanation)Say, total unit of work = 1

X, working alone, can complete in 1 day = \(\frac{1}{10}\) amount of work

Y, working alone, can complete in 1 day = \(\frac{1}{20}\) amount of work

Z, working alone, can complete in 1 day = \(\frac{1}{30}\)amount of work

Therefore, all working together, in 1 day they complete = [\(\frac{1}{10}\) + \(\frac{1}{20}\)\(\) + \(\frac{1}{30}\)] amount of work.

For 2 days, all three of them worked together, therefore total work completed in 2 days = 2 * [\(\frac{1}{10}\)\(\) + \(\frac{1}{20}\) + \(\frac{1}{30}\)]

Let us suppose that after the 2nd day (i.e 3rd day onwards), when X and Y have left, Z takes 'd' number of days to complete the remaining work.

The final equation we get is:

\(2 * [\frac{1}{10} + \frac{1}{20} + \frac{1}{30}] + d * [\frac{1}{30}] = 1\) ----------------

(work completed + remaining work = total work)Solving the equation, we get:

\(\frac{11}{30} + \frac{d}{30} = 1\)

d = 19 days ----- Answer.