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X and Y can d...

Question

X and Y can do a piece of work in 20 days and 12 days respectively. X started the work alone and then after 4 days Y joined him till the completion of the work. How long did the work last?

6 days

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10 days

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15 days

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20 days

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16 days

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Solution

The correct option is B
10 days

Work done by x in 4 days $= (\frac{1}{20} \times 4) = \frac{1}{5}$

Remaining work $= (1 - \frac{1}{5}) = \frac{4}{5}$

(X+Y)’s 1 day’s work = $(\frac{1}{20} + \frac{1}{12}) = \frac{8}{60} = \frac{2}{15}$

So, $\frac{4}{5}$ work will be done by X and Y in $(\frac{15}{2} \times \frac{4}{5})$ = 6 days

Hence, total time taken $= (6 + 4) d a y s = 10 d a y s$

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$A$ can do a piece of work in $25$ days, which $B$ alone can do in $20$ days. A started the work and was joined by $B$ after $10$ days. The work lasted for
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X and Y together can complete a work in 6 days. X alone can complete it in 10 days. If Y does the work only for half a day daily, then in how many days will X and Y together complete the work?

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X and Y can do a piece of work in 20 days and 12 days respectively. X started the work alone and then after 4 days Y joined him till the completion of the work. How long did the work last?

The correct option is B 10 days Work done by x in 4 days =(120×4)=15 Remaining work =(1−15)=45 (X+Y)’s 1 day’s work = (120+112)=860=215 So,45 work will be done by X and Y in (152×45) = 6 days Hence, total time taken =(6+4) days=10 days