A husband and wife, started painting their house, but husband left painting 5 days before the completion of the work. How many days will it take to complete the work, which the husband alone would have completed in 20 days and wife in 15 days?
To solve this problem, At first I have evaluated the portion of the work done by husband.
Which is: $15 \times \frac{1}{20} =\frac{3}{4}$ as I thought total work is $1$.
Now, work done = time required $\times$ work rate.
Where the wrong, I have done?
now for the wife, she will do the portion of the work, $(1 - \frac{3}{4} = \frac{1}{4})$
and therefore, $$1/4 = time \times \frac1{15}$$ then the time is = $\frac{15}{4}$
Answer
Which is: $15 \times 120 = \frac{3}{4}$ as I thought total work is 1.
I think this is wrong because they both work together, they do $(\frac{1}{15} + \frac{1}{20}) = \frac{7}{60}$ of the whole in a day. So, if husband leaves $5$ days early, work left = $5 \times \frac{7}{60} = \frac{7}{12}$. instead of $\frac{1}{4}$ that you used.
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