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Question

Suppose $$A$$ can do a piece of work in $$14$$ days while $$ B$$ can do it in $$21$$ days. They begin together and worked at it for $$6$$ days. Then $$A$$ fell ill and $$B$$ had to complete the work alone. In how many days was the work completed?


Solution

$$A$$ can do a piece of work in $$14$$ days.
Therefore $$A's$$ one days work will be $$\dfrac{1}{14}$$
$$B$$ can do a piece of work in $$21$$ days.
Therefore $$B's$$ one days work will be $$\dfrac{1}{21}$$
If $$A$$ and $$B$$ together work then their one days work will be
$$\Rightarrow \dfrac{1}{A}+\dfrac{1}{B}=\dfrac{1}{14}+\dfrac{1}{21}$$

$$\Rightarrow \dfrac{1}{A}+\dfrac{1}{B}=\dfrac{3+2}{42}$$ taking LCM

$$\Rightarrow \dfrac{1}{A}+\dfrac{1}{B}=\dfrac{5}{42}$$

$$A$$ and $$B$$ worked for $$6$$ days
Therefore their $$6$$ days work is $$\dfrac{5}{42}\times 6=\dfrac{5}{7}$$
Amount of work completed is $$\dfrac{5}{7}$$
Work to be completed is $$1-\dfrac{5}{7}=\dfrac{7-5}{7}=\dfrac{2}{7}$$
Therefore $$\dfrac{2}{7}$$ work has to be completed by $$B$$ only.
$$=\dfrac{2}{7}\times 21$$
$$=2\times 3=6\ days$$
Work will be completed in $$6$$ days


Mathematics

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