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Question

Two workers, working together, completed the job in $$5$$ days. If one worker worked twice as fast and the other worked half as fast, they would have to work for $$4$$ days. How much time would it take the first worker to do the job if he worked alone.?


Solution

Let the two workers finish the work in $$x$$ hours & $$y$$ hours.
$$\dfrac { 1 }{ x } +\dfrac { 1 }{ y } =\dfrac { 1 }{ 5 } \quad \longrightarrow (I)$$
If $$x$$ is $$2x$$
& $$y$$ is $$\dfrac{y}{2}$$.

$$\therefore \quad \dfrac { 1 }{ 2x } +\dfrac { 2 }{ y } =\dfrac { 1 }{ 4 } \quad \longrightarrow (II)$$
Solving (I) & (II)
$$5(x+y)=xy$$
$$\underline { 4\left( y+4x \right) =2xy } $$
$$5x+5y=xy$$  $$\times 2$$
$$\underline { 8x+2y=xy\quad \times 5 } $$
$$10x+10y=2xy$$
$$\underline { 40x+10y=5xy } $$
$$-30x=-3xy$$

      $$y=10$$ hours
      $$\underline { x=10\quad hours } $$

Mathematics

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