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(A) Let the time taken by $Y$ to complete the work be $T_Y$ days.Since $X$ is three times as fast as $Y$,the time taken by $X$ $(T_X)$ will be $\frac{

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(A) Let the time taken by $Y$ to complete the work be $T_Y$ days.Since $X$ is three times as fast as $Y$,the time taken by $X$ $(T_X)$ will be $\frac{1}{3}$ of the time taken by $Y$,so $T_X = \frac{T_Y}{3}$.Given that $X$ completes the work $40 	ext{ days}$ faster than $Y$,we have $T_Y - T_X = 40$.Substituting $T_X = \frac{T_Y}{3}$,we get $T_Y - \frac{T_Y}{3} = 40$.$\frac{2T_Y}{3} = 40 \implies T_Y = 60 	ext{ days}$.Therefore,$T_X = \frac{60}{3} = 20 	ext{ days}$.Work done by $X$ in $1 	ext{ day} = \frac{1}{20}$.Work done by $Y$ in $1 	ext{ day} = \frac{1}{60}$.Work done by $(X+Y)$ in $1 	ext{ day} = \frac{1}{20} + \frac{1}{60} = \frac{3+1}{60} = \frac{4}{60} = \frac{1}{15}$.Thus,$X$ and $Y$ together will complete the work in $15 	ext{ days}$.

$X$ is three times as fast as $Y$ and is able to complete the work in $40 \text{ days}$ less than $Y$. Then the time in which they can complete the work together is (in $\text{days}$)

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