A and B together can do a piece of work in 30 | vedclass

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Question

(A) Let the total work be the $LCM$ of $30$ and $20$,which is $60$ units.Efficiency of $(A + B) = \frac{60}{30} = 2$ units/day.Efficiency of $(B + C)

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(A) Let the total work be the $LCM$ of $30$ and $20$,which is $60$ units.Efficiency of $(A + B) = \frac{60}{30} = 2$ units/day.Efficiency of $(B + C) = \frac{60}{20} = 3$ units/day.Let $A, B, C$ be the efficiencies of $A, B, C$ respectively.$A + B = 2$ $(i)$$B + C = 3$ (ii)Total work done = $5A + 15B + 18C = 60$.We can rewrite this as $5(A + B) + 10B + 18C = 60$.Substitute $(A + B) = 2$:$5(2) + 10B + 18C = 60 \Rightarrow 10 + 10B + 18C = 60 \Rightarrow 10B + 18C = 50 \Rightarrow 5B + 9C = 25$ (iii).From (ii),$B = 3 - C$. Substitute this into (iii):$5(3 - C) + 9C = 25 \Rightarrow 15 - 5C + 9C = 25 \Rightarrow 4C = 10 \Rightarrow C = 2.5$ units/day.Time taken by $C$ alone = $\frac{	ext{Total Work}}{	ext{Efficiency of } C} = \frac{60}{2.5} = 24$ days.

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