Explanation
In 12 hours, they are at right angles 22 times.$$\therefore $$In 24 hours, they are at right angles 44 times.
Option D is correct.
$$A$$ can finish a work in $$24$$ days, $$B$$ in $$9$$ days and $$C$$ in $$12$$ days. $$B$$ and $$C$$ start the work but are forced to leave after $$3$$ days. The remaining work was done by $$A$$ in:
Work efficiency of $$34$$ men in $$8$$ days working $$9$$ hours a day is $$\dfrac{2}{5}$$th
Hence, work efficiency of $$x$$ men to finish rest of the work in $$6$$ days working $$9$$ hours a day = $$1 - \dfrac{2}{5} = \dfrac{3}{5}$$
$$\therefore$$ By direct proportionality,
$$\dfrac{34\times8\times9}{\dfrac{2}{5}} =\dfrac{x\times6\times9}{ \dfrac{3}{5}}$$
$$\therefore 34\times72\times\dfrac{3}{5} = x\times54\times\dfrac{2}{5}$$
$$\therefore x = \dfrac{34\times72\times3\times5}{54\times2\times5}$$
$$\therefore x = 68$$
Hence extra men required = $$68 - 34 = 34$$men
So, the answer is option C
(A+B+C)'s one days's work$$=\cfrac{1}{10}$$
So (A+B+C)'s 4 day's work$$=\cfrac{4}{10}=\cfrac{2}{5}$$
Remaining part of the work$$=1-\cfrac{2}{5}=\cfrac{3}{5}$$
$$\because \cfrac{3}{5}$$part of the work done by (B+C)= $$10$$ days
$$\therefore $$ 1 work done by (B+C)$$=\cfrac{10}{\cfrac{3}{5}}=\cfrac{50}{3} Days$$
$$\therefore$$A alone work$$=\cfrac{1}{10}-\cfrac{3}{50}=\cfrac{1}{25}$$
Hence A alone complete the work in $$25$$ days.
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