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 = 34men
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
\thereforeA alone work=\cfrac{1}{10}-\cfrac{3}{50}=\cfrac{1}{25}
Hence A alone complete the work in 25 days.
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