A) \[4\frac{2}{7}\text{days}\]
B) 30 days
C) \[8\frac{4}{7}\text{days}\]
D) \[\frac{7}{10}\text{days}\]
Correct Answer: C
Solution :
(A + B)'s day work \[=\frac{1}{20}\] |
(B + C)'s 1 day work \[=\frac{1}{10}\] |
(C + A)'s 1 day work \[=\frac{1}{12}\] |
2 (A + 5 + C)'s 1 day work \[=\frac{1}{20}+\frac{1}{10}+\frac{1}{12}\] |
\[=\frac{3+6+5}{60}=\frac{14}{60}=\frac{7}{30}\] |
\[\therefore \] (A+B+C)'s 1 day work\[=\frac{7}{30}\times \frac{1}{2}=\frac{7}{60}\] |
Hence, A, B and C can finish the work together in\[\frac{60}{7}\]days or \[8\frac{4}{7}\]days. |
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