A) \[3\frac{1}{7}\]days
B) \[3\frac{3}{7}\,\,days\]
C) \[5\frac{5}{47}\,\,days\]
D) \[4\frac{4}{9}\,\,days\]
E) \[3\frac{3}{17}\,\,days\]
Correct Answer: C
Solution :
According to the question, LCM of 8, 6 and 10 - 120 \[\therefore \] (A + B) can. do \[\left( \frac{820}{8} \right)\] 15 units/day Similarly, (B + C) can do \[\left( \frac{120}{6} \right)\]20 units/ day and (C + A) can do\[\frac{120}{10}\]= 12 units day Total efficiency = 15 + 20 + 12 \[\frac{47}{2}\] (A + B + C) together will complete the whole work in \[\frac{120\times 2}{47}=\frac{240}{47}=5\frac{5}{47}\] days Another method: (A + B) can do the work in \[\frac{1}{8}\] days (B + C) can do the work in\[\frac{1}{6}\] days (C + A) can do the work in \[\frac{1}{10}\] days 2(A + B+ C) together do the work \[=\frac{1}{8}+\frac{1}{6}+\frac{1}{10}=\frac{15+20+12}{120}=\frac{47}{120}\] (A + B + G) can do the whole work in \[\frac{120\times 2}{47}=5\frac{5}{47}\] days
You need to login to perform this action.
You will be redirected in
3 sec
