A) \[7\frac{7}{20}days\]
B) \[8\frac{8}{21}days\]
C) \[8\frac{9}{20}days\]
D) \[6\frac{6}{17}days\]
E) None of these
Correct Answer: B
Solution :
Work done by A and C in 6 days \[=6\times \left( \frac{1}{36}+\frac{1}{42} \right)=\frac{13}{42}\] Work done by A and C in last 10 days \[=10\times \left( \frac{1}{36}+\frac{1}{42} \right)=\frac{65}{126}\] \[\therefore \]Remaining work which is done by B \[=1-\left( \frac{13}{42}+\frac{65}{126} \right)=\frac{11}{63}\] Time taken by B on the work \[=\frac{11}{63}\times 48=8\frac{8}{21}days\] Method II. LCM of 36, 48 and 42 = 1008 units A can do \[\frac{1008}{36}=28\] units/dat B can do \[\frac{1008}{48}=21\]units/day C can do \[\frac{1008}{42}=24\]units/day Now, in (6+10) days (A+C) can do \[=16\times 52=832\,\,units\] \[\therefore \] Remaining units done by B = 1008 - 832 = 176 \[\therefore \] B?s work for \[\frac{176}{21}=8\frac{8}{21}days\]You need to login to perform this action.
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