A) \[3\frac{5}{22}\]days
B) \[5\frac{7}{21}\]days
C) \[3\frac{7}{24}\]days
D) \[4\frac{21}{22}\]days
E) \[6\frac{2}{27}\]days
Correct Answer: D
Solution :
\[(10\times 20)M=(15\times 12)W\] \[=(22\times 16)C\] or,\[200M-180W=352C\] or,\[50M=45W=88C\] Work done by \[9\] women and \[14\] children in \[7\] days\[=(9W+14C)\times 7\] \[=\left( \frac{10}{9}\times 9M+\frac{25}{44}\times 14M \right)\times 7\] \[=\left( 10M+\frac{175}{22}M \right)\times 7=\left( \frac{220M+175}{22} \right)\times 7\] \[=\frac{395}{22}M\times 7\] \[\therefore \]Remaining work\[=(10\times 20)M-\frac{2765}{22}M\] \[=\frac{(4400-2765)M}{22}=\frac{1635}{22}M\] \[\therefore \]15 men can complete it in\[\frac{1635}{22\times 15}\] \[=\frac{1635}{330}=\frac{109}{22}days=4\frac{21}{22}days\] Another Method: \[15W-12days-1work\] \[\Rightarrow \]\[9W-7days-\frac{9\times 7\times 1}{15\times 12}=\frac{7}{20}\](Using\[{{M}_{1}}{{D}_{1}}{{W}_{2}}={{M}_{2}}{{D}_{2}}{{W}_{1}})\] 14 children - 7 days\[-\frac{14\times 7\times 1}{22\times 16}\] Remaining work \[=1-\left( \frac{7}{20}+\frac{49}{176} \right)=1-\frac{553}{880}=\frac{327}{880}\] \[\therefore \] \[15M\] do the remaining work in\[\frac{10\times 20}{15}\times \frac{327}{880}\] \[=\frac{109}{22}=4\frac{21}{22}days\]
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