A) 10
B) 14
C) 19
D) 41
Correct Answer: D
Solution :
1 man?s 1 day work \[=\frac{1}{3}\] 1 women?s 1 day?s work \[=\frac{1}{4}\] 1 boys 1 day?s work\[=\frac{1}{12}\] 1 man?s\[\frac{1}{4}\] day?s work\[=\left( \frac{1}{3}\times \frac{1}{4} \right)=\frac{1}{2}\] 1 woman?s \[\frac{1}{4}\]day?s work\[=\left( \frac{1}{4}\times \frac{1}{4} \right)=\frac{1}{16}\] 1 boy?s\[\frac{1}{4}\] day?s work\[=\left( \frac{1}{12}\times \frac{1}{4} \right)=\frac{1}{48}\] Let (1 man+1 women + x boy)?s work = 1 \[\Rightarrow \,\] \[\frac{1}{12}+\frac{1}{16}+\frac{x}{48}=1\] \[\Rightarrow \,\] \[7+x=48\] \[\Rightarrow \,\] \[x=41\] \[\therefore \]41 boy must assist 1 man and 1 women.
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