Question

3 men, 5 women and 7 children finish a job in 40 days; 6 men, 10 women and 18 children finish it in 18 days. In how many days can 9 men, 15 women and 5 children finish the same job?

• A. \(17\tfrac{17}{18}\)
• B. \(18\tfrac{18}{19}\)
• C. \(18\tfrac{17}{18}\)
• D. \(19\tfrac{18}{19}\)

Hint:

Look for a way to combine the two work-rate equations by scaling one of them so that the man and woman terms cancel out directly.

Answer 1

The Correct Option is B

Step 1: Let daily work rates be $m,w,c$ for a man, woman and child. Given $3m+5w+7c=\frac{1}{40}$ ...(i) and $6m+10w+18c=\frac{1}{18}$ ...(ii).
Step 2: Multiply (i) by 2: $6m+10w+14c=\frac{1}{20}$.
Step 3: Subtract this from (ii): $(6m+10w+18c)-(6m+10w+14c)=4c=\frac{1}{18}-\frac{1}{20}=\frac{10-9}{180}=\frac{1}{180}$, so $c=\frac{1}{720}$.
Step 4: From (i): $3m+5w=\frac{1}{40}-7c=\frac{18}{720}-\frac{7}{720}=\frac{11}{720}$.
Step 5: Required daily rate for 9 men, 15 women, 5 children: $9m+15w+5c=3(3m+5w)+5c=3\left(\frac{11}{720}\right)+\frac{5}{720}=\frac{33+5}{720}=\frac{38}{720}=\frac{19}{360}$.
Step 6: Time taken $=\dfrac{360}{19}=18\tfrac{18}{19}$ days.
\[\boxed{18\tfrac{18}{19}\text{ days}}\]