Question : A man does double the work done by a boy at the same time. The number of days that 3 men and 4 boys will take to finish a work that can be done by 10 men in 8 days is:
Option 1: $4$
Option 2: $16$
Option 3: $7\frac{3}{11}$
Option 4: $8\frac{4}{5}$
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Correct Answer: $16$
Solution :
Let the efficiency of 1 man and 1 boy be m and b respectively.
m = 2b
m : b = 2 : 1
We know that, work = efficiency × time
So, work done by 10 men in 8 days = 2 × 10 × 8 = 160
Total efficiency of 3 men and 4 boys = (3 × 2) + (4 × 1) = 10
$\therefore$ Days required = $\frac{\text{Total work}}{\text{Efficiency}}$ = $\frac{160}{10}$ = 16
Hence, the correct answer is $16$.
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