Time and Work Practice Test 2 with solutions

15 men's 10 days' work = 10 × $\frac{1}{210}$ = $\frac{1}{21}$
Remaining work = 1 - $\frac{1}{21}$ = $\frac{20}{21}$
Now, (15 men + 15 men having double efficiency)'s 1 day's work = $\frac{1}{210}+\frac{2}{210}$
= $\frac{3}{210}$ = $\frac{1}{70}$
Hence, number of days = $\frac{\frac{20}{21}}{\frac{1}{70}}$ = $\frac{20}{21}\times 70$ = $\frac{200}{3}$ = $66\frac{2}{3}$ days
So, required number of total days = 10 + $66\frac{2}{3}$ = $76\frac{2}{3}$ days