250 men can finish a work in 20 days working 5hours aday. To finish the with in 10 days working 8 hours a day,the minimum number of men required is
M1 D1 T1 = M2 D2 T2
given M1 = 250 men
D1 = 20 days
T1 = 5 hours
D2 = 10 days
T2 = 8 hours
M2 = ?
substituting
$$ 250 \times 20 \times 5 = M2 \times 10 \times 8 $$
solving M2 = $$ \frac{250 \times 20 \times 5}{10 \times 8} = 312.5 $$
312.5 approximately equal to 313
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