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Two coal loading machines, each working 12 hours per day for 8 days, handle 9000 tonnes of coal with an efficiency of 90%; while 3 other coal loading machines at an efficiency of 80% are set to handle 12000 tonnes of coal in 6 days. How many hours per day should each machine
work?
Using, $$\frac{M_1D_1H_1}{W_1}$$ $$=\frac{M_2D_2H_2}{W_2}$$
Let $$t$$ hours per day each machine should work
=> $$\frac{2\times8\times12}{9000}\times90=\frac{3\times6\times t}{12000}\times80$$
=> $$16\times12=12t$$
=> $$t=16$$
$$\therefore$$ The machine should work 16 h/day
=> Ans - (B)
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