If 6 engines consume 15 metric tones of coal when each is running 9 hours a day, how much coal will be required for 8 engines, each running 12 hours a day, if being given that 3 engines of former type consume as much as 4 engines of latter type?

None of these

Answer : Option A

Explanation :

3e = 4E

${ m{e}} = frac{4}{3}{ m{E}}$

$frac{{{{ m{M}}_1}{{ m{H}}_1}}}{{{{ m{W}}_2}}} = frac{{{{ m{M}}_2}{{ m{H}}_2}}}{{{{ m{W}}_2}}}$

$frac{{8 imes 9}}{{15}} = frac{{8 imes 12}}{{{W_2}}}$

W₂ = 20

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