Ratio And Proportion If 8 engines consume 12000 kgs. of coal when each running 10
hours per day, how much coal is required for 7 engines out
which 5 are running 12 hours a day and the rest 6 hours a day?​

11200 for 7 engines for a total of 80 hours

Step-by-step explanation:

You have 8 engines burning 12000 .for 10 hours.

Divide 10 hours into 12000=1200 PER hour.

1200 ÷8 engines = 140 per hour per engine.

140 × # of hours (12)=1680 for 12 hours per engine.

# of engines × 1680 for 12 hour/per engine

5×1680 = 8400 of fuel for 12 hours/5 engine

Then you have 2 that runs 10 hours

2 × 140 per hour per engine =280 × 10 =2800 for 2 engines for 10 hours

8400. + 2800= 11200 of fuel for 6 engines

what?

step-by-step explanation:

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