Question

If 5 engines consume 6 metric tonnes of coal when each is running 9 hours a day, how much coal (in metric tonnes) will be needed for 8 engines, each running 10 hours a day, it being given that 3 engines of the former type consume as much as 4 engines of latter type?

Answer 1

Answer:

8 metric tonnes

Step-by-step explanation:

Given If 5 engines consume 6 metric tonnes of coal when each is running 9 hours a day, how much coal (in metric tonnes) will be needed for 8 engines, each running 10 hours a day, it being given that 3 engines of the former type consume as much as 4 engines of latter type?

Now first coal consumed by one engine = 6/5 x 9 = 2/15 m tonnes per hour

3 engines of first consume 4 engines of second one.

So 2 = 3/4 x 2/ 15 = 1/10 metric tonnes per hour.

Coal consumed by 8 engines of second one running 10 hours a day is 1/10 x 8 x 10 = 8 metric tonnes.

                 OR

We can do it as

3 former type requires 4 engines

1 former type requires 4/3

So 5 former type requires 4/3 x 5

So

6 / 4/3 x 5 x 9 = X / 8 x 10

x/2 x 2 = 2

or x = 8 m tonnes