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Question:

If 9 engines consume 12 metric ton of coal, when each is working 8 hours day, how much coal will be required for 4 engines, each running 13 hours a day, it being given that 3 engines of former type consume as much as 4 engines of latter type? (in metric tons)

Explanation:

Let 3 engines of former type consume 1 unit in 1 hour.

Then, 4 engines of latter type consume 1 unit in 1 hour.

Therefore 1 engine of former type consumes (1/3) unit in 1 hour.

1 engine of latter type consumes (1/4) unit in 1 hour.

Let the required consumption of coal be x units.

Less engines, Less coal consumed (direct proportion)

More working hours, More coal consumed (direct proportion)

Less rate of consumption, Less coal consumed (direct proportion)

⇒ Coal consumption ∝ engines × working hours/day × rate of consumption

⇒ $\frac{(C o a l c o n s u m p t i o n)}{(e n g i n e s \times w o r k i n g h o u r s p e r d a y \times r a t e o f c o n s u m p t i o n)}$ = constant

⇒ $\frac{12}{9 \times 8 \times \frac{1}{3}} = \frac{x}{4 \times 13 \times \frac{1}{4}}$

⇒ x = 26

Hence, the required consumption of coal = 26/4 = 6.5 metric ton.

Hence, 6.5.

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