Six engines of a certain (former) type consume 24 metric tonnes of coal when each engine works for 8 hours per day. If two engines of this former type consume as much coal as three engines of a latter type, how many metric tonnes of coal will be required for 9 engines of the latter type, each running for 13 hours?

Introduction / Context:
This question is a variation of a time and work type problem but with machines and fuel consumption instead of human workers and jobs. We are given how much coal a group of engines of one type consumes in a certain number of hours, and a relationship between the consumption of engines of two different types. The task is to determine the coal needed for a different number of engines of the second type working for a different number of hours.

Given Data / Assumptions:

• Six engines of the former type consume 24 metric tonnes of coal when each works for 8 hours.
• Two engines of the former type consume as much as three engines of the latter type for the same time.
• We want coal required for 9 engines of the latter type, each running 13 hours.
• Consumption is directly proportional to the number of engines and to the hours they work.
• All engines of a given type have identical consumption rates.

Concept / Approach:
First, we find the coal consumption per engine per hour for the former type. Then we use the equivalence condition two former engines equal three latter engines to obtain the per engine per hour consumption for the latter type. Once we know that, we multiply by the number of latter type engines and the hours they run to find the total coal consumption required.

Step-by-Step Solution:
Total coal consumed by six former type engines in 8 hours = 24 metric tonnes.Total engine hours for former type = 6 engines * 8 hours = 48 engine hours.Consumption per engine hour for former type = 24 / 48 = 0.5 metric tonnes per engine hour.We are told that 2 former type engines consume as much coal as 3 latter type engines for the same time.Let cF be former per engine hour, cL be latter per engine hour. Then 2 * cF = 3 * cL.Substitute cF = 0.5: 2 * 0.5 = 3 * cL, so 1 = 3 * cL.Therefore, cL = 1 / 3 metric tonne per engine hour for the latter type.Now compute usage for 9 latter type engines running 13 hours.Total engine hours for latter type = 9 * 13 = 117 engine hours.Total coal required = 117 * (1 / 3) = 39 metric tonnes.

Verification / Alternative check:
We can cross check the equivalence condition. If cL = 1 / 3, then three latter engines in 1 hour use 3 * 1 / 3 = 1 metric tonne. Two former engines in 1 hour use 2 * 0.5 = 1 metric tonne. This exactly matches the given relationship. Thus the derived latter type rate is consistent, and multiplying by 117 engine hours to get 39 metric tonnes is correct.

Why Other Options Are Wrong:
Values like 45, 34 or 55 metric tonnes are not proportional to the correct per engine hour consumption and total engine hours. For example, 45 tonnes would correspond to a larger per engine hour rate than 1 / 3 and would violate the given equivalence between the two types of engines. Thirty four tonnes is too small and would imply a latter type engine is more efficient than implied by the relation. Only 39 metric tonnes satisfies all the constraints.

Common Pitfalls:
Some learners treat the 24 metric tonnes as a daily figure without computing per engine hour consumption. Others misinterpret the equivalence condition and set 2 former engines equal to 3 latter engines with inverse ratios. It is essential to first reduce everything to per engine hour consumption and then carefully apply proportional reasoning to the new situation.

Final Answer:
The coal required for 9 engines of the latter type running for 13 hours is 39 metric tonnes.