Engine consumption comparison: 9 engines of type A consume 24 tonnes of coal when each runs 8 h/day. If 3 engines of type A consume as much as 4 engines of type B, how much coal is required for 8 engines of type B each running 13 h/day?

Introduction / Context:
This question involves two engine types with a stated consumption equivalence. By first finding consumption per engine-hour for type A and then using the given equivalence to infer type B’s consumption, we can compute the required coal for type B in the new schedule.

Given Data / Assumptions:

• Type A: 9 engines * 8 h/day consume 24 tonnes ⇒ per engine-hour for A is constant
• Equivalence: 3 A-engines consume as much as 4 B-engines (same time)
• Target: 8 B-engines running 13 h/day; find total coal

Concept / Approach:
Let a be tonnes per engine-hour for A; then 9*8*a = 24 ⇒ a = 1/3. From 3A ≡ 4B, per engine-hour for B is b = (3/4)*a = 1/4. Then compute 8*13*b for the required total consumption.

Step-by-Step Solution:

Verification / Alternative check:
Equivalent A-engines for 8 B-engines: 4B ≡ 3A ⇒ 8B ≡ 6A. Consumption via A-rate: 6 * 13 * (1/3) = 26 tonnes—same result, confirming consistency.

Why Other Options Are Wrong:

• 20, 23: underestimate consumption given the derived 1/4 tonne per engine-hour for B.
• 39: overestimates; requires higher per-hour usage than implied by the equivalence.
• 24: still under the computed requirement of 26.

Common Pitfalls:
Applying 3A = 4B backwards (treating B as more consumptive), or forgetting to convert to per engine-hour before scaling.

Final Answer:
26 metric tonnes