What is the minimum number of times one needs to measure to get 298 litres of water from a tank, if the measuring cylinders have capacities 1 litre, 6 litres, 25 litres and 100 litres?

This problem involves mathematical optimization and is fundamentally related to the classic Water Jug Problem, solvable via Linear Diophantine equations. In such puzzles, measuring allows both additive (drawing from the tank) and subtractive (pouring back) operations.

Why Option B is correct: To minimize the number of measurements, we must minimize the total sum of operations (|a| + |b| + |c| + |d|) in the equation 100a + 25b + 6c + 1d = 298. Instead of relying strictly on addition, the mathematically optimal strategy is to overshoot the target and subtract the difference:

1. Draw three 100-litre measures from the tank: 3 × 100L = 300L (3 operations).
2. Pour back two 1-litre measures into the tank: 300L - (2 × 1L) = 298L (2 operations). Total measurements = 3 + 2 = 5. Thus, 5 is the true mathematical minimum.

Why the other options are incorrect:

• Option A (4): Reaching 298L in exactly 4 steps would require an impossible combination, such as drawing three 100L measures and subtracting one 2L measure (3 × 100L - 1 × 2L). Since no 2L cylinder is provided, this cannot be done.
• Option C (9): This represents a mathematically suboptimal measurement sequence rather than the required minimum.
• Option D (13): This represents the traditional 'Greedy Algorithm' or 'Coin Change' approach using strictly additive measurements (2×100L + 3×25L + 3×6L + 5×1L = 13). It is incorrect because it fails to account for the efficiency of subtractive operations.

Takeaway: In aptitude-based capacity and measurement puzzles, always account for both forward (addition) and backward (subtraction) operations. Overshooting your target and stepping backward is often the key to minimizing the total number of steps.