Three liquids of 403 L petrol, 456 L diesel, and 496 L motor oil are to be filled into bottles of equal capacity, with each bottle completely full and no mixing across types. What is the least possible number of bottles required?

Introduction: Packing different liquids into equal-sized bottles without mixing requires that bottle capacity be a common divisor of each total volume. To minimize the number of bottles, we maximize the bottle capacity, which equals the highest common factor of the volumes.

Given Data / Assumptions:

• Volumes: 403 L, 456 L, 496 L
• Equal bottle capacity
• Each bottle filled completely, no mixing of liquids

Concept / Approach: Bottle size = HCF(403, 456, 496). Total bottles = 403 / size + 456 / size + 496 / size.

Step-by-Step Solution:

Verification / Alternative check: Since the greatest common divisor is 1, no larger equal capacity is possible that divides all three volumes exactly. Therefore 1 L is forced, giving 1355 bottles.

Why Other Options Are Wrong: 34, 44, 46, and 54 do not equal the computed minimum 1355, so the correct choice is the catchall None of these.

Common Pitfalls: Attempting to use pairwise common divisors that do not divide the third volume, which would break the equal capacity rule.

Final Answer: None of these