A drink vendor has 80 liters of Mazza, 144 liters of Pepsi, and 368 liters of Sprite. He wants to pack them in cans such that: • Each can has the same capacity (in liters), • No can contains a mix of two different drinks, and • All the drink is packed completely with no leftover. What is the least number of cans required?

Introduction / Context:
This question is very similar in structure to another HCF-based packing problem. To minimize the number of cans, the capacity of each can should be the largest possible value that divides each of the three drink quantities exactly. Once that capacity is known, we count how many cans are needed for each drink and add them together.

Given Data / Assumptions:

• Mazza = 80 liters
• Pepsi = 144 liters
• Sprite = 368 liters
• All cans have equal capacity
• No mixing of drinks in the same can, and no leftovers

Concept / Approach:
If the can capacity is x liters, then x must divide 80, 144, and 368. The maximum such x is the HCF of the three volumes:
x = HCF(80, 144, 368)
The least number of cans is then computed by dividing each volume by x and summing the results.

Step-by-Step Solution:
Step 1: Compute gcd(80, 144).144 mod 80 = 64.80 mod 64 = 16.64 mod 16 = 0, so gcd(80, 144) = 16.Step 2: Compute gcd(16, 368).368 ÷ 16 = 23 exactly, so 368 mod 16 = 0.Therefore HCF(80, 144, 368) = 16, so can capacity x = 16 liters.Step 3: Number of cans of each drink:Mazza: 80 / 16 = 5 cans.Pepsi: 144 / 16 = 9 cans.Sprite: 368 / 16 = 23 cans.Step 4: Total cans = 5 + 9 + 23 = 37.

Verification / Alternative check:
Using a smaller can size that also divides all three volumes (for example, 8 liters) yields more cans: 80/8 + 144/8 + 368/8 = 10 + 18 + 46 = 74, which is larger than 37. Since 16 is the greatest common divisor, it gives the minimum possible number of cans.

Why Other Options Are Wrong:
35, 36, 38, 46: These totals do not correspond to the distribution obtained when using the maximum possible can capacity of 16 liters. Any different total implies a smaller can size or an invalid capacity that does not divide all three volumes exactly.

Common Pitfalls:
Using LCM instead of HCF, which is not appropriate for this type of packing problem.Choosing a can size that divides some but not all of the drink volumes.Forgetting to add the cans needed for each drink after determining the capacity.

Final Answer:
37