A drink vendor has 368 litres of Maaza, 80 litres of Pepsi, and 144 litres of Sprite.\nHe wants to pack these drinks in cans so that:\n• each can contains the same whole number of litres of a single drink, and\n• no can contains a mixture of two drinks.\nWhat is the least number of cans required?

Introduction:
This question uses the concepts of highest common factor (H.C.F.) and divisibility in a real-life context. It checks whether you can identify a common capacity for containers such that the total number of containers used is minimised while volumes remain whole numbers.

Given Data / Assumptions:

• Maaza: 368 litres.
• Pepsi: 80 litres.
• Sprite: 144 litres.
• All cans must have the same capacity in litres.
• Each can is filled with only one kind of drink.
• We want the least possible number of cans.

Concept / Approach:
To minimise the total number of cans, we need each can to have the greatest possible capacity that divides evenly into each of the three drink quantities. This capacity is the H.C.F. of 368, 80, and 144. Once we know the capacity, the number of cans for each drink is volume divided by capacity, and the total is the sum of these counts.

Step-by-Step Solution:
Step 1: Find HCF(368, 80, 144). Step 2: HCF(368, 80) = 16. Step 3: HCF(16, 144) = 16. Step 4: Therefore, maximum can capacity = 16 litres. Step 5: Number of cans for Maaza = 368 / 16 = 23. Step 6: Number of cans for Pepsi = 80 / 16 = 5. Step 7: Number of cans for Sprite = 144 / 16 = 9. Step 8: Total cans required = 23 + 5 + 9 = 37.

Verification / Alternative check:
If we used a smaller common divisor, such as 8 litres per can, the number of cans would be: 368 / 8 = 46, 80 / 8 = 10, 144 / 8 = 18; total = 74 cans. This is clearly more than 37, so 16 litres per can is better. Because 16 is the H.C.F., no larger capacity works for all three volumes, so 37 is indeed the minimum number of cans.

Why Other Options Are Wrong:
47, 46, 41, 35: These totals would correspond to either smaller can sizes or non-integer volumes per can, violating the conditions. They do not arise from using the true H.C.F. as the capacity.

Common Pitfalls:
Learners sometimes choose the least common multiple or mistakenly use any common divisor instead of the greatest one. Others forget that the can capacity must divide all three quantities exactly, leading to fractional cans, which are not allowed.

Final Answer:
The least number of cans required is 37.