Milk container capacity from fractional fills and transfers A container was 4/5 full. After removing 12 bottles and then adding 8 bottles, it became 3/4 full. How many bottles can the container hold in total?

Introduction / Context:
This is a classic capacity-and-transfer problem. The container’s capacity in “bottles” is unknown, yet fractional fullness and discrete bottle moves allow determining the total capacity using algebra with fractions and integers.

Given Data / Assumptions:

• Total capacity = B bottles.
• Initial volume = (4/5) * B.
• Remove 12 bottles; add 8 bottles.
• Final volume = (3/4) * B.

Concept / Approach:
Track the number of bottles through each operation. Since “remove 12” then “add 8” is a net change of −4 bottles, write the final equality in terms of B and solve a simple linear equation in B.

Step-by-Step Solution:
Initial amount = (4/5)B.After removing 12: (4/5)B − 12.After adding 8: (4/5)B − 12 + 8 = (4/5)B − 4.This equals the final fullness: (3/4)B.Therefore (4/5)B − 4 = (3/4)B ⇒ (4/5 − 3/4)B = 4.Compute the fraction: 4/5 − 3/4 = 16/20 − 15/20 = 1/20.So (1/20)B = 4 ⇒ B = 80.

Verification / Alternative check:
4/5 of 80 is 64. Remove 12 ⇒ 52. Add 8 ⇒ 60. 3/4 of 80 is 60. Everything matches exactly.

Why Other Options Are Wrong:
40, 30, 90: Plugging any of these into the equation fails to satisfy the final fullness condition.

Common Pitfalls:
Forgetting that remove then add is a net −4; mishandling fractional subtraction (4/5 − 3/4). Always reduce to a single equation in B and solve carefully.

Final Answer:
80