B would take 10 hours more than A to complete a certain task if each of them worked alone. Working together, A and B can complete the entire task in 12 hours. How many hours would B take to complete 50% of the task if he works alone?

Introduction / Context:
This Time and Work problem involves two workers, A and B, whose individual times are related by a fixed difference. We are told that B takes 10 hours longer than A to complete a task alone, and that together they complete the task in 12 hours. We must first determine B's individual time for the full task and then find how long B alone would take to complete half of the work. This requires setting up and solving a quadratic equation in terms of A's time or B's time.

Given Data / Assumptions:

• A's time to complete the task alone = x hours (unknown).
• B's time to complete the task alone = x + 10 hours.
• Working together, A and B finish the whole task in 12 hours.
• Both A and B work at constant rates.
• Total work is considered as 1 unit.

Concept / Approach:
Since time and efficiency are inversely related, we express A's rate as 1 / x and B's rate as 1 / (x + 10). The combined rate of A and B is thus 1 / x + 1 / (x + 10). Given that they complete the task in 12 hours, their combined rate must equal 1 / 12. Solving this equation gives the value of x (A's time). From x, we obtain B's time as x + 10 hours. To find the time B takes for 50% of the work, we divide half the work by B's rate.

Step-by-Step Solution:
Let A alone take x hours. Then B alone takes x + 10 hours. A's rate = 1 / x units per hour. B's rate = 1 / (x + 10) units per hour. Combined rate = 1 / x + 1 / (x + 10). Given that together they complete the work in 12 hours, so combined rate = 1 / 12. Thus, 1 / x + 1 / (x + 10) = 1 / 12. Combine the left side: (x + 10 + x) / [x(x + 10)] = 1 / 12. (2x + 10) / [x(x + 10)] = 1 / 12. Cross-multiplying: 12(2x + 10) = x(x + 10). 24x + 120 = x^2 + 10x. Rearrange: x^2 + 10x - 24x - 120 = 0 ⇒ x^2 - 14x - 120 = 0. Solve the quadratic: x^2 - 14x - 120 = 0. This factors as (x - 20)(x + 6) = 0. Since time cannot be negative, x = 20 hours. So, A alone takes 20 hours and B alone takes 20 + 10 = 30 hours. B's rate = 1 / 30 units per hour. To complete 50% (i.e., 1 / 2) of the work, time taken by B = (1 / 2) / (1 / 30) = 15 hours.

Verification / Alternative check:
Check the combined time: A's rate is 1 / 20, B's rate is 1 / 30. Combined rate = 1 / 20 + 1 / 30 = (3 + 2) / 60 = 5 / 60 = 1 / 12. Therefore, together they take 12 hours, which matches the given condition. For half the work, B at 1 / 30 units per hour needs 15 hours, since in 15 hours he does (15 / 30) = 1 / 2 of the job. Everything is consistent.

Why Other Options Are Wrong:
Option A (30 hours) is the time B takes to finish the entire task, not just 50% of it. Option C (20 hours) or Option D (10 hours) do not correspond to half the work at B's actual rate of 1 / 30; they yield different fractions of the total work (2/3 and 1/3 respectively). Only 15 hours gives exactly half the job completed by B alone.

Common Pitfalls:
One common mistake is to assume that if together they take 12 hours and B is slower, B must automatically take 24 hours for the whole work, which is not correct because of the specific 10-hour difference from A. Another error is to forget that half the work means dividing by the rate while using 1 / 2 as the numerator, not confusing it with doubling the time incorrectly. Careful handling of the quadratic equation and the half-work calculation is essential.

Final Answer:
B would take 15 hours to complete 50% of the task working alone.