A and B can together complete a piece of work in 10 days. If A works at twice his usual efficiency and B works at an efficiency that is one-third less than his usual efficiency, then the same work is finished in 6 days. In how many days can A alone and B alone complete the entire work respectively?

Introduction / Context:
This is a classic time and work question that tests your understanding of individual work rates and how changes in efficiency affect the overall time required to complete a job. Two workers, A and B, can finish a task together in a known number of days, and then their efficiencies are altered. From the new completion time, we have to reverse engineer their original individual times to complete the same work alone.

Given Data / Assumptions:

The whole work is considered as 1 unit of job.
Original time when A and B work together is 10 days.
When A works at twice his usual efficiency and B works at one-third less (that is two-thirds) of his usual efficiency, the work finishes in 6 days.
Work rates are assumed constant over time, and there are no breaks or interruptions.

Concept / Approach:
We use the concept that Work = Rate * Time and that individual rates add up when people work together. Let the original daily work rates of A and B be a and b respectively. Then their combined rate is a + b. From the given conditions, we will form two linear equations in terms of a and b and solve those equations to find each worker's rate and hence their individual times to complete the work alone.

Step-by-Step Solution:
Let total work = 1 unit. Since A and B together finish the work in 10 days, (a + b) * 10 = 1, so a + b = 1/10. When A works at twice his efficiency, his new rate is 2a. When B works at one-third less efficiency, his new rate is (2/3)b. Together in the modified scenario they complete the same 1 unit of work in 6 days, so (2a + (2/3)b) * 6 = 1, hence 2a + (2/3)b = 1/6. We now solve the system: a + b = 1/10 and 2a + (2/3)b = 1/6. From a + b = 1/10, we get a = 1/10 - b. Substitute in the second equation: 2(1/10 - b) + (2/3)b = 1/6. This gives 1/5 - 2b + (2/3)b = 1/6, so 1/5 - (4/3)b = 1/6. Rearranging, (4/3)b = 1/5 - 1/6 = (6 - 5)/30 = 1/30, so b = (1/30) * (3/4) = 1/40. Then a = 1/10 - 1/40 = 4/40 - 1/40 = 3/40. Time taken by A alone = 1/a = 1 / (3/40) = 40/3 days. Time taken by B alone = 1/b = 1 / (1/40) = 40 days.

Verification / Alternative check:
Check the original combined time: a + b = 3/40 + 1/40 = 4/40 = 1/10, so together they take 10 days, which matches the given data. Now check the modified rates: 2a + (2/3)b = 2*(3/40) + (2/3)*(1/40) = 6/40 + 2/120 = 6/40 + 1/60 = 9/60 + 1/60 = 10/60 = 1/6, which means 6 days for the whole job. Both conditions are satisfied, so the solution is consistent.

Why Other Options Are Wrong:
Option 20/3 days, 20 days gives a combined rate that does not equal 1/10 and fails the second condition as well.
Option 30 days, 20/3 days swaps and distorts the time values and does not satisfy either equation for the modified efficiencies.
Option 50/3 days, 25 days leads to slower workers and predicts a combined time greater than 10 days, so it contradicts the problem statement.
Option 30 days, 30 days assumes equal speeds, which again does not match the linear system formed from the given conditions.

Common Pitfalls:
A common mistake is to directly scale times instead of rates when efficiency changes. Another mistake is to misinterpret "one-third less" as one-third of the original instead of reducing the rate to two-thirds of the original. Sign errors while solving simultaneous equations are also frequent. Always define rates carefully and form equations systematically.

Final Answer:
Therefore, A alone completes the work in 40/3 days and B alone completes the work in 40 days, so the correct option is 40/3 days, 40 days.