The ratio of efficiency of A to C is 5 : 3. The ratio of the number of days taken by B to C is 2 : 3. A takes 6 days less than C to complete a work, when each works individually. B and C start the work together and leave after working for 2 days. How many days will A alone take to finish the remaining work?

Introduction / Context:
This is a multi step time and work problem involving efficiency ratios and days ratios. You are given relationships between the efficiencies and times of A, B and C, along with an absolute difference between days of A and C. After some initial joint work by B and C, A finishes the remainder. You must compute the number of days A takes for that remaining part.

Given Data / Assumptions:

• Efficiency ratio A : C = 5 : 3.
• Days ratio B : C = 2 : 3 (that is, B takes 2 parts, C takes 3 parts).
• A takes 6 days less than C to do the work individually.
• B and C work together for 2 days and then stop.
• A completes the remaining work alone.

Concept / Approach:
Efficiency is inversely proportional to days taken. Use the given efficiency ratio to relate days of A and C, and then apply the 6 day difference. Similarly, use the B : C days ratio to find B's time. Once you know individual days for A, B and C, convert them into daily work rates. Compute total work done in the first 2 days by B and C, and then let A finish the rest. The time A uses at the end is exactly what the question asks for.

Step-by-Step Solution:
Let T_A and T_C be days taken by A and C alone. Efficiency ratio A : C = 5 : 3 implies work rate ratio = 5 : 3. Since rate is inversely proportional to time, T_A : T_C = 3 : 5. We are told T_C - T_A = 6 days. Let T_A = 3k and T_C = 5k for some k. Then 5k - 3k = 6, so 2k = 6 and k = 3. Hence T_A = 3 * 3 = 9 days and T_C = 5 * 3 = 15 days. Days ratio B : C = 2 : 3 implies T_B : T_C = 2 : 3. So T_B = (2/3) * T_C = (2/3) * 15 = 10 days. Now compute daily work rates (assuming total work = 1 unit). A's rate = 1 / 9 per day, B's rate = 1 / 10 per day, C's rate = 1 / 15 per day. B and C work together for 2 days, so work done = 2 * (1/10 + 1/15). 1/10 + 1/15 = (3 + 2) / 30 = 5 / 30 = 1 / 6. Thus work completed in 2 days = 2 * (1 / 6) = 1 / 3. Remaining work = 1 - 1 / 3 = 2 / 3. A alone now works at rate 1 / 9 per day. Time for A to finish remaining work = (2 / 3) / (1 / 9) = (2 / 3) * 9 = 6 days.
Verification / Alternative check:
Total work = 1: B and C did 1/3; A does 2/3. If A did the entire work alone, he would need 9 days, which is greater than 6, so the partial work reasoning is consistent.
Why Other Options Are Wrong:
4.5, 5 or 9 1/3 days would not match the calculated remaining work fraction 2/3 at A's rate. Inserting them into rate * time would give either less than or more than 2/3 of the work.
Common Pitfalls:
Some test takers misinterpret efficiency ratio as days ratio directly, which is incorrect. Others handle the 6 day difference wrongly by assigning k values without carefully resolving equations. Always link efficiency ratios to inverse time ratios and solve the system systematically.
Final Answer:
A takes 6 days to finish the remaining work.