Pavan and Sravan can complete a piece of work under two different efficiency arrangements. If Pavan works with his usual efficiency and Sravan works at twice his usual efficiency, they complete the work in 25 days. If Pavan works at twice his usual efficiency and Sravan works at his usual efficiency, they complete the work in 20 days. In how many days would Sravan alone, working with his usual efficiency, complete the work?

Introduction / Context:
Here we have a time and work problem involving variable efficiencies. The same two workers, Pavan and Sravan, work under two different efficiency arrangements, leading to two different completion times for the same job. Using these two conditions, we can solve for the individual daily work rates of each person and then determine how long Sravan alone would take to finish the work at his normal efficiency.

Given Data / Assumptions:

• Case 1: Pavan at normal efficiency, Sravan at twice his normal efficiency, time = 25 days.
• Case 2: Pavan at twice his normal efficiency, Sravan at normal efficiency, time = 20 days.
• Both arrangements complete the same amount of work.
• Efficiencies remain constant within each scenario.
• We need Sravan's time working alone at his normal efficiency.

Concept / Approach:
Let P and S be the normal daily work rates of Pavan and Sravan respectively. In the first scenario, the combined rate is P + 2S, and in the second scenario, it is 2P + S. Using Work = Rate * Time, we express the same total work using both scenarios and form two equations. After solving these simultaneous equations for P and S, we can find the time taken by Sravan alone as 1 / S.

Step-by-Step Solution:
Let Pavan's normal daily work rate be p units. Let Sravan's normal daily work rate be s units. Total work W is the same in all cases. Case 1: (p + 2s) * 25 = W ⇒ p + 2s = 1 / 25. Case 2: (2p + s) * 20 = W ⇒ 2p + s = 1 / 20. Solve the system: p + 2s = 1 / 25 …(1) 2p + s = 1 / 20 …(2) From (1), p = 1 / 25 − 2s. Substitute in (2): 2(1 / 25 − 2s) + s = 1 / 20. ⇒ 2 / 25 − 4s + s = 1 / 20. ⇒ 2 / 25 − 3s = 1 / 20. ⇒ −3s = 1 / 20 − 2 / 25. Compute 1 / 20 − 2 / 25 = (5 / 100 − 8 / 100) = −3 / 100. So −3s = −3 / 100 ⇒ s = 1 / 100. Therefore Sravan's rate = 1 / 100 of the work per day. Time by Sravan alone = 1 / s = 100 days.

Verification / Alternative check:
Using s = 1 / 100, we can compute p from (1): p + 2 * (1 / 100) = 1 / 25 ⇒ p + 1 / 50 = 1 / 25 ⇒ p = 1 / 25 − 1 / 50 = 1 / 50. Check Case 2: 2p + s = 2 * (1 / 50) + 1 / 100 = 1 / 25 + 1 / 100 = 4 / 100 + 1 / 100 = 1 / 20, which matches equation (2). This confirms that the values of p and s are correct.

Why Other Options Are Wrong:
50 days would require Sravan's rate to be 1 / 50, which would not satisfy the given relations. 70 or 35 days correspond to other incorrect combinations of p and s. Only 100 days is consistent with both scenarios and the simultaneous equations.

Common Pitfalls:
The main pitfalls include miswriting the equations, interchanging the coefficients of p and s, or making small errors when subtracting the fractions 1 / 20 and 2 / 25. Systematic use of fractional arithmetic or a common denominator avoids these mistakes.

Final Answer:
Sravan, working alone at his normal efficiency, will complete the work in 100 days.