Comparative Efficiency with Time Difference (Hours) Romil is twice as efficient as Sanjay and finishes the work 4 hours sooner than Sanjay. If both work together from the start, in how many hours will they finish the work?

Introduction / Context:This problem links relative efficiency (twice as efficient) with a time difference. Solve for each person’s solo time first, then add rates to get the combined finishing time.

Given Data / Assumptions:

• Romil's efficiency = 2 * Sanjay's efficiency.
• Romil finishes 4 hours earlier than Sanjay.
• Constant rates throughout.

Concept / Approach:If Sanjay's time is t hours, Romil's time is t/2 (twice as efficient ⇒ half the time). Given t/2 = t − 4 allows solving for t. Then compute the combined rate.

Step-by-Step Solution:t/2 = t − 4 ⇒ t − t/2 = 4 ⇒ t/2 = 4 ⇒ t = 8 hours (Sanjay).Romil's time = t/2 = 4 hours.Combined rate = 1/8 + 1/4 = 3/8 job/hour.Together time = 1 / (3/8) = 8/3 hours ≈ 2.667 hours.

Verification / Alternative check:Romil being faster and both working together should yield a time less than 4 hours, which 8/3 satisfies.

Why Other Options Are Wrong:3 h or 4/3 h do not match the combined rate; 8 h is Sanjay’s solo time; 5 h is arbitrary.

Common Pitfalls:Misreading “twice as efficient” as “2 hours less” or subtracting times rather than adding rates to combine efforts.

Final Answer:8 / 3 h