A contractor employed a certain number of workers to finish a road within a scheduled time. After some time, he realised that at the current rate the work would be delayed by three-fourths of the scheduled time. He immediately doubled the number of workers and managed to finish the road exactly on schedule. What percentage of the work had been completed before he increased the number of workers?

Introduction / Context:
This is a conceptual time-and-work puzzle about planning and adjusting workforce. The contractor initially underestimates the number of workers needed and later doubles the workforce to avoid a large delay. You have to determine what fraction of the total work had been finished before the adjustment, expressed as a percentage.

Given Data / Assumptions:

• Scheduled completion time for the road = S (unknown) days.
• Initial number of workers = N (unknown).
• At some checkpoint, the contractor realises the work would be delayed by three-fourths of S (that is, total time would become S + 3S/4 = 7S/4) if he keeps N workers.
• He then doubles the workforce to 2N and finishes exactly in S days from the start.
• We assume constant work rate per worker and no other interruptions.

Concept / Approach:
The total quantity of work is fixed. We express it in terms of the original workforce N and the time that would have been needed without any change (7S/4). We then equate that with the actual scenario where the contractor works with N workers for some time and 2N workers for the remaining time but completes in S days. That gives us the time of switching and the fraction of work done by that moment.

Step-by-Step Solution:
Let total work = W units. Original workforce = N workers, each with unit rate per worker. If no change is made, total time needed at rate N is 7S/4 days. So W = N * (7S / 4). In reality, the contractor uses N workers for t days and then 2N workers for S - t days, finishing in S days. Actual work done: W = N * t + 2N * (S - t) = N t + 2N S - 2N t = 2N S - N t. Equate both expressions for W: N * (7S / 4) = 2N S - N t. Divide both sides by N: 7S / 4 = 2S - t. Solve for t: t = 2S - 7S / 4 = (8S / 4 - 7S / 4) = S / 4. So the workforce was doubled after t = S / 4 days from the start. Work done up to that time with N workers = N * (S / 4). Total work W = N * (7S / 4). Fraction of work completed = [N * (S / 4)] / [N * (7S / 4)] = 1 / 7. Convert to percentage: 1 / 7 = 14 2/7 % approximately.
Verification / Alternative check:
Remaining work fraction after S / 4 days = 6/7 of W. Rate after doubling = 2N, so time needed for remaining work = (6/7 W) / (2N). Since W = N * 7S / 4, substitute: (6/7 * N * 7S / 4) / (2N) = (6S / 4) / 2 = 3S / 4. Total time = S / 4 + 3S / 4 = S, so it matches the scheduled time.
Why Other Options Are Wrong:
10% is less than 1/7 and does not satisfy the time algebra. 20% equals 1/5, again inconsistent with the equation derived from actual and projected times. "Can not be determined" is incorrect because the system of equations yields a unique fraction independent of actual S and N values.
Common Pitfalls:
Some students misinterpret "delayed by three-fourths of the scheduled time" as a fraction of remaining time, not of S. Others assume an explicit numerical S value without realising that it cancels out in the ratio. The key is to treat S and N as symbols and recognise that only the fraction of work completed matters.
Final Answer:
The work completed before increasing the number of workers was 1/7 of the total work, that is 14 2/7 %.