Time to weave a fixed length at a constant per-second rate: An industrial loom weaves 0.128 meters of cloth every second at a steady rate. Approximately how many seconds are required for the loom to weave 25 meters of cloth in total?

Introduction / Context:
Speed–time–distance style problems in manufacturing are identical to motion problems. Here, “distance” is cloth length, “speed” is meters per second, and “time” is what we need, given a fixed target length. We will use the formula time = length / rate and round to the nearest reasonable integer second as the problem asks for an approximate value.

Given Data / Assumptions:

• Weaving rate = 0.128 m/s
• Target length = 25 m
• Rate is constant; ignore startup/slowdown effects.

Concept / Approach:
Time = distance / speed. Substitute the given length and rate. The result may be fractional; choose the closest option. No unit conversions are needed because both are in meters and seconds.

Step-by-Step Solution:
t = 25 / 0.128 secondst = 195.3125 secondsApproximate time ≈ 195 seconds

Verification / Alternative check:
Reverse check: 195 s * 0.128 m/s = 24.96 m, which is very close to 25 m and appropriate for the approximation requested.

Why Other Options Are Wrong:
184 and 190 are low; 197 is slightly high but not as close as 195; 205 is too high for the exact computation of 195.3125 s.

Common Pitfalls:
Inverting speed incorrectly (using 0.128 / 25), or rounding too early, which may push the choice away from the closest correct approximation.

Final Answer:
195