I’m stuck on setting up proportional relationships when two things change at once. Example: A crew of 4 workers lays 240 m of cable in 6 hours. If the rate per worker is constant, how long should 7 workers take to lay 500 m?
I know time should increase with distance and decrease with number of workers, so t is proportional to distance and inversely proportional to workers. But when I try to write a single proportion I keep flipping ratios. I wrote t/6 = (500/240) × (4/7). Then I second-guessed myself and wondered if it should be (7/4) instead because more workers means less time.
My other attempt was to use a unit rate per worker: r = 240 / (6 × 4) meters per hour per worker, so t = 500 / (7r). That feels safer, but I want to be sure I’m not smuggling in an assumption.
Can someone confirm the correct way to combine these (direct with distance, inverse with workers) into one equation, and show a clean algebraic justification? Also, is there a quick sanity check to tell whether I inverted a ratio (for example, by testing what happens if I double the workers but keep the distance the same)?
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3 Responses
You’ve got the right idea: combine the effects by t ∝ distance and t ∝ 1/workers, so t = k·(d/n). From the given job, k = (6·4)/240 = 1/10, hence t = (1/10)·(500/7) = 50/7 ≈ 7.14 hours. This matches your scaling equation t/6 = (500/240)·(4/7): 500/240 (>1) stretches time because there’s more to do, while 4/7 (<1) shrinks time because there are more workers, so the directions check out. Your unit-rate method is the same reasoning in another form: each worker lays 240/(6·4) = 10 m per hour, so 7 workers lay 70 m per hour and need 500/70 = 50/7 hours. A quick sanity check: if you keep distance at 240 m and double workers from 4 to 8, time should halve from 6 h to 3 h; any setup that makes it increase (e.g., 6·(7/4)) signals you flipped a ratio.
Don’t overthink it: t = k*(distance/workers); from 6 = k*(240/4) get k = 0.1, so t = 0.1*(500/7) = 50/7 ≈ 7.14 h, and as a check, since t is inversely proportional to workers, doubling workers halves time-so t/6 = (500/240)*(4/7) is the right way round.
Hope this helps!
Think of time as a stretchy rubber band: it lengthens with distance and tightens when more workers tug on it, so t varies like t = k·(d/n). Use the original job to find the constant: 6 = k·(240/4) = 60k, so k = 0.1. Then the new time is t = 0.1·(500/7) = 50/7 ≈ 7.14 hours. Your proportion t/6 = (500/240)·(4/7) is exactly this in ratio form, and 4/7 is the right way round-if you double the workers with the distance fixed, that factor halves and so does t, which is a nice built-in sanity check. The unit-rate route r = 240/(6·4) = 10 m per hour per worker and t = 500/(7r) is the same algebra wearing a different hat. For a tidy refresher on mixing direct and inverse variation, see https://www.purplemath.com/modules/variatn.htm