I’m practicing inverse proportion and I thought I had it: more workers means less time, so workers × time = constant for the same job. That feels super clean and satisfying. But I hit a word problem that made me second-guess everything.
Example: “6 machines make 480 parts in 5 hours. How long would 8 machines take to make 720 parts?” My brain says time is inversely proportional to machines… but also directly proportional to how many parts we need. So is this still an inverse proportion situation, or is it a combo of direct and inverse at the same time? I tried writing T ∝ 1/M at first, but then the output changed and that seems to break the ‘constant product’ idea.
I tried switching to rates: let r be parts per machine per hour, so M × r × T = total parts. That seems more sensible, but I’m not sure if I’m overcomplicating it or if that’s actually the right lens for “inverse proportion” problems. I also tried doing a quick proportion like 6/8 = T/? and immediately got tangled because the total parts weren’t the same-so I think I just did something illegal there.
What’s the quick way to detect when a situation is truly inverse proportion (like a pure hyperbola T = k/M) versus when I need to bring in another varying thing (like output or distance) and treat it differently? Also, as a follow-up: if I graphed T vs M for a set of targets (e.g., 480 parts vs 720 parts), is it basically a family of hyperbolas with different constants, or am I thinking about that wrong?
Why I’m stuck: I keep mixing up “inverse proportion” with “just take the reciprocal of the scaling factor,” and I’m not sure when I’m allowed to keep the product constant and when the ‘constant’ is secretly changing because something else changed. Any tips to keep my logic straight here?
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3 Responses
You’ve got the right instinct: “inverse proportion” only holds when the job size and the per-machine rate are fixed. The clean rule is: hold everything else constant, then see how one thing scales when you change the other. For work problems, the backbone is Work = (machines) × (rate per machine) × (time). In your example, 6 machines make 480 parts in 5 hours, so the per‑machine rate is r = 480/(6×5) = 16 parts/hour. Then time for 8 machines to make 720 parts is T = 720/(8×16) = 720/128 = 45/8 = 5.625 hours (about 5 h 37.5 min). If you like quick scaling: more machines cuts time by the same factor (6→8 means multiply time by 6/8), more parts stretches time by the same factor (480→720 means multiply time by 720/480), so 5 × (6/8) × (720/480) = 5 × (3/4) × (3/2) = 45/8. So yes, it’s “direct in parts, inverse in machines” at the same time: T ∝ parts and T ∝ 1/machines, with T = (parts)/(r×machines). Graph-wise, for a fixed target (and fixed r), T vs M is a hyperbola; different targets give you a family of hyperbolas with different constants. The only time this breaks is when something else isn’t constant-rates change with crowding, setup time, fatigue, etc.-then the product isn’t constant and it’s not pure inverse anymore. Want to try one with a fixed setup time to see how the curve stops being a neat hyperbola?
I like to check “what stays fixed?” before deciding on inverse proportion. If the job size (number of parts) and the per-machine productivity are fixed, then workers × time = constant and you truly have T ∝ 1/M. But the moment the job size changes, that “constant” can’t stay the same anymore. A cleaner lens is the rate model: let r be parts per machine per hour. Then total parts P = M × r × T, so T = P/(M r). From this you can read off two facts: for a fixed target P, time is inversely proportional to machines; for a fixed crew M, time is directly proportional to the target. If both P and M change at once, it isn’t inverse proportion anymore in the strict sense (because the product M·T doesn’t stay the same), so you shouldn’t expect one single hyperbola to cover all those changes.
Worked example: “6 machines make 480 parts in 5 hours. How long would 8 machines take to make 720 parts?” First find r from the first scenario: 6 × r × 5 = 480 ⇒ r = 480/30 = 16 parts per machine per hour. Then T = P/(M r) = 720/(8 × 16) = 720/128 = 5.625 hours (that’s 5 hours 37.5 minutes). A quick proportional shortcut says the same thing: scale time by “how much more work” and “how much more help,” so T_new = T_old × (P_new/P_old) × (M_old/M_new) = 5 × (720/480) × (6/8) = 5.625.
On graphs: for each fixed target P you get T vs M as a hyperbola T = k/M with k = P/r. Changing the target changes k, so you get a whole family of “shifted” hyperbolas stacked higher or lower depending on P (doubling P doubles k). That’s why M·T is only constant along one curve at a time-once you change P, the constant changes too. A quick detection tip I use: ask “if I freeze everything except M and T, does M·T stay the same?” If yes, it’s pure inverse proportion; if the output or speed is also moving, then it’s really the combined relation T ∝ P/M instead.
Think “what’s held fixed?” If the size of the job is fixed, then time is inversely proportional to machines: T ∝ 1/M, so M·T is constant. If the job size (parts) can change, the right model is joint proportionality: T ∝ P/M. Equivalently, with a constant per-machine rate r (parts per machine-hour), you have M·r·T = P, so T = P/(M r). That shows the “constant” in M·T = constant is actually P/r, so it only stays constant when P is fixed. Quick scaling test: multiply machines by s and parts by t; then time scales by t/s (pure inverse is the special case t = 1). For your numbers, from 6 machines making 480 parts in 5 hours, r = 480/(6·5) = 16 parts per machine-hour. Then for 8 machines making 720 parts, T = 720/(8·16) = 720/128 = 5.625 hours = 5 h 37.5 min. Graphically, for each fixed target P you get a hyperbola T = (P/r)·(1/M); changing P just changes the hyperbola’s constant. Hope this helps!