I’m getting tangled up with compound measures and when to add, average, or something else. It feels like I’m trying to blend two smoothies and just guessing the recipe. I keep telling myself “average it!” and then everything falls apart.
Example 1 (speed): I walk 3 km to the shop at 4 km/h and 3 km back at 6 km/h. What’s my overall average speed for the whole 6 km? My first instinct is to just average the speeds: (4 + 6) / 2 = 5 km/h. That seems neat and tidy, but I’m not confident it’s right.
Example 2 (flow rate): A 240 L tank is being filled. I use a hose that does 12 L/min for the first 10 minutes, then I switch to a faster hose that does 18 L/min until it’s full. I tried doing 12 + 18 = 30 L/min, then 240 / 30 = 8 minutes, and finally adding the first 10 minutes to get 18 minutes total. I’m pretty sure that’s nonsense because I didn’t actually run both hoses at the same time.
Could someone explain, in plain terms, how to properly combine these rates? Like, when do I add things, when do I average, and what exactly should I be averaging (speeds, times, volumes…)? I keep tripping over my own shoelaces with the units, too, so any pointers there would help.
Any help appreciated!
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3 Responses
I get tangled in this too, so my go-to rule is: average rate = total amount divided by total time, and only add rates when things run in parallel at the same time. For the walk, you cover 3 km at 4 km/h (that’s 0.75 h) and 3 km at 6 km/h (0.5 h), so the total time is 1.25 h for 6 km; average speed is 6 ÷ 1.25 = 4.8 km/h, not the simple average 5. That “don’t just average the speeds” thing is because the slower part eats more time. For the tank, do it piece by piece: first 10 minutes at 12 L/min gives 120 L; 120 L remain, which the 18 L/min hose fills in 120 ÷ 18 = 6.666… minutes, so the total time is about 16 minutes 40 seconds. If you want an overall average fill rate afterward, it’s 240 L divided by that total time, about 14.4 L/min. Big picture: add rates only when hoses or workers run together; if they run one after the other, add times. And if you’re ever unsure, chase the units: distance = speed × time and volume = rate × time, so your final step should look like “total amount ÷ total time.” Want to try a twist: what if the outward and return legs took the same time instead of the same distance-what would the average speed be then?
A reliable way to combine rates is to return to the definition rate = quantity ÷ time and build the total time and total quantity step by step. Example 1: you travel 3 km at 4 km/h, which takes 3/4 = 0.75 hours, and 3 km at 6 km/h, which takes 3/6 = 0.5 hours; total distance is 6 km and total time is 1.25 hours, so the overall average speed is 6 ÷ 1.25 = 4.8 km/h. It is not the simple average (5 km/h) because you spend more time at the slower speed; with equal distances, the correct “average of speeds” is the harmonic mean, which here also gives 4.8. Example 2: in the first 10 minutes at 12 L/min you fill 12 × 10 = 120 L, leaving 120 L; at 18 L/min that takes 120 ÷ 18 = 6 2/3 minutes, so the total fill time is 10 + 6 2/3 = 16 2/3 minutes. Adding rates only makes sense when they act at the same time on the same task; if both hoses ran together the rate would be 12 + 18 = 30 L/min and a 240 L tank would then take 240 ÷ 30 = 8 minutes. As a rule: for sequential phases, add times (or compute each phase’s quantity and time, then sum appropriately) and finish with overall rate = total quantity ÷ total time; for concurrent agents, add their rates. For averaging speeds, use the arithmetic mean only when the times at each speed are equal; with equal distances, use the harmonic mean. A quick unit check helps: time = distance ÷ speed or volume ÷ flow, and any “average rate” must be total quantity ÷ total time.
I feel this! Rates are like smoothie recipes: you can’t just average strawberries and bananas unless you know how many scoops of each you actually blended. The safe north star is “overall rate = total stuff ÷ total time.” So for the walk: total distance is 6 km; time out is 3/4 hour (3 km at 4 km/h) and time back is 1/2 hour (3 km at 6 km/h), so total time is 1.25 hours and the average speed is 6 ÷ 1.25 = 4.8 km/h-not 5. A handy (slightly nerdy) view is to average “time per km”: 15 min/km and 10 min/km over equal distances give 12.5 min/km on average, which flips back to 4.8 km/h. For the tank: first 10 minutes at 12 L/min gives 120 L; you’ve got 120 L left, and the 18 L/min hose takes 120 ÷ 18 = 6 2/3 minutes; total time is about 16 2/3 minutes, and the overall average fill rate is 240 ÷ 16 2/3 ≈ 14.4 L/min (closer to the slower hose, since you spent more time with it). You only add rates when they run at the same time (two hoses together), not one after another; with sequential chunks, you add the times, not the rates. My rule of thumb-though I admit I sometimes second-guess it-is “average the thing that’s naturally per unit of what stays the same”: with equal distances, average the time-per-distance; with time chunks, the overall rate is like a time-weighted average of the segment rates (I might be oversimplifying that, but it keeps my shoelaces untangled). A nice walkthrough is here: https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:rational-expressions-equations/x2f8bb11595b61c86:rate-word-problems/a/average-rate-problems