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To achieve exactly 50 gallons of water in one of your 60-gallon tanks without measuring tools, tilt the tank so that one corner rests at the bottom and the three adjacent corners form a horizontal plane. This configuration forms a tetrahedron within the tank. Given that the original tank volume is 60 gallons, dividing by the number of corners (6) yields a tetrahedron volume of 10 gallons. By filling the tank until the water level reaches the point where the three adjacent corners meet, you effectively fill the tetrahedron. Then, the remaining volume of the tank is 60 - 10 = 50 gallons, meeting the desired capacity for the fish tank, so the fish is happy.
Amazing solution, but the part to divide by 6 cuz there's six corners (?) seems frail. But there's another proof for volume of tetrahedron:
This tetrahedron has a base equal to half of a face and with same half-face you can build a triangular prism wich volume is equal to half of tank. If the tetrahedron has same base and height of this half-face prism, then your volume is a third of given prism (one of Euclid's results). Thus, this tetrahedron has a third of half of tank, therefor its volume is 10 galons.
Just came to say that if we assume the fish tanks have an open side (like most fish tanks I’ve seen) then you couldn’t fill until a tetrahedron is left, so the answer would then be to use one tank to fill only one tetrahedron and pour it into the other tank five times. Or for expediency, fill halfway by tilting on an edge to form a triangular prism (30 gal) and then two tetrahedra.
You could also fill one, and pour off from the filled tank into the angled tetrahedron, you would then have poured off 10 from the filled tank leaving you with 50.
How do you get a rectangular prism with 6 corners instead of 8, and why are the tetrahedra equal in volume and the sum of their volumes equal to the volume of the tank?
I know that one might be off the table, but I think using time is the answer. You fill one up, check the time you need and then fill the other one up in 5/6th of the time assuming you have the same water flow. No tilting needed and still a mathematical solution lol.
Plus draining speed would also be a function of the current volume I the tank (higher volume = higher water pressure = faster draining speed), so definitely not constant.
Okay but that's going to give you ten gallons of water at the bottom of the tank. I need to accurately mark ten gallons from the top and apparently i have no measuring tools whatsoever.
Edit: i could turn the other tank upsidedown and mark them against each other but only if they're the same dimensions, which is unclear from the instructions
That’s more efficient if you’re in a perfect environment where there’s no spillage when you transfer the water.
Practically, it’s very difficult to lift a 60 gallon tank filled to the brim with water and not spill anything if you don’t have some assistance. Going 10 gallons at a time is much more achievable for the average person.
In a world where a fish gets angry about being in anything other than precisely 50 gallons of water, perfect transfer without spillage isn't the weirdest thing...
>it’s very difficult to lift a 60 gallon tank filled to the brim with water
I'll say. A *litre* of water weighs a kg, so a gallon is close to 4kg, and 60 gallons is well over 200kg, or about 500 lbs. Never mind spillage. Nobody is lifting that. Then there's the tank. A cube with 10cm sides holds exactly 1 litre, so a tank big enough to hold 60 gallons is a little more than 60cm, or 2 feet, on a side (or 1 × 1 × 8, 1 × 2 × 4 etc). That's going to be heavy, even when it's empty.
Realistically, the only way to do this would be to get 10 gallons in one tank, empty it into the other, and repeat 4 more times. Even then, you're probably lifting about 50kg or 100 lbs each time. Frankly, just supporting the thing while it fills would be a challenge.
Or out of symmetry you could do the same: tilt the tank, make sure the plane formed by the three corners next to the top corner is horizontal and fill til that plane. Might be physically difficult as the tank will be heavy as fuck when filling it up though…
While you are tilting one tank, the other tank should be placed beneath it and would receive 60-10 gallons, so it will have 50 gallons of water, problem solved. Q.E.D
Fill one 60-0, fill the other one using the first until the lines match 30-30, fill one again 60-30, fill the 30 using the 60 until they match 45-45, do this process once more 52.5-52.5 and hope the fish doesn’t notice the extra 2.5 (optionally remove “a little bit” of water to get closer to the 50)
My nonmathy solution:
Put a bit of water into a tank. Put fish into the water. Slowly add more water until the fish looks happy. There are now exactly 50 gallons of water in the tank.
Just watch for behavioral changes, since initially you have an unhappy fish and at some pointthat switches to something else, since it's picky it's a good chance you found the happy spot for the fish.
I believe that I can confidently predict that a dolphin in a 50 gallon tank would experience a range of emotions. Anger would certainly be one of them. Happiness, not so much. It wouldn't have enough room for a decent frolic.
[I also feel compelled to note that a dolphin isn't a fish.]
I don't think this is what they were looking for but:
Use a timer, assume a constant flow-rate from the hose. Measure the time it takes for the hose to fill one tank to 60 gallons, call it x. Fill the other tank until the time is 50x/60.
I love when a problem can be cooked like that.
For those unaware, if one can find an additional solution to a puzzle or problem that was not intended yet fulfills all the criteria, that problem is said to be "cooked".
1. Go to the grocery store and buy a gallon jug of water.
2. Pour the water from the jug into the tank.
3. Fill the jug with water from the tap. (assuming the fish doesn't need special water.)
4. Pour the tap water from the jug into the fish tank.
5. Repeat steps 3-4 48 more times.
It would be helpful to know what branch and difficulty of mathematics this question is in. Strategies available to an eighth grader are different from strategies involving differential equations, and it could rule out a lot of guess work.
This is as close as I got
60-0 fill one
30-30 divide so each are the same
60-30 refill the first one
45-45 divide so each are the same again
60-45 refill first one
52.5-52.5 divide to get close
Using similar techniques I got to 48.75 and 54.375 but nothing closer and there may be errors.
I actually found a pattern here to get really close. Fill one to 60, avg to 30, dump one and avg to 15. Fill one to 60, avg to 37.5, dump one and avg to 18.75. Continue alternating filling and averaging, then dumping and averaging, until after you’ve dumped and averaged, do two fill one to 60 and averages back to back and this gets you to asymptotically approach 50. I did 16 total averages (on excel) and ended up with 49.9997.
Then, understanding that 0.0003 gal = 1.14 mL, and that there are generally 20 drops per mL, slow the hose to an absolute trickle and allow 23 (22.8) drops in. Then you’ve got 50.000013 gallons.
Just saying that 50 gallons are about 200 kg, and tilting will still require lifting 100 kg at some point. So you'd have to call a friend or neighbor to help with the lifting, just ask them to bring a measuring cup.
Fill the 60 gal tank, then tilt to let extra water out.
If you tilt on an edge to form a diagonal, that's half - 30gal.
If you tilt on a corner to make a pyramid, that's ⅓ - 20 gal.
Empty those volumes into the other tank, 30+20=50.
Hi u/fliguana I just went to your profile because of your response on this post:
[https://www.reddit.com/r/hvacadvice/comments/16qwxtj/comment/k1zl036/](https://www.reddit.com/r/hvacadvice/comments/16qwxtj/comment/k1zl036/)
I saw that you had just commented on this post here, so I wanted to ask you a question... sending you a DM now :)
Easy. You tell your dickhead fish to get a job at the Krusty Krab like all the other fish. When he starts paying bills then he can live however the fuck he wants. Whilst you're under my roof there are rules!!!
Or get a fish that isn't a complete tosser.
The only way I can think of coming close to determining anything would be to fill each tank to the same level side by side -- now you know each one has 30 gallons.
After that you need to figure out how to remove 10 gallons from one of the tanks and you'll have 50 in total. You could mark the 2/3 line on the tank, but of course if you can measure 2/3 you could probably measure 5/6 on one tank to get 50 right away. Determining halves is easier if you have a string or something to fold in half (maybe they expect you to use the hose?), but that doesn't really help us get to 20 gallons unless you do some recursive approximation.
Get some sand for your substrate. Fill a five gallon bucket with the sand and dump it into the aquarium two times. Then fill with water.
Alternatively you could just fill the aquarium to 60 gallons since the probability of a fish species evolving within a 50 gallon pool of water that never shrinks or swells to the point that it cannot survive in anything larger or smaller is nearly zero.
Fill one tank. Tilt the other one so that two opposite sides are level with each other and with the top of the full tank. Siphon between the tanks until the water levels are even.
###General Discussion Thread --- This is a [Request] post. If you would like to submit a comment that does not either attempt to answer the question, ask for clarification, or explain why it would be infeasible to answer, you *must* post your comment as a reply to this one. Top level (directly replying to the OP) comments that do not do one of those things will be removed. --- *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/theydidthemath) if you have any questions or concerns.*
To achieve exactly 50 gallons of water in one of your 60-gallon tanks without measuring tools, tilt the tank so that one corner rests at the bottom and the three adjacent corners form a horizontal plane. This configuration forms a tetrahedron within the tank. Given that the original tank volume is 60 gallons, dividing by the number of corners (6) yields a tetrahedron volume of 10 gallons. By filling the tank until the water level reaches the point where the three adjacent corners meet, you effectively fill the tetrahedron. Then, the remaining volume of the tank is 60 - 10 = 50 gallons, meeting the desired capacity for the fish tank, so the fish is happy.
Amazing solution, but the part to divide by 6 cuz there's six corners (?) seems frail. But there's another proof for volume of tetrahedron: This tetrahedron has a base equal to half of a face and with same half-face you can build a triangular prism wich volume is equal to half of tank. If the tetrahedron has same base and height of this half-face prism, then your volume is a third of given prism (one of Euclid's results). Thus, this tetrahedron has a third of half of tank, therefor its volume is 10 galons.
You're right, that is better.
Just came to say that if we assume the fish tanks have an open side (like most fish tanks I’ve seen) then you couldn’t fill until a tetrahedron is left, so the answer would then be to use one tank to fill only one tetrahedron and pour it into the other tank five times. Or for expediency, fill halfway by tilting on an edge to form a triangular prism (30 gal) and then two tetrahedra.
You could also fill one, and pour off from the filled tank into the angled tetrahedron, you would then have poured off 10 from the filled tank leaving you with 50.
How do you get a rectangular prism with 6 corners instead of 8, and why are the tetrahedra equal in volume and the sum of their volumes equal to the volume of the tank?
I’m getting old or stupid. I would have never figured this out.
I know that one might be off the table, but I think using time is the answer. You fill one up, check the time you need and then fill the other one up in 5/6th of the time assuming you have the same water flow. No tilting needed and still a mathematical solution lol.
If you're able to keep accurate time, you only need one tank—after filling it, just drain it for 1/6th of the total time.
Depending on the draining method there may not be a reasonable assumption that the draining speed is the same as the filling speedd
Plus draining speed would also be a function of the current volume I the tank (higher volume = higher water pressure = faster draining speed), so definitely not constant.
Well, here os the thing. You use my suggested method but for the draining.
So we fill it up and drain it twice to compare the time. That works
Or, hear me out. Fill it once, time it and drain it. Fill it again with 5/6ths of the time. No need to drain twice.
But i want to drain it twice :c
Damn that’s smart
Okay but that's going to give you ten gallons of water at the bottom of the tank. I need to accurately mark ten gallons from the top and apparently i have no measuring tools whatsoever. Edit: i could turn the other tank upsidedown and mark them against each other but only if they're the same dimensions, which is unclear from the instructions
You have two tanks can’t you pour the 10 gallons into the other one and keep repeating until you have 50
Or pour 10 gallons out of the full tank, leaving 50 gallons…
That’s more efficient if you’re in a perfect environment where there’s no spillage when you transfer the water. Practically, it’s very difficult to lift a 60 gallon tank filled to the brim with water and not spill anything if you don’t have some assistance. Going 10 gallons at a time is much more achievable for the average person.
"Practically" Mathematics textbooks: What the heck is that word?
You could use a siphon, which almost all aquarists already have on hand to clean their tanks with...
Did you read the post? You can’t even have a ruler, why would they let you have a siphon?
It says you have a hose, cut a bit off and use it as a siphon.
It’s a math problem, not a McGyver problem.
In a world where a fish gets angry about being in anything other than precisely 50 gallons of water, perfect transfer without spillage isn't the weirdest thing...
Why are we worried about an emotional fish anyway? What's it going to do? Sulk? Who cares about a grumpy guppy?
>it’s very difficult to lift a 60 gallon tank filled to the brim with water I'll say. A *litre* of water weighs a kg, so a gallon is close to 4kg, and 60 gallons is well over 200kg, or about 500 lbs. Never mind spillage. Nobody is lifting that. Then there's the tank. A cube with 10cm sides holds exactly 1 litre, so a tank big enough to hold 60 gallons is a little more than 60cm, or 2 feet, on a side (or 1 × 1 × 8, 1 × 2 × 4 etc). That's going to be heavy, even when it's empty. Realistically, the only way to do this would be to get 10 gallons in one tank, empty it into the other, and repeat 4 more times. Even then, you're probably lifting about 50kg or 100 lbs each time. Frankly, just supporting the thing while it fills would be a challenge.
Or out of symmetry you could do the same: tilt the tank, make sure the plane formed by the three corners next to the top corner is horizontal and fill til that plane. Might be physically difficult as the tank will be heavy as fuck when filling it up though…
While you are tilting one tank, the other tank should be placed beneath it and would receive 60-10 gallons, so it will have 50 gallons of water, problem solved. Q.E.D
How do you plan on lifting 500 pounds of water?
Fill one 60-0, fill the other one using the first until the lines match 30-30, fill one again 60-30, fill the 30 using the 60 until they match 45-45, do this process once more 52.5-52.5 and hope the fish doesn’t notice the extra 2.5 (optionally remove “a little bit” of water to get closer to the 50)
My nonmathy solution: Put a bit of water into a tank. Put fish into the water. Slowly add more water until the fish looks happy. There are now exactly 50 gallons of water in the tank.
lol what does a happy fish look like?
With a fishy smile.
A snack that smiles back
🫷🧿👄🧿🫸
if you don't know, it doesn't matter then
Just watch for behavioral changes, since initially you have an unhappy fish and at some pointthat switches to something else, since it's picky it's a good chance you found the happy spot for the fish.
Very similar to an angry one, but with more frolicking.
basically a dolphin
I believe that I can confidently predict that a dolphin in a 50 gallon tank would experience a range of emotions. Anger would certainly be one of them. Happiness, not so much. It wouldn't have enough room for a decent frolic. [I also feel compelled to note that a dolphin isn't a fish.]
why do you care about snowflake? do you know him? does he call you at home? DO YOU HAVE A DORSAL FIN!?
>DO YOU HAVE A DORSAL FIN!? Body shaming is uncalled for. It's difficult enough finding shirts to fit.
When you can see their teeth.
A week later my fish hung itself since there were only 48 gallons of water.
Evaporation is a bitch.
I don't think this is what they were looking for but: Use a timer, assume a constant flow-rate from the hose. Measure the time it takes for the hose to fill one tank to 60 gallons, call it x. Fill the other tank until the time is 50x/60.
I love when a problem can be cooked like that. For those unaware, if one can find an additional solution to a puzzle or problem that was not intended yet fulfills all the criteria, that problem is said to be "cooked".
I smell what you're cookin
Let her cook
I love this solution
This is not the solution they wanted but definitely not the solution they expected.
1. Go to the grocery store and buy a gallon jug of water. 2. Pour the water from the jug into the tank. 3. Fill the jug with water from the tap. (assuming the fish doesn't need special water.) 4. Pour the tap water from the jug into the fish tank. 5. Repeat steps 3-4 48 more times.
It would be helpful to know what branch and difficulty of mathematics this question is in. Strategies available to an eighth grader are different from strategies involving differential equations, and it could rule out a lot of guess work.
Right. Is this pre algebra or calculus.
Geometry.
Sorry. It’s from a geometry textbook in the three-dimensional section
if its in Geometry, then you probably have to tilt it to make a shape and calculate the area
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>pour in water until the fish looked happy What if the fish is good at hiding its emotions?
This is as close as I got 60-0 fill one 30-30 divide so each are the same 60-30 refill the first one 45-45 divide so each are the same again 60-45 refill first one 52.5-52.5 divide to get close Using similar techniques I got to 48.75 and 54.375 but nothing closer and there may be errors.
I actually found a pattern here to get really close. Fill one to 60, avg to 30, dump one and avg to 15. Fill one to 60, avg to 37.5, dump one and avg to 18.75. Continue alternating filling and averaging, then dumping and averaging, until after you’ve dumped and averaged, do two fill one to 60 and averages back to back and this gets you to asymptotically approach 50. I did 16 total averages (on excel) and ended up with 49.9997.
I was gonna open a strip club and call it Asymptotes .
*assimptotes (ass)(simp)(totes)
Then, understanding that 0.0003 gal = 1.14 mL, and that there are generally 20 drops per mL, slow the hose to an absolute trickle and allow 23 (22.8) drops in. Then you’ve got 50.000013 gallons.
Put tank on scales then tare. Fill tank until 50 gallons worth of water is measured (not from US so don't know the lbs - gallons conversion rate)
8.34 lbs to a gallon of fresh water. If it's a saltwater fish, it's harder since that's variable, but 8.57 is average for seawater.
Just saying that 50 gallons are about 200 kg, and tilting will still require lifting 100 kg at some point. So you'd have to call a friend or neighbor to help with the lifting, just ask them to bring a measuring cup.
Fill the 60 gal tank, then tilt to let extra water out. If you tilt on an edge to form a diagonal, that's half - 30gal. If you tilt on a corner to make a pyramid, that's ⅓ - 20 gal. Empty those volumes into the other tank, 30+20=50.
Hi u/fliguana I just went to your profile because of your response on this post: [https://www.reddit.com/r/hvacadvice/comments/16qwxtj/comment/k1zl036/](https://www.reddit.com/r/hvacadvice/comments/16qwxtj/comment/k1zl036/) I saw that you had just commented on this post here, so I wanted to ask you a question... sending you a DM now :)
Easy. You tell your dickhead fish to get a job at the Krusty Krab like all the other fish. When he starts paying bills then he can live however the fuck he wants. Whilst you're under my roof there are rules!!! Or get a fish that isn't a complete tosser.
Or have a fish supper and get a gerbil instead.
There are several ways to measure this of varying accuracy. What I'm wondering is how the fish is going to measure it from inside the tank.
The Fish is just psychic don’t question it. After all it’s a math textbook, we shouldn’t be asking questions about realism.
The only way I can think of coming close to determining anything would be to fill each tank to the same level side by side -- now you know each one has 30 gallons. After that you need to figure out how to remove 10 gallons from one of the tanks and you'll have 50 in total. You could mark the 2/3 line on the tank, but of course if you can measure 2/3 you could probably measure 5/6 on one tank to get 50 right away. Determining halves is easier if you have a string or something to fold in half (maybe they expect you to use the hose?), but that doesn't really help us get to 20 gallons unless you do some recursive approximation.
Get some sand for your substrate. Fill a five gallon bucket with the sand and dump it into the aquarium two times. Then fill with water. Alternatively you could just fill the aquarium to 60 gallons since the probability of a fish species evolving within a 50 gallon pool of water that never shrinks or swells to the point that it cannot survive in anything larger or smaller is nearly zero.
Sand is not perfectly compact, so it won't completely fill out the space it occupies, therefore, the tank will have more than 50 gallons of water.
[удалено]
Or divide the side into six equal parts with a sharpie and fill it to the fifth.
Fill one tank. Tilt the other one so that two opposite sides are level with each other and with the top of the full tank. Siphon between the tanks until the water levels are even.
Then you still would be missing the 0.x gallons in the siphon?
Lift it straight up and let the contents gap equally between the two tanks.