If you have One hundred pounds of water counting the weight of the container on a scale reading 100lbs, and you drop a live 10 lb northern into it, What does the scale read? 
Theres actually an answer to this one.
If you’ve heard this before stay quiet. I want to hear the theories.
First one to get it right with the proper explanation gets a big ego.
I’d like to be the first to say that a scale can’t read 
But as for the reading, I would think as long as the northern is totally submerged (since it is alive) and no water is displaced from the container….it would weigh 110 #’s. Those are my observations/guesses.
Chris
Ok, I am going to go ultra obvious on this as I have spent quite a bit of time think about this. I may have some more guesses for you if this one is not correct. Let me ask one more question if the below answer is not correct…Do I have to answer with a specific answer in LBS?
My First Answer:
The scale reads how much the container, water and fish weigh after the fish was added…
Best to start dumb and get more complicated.
Thanks Matt this is keeping me entertained… 
O.K. If the water is not displaced the weight would still be 100# because the northerns swim bladder keeps him neutrally boyuant, therefor adding no weight to the water!
I think!
i think 100lb because the fish took up 10 lb of water that it diplaced
The scale reads weight in pounds
210 pounds??? 
100 lb tank + 10 lb fish = 110 lb fish tank. That’s my theory. 
I think Jake has it
All BS aside, The scale reads weight in pounds. No tricks here. The 100# of water and the 10 pound fish are all in the container with nothing leaving the container.
WOOHOO!!! I GOT IT!!!
Time for a 
I’ll go with FishOn’s theory.
Barry
Wait…. what is the specific gravity of Northern?
Barry
I will split the difference of the neutrally bouyant and obvious theory and go with 105#.
I am going with Bigfife.
My head hurts.
I am old so please stop confusing me
Ron
That depends on how much water did the northern drank since hes been submerged in the water. 
Got out my bong and thought about this. If you have 100lbs of water counting the container, the container must be water also. Throw in a live northern and it will probably swim away and the scale will still read 100 lbs.
Well pack that baby up again and keep thinking
I thought about it and changed my theory. If the fish is alive and is neutrally bouyant, the tank will still weigh 100lbs, but if the fish dies and floats to the top or sinks to the bottom, the tank will weigh 110 lbs. The fish does not weigh anything if it is alive and neutrally bouyant, otherwise it would sink to the bottom. Sidenote, 100 lbs of water is roughly 12 gallons. If you consider the weight of the tank, it’s probably less than 12 gallons. That’s a pretty small tank to keep a 10lb northern in. Will it even fit in there? Probably, but I’m just nit picking.
I tried the scientific approach. Took a big Rubbermaid container in the bathroom, put it on the scale, and filled it with water till it got around 100 pounds. Couldn’t find a 10 pound northern, but did manage to snag my wife’s 13 pound cat. Figured that was close enough for this experiment. Tossed the cat in the water and lo and behold, learned two new things, and reinforced one I already knew. The one I knew was cats don’t really like water. I can now personally attest to that one. It’s a fact. New item number one. Wives do NOT like soaking wet cats running around the house. 
Honestly, what’s the big deal? That’s what towels are for. New item number two. A 13 pound long haired cat will absorb approximately 1.8 pounds of water in 3 seconds and deposit it around the bathroom in less than one second. 
Unfortunately, the cat didn’t hang around long enough for me to get an answer to the original question. 
Good one 
it would depend on how many onions and spices you put in there. E=mc2!
I’m working on the answer but so far this is all I have. Load one for me Putz.
c2 = a2 + b2 – a·sin(B)·b·sin(A) – a·sin(B)·b·sin(A) + 2ab·cos(A)·cos(B)
= a2 + b2 + 2ab(cos(A)cos(B) – sin(A)sin(B))
= a2 + b2 + 2ab(cos(A + B))
= a2 + b2 + 2ab(cos(180o – C))
= a2 + b2 – 2ab(cos(C))
Sgt. Rock, You are either on something, or on to something.
I say 105. Every 10lb northern I’ve caught actually weighs closer to 5lbs.
I fish, therefore I lie.
Quote:
I say 105. Every 10lb northern I’ve caught actually weighs closer to 5lbs.
I fish, therefore I lie.
Its gotta to be 110#… A cooler of water and ice certainly feels heavier than the water by itself. You can’t make 10 #lbs disappear…it had to be displaced somewhere.
And its a tie between Fish-on and Mile. However Mile gave the proper explanation. If the fish is alive its weightless in the water. If you don’t get it, think of how much a helicopter weighs hovering……0. Putting the live fish in the tank is the same thing as tying a rock to a string and dropping it in the water and letting it hang there;The scale wouldn’t read different in this scenario. 
Now put your bongs away!
If the fish was lying on the bottom of the tank the scale would read more but not 10# more. Everything has a bouyancy force in or underwater. If the fish floated to the top Only the portion of the fish above water would show on the scale.
Nice work, we should have a riddle forum. These kinds of things drive me nuts. It took me heavy thinking for one hour while I was drunk at a family get together. It was one those, “well you’re so smart, figure this one out” I almost had to call putz and bring my own munchies.
You couldnt be more wrong. The answer is 110 pounds.
This is the old “would you rather have 10 pounds of bricks or 10lbs of feathers fall on your head”, answer, it doesnt matter.
Adding a 10lb northern, floating or not, would still add 10lbs to the scale. Following you logic, what happens to taking 100,000 of water on a scale and dropping a boat it? The water displaces, but the actual weight increases. Ice floats, but if you have 6oz of water and add ice, it gets heavier. A northern, rock, piece of foam, wood or anything will add weight.
For proof-look into the law of conservation of mass.
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