The fundamental problem with this word problem and its formula is that the tank never gets empty. So there is a discrepancy between theory and practice. It reminds me of the story of the hare and the turtle. They agreed to have a race against each other and because the hare could run much faster than the turtle, the turtle was given some meters ahead before they started the race. And then each time the hare covered half the distance between between him and the turtle, and then half of half that distance again and so on. So the hare never could pass or even match the turtle and therefore the turtle won the race, much to the surprise of the hare. But in reality it obviously works differently, even though it seems acceptable theoretically. :)
But it's funny how you redefined "t" later in the problem. It started out representing the amount in a full tank and then it ended up being the number of strokes.
A comment from the UK about language. I have never heard of a quarter (i.e. 25% or 0.25 or 1/4) described as "one-fourth", and we would never put "th" after the 1/4.
3/4^n or 3/4 ^8 * 100%. I think I was right. My logic was different from yours, but I think it basically was the same. Being lazy and not having a pencil handy, I did not do the math. Therefore, I get 50%?? On test.
Given: Enclosed tank starts with full supply of air. Attached pump removes 1/4 (25%) of remaining air each stroke. I.E. N : number of strokes A : fraction of remaining air This very much paralles compounded interest... I.E. a sum of money is deposited and every year earns 5% interest... So there's interest upon interest. With the tank/pump the first stroke removes 25% yielding 75%... 0 75 × 100 = 75% remaining. The next stroke removes 25% of 75% for remainder of 56.25%. Third: 25% of 56.25% = 42.19% create a formula..... and this will result in a value raised to a power. Actually... it simply reduces to A = (0.75)^N so A = (0.75)^8 = 10.0% And realize.... the tank is NEVER completely empty. It approaches 0 but is always greater than... Asymptotic?
Can we conclude that algebra not always gives us the result reality offers? If the tank were filled with milk for instance, the pump should empty the tank completely. But this formula learns us that this is not the case :)
If 1/4 of the air that remains in the tank is removed with each stroke that means that 3/4 of the air remains, so after 8 strokes that would be (3/4) x (3/4) eight times. 3/4 = 0.75 and so here we go 0.75 ... 0.5625 ... 0.4218 ... 0.3164 ... 0.3273 ... 0.1779 ... 0.1334 ... 0.1001 and that's 8 times so it's 0.1001 x 100 10 %
btw, in this video only, my view of your screen on iphone and computer, even at expanded view, does not show the last line of writing on yours, just fyi.
I guess this problem out if reach for me ! I have an air compressor and if I pump air my car tire ms and each time i pump air and each time remove 1/4 of the air its going to be empty after 4 pumps. This sounds like math politicians would use.
A car tires shape can vary with change in volume and it is not a good analogy for this problem. In this problem, think of the air being pumped out of a container with fixed sides. Then after each stroke, the remaining air would expand to fill up the volume of the tank.
You are on a word problem kick right now …Word problems strikes fear in many many math students LOL I tell them you have to slow downnnnnnn and take a deep breath And slowly think about it Read it 3 to 4 times .So one can understand what is going on. The fear goes away .and with not to much help or none at all they can solve it by themselves
Hydraulics deal with liquids. This problem deals with gasses because it is stated that a quarter of the AIR is removed. The remaining air expands to fill the volume of the tank after each stroke. This is a discussion about a vacuum pump.
Nothing excuses poorly written English but I understood what the question meant. If I saw that question written as such in a student's paper, I'd berate the examiner.
2 is full tank capacious, then all zero is remanined , but equation is not needed , 1/4th eight times is 2 only, so, tank exhsusted , but 1/4 will be there ✌😊✌, And equation for 1/2 Is X-8×1/4 Or, 4x-2 X = 2/4 = 1/2 , its general , but the real value for initial is 2 , Cause if initial was half, then post 8 strokes , it will be 1/2- eight times 1/4 = 1/2 × 8 -1/4 = 1/2 -2 = 1.5 remaining . But that is not possible, nor initial caould be 1/4 Then 1/4-2 = 2 remaining , that as nothing consumed . But if initial is 2 then 2-2 = 0 remaining . These cant be backed by maths its practical iq needed , cause onevequation has different yields from diffetent approach to it, but with Bodmas its practical , not any other simplification or algebra method ✌😂✌😊✌😂✌
No it is 1/4 of what is left after the last stroke,but an air pump is not a good tangible example .A sump pump working on some container would make more sense
The fundamental problem with this word problem and its formula is that the tank never gets empty. So there is a discrepancy between theory and practice. It reminds me of the story of the hare and the turtle. They agreed to have a race against each other and because the hare could run much faster than the turtle, the turtle was given some meters ahead before they started the race. And then each time the hare covered half the distance between between him and the turtle, and then half of half that distance again and so on. So the hare never could pass or even match the turtle and therefore the turtle won the race, much to the surprise of the hare. But in reality it obviously works differently, even though it seems acceptable theoretically. :)
Nice word problem.
"Of" is a code word for multiply, most often used with fractional multipliers. 1/4 of 60 is 15, because 1/4 * 60 is 15.
But it's funny how you redefined "t" later in the problem. It started out representing the amount in a full tank and then it ended up being the number of strokes.
Yeah, V=volume; t=times(it happened)
A comment from the UK about language. I have never heard of a quarter (i.e. 25% or 0.25 or 1/4) described as "one-fourth", and we would never put "th" after the 1/4.
It's fine, we never expect the British to get English correct anyway. 😂
3/4^n or 3/4 ^8 * 100%. I think I was right. My logic was different from yours, but I think it basically was the same. Being lazy and not having a pencil handy, I did not do the math. Therefore, I get 50%?? On test.
Given:
Enclosed tank starts with full supply of air.
Attached pump removes 1/4 (25%) of remaining air each stroke.
I.E. N : number of strokes
A : fraction of remaining air
This very much paralles compounded interest... I.E. a sum of money is deposited and every year earns 5% interest... So there's interest upon interest.
With the tank/pump the first stroke removes 25% yielding 75%... 0 75 × 100 = 75% remaining. The next stroke removes 25% of 75% for remainder of 56.25%. Third: 25% of 56.25% = 42.19%
create a formula..... and this will result in a value raised to a power.
Actually... it simply reduces to
A = (0.75)^N
so
A = (0.75)^8
= 10.0%
And realize.... the tank is NEVER completely empty. It approaches 0 but is always greater than... Asymptotic?
Can we conclude that algebra not always gives us the result reality offers? If the tank were filled with milk for instance, the pump should empty the tank completely. But this formula learns us that this is not the case :)
If an open system, then 100%, if a closed system. Then 100%*(1-1/4)^8= 10.0113%, like time value of money with a negative interest rate
That’s how I did it - except I didn’t multiply it by 100%, because that’s just multiplying the rest times 1 (which doesn’t change anything).
@@thetooginator153 it outputs the answer in %, as requested by the question. Thx
Air left = 10% of original. (approx.) exact 6561/65536
If 1/4 of the air that remains in the tank is removed with each stroke that means that
3/4 of the air remains, so after 8 strokes that would be (3/4) x (3/4) eight times.
3/4 = 0.75 and so here we go 0.75 ... 0.5625 ... 0.4218 ... 0.3164 ... 0.3273 ... 0.1779
... 0.1334 ... 0.1001 and that's 8 times so it's 0.1001 x 100 10 %
btw, in this video only, my view of your screen on iphone and computer, even at expanded view, does not show the last line of writing on yours, just fyi.
A little over 10 percent if each stroke of the pump leaves 75 percent of what remained in the tank after the previous stroke.
Yes,but I think a sump pump emptying something would be a better example...
never came up with an answer this problem was unsolvable.
This video should have 10 pct as long too and it could have been understood
I guess this problem out if reach for me ! I have an air compressor and if I pump air my car tire ms and each time i pump air and each time remove 1/4 of the air its going to be empty after 4 pumps. This sounds like math politicians would use.
A car tires shape can vary with change in volume and it is not a good analogy for this problem. In this problem, think of the air being pumped out of a container with fixed sides. Then after each stroke, the remaining air would expand to fill up the volume of the tank.
why round up ??? I got 9.8037
3/4😌
6
You are on a word problem kick right now …Word problems strikes fear in many many math students LOL I tell them you have to slow downnnnnnn and take a deep breath And slowly think about it Read it 3 to 4 times .So one can understand what is going on. The fear goes away .and with not to much help or none at all they can solve it by themselves
ten.
I design hydraulic systems, please explain how this is practically possible.
Hydraulics deal with liquids. This problem deals with gasses because it is stated that a quarter of the AIR is removed. The remaining air expands to fill the volume of the tank after each stroke. This is a discussion about a vacuum pump.
(3/4)^8, obviously (multiplied by 100 to get it in percent); how can this video possibly be almost 14 minutes long?
Yes but the prof is nice .. about 3/4 to the 8th power should have been the length of the video ..
Nothing excuses poorly written English but I understood what the question meant. If I saw that question written as such in a student's paper, I'd berate the examiner.
I'm guessing the desired answer is approx. 10% but a real pump the answer would be 0% after 4 strokes.
Unsure about zero, now I'll watch the vid.
Yes in real life pumps will such equivalent volume at each stroke. This can really confuse kids later on in practical applications.
This again is a false answer, like the 1/5 air tank one.
If you have a 100l tank 4 strokes and its gone.
2 is full tank capacious, then all zero is remanined , but equation is not needed , 1/4th eight times is 2 only, so, tank exhsusted , but 1/4 will be there
✌😊✌,
And equation for 1/2
Is
X-8×1/4
Or, 4x-2
X = 2/4 = 1/2 , its general , but the real value for initial is 2 ,
Cause if initial was half, then post 8 strokes , it will be 1/2- eight times 1/4 = 1/2 × 8 -1/4
= 1/2 -2
= 1.5 remaining . But that is not possible, nor initial caould be 1/4
Then
1/4-2 = 2 remaining , that as nothing consumed .
But if initial is 2 then 2-2 = 0 remaining . These cant be backed by maths its practical iq needed , cause onevequation has different yields from diffetent approach to it, but with Bodmas its practical , not any other simplification or algebra method ✌😂✌😊✌😂✌
No it is 1/4 of what is left after the last stroke,but an air pump is not a good tangible example .A sump pump working on some container would make more sense
3/4
0%