Easy math puzzles
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53 members have voted
August 19th, 2026 at 6:12:07 PM
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Quote: DogHandQuote: WizardYou have a 100-gallon tub full of pure water. Someone pours two gallons of salt into it, which is immediately absorbed. Pure water flows into the tank at a rate of five gallons per minute. Meanwhile, water leaves a hole in the bottom at the same rate. How long will it take for the water to contain 0.1 gallons of salt?
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Wiz,
Here is my solution:
Dog Hand
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I have a slightly difference answer.
At time t=T min, salt=0.1 gal, water=100 gal.
At time t=x, salt=y. We have dy/dx=-5y/100; therefore, the general solution to the differential equation is
𝑦=A* exp(-x/20)=2*exp(-x/20).
Where we have used: at time t=0 min, salt=2 gal, water=100 gal. Therefore
x=-20 ln(y/2)= -20 ln (0.1/2) = 59.915 min.
August 19th, 2026 at 7:18:57 PM
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Quote: DogHandQuote: WizardYou have a 100-gallon tub full of pure water. Someone pours two gallons of salt into it, which is immediately absorbed. Pure water flows into the tank at a rate of five gallons per minute. Meanwhile, water leaves a hole in the bottom at the same rate. How long will it take for the water to contain 0.1 gallons of salt?
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Wiz,
Here is my solution:
Dog Hand
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I agree based on your interpretation of the question. I meant to say the initial 2 gallons of walt was immediately absorbed, like a sponge, into the pure water. Thus, the starting volume of saltwater was still 100 gallons. However, I'll take the blame for the ambiguous wording and give you credit. This one was beer worthy.
"No great mind has ever existed without a touch of madness." -- Aristotle
August 19th, 2026 at 8:13:45 PM
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In chemistry, volumes are not additive. The volume of the final solution will be between 100.8 and 101.5 gallons, depending on whether the 2 gallons of salt are measured as a loose powder or a solid block.
Gambling is a math contest where the score is tracked in dollars. Try not to get a negative score.
October 7th, 2026 at 7:17:52 PM
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There is a city with a square grid of three streets running east to west and 3 running north to south, creating 4 blocks. A drunk is at the north west corner of the grid. His home is at the south east corner. Every turn he walks in a random direction, all equally likely, of all available streets to him. How many turns will it take, on average, to get home?
"No great mind has ever existed without a touch of madness." -- Aristotle
October 7th, 2026 at 11:05:05 PM
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13
?
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It’s all about making that GTA
October 8th, 2026 at 4:22:43 AM
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Quote: Ace213
?
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I disagree, but could be wrong.
"No great mind has ever existed without a touch of madness." -- Aristotle
October 8th, 2026 at 7:18:16 AM
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44 4/7
Let T(r,c) be the number of turns it will take to get from (r,c) to (3,3); the solution is T(0,0)
Because of symmetry, T(r,c) = T(c,r), which reduces the number of equations involved
T(3,3) = 0
T(3,2) = 1 + 1/3 T(2,2) + 1/3 T(3,1) {+ 1/3 T(3,3)}
T(3,1) = 1 + 1/3 T(3,2) + 1/3 T(2,1) + 1/3 T(3,0)
T(3,0) = 1 + 1/2 T(3,1) + 1/2 T(2,0)
T(2,2) = 1 + 1/2 T(3,2) + 1/2 T(2,1)
T(2,1) = 1 + 1/4 T(3,1) + 1/4 T(2,2) + 1/4 T(1,1) + 1/4 T(2,0)
T(2,0) = 1 + 1/3 T(3,0) + 1/3 T(2,1) + 1/3 T(1,0)
T(1,1) = 1 + 1/2 T(2,1) + 1/2 T(1,0)
T(1,0) = 1 + 1/3 T(2,0) + 1/3 T(1,1) + 1/3 T(0,0)
T(0,0) = 1 + T(1,0)
T(1,0) = 1 + 1/3 T(2,0) + 1/3 T(1,1) + 1/3 (1 + T(1,0))
2/3 T(1,0) = 4/3 + 1/3 T(2,0) + 1/3 T(1,1)
T(1,0) = 2 + 1/2 T(2,0) + 1/2 T(1,1)
This is 8 equations in 8 unknowns; I won't bother with the details of solving them here, but T(1,0) = 305/7, so T(0,0) = 23 + 1 = 312/7
Because of symmetry, T(r,c) = T(c,r), which reduces the number of equations involved
T(3,3) = 0
T(3,2) = 1 + 1/3 T(2,2) + 1/3 T(3,1) {+ 1/3 T(3,3)}
T(3,1) = 1 + 1/3 T(3,2) + 1/3 T(2,1) + 1/3 T(3,0)
T(3,0) = 1 + 1/2 T(3,1) + 1/2 T(2,0)
T(2,2) = 1 + 1/2 T(3,2) + 1/2 T(2,1)
T(2,1) = 1 + 1/4 T(3,1) + 1/4 T(2,2) + 1/4 T(1,1) + 1/4 T(2,0)
T(2,0) = 1 + 1/3 T(3,0) + 1/3 T(2,1) + 1/3 T(1,0)
T(1,1) = 1 + 1/2 T(2,1) + 1/2 T(1,0)
T(1,0) = 1 + 1/3 T(2,0) + 1/3 T(1,1) + 1/3 T(0,0)
T(0,0) = 1 + T(1,0)
T(1,0) = 1 + 1/3 T(2,0) + 1/3 T(1,1) + 1/3 (1 + T(1,0))
2/3 T(1,0) = 4/3 + 1/3 T(2,0) + 1/3 T(1,1)
T(1,0) = 2 + 1/2 T(2,0) + 1/2 T(1,1)
This is 8 equations in 8 unknowns; I won't bother with the details of solving them here, but T(1,0) = 305/7, so T(0,0) = 23 + 1 = 312/7
Edit: I read the problem wrong - I had 4 streets in each direction
Last edited by: ThatDonGuy on Oct 8, 2026
October 8th, 2026 at 7:32:24 AM
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I'm slightly confused by the "three roads" puzzle, especially given the above solution, as my interpretation of the puzzle was different.
If there are three roads across and three roads down, then do I assume the drunk is walking down the middle of these roads, and hence there are only nine junctions/places where they can change direction.
Thus the drunk could, if lucky, go $ABEZ but also $A$CFGZ etc.
This would imply the points are (0,0)(1,0)(2,0)(0,1)...(2,2).
Or is the puzzle using the side-walks so has (0,0) thru (3,3).
If there are three roads across and three roads down, then do I assume the drunk is walking down the middle of these roads, and hence there are only nine junctions/places where they can change direction.
$AB
CDE
FGZ
This would imply the points are (0,0)(1,0)(2,0)(0,1)...(2,2).
Or is the puzzle using the side-walks so has (0,0) thru (3,3).
October 8th, 2026 at 7:45:34 AM
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18
Last edited by: ThatDonGuy on Oct 8, 2026
