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?
link to original post
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.
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