Tag Archives: mixture problem

Two Containers Mixing Puzzle

This is a slightly different type of a mixture problem from Dan Griller.

“Two containers A and B sit on a table, partially filled with water.  First, 40% of the water in A is poured into B, which completely fills it.  Then 75% of the water in B is poured into A, which completely fills it.  80% of the water in A is poured into B, which completely fills it.  Calculate the ratio of the capacity of container A to the capacity of container B, and the fraction of container A that was occupied by water at the start.”

See the Two Containers Mixing Puzzle

Al the Chemist II

This is the second part of the problem from Raymond Smullyan in the “Brain Bogglers” section of the 1996 Discover magazine.

“On another occasion, Al made a mixture of water and wine. There was more water than wine—in fact, the excess of water over wine was equal to one-fourth the quantity of wine. Al then added 12 ounces of wine, at which point there was one ounce more of wine than water.

According to another version of the story, before Al added the 12 ounces of wine, he first boiled off 12 ounces of water (the net effect being that he replaced 12 ounces of water with wine), and again there was one more ounce of wine than water.

Would there be more mixture present at the end of the first version or the second?”

I found this statement a tad ambiguous with the result that I found two possible solutions: the one Smullyan gave and another, surprising one.

See Al the Chemist II

Milk Mixing Puzzle

This is a classic example of a mixture problem from Dan Griller that recalls my agonies of beginning algebra.

“In Cauchy Village, full fat milk has 3.5% fat content, semi-skimmed milk has a 1.5% fat content, and skimmed milk has a 0.2% fat content. How many liters of full fat milk must be added to 100 liters of skimmed milk to produce semi-skimmed milk?”

See the Milk Mixing Puzzle

Wine Into Water Problem

Here is a challenging problem from the 1874 The Analyst.

“A cask containing a gallons of wine stands on another containing a gallons of water; they are connected by a pipe through which, when open, the wine can escape into the lower cask at the rate of c gallons per minute, and through a pipe in the lower cask the mixture can escape at the same rate; also, water can be let in through a pipe on the top of the upper cask at a like rate. If all the pipes be opened at the same instant, how much wine will be in the lower cask at the end of t minutes, supposing the fluids to mingle perfectly?

—  Communicated by Artemas Martin, Mathematical Editor of Schoolday Magazine, Erie, Pennsylvania.”

I found the problem in Benjamin Wardhaugh’s book where he describes The Analyst:

“Beginning in 1874 and continuing as Annals of Mathematics from 1884 onward, The Analyst appeared monthly, published in Des Moines, Iowa, and was intended as “a suitable medium of communication between a large class of investigators and students in science, comprising the various grades from the students in our high schools and colleges to the college professor.” It carried a range of mathematical articles, both pure and applied, and a regular series of mathematical problems of  varying difficulty: on the whole they seem harder than those in The Ladies’ Diary and possibly easier than the Mathematical Challenges in the extract after the next. Those given here appeared in the very first issue.”

I tailored my solution after the “Diluted Wine Puzzle”, though this problem was more complicated.  Moreover, the final solution must pass from discreet steps to continuous ones.

There is a bonus problem in a later issue:

“19.  Referring to Question 4, (No. 1): At what time will the lower cask contain the greatest quantity of wine?

—Communicated by Prof. Geo. R. Perkins.”

See the Wine Into Water Problem

Three Jugs Problem Redux

I was sifting back through some problems posed by Presh Talwalkar on his website Mind Your Decisions, when I found another 3 Jugs problem, which was amenable to the skew billiard table solution from my earlier Three Jugs Problem. Here is his statement:

“A milkman carries a full 12-liter container. He needs to deliver exactly 6 liters to a customer who only has 8-liter and a 5-liter containers. How can he do this? No milk should be wasted: the milkman needs to leave with 6 liters of milk. Can he measure all amounts of milk from 1 to 12 (whole numbers) in some container?”

I also believe I found a case where Talwalkar’s solution to the last question needs revision. See the Three Jugs Problem Redux.

Diluted Wine Puzzle

This was a rather intricate puzzle from Presh Talwalkar. I found his solution a bit hard to follow, so I tried for a clearer presentation.

“A servant has a method to steal wine. He removes 3 cups from a barrel of wine and replaces it with 3 cups of water. The next day he wants more wine, so he does the same thing: he removes 3 cups from the same barrel (now with diluted wine) and replaces it with 3 cups of water. The following day he repeats this one more time, so he has drawn 3 times from the same barrel and has poured back 9 cups of water. At this point the barrel is 50% wine and 50% water. How many cups of wine were originally in the barrel? ”

See the Diluted Wine Puzzle.

Two Pints of Cider

This is another problem from the defunct Wall Street Journal Varsity Math Week column.

“Team member Janice recently visited the U.K. and poses this puzzle to her teammates: You have three containers that can hold exactly 15, 10 and 6 pints. The 15-pint container starts full of cider. You want to measure out exactly 2 pints of cider, drink it all, and end with an empty 15-pint container and 8 and 5 pints of cider in the other two containers. What transfers should you make to accomplish this?”

The solution is based on my Three Jugs Problem. See Two Pints of Cider.

Three Jugs Problem

Years ago (1967) I read about an interesting solution to the three jugs problem in a book by Nathan Court which involved the idea of a billiard ball traversing a skew billiard table with distributions of the water between the jugs listed along the edges of the table. The ball bounced between solutions until it ended on the desired value. I thought it was very clever, but I really did not understand why it worked. Later I figured out an explanation, which I present here. See the Three Jugs Problem.