January 1978 Archives

What can I say; I think math is way cool.  Always Have.  Always Will.  Like when I start with X only to find out that it's no longer equal to itself.  Kinda ...

Let x = 0.999... (to ∞)

Now we can multiply this equation by 10 to get

10x = 9.999... (to ∞)

Now we have two equations and so far OK.  But let's subtract them:

     10x  = 9.999...

-      x   = 0.999...

________________

        9x = 9.0  which of course simplifies to x = 1.

And "1" is not the number I started with, which was 0.999... (to ∞) !!!

Or does 1 really equal the first number after all: 1 = 0.999... ???

Infinity is a mischievous concept and there's no telling what it's up to!

Quote

"Don't just learn the tricks of the trade. Learn the trade." James Bennis

Fun Facts to Know and Share

111,111,111  X  111,111,111 = 12,345,678,987,654,321

Bumper Stickers that caught my eye

If you're not part of the solution, you're part of the precipitate.

What if the hokey pokey is really what it's all about?

ALGEBRA PROBLEMS                  

Problem 1:

An advertisement for an orange drink claims that the drink contains 10% orange juice. How much pure orange juice would have to be added to 5 quarts of the drink to obtain a mixture containing 40% orange juice?

Solution 1:

Hint #1:

a)  Read and think about the word problem, in English, until you are satisfied you understand it.

b)   Define your Lets.  1-What don't you know? / What are they asking you to find?

c)  What units is the unknown in?

Let X = quarts of pure OJ to be added

Hint #2:

a)  What did we start with?  How much did we start with?  How do you know this?

b)   We start with 5 quarts of orange drink, which is 10% OJ, or [0.10(5)] quarts of OJ.

c)  What must we end with?  How much must we end with? How do you know this?

d) We must end with a volume of the initial 5 quarts AND the quarts added, or (5 + X) quarts of orange drink.

The volume of orange drink we end up with must be 40% OJ,

 \is the symbol used for "therefore".

\ Initial quarts + pure quarts added = result, or better yet; Initial OJ + Pure OJ = Final OJ

0.10(5) + 1.00(X) = 0.40(5 + X)

The rest is left as an exercise for the reader.

Problem 2:

A cask is filled with 45 gallons of wine. Nine gallons are removed and the cask is refilled with water. The nine gallons of the mixture are removed and the cask is again refilled with water. What is the ratio of water to wine in the final mixture?

Solution 2:

Hint #1:

a)  Our "Let", or more appropriately, our "Unknown Variable Assignment", comes straight from the question because it asks "What is the ratio ...".

b)   There are 3 significant bits of information here: 1) A volume (gallons) of final mixture, which is 2) A final gallons of water, and 3) A final gallons of wine.  Even though these are 3 things, if we know any 2 of them, we can calculate the 3^rd. 

c)  Out of the above 3, experience and thought direct me to assign unknown variables to final gallons of pure wine and pure water.

Let X = gallons of pure water in final mixture

Let Y = gallons of pure wine in final mixture

Hint #2:

a)  This is what I consider a "Trick" question because there's something funny about it.  There are actually 2 mixtures being made instead of our usual 1.

b)   We can't make the (2^nd) final mixture until after we've made the 1^st mixture.

c)  Here's an English description of what is going on, and what needs to be going on. There are 5 steps:

d) 1) We start with 45 gallons of pure wine. 

e)  2) We remove 9 gallons of pure wine.

f)  3) We add 9 gallons of water, creating mixture #1.

g) 4) We remove 9 gallons of mixture #1. 

h)   5) We add 9 gallons of water.

1^st Mixture:

45   is the gallons of wine we start with.

 9    is the gallons of wine we remove.

 9    is the gallons of water we add.

\ 36 gallons wine / 9 gallons water = ratio of wine to water in mixture #1

This means that for every gallon of mixture #1 we remove, we are removing 4 parts wine and 1 part water; i.e. 80% wine and 20% water.

\ 0.80(9) is the gallons of wine removed from mixture #1

2^nd and final mixture.

45   is the gallons of mixture #1 we start with.

 9    is the gallons of mixture #1 we remove.

 9    is the gallons of water we add.

Now we can make our 1^st equation: Initial wine - wine removed - wine in mixture #1 we removed = wine in final fluid.

\ Y = 45 initial gallons wine - 9 gallons wine - 0.80(9) gallons wine

The rest is left as an exercise for the reader.