## Sunday, April 23, 2017

### Tricky Algebra Mathcounts National Questions: Counting Backwards

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#24 1993 National Sprint: Bottle A contains more Diet Coke than Bottle B. Now do the following:
a. Pour from Bottle A into B as much Diet Coke as B already contains.
b. Pour from B into A as much Diet Coke as A now contains.
c. Pour from A into B as much Diet Coke as B now contains.
Both bottles now have 64 ounces. How many more ounces were in A than in B at the beginning?

#30: 1993 National Sprint: Auggie spent all of his money in 5 stores. In each store, he spent \$1 more than one-half of what he had when he went in. How many dollars did Auggie have when he entered the first store?

#25: 1998 AMC-8  Three generous friends, each with some money, redistribute the money as follow: Amy gives enough money to Jan and Toy to double each amount has. Jan then gives enough to Amy and Toy to double their amounts. Finally, Toy gives enough to Amy and Jan to double their amounts. If Toy had 36 dollars at the beginning and 36 dollars at the end, what is the total amount that all three friends have?

Solution I: Use Algebra:
#24: Let A contains x ounces and B contains y ounces and x > y (given).
After first pouring, A has (x - y) ounces left and B has 2y ounces (double the original amount)
After second pouring, A has ( 2x - 2y)(double the amount) ounces and B has (3y - x) ounces left.
After third pouring, A has (3x - 5y) ounces left and B has (6y - 2x) (double the amount)
3x - 5y = 64   times 2 for each terms      6x - 10y = 128  ----equation 3
6y - 2x = 64   times 3 for each terms      18y - 6x = 192  ---- equation 4
equation 3 + equation 4 and you have 8y = 320 and y = 40 ; Plug in any equation and you get x = 88
88 - 40 = 48 ounces

Solution II: Solving it backwards:
At the end,both A and B have 64 ounces, which is after same amount of Diet Coke being pour from A to B.
Thus before action C, A has 64 + half of 64 = 96 ounces and B has 32 ounces. [Make sure you understand this]
With the same reasoning, before action B, B has 32 + half of 96 = 80 oz. and A has 48 ounces.
Again, use the same strategy, you have before action A, A has 48 + half of 80 = 88 and 40.
The difference is 88 - 40 = 48 oz.

Solution I: Use Algebra
#30: Let Auggie had x dollars at the beginning. At the first store, he would spent 1 + (x/2) = (2+x)/2 and would have x - (2 +x)/2 = (x-2)/2 left
At the second store, he would spend 1 + (x-2)/4 and would have (x-2)/2 - 1 - (x-2)/4 or (x-6)/4 left
At the third store, he would spend 1 + (x-6)/8 and would have (x-14)/8 left
It looks like there's a pattern. At the fourth store, he would spend (x-30)/16
and at the 5th store he would spent (x-62)/32 = 0 so x - 62 = 0 and x = 62 dollars

Solution II: Work backwards
Since Auggie spent all his money at the 5th store. If there are x dollars left before he spent the money all at the 5th store. You can set up the equation such as this:  x = 1 + 1/2 of x (according to the given)
So at the 5th store, he had 2 dollars.
Use the same strategy, if he had y dollars before he spent the money at the 4th store, he had
y = 1 + 1/2 of y + 2 ; y = 6
Use the same method, Aggie had 16 before he spent at the 3rd store, 30 before the 2nd store and finally,
62 dollars at the beginning.

#25: The total sum of what Amy, Jan, and Toy have stay constant so use Toy's amount to solve this problem.

Amy          Jan         Toy
?               ?             36
First round Amy gave Jan and Toy double the amount of what each of them has, so
Amy           Jan         Toy
?                ?            72
Second round Jan gave Amy and Toy double the amount of what each of them has, so
Amy           Jan         Toy
?                 ?            144
Third round Toy gave Amy and Jan double the amount of what each of them has an at the end Toy has 36 dollars        Amy           Jan            36
That means that  at the second round, Amy + Jan = 144 - 36 = 108 dollars.
So they total have 108 + 144 = 252 dollars.