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.                                                                                                    
 

2013 Mathcounts State Harder Problems

 You can download this year's Mathcounts state competition questions here.

Trickier 2013 Mathcounts State Sprint Round questions :
Sprint #14:  
From Varun: 
Assume the term "everything" refers to all terms in the given set.
1 is a divisor of everything, so it must be first.
Everything is a divisor of 12, so it must be last.
The remaining numbers left are 2, 3, 4, and 6.
2 and 3 must come before 6, and 2 must come before 4.
Therefore, we can list out the possibilities for the middle four digits:
2,3,4,6
2,3,6,4
3,2,4,6
3,2,6,4
2,4,3,6
There are 5 ways--therefore 5 is the answer.

From Vinjai:
Here's how I did #14:
First, notice that 1 must be the first element of the set and 12 must be the last one.
So that leaves only 2,3,4,6 to arrange.
We can quickly list them out.
The restrictions are that 2 must be before 4, 3 must be before 6, and 2 must be before 6:
2,3,4,6          3,2,4,6          4,2,3,6            6,2,3,4
2,3,6,4          3,2,6,4          4,2,6,3            6,2,4,3
2,4,3,6          3,4,2,6          4,3,2,6            6,3,2,4
2,4,6,3          3,4,6,2          4,3,6,2            6,3,4,2
2,6,3,4          3,6,2,4          4,6,2,3            6,4,2,3
2,6,4,3          3,6,4,2          4,6,3,2            6,4,3,2

Only the bold ones work. So, the answer is 5.

#17: Common dimensional change problem
\(\overline {ZY}:\overline {WV}=5:8\) -- line ratio
The volume ratio of the smaller cone to the larger cone is thus \(5^{3}: 8^{3}\).
The volume of the frustum is the volume of the larger cone minus the volume of the smaller cone
= \(\dfrac {8^{3}-5^{3}} {8^{3}}\times \dfrac {1} {3}\times 8^{2}\times 32\times \pi\) = 516\(\pi\)

More problems to practice from Mathcounts Mini 

#24:  The answer is \(\dfrac {1} {21}\).

#28: Hats off to students who can get this in time !! Wow!!
From Vinjai:

For #28, there might be a nicer way but here's how I did it when I took the sprint round:

# 4's     # 3's     # 2's       # 1's     # ways

   1         2          0            0          3

   1         1          1            1          24

   1         1          0            3          20

   1         0          3            0          4

   1         0          2            2          30

   1         0          1            4          30

   0         2          2            0          6

   0         2          1            2          30

   0         2          0            4          15

   0         1          3            1          20

   0         1          2            3          60

   0         0          3            4          35

TOTAL: 277

#29: 

\(\Delta ADE\) is similar to \(\Delta ABC\)

Let the two sides of the rectangle be x and y (see image on the left)

\(\dfrac {x} {21}=\dfrac {8-y} {8}\)
x =\(\dfrac {21\left( 8-y\right) } {8}\)

xy =  \(\dfrac {21\left( 8-y\right) } {8}\)  * y = \(\dfrac {-21y\left( y-8\right) } {8}\) =
\(\dfrac {-2l\left( y-4\right) ^{2}+21\times 16} {8}\)

From the previous equation you know when y = 4, the area \(\dfrac {21\times 16} {8}\)is the largest. The answer is 42. 





Here is a proof to demonstrate that the largest area of a rectangle inscribed in a triangle is
half of the area of that triangle.

#30:
Solution I :
If (x, y) are the coordinates of the center of rotational points, it will be equal distance from A and A' as well as from B and B'.
Use distance formula, consolidate/simplify and solve the two equations, you'll get the answer (4, 1).

Solution II:

How to find the center of rotation from Youtube.

From AoPS using the same question

To sum up:
First, connect the corresponding points, in this case A to A' and B to B'.
Second, find the equation of the perpendicular bisector of line \(\overline {AA'}\), which is
y =  - x + 5
and \(\overline {BB'}\), which is y = 5x - 19
The interception of the two lines is the center of rotation.
The answer is (4, 1).

2013 Mathcounts Target :
#3:
RT = D, unit conversions and different rates are tested here:

Make Joy's rate (speed) uphill be x m/s, his downhill speed be 2x m/s.

It takes Greg 3000 seconds (time) to reach the starting point and that is also what it takes Joy to
ride up to the hill and down to the same point.

\(\dfrac {7000} {x}+\dfrac {10000} {2x}=3000\) \(\rightarrow\) x = 4 m/s 

#8:

Using "finding the height to the hypotenuse".( click to review)

 you get \(\overline {CD}=\dfrac {7\times 24} {25}\).

Using similar triangles ACB and ADC, you get  \(\overline {AD}=\dfrac {576} {25}\).
[\(\dfrac {24} {x}=\dfrac {25} {24}\)]

Using angle bisector (click to review),

you have \(\overline {AC}:\overline {AD}=\overline {CE}:\overline {ED}= 24: \dfrac {576} {25}\) = 600 : 576 = 25 : 24

\(\rightarrow\)\(\overline {ED}= \overline {CD}\times \dfrac {24} {24+25}\) = \(\dfrac {7\times 24} {25}\times \dfrac {24} {24+25}\) = \(\dfrac {576} {175}\)

2013 Mathcounts Nationals Sprint #29 and similar problem hints

Mathcounts problems : state/nationals level

2013 Mathcounts Natinals Sprint # 28

2013 Mathcounts National Sprint #28 : In right triangle ABC, shown here,  line AC = 5 units and line BC = 12 units. Points D and E lie on  line AB and line BC respectively, so that line CD is perpendicular to line AB and E is the midpoint of line BC. Segments AE and CD intercept at point F. What is the ratio of AF to FE ? Express your answer as a common fraction.

                                                     Solution I : Using similar triangles

                                                        Solution II : Use Mass Point Geometry

2013 Mathcounts School and Chapter Harder Problems

You can now download and discuss with your friends this year's school and chapter problems.
Here is the link to the official Mathcounts website.

Some more challenging problems from this year's Mathcounts school/or chapter problems.

2013 school team #10 : Three concepts are testing here :
Hint: 
a. If you get rid of the remainder, the numbers will be evenly divided into 192, so you are looking at
those factors of 192 - 12 = 180

b. To leave a remainder of 12, those factors of 180 that are included in the Set must be smaller than 12, otherwise, you can further divide it.

c. To find the median, make sure to line up the numbers from the smallest to the largest and find the middle numbers. If there are even numbers of factors larger than 12, average the middle two. Otherwise, the middle number is the answer.

\(180=2^{2}\times 3^{2}\times 5 \) so there are (2 + 1) (2 + 1) (1 + 1) = 18 factors

The list on the left side gives you the first 9 and if you times those numbers with "5", you get 9 other factors,which are 5, 10, 20, 15, 18, 60, 45, 90 and 180.

Discard the factors that are smaller or equal to 12 and list all the other factors in order and find the median.

The answer is "36".



2013 Chapter Sprint:
#21: Dimensional change problem : The height of the top pyramid is \(\dfrac {2} {3}\) of the larger
pyramid so its volume is \(\left( \dfrac {2} {3}\right) ^{3}\) of the larger pyramid.

\(\left( \dfrac {2} {3}\right) ^{3}\times \dfrac {1} {3}\times \left( \dfrac {36} {4}\right) ^{2}\times 12
= \) \(96  cm^{3}\)

# 24:  According to the given:   \(xyz=720\)   and   \(2( xy+yz+zx)= 484 \) so
\(( xy+yz+zx )= 242\)

Since x, y and z are all integers, you factor 720 and see if it will come up with the same x, y and z values
for the second condition.

Problem writer(s) are very smart using this number because the numbers "6", "10", "12" would give you
a surface area of 252. (not right)

The three corrrect numbers are "8", "9", and "10" so the answer is \(\sqrt {8^{2}+9^{2}+10^{2}}=\) \(7\sqrt {5}\)

#25: Geometric probability: Explanations to similar questions and more practices below. 

Probability with geometry representations form Aops.

Geometric probability from "Cut the Knots".

#26: This one is similar to 2002 AMC-10B #21, so try that question to get more practices. 
2002 AMC-10B #21 link 

#27:
\(\dfrac {1} {A}+\dfrac {1} {B}=\dfrac {1} {2}\)
\(\dfrac {1} {B}+\dfrac {1} {C}=\dfrac {1} {3}\)
\(\dfrac {1} {C}+\dfrac {1} {A}=\dfrac {1} {4}\)
Add them up and you have  \(2 * (\dfrac {1} {A}+\dfrac {1} {B}+ \dfrac {1} {C})=\dfrac {13} {12}\)

\((\dfrac {1} {A}+\dfrac {1} {B}+ \dfrac {1} {C})=\dfrac {13} {24}\)

\(\dfrac {1} {\dfrac {1} {A}+\dfrac {1} {B}+\dfrac {1} {C}} = \)\(\dfrac {{24}} {13}\) hours

#28: Hint : the nth triangular number is the sum of the first "n" natural numbers and \(\dfrac {n\left( n+1\right) } {2}\) is how you use to find the sum.
From there, you should be able to find how many numbers will be evenly divided by "7".

#29 : Circle questions are very tricky so make sure to find more problems to practice accuracy.

#30 :  
Solution I: 
Read the solution that is provided by Mathcounts.org here.
Solution II:
Case 1 : \(x-1 > 0\rightarrow x > 1\) Times ( x - 1) on both sides and you have
\(x^{2}-1>8\) so x > 3 or x < -3 (discard)

Case 2: \(x-1 < 0\) so \(x < 1\) \(\rightarrow x^{2}-1 < 8\) [You need to change the sign since it's negative.]-3 < x < 3. Combined with x < 1 you have the range as -3 < x < 1
The answer is 60%.

                                               2013 Mathcounts Target #7 and 8: 

Target question #8 is very similar to 2011 chapter team #10. 
It just asks differently.   
Read the explanations provided on the Mathcounts official website.
They are explained very well.
Let me know if there are other easier ways to tackle those problems.

Hope this is helpful !! Thanks a lot !! Good luck on Mathcounts state.