Mathcounts Prep -- Number Sense

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Problems: (Solutions below)

#1: 2005 Chapter Team-- A standard deck of playing cards with 26 red cards and 26 black cards is split into two piles, each having at least one card. In pile A there are six times as many black cards as red cards. In pile B, the number of red cards is a multiple of the number of black cards. How many red cards are in pile B?

#2:  2000 State sprint #30. Joe bought a pumpkin that cost cents more per pound than his sister's. Together, the two pumpkins weighed pounds, but Joe's pumpkin was heavier. Joe paid dollars and his sister paid cents. How many pounds did Joe's pumpkin weigh?











 

Solutions :

#1: You know the total cards in pile A is a multiple of 7 because there are six times as many black cards as the red cards. (given)

6 Black, 1 Red on pile A gives you 20 Black and 25 Red cards on pile B. (doesn't work)

12 Black and 2 Red cards on pile A gives you 14 Black and 24 Red cards on pile B. (doesn't work)

18 Black and 3 Red cards on pile A gives you 8 Black and 23 Red cards on pile B. (doesn't work)

24 Black and 4 Red cards on pile A gives you 2 Black and 22 Red cards on pile B. Yes!!

The answer is 22 Red cards.

#2: 

Solution I : 

Let x dollars be the cost per pound for Joe's sister's pumpkin and x + .1 dollars are the cost per pound for Joe's pumpkin. Since the pounds of each pumpkin is the cost cost per pound, we have

. Solving, . Cost per pumpkin for Joe is , so

Solution II:  Make a list:

Joe's sister        Joe

1 lb.                 19 lb.       (doesn't work since 19 x 58 cents are too much)

2 lbs.                18 lbs      (doesn't work)

3 lbs.                17 lbs      (No)17 x (48/3 + 10) = 442 (still too much)

4 lbs                 16 lbs      16 x (48/4 + 10)= 352 (yes)

The answer is 16 pounds.

This Week's Work : Week 44 - for Inquisitive Young Mathletes

From Mathcounts Mini : More Constructive Counting

Also try the warm-up, retry the problems in the video on your own and spend some time
pondering on those follow-up, harder problems.

#7 and 9 are more challenging.

Review 2014 harder school round problems, especially the "Harvey" method.

Try the first 8 2012 AMC-10 B questions again and see if you have the speed.

Have fun problem solving !! Mrs. Lin

This Week's Work : Week 43 - for Inquisitive Young Mathletes

From Mathcounts Mini : Using Similarity to Solve Geometry Problems

Also try the warm-up, retry the problems in the video on your own and spend some time
pondering on those follow-up, harder problems.

#4, 5, 6 are state level problems. #7 is challenging.

From Mathcounts Mini : Tangent Segments and Similar Triangles

#8 is state/national level.

Have fun problem solving !! Mrs. Lin

Problem Solving is Fun

There is a goat, says the question. The goat is tethered to one edge of a barn, on a rope 10 meters long. If the barn is 9 meters long and 5 meters wide, and the goat cannot enter the barn, what is the total area that the goat can cover?

To my eleven-year-old mind, this question brought up countless other questions: what, for example, would inspire the goat to traverse as much distance as possible in the first place? Would it not be more sensible for the goat to double 'round and chew off the rope, then escape the tyranny of the barn and traipse off to whatever happy meadows were certainly waiting in store? What exactly is the farmer hiding in the barn that is so important that he will not even allow an innocent goat to see it?

The problem with the stereotypical math questions posed in high school is that they all look something like this: If x^2 - 6x + 5 = 0, what is x? Such a question gives no incentive to solve it. There is no sympathy for 'x', no tears shed over the tragedy of a beleaguered animal. All the question does is to continue to drill factoring and quadratics into brains already well-tired of such things.

Stemming from this is my proposition on how to make math 'cool'. Rather than the continuous drilling of formulas and equations, problems in math classes should be set up like the problems offered on standard math competitions such as the MathCounts programs and the AMCs. The goat question is a perfect example - it provides the blending of a possible real-life situation (if any mathematicians are striving to become farmers with curious goats) with the concept of sectors of circles. Schools offer the basics; almost all high school-students know that the area of a circle is "pi-r-squared". However, I have offered the goat question - a standard on several MathCounts state tests - to a roomful of honors-level high-schoolers during my time as MathLeague coach, and been rewarded with a bunch of blank stares.

"We don't know how to do that," say my teammates. "Why is there a goat? The goat is confusing us!"

And here's the problem - schools force the memorization of formulas, but do nothing to teach applied math. In order to really understand math, students should be offered questions which challenge their imagination and understanding. It's not hard, either - after explaining the goat question once, all of the MathLeague students present at the lecture could solve any similar questions; all they needed was a moment's exposure to competitional math.

Additionally, several of the - well, granted, not-very-mature - students suddenly caught a glimpse of how I, in all my madness, could call math 'fun'. "This is kind of cool," one boy admitted, after a ten-minute discussion on how we could find the number of diagonals that a convex pentagon - whimsically named DUCKS - had. "I like ducks."

So it's easy, really, to make math fun. Amuse the students - put math in situations they've never seen before, with not only goats, but ducks, and cows, and even once, to the chagrin of our MathLeague supervisor who had already been losing faith in the human race, a llama. Offer new ways to play with probability - probability as used in gambling, for example, which tends to amuse students to no end.

And, finally, challenge the kids, because that's the only way they're going to go from memorizing formulas to applying them. After all, my teammates can now say quite confidently each time they see a question dealing with circles, "I've got your goat."

2014 Mathcounts State Prep : Inscribed Circle Radius and Circumscribed Circle Radius of an equilateral triangle

#1: P is the interior of equilateral triangle ABC, such that perpendicular segments from P to each of the sides of triangle ABC measure 2 inches, 9 inches and 13 inches. Find the number of square inches in the area of triangle ABC, and express your answer in simplest radical form.

#2: AMC 2007-B: Point P is inside equilateral  ABC. Points Q, R, and S are the feet of the
perpendiculars from P to AB, BC, and CA, respectively. Given that PQ = 1,PR = 2, and PS = 3, what is AB ?
#3:

This is an equilateral triangle. If the side is "S", the length of the in-radius would be\(\dfrac {\sqrt {3}} {6}\) of  S (or \(\dfrac {1} {3}\)of the height) and the length of the circum-radius would be\(\dfrac {\sqrt {3}} {3}\) of  S (or \(\dfrac {2} {3}\)of the height).

You can use 30-60-90 degree special right triangle angle ratio to get the length of each side as well as the height.

Solution I: Let the side be "s" and break the triangle into three smaller triangles.

\(\dfrac {9+13+2} {2}\)= 12s (base times height divided by 2)= \(\dfrac {\sqrt {3}} {4}\times s^{2}\)

s = 16√ 3 

Area of the triangle = 192√ 3



                                           
 

Solution II: Let the side be "s" and the height of the equilateral triangle be "h"

24* s (by adding 9, 2 and 13 since they are the height of each smaller triangle)  = s*h 
(Omit the divided by 2 part on either side since it cancels each other out.)

h = 24
Using 30-60-90 degree angle ratio, you get \(\dfrac {1} {2}\) s = 8√ 3  so  s = 16√ 3 

Area of the equilateral triangle = \(\dfrac {24\times 16\sqrt {3}} {2}\) = \(192\sqrt {3}\) 

#2: This one is similar to #1, the answer is 4√ 3