# Thread: Help with story problems PLEASE!

1. ## Help with story problems PLEASE!

1. A gambler sets up a game in which she puts four $10 bills and one$20 bill in one box. Blindly grab one of the bills and keep it. To break even on average, how much must the gambler charge the player to win?

2. Brittany pays $4.00 for a square piece of wood, which she makes into a stop sign by cutting the corners off. What is the cost of the wasted parts? Any help would be greatly appreciated! 2. To break even means you are left with at the end the same amount that you started with. So to break even, the person would need to gain in fees, what they could lose in dollars. If you set up an equation relating the number of X people paying Y dollars to play, you can figure this out. For the second question lets think about the shape of a square with a hexagon on the inside. The sides of a hexagon are all equal, and each angle measure 120 degrees yes? Using this information I can label these sides of my hexagram, and figure out the area (general area since we have no dimensions here) of ONE cut off corner. Hopefully this helps. 3. Still a little confused! 4. On the first one: How much has the woman put up of her own cash? What would be an equation that defined her profits? On the second one, have you drawn a diagram? Do you know how to draw a hexagon, and what its properties are? In this case the hexagon has all sides equal, with the internal angle having degree measure of 120. Thus, if you inscribe a hexagon inside of a square, the four "corner" will all be equilateral triangles. 5. Hello, epetrik! 1. A gambler sets up a game in which she puts four$10 bills and one $20 bill in one box. A player randomly tales one of the bills and keep it. To break even on average, how much must the gambler charge the player to play? The player will draw: . $\begin{array}{c}\text{a \10-bill }\frac{4}{5}\text{ of the time} \\ \\[-4mm] \text{a \20-bill }\frac{1}{5}\text{ of the time} \end{array}$ In five games, the gambler can expect to pay out$10, $10,$10, $10,$20
. . a total of $60 over 5 game. That is, he can expect to pay out an average of $\frac{\60}{5} \:=\:\12$ per game. To break even, the gambler should charge$12 per game.

2. Brittany pays $4.00 for a square piece of wood, which she makes into a stop sign by cutting the corners off. What is the cost of the wasted parts? A stop sign is tranditionally a regular octagon. Code:  : - - - x - - - : - * - - * * * * - - * : | * * | : | * * | : * * : * * x * * : * * : * * : | * x-2a * | a : | * * | - * - - * * * * - - * : a : x-2a: a : The square board is $x\text{-by-}x$ feet. . . She paid$4 for $x^2$ square feet of wood.

Let the corner pieces be $a\text{-by-}a.$
Then the side of the octagon is $x-2a.$

At the lower-right, we have an isosceles right triangle
. . with equal sides $a$ and hypotenuse $x-2a$.

Pythagorus: . $(x-2a)^2 \:=\:a^2 + a^2 \quad\Rightarrow\quad 2x^2 - 4xa + x^2 \:=\:0$

Quadratic Formula: . $a \;=\;\frac{4x \pm\sqrt{8x^2}}{4} \;=\;\left(\frac{2\pm\sqrt{2}}{2}\right)x$
Since $a < x$, we have: . $a \:=\:\frac{2 - \sqrt{2}}{2}\,x$

The area of the four triangles is: . $A \;=\;2a^2 \;=\;2\bigg[\frac{2-\sqrt{2}}{2}\,x\bigg]^2 \;=\;(3-2\sqrt{2})x^2$ square feet

The fraction of wasted wood is: . $\frac{(3-2\sqrt{2})x^2}{x^2} \:=\:3-2\sqrt{2}$

The cost of the wasted wood is: . $(3-\sqrt{2})(\4) \;\approx\;\0.68$

6. Ha. Where did I get a hexagon?