Problem Set Four

Second problem solving assignment. Yay math!

1) Question:
The price of each item at the Gauss Gadget store has been reduced by 20% from it's original price. An MP3 player has a sale price of $112.00. What would the same MP3 player sell for if it was on sale for 30% off it's original price?

2) Answer:
I'll just break into equations.

To begin, we should find the original price.
To find the original price, I'll divide $112.00 by 80, since that's the percent of the full price that is left (100% - 20% = 80% price, which means, x - y = $112.00).
$112.00/80 = $1.40. That means that 1% of the product is the equivalent of $1.40.
To find out the original price, we have to find out how much the 20% that was removed is worth in dollars. 20%*$1.40 = $28.00.
Now we can get the original price. As I have previously stated, 100% - 20% = 80%, so 80% + 20% = 100%. In other words, $112.00 + $28.00 = x.
The original, full price, x, is $140.00. (You can also find this by taking $1.40 and multiplying it by 100%).

Now we're looking for a 30% sale price. The original price remains constant.
So, the equation we're working with is: Original Price - 30% off = Sale Price, or $140.00 - z = a, a being the 30%-off sale price.
We can find the 30% off by using the same method as we did for finding 20%. Take $1.40 and multiply it by 30 to get the 30%. The answer is $1.40*30% = $42.00.
The amount saved with the 30% off sale is $42.00. Referring back to the equation we had earlier, that means $140.00 - $42.00 = a. $140.00 - $42.00 = $98.00.

With a 30% off sale, the MP3 player would cost $98.00.

In summary, and if you'd rather just get it all out of the way and see a dumbed-down version of the same question:
i) Take $112.00 and divide it by 80. This is because $112.00 is 80% of the original cost. You will get an answer of $1.40.
ii) Multiply $1.40 by 100 to get the full price. You will get an answer of $140.00.
iii) Multiply $1.40 by 30 to get the amount off for the 30%-off sale price. You will get an answer of $42.00.
iv) Take the full price ($140.00) and subtract the 30% sale ($42.00). You will get an answer of $98.00.

3) I wasn't originally going to do this question, but Margo brought it up so I used it. I dislike working with finding percentages of numbers, but this one turned out to be easier than I remember with percentages. So, I guess I can say I like it because I felt I am understanding percentages more.

4) I learned more about using percentages through this question, and finding amounts based on percentages that are based on amounts (make sense of that!). That's really the main thing, since I'm normally not so good with working with percentages.

The end.

Problem Set Three

First problem solving assignment. Here we go!

1) Question:
An equilateral triangle has a side length of 20. If a square has the same perimeter as the triangle, what's the area of the square?

2) Solution:
Okay, so. First we should get the shapes down.
The triangle is an equilateral. That means that each side is the same length. The square is a square -- four sides, same length for each side.
Now, each side of the triangle is 20 in length (no unit of measurement is given). There are three sides to a triangle. Three multiplied by twenty equals sixty (3*20= 60). The triangle has a perimeter of 60.
The triangle and the square share the same perimeter. That means that each side of the square must be sixty divided by four (60/4), since there are four sides.
Each side of the square, therefore, must be fifteen.
But we aren't just looking for the side length of the square. The question asks for the area.
Area equals length multiplied by width (a=l*w), or in this case, as with all squares, side length multiplied by side length (a=s*s, or a=s^2). In this problem, the area of the square equals fifteen multiplied by fifteen (a=15*15).
In answer to the question, the area of the square is 225.

3) I like this question because it shows the relationship shapes have with each other, and how those relationships can be beneficial to finding the solutions to problems that we may encounter in our work.

4) Through this exercise, I have learned that it's always important to look for relationships within the question, specific data that is relative to two or more distinct parts, that may serve as a basis to find all the needed information.

The end.

Start.

A journal for Math?
...Interesting?

Well. We'll see what happens. Stuff about math and problem solving and equations.
Hopefully, nothing about quadratic functions. =.="

Can you tell I'm just filling this with text for the sake of filling this with text?

Well.
Now you do.