# Physics Beginner!

1. Jun 1, 2008

### SBains88

1. A poster advertising a student election candidate is too large according to the election rules. The candidate is told she must reduce the length and width of the poster by 15.0%. By what percentage will the area of the poster be reduced?

Im not sure how to go about this problem. Im just beginning in physics and am kind of confused. If anyone can help much would be appreciated!

2. Jun 1, 2008

### konthelion

First, ask yourself what shape is the poster and what is the area of that shape?

3. Jun 1, 2008

### SBains88

The shape itself is most likely a rectangle and the area is not given. Im not sure how estimate that. That is why this is somewhat confusing I feel like there is a lack of information in the question

4. Jun 1, 2008

### dynamicsolo

You don't need to know the actual area of the poster, but just how you would calculate the area. What is the formula?

Now if you reduced the length and the width each by 15%, how would the new length and width compare with the original values? What would you find for the new area?

5. Jun 1, 2008

### SBains88

The formula is LxW. I would think to find the new formula u would substract the percentage that is being taking off. So (L-15%)(W-15%)= new area

6. Jun 1, 2008

### konthelion

Ah, you can not do it like that. Example, Let's suppose length = 100. Then 100-15%=100-.15= 99.85 which is not equal to 85%(i.e. 15% off) of 100 which is 85.

If you do it like that then It is L-.15*(blank). What is blank? From then, you can simplify it further.

7. Jun 1, 2008

### SBains88

Im not sure what you mean. What is the blank suppose to indicate

8. Jun 1, 2008

### konthelion

It's some variable kind of like x or like a

If you are taking 15% off, then L-.15*x. What is x? Then simplify this expression.

9. Jun 1, 2008

### konthelion

I'll make it simple. L-.15L = L(1-.15) .85L (which can be interpreted as 85% of L). This can be done with W as well. After you've done that, find the new area in terms of L and W. Then you should be able to compute the percentage off from the original area.

10. Jun 1, 2008

### SBains88

Is x the percentage that comes off of 100%? So would it be 85%?

11. Jun 1, 2008

### konthelion

I was just using x to test what you've already know about percentage. In this case, x=L. If you are taking a percentage of a number, so you must convert that percentage into a decimal and multiply by that number. Since you are taking 15% off, then you must subtract from the original number
i.e. number - .15*number which is the same as .85* number

12. Jun 1, 2008

### SBains88

so would the width and the length come out to have the same number? .85L and .85W? How would I compute the existing area based off the original area without have containing a variable in the problem?

13. Jun 1, 2008

### konthelion

No, you can not assume that L=W. You don't need to know the numerical value... for the area nor the width or length. You are figuring out the percentage change in area.
Old Area = LxW.
Okay, good. Since you've now know that the New Area=(.85L)(.85W), simplify this equation. Use basic algebra. Then, use the definition of a percentage change and find it.

i.e. $$change_{area} = \frac{Area_{new}-Area_{old}}{Area_{old}}*100$$

I'm sleepy. You can figure this out by yourself. It's straightforward.

Last edited: Jun 1, 2008