When some quantity changes, such as a price or the amount of students, we can measure either the absolute change ("The price increased by $5" or "There were 93 less students this year"), or the percent change.
In percent change, we express WHAT PART of the original quantity the change was.
For example, if a gadget costs $44 and the price is increased by $5, we measure the percent change by first considering WHAT PART $5 is of $44. Of course the answer is easy: it is 5/44 or five forty-fourths parts.
To make it percent change, however, we need to express that part using hundredths and not 44th parts. this happens to be easy, too. As seen in my previous post, you COULD make a proportion to find out how many hundredths 5/44 is:
5/44 = x/100
To solve this, you simply go 5/44 x 100, which is easy enough to remember in itself. In fact, this is the rule often given: you compare the PART to the WHOLE using division (5/44), and multiply that by 100.
There were 568 students one year, and 480 the next year. By how may percent did the student population decrease?
You first calculate the absolute change, which is 568 − 480 = 88. Then we find what part of the original population is 88 (it is 88/568), and express that using hundredth parts (percents):
88/568 x 100 = 15.49%
The student population decreased by 15.49%.
Often we are given the opposite problem: we know the percent of change and the original situation, and are asked about the new situation.
The price was $4.55 and increased by 14.78%. What is the new price?
Here, we'd need to find the price increase, or the absolute change in price first. We know the percent part of the total (it is 14.78/100) and the total amount, so multiplying those we get the part as a dollar-amount: 14.78/100 x $4.55 = $0.67249. So this is the increase. To find the new price, add the increase to the original: $4.55 + $0.67249 = $5.22249 = $5.22.
Instead of multiplying by 14.78/100, it is far quicker to multiply by 0.1478 — or to change the percent-amount 14.78% to a decimal 0.1478 and multiply by it.
And, since in the end we need to add the original total, the whole calculation looks like this:
0.1478 × $4.55 + $4.55
Here, using distributive property we can make it look like this:
= $4.55 (0.1478 + 1) = $4.55(1.1478) = 1.1478 × $4.55
So it can all be done in one multiplication. Instead of multiplying by the decimal 0.1478, you add 1 to it before multiplying.
Then one more possible problem type is that you know the percent of change and the actual change amount (absolute change), and are asked the original and/or the new total.
The price increased by 13%, or by $10.14. What was the original price?
Let the original price be p. Then you can build an equation based on the idea that the price increase is 13% of p:
0.13p = $10.14 or 13/100p = $10.14
p = $10.14/0.13 = $78.
The new price would be found by adding.
Some other lessons to read are below:
Percent Of Change - Lesson and Problems
General Increase and Decrease Examples from Purplemath.com
Percent of change calculator - enter the original and the changed quantities, and it calculates the percent of change.
PERCENT INCREASE OR DECREASE lesson from TheMathPage.com
Selasa, 28 Agustus 2007
Sabtu, 25 Agustus 2007
A free download of a digital Algebra 1 book
Kinetic Books Algebra 1 looks really interesting! It is not really just a book, but software, or a digital interactive textbook.
It contains text, interactive problems and activities, and a scoring system all on the computer. Students can get step-by-step assistance in the form of audio hints and one-click access to relevant examples.
See a demo here. But the best is that the company Kinetic Books is even offering a free download of the product till September 30!
That really sounds fantastic, so if you have algebra 1 student(s), don't fail to take advantage of this tremendous offer.
It contains text, interactive problems and activities, and a scoring system all on the computer. Students can get step-by-step assistance in the form of audio hints and one-click access to relevant examples.
See a demo here. But the best is that the company Kinetic Books is even offering a free download of the product till September 30!
That really sounds fantastic, so if you have algebra 1 student(s), don't fail to take advantage of this tremendous offer.
Jumat, 24 Agustus 2007
Equation wizard
Back last spring I promised I'd write something about this tool, so here goes.
Equation Wizard is a software, a tool, that solves first, second, third, and fourth degree equations, simplifies expressions, and calculates values of complex expressions.
I had my assistant use it when checking and making answers to my Algebra 1 worksheets.
Based on our experience, the tool works really well and was useful, for example with rational expressions, or checking answers to equations.
The two features I was missing were:
1) The ability to solve (even simple) systems of equations. There's quite a bit of work when solving a bunch of these by hand!
2) The ability to give exact roots (in our case to second-degree equations). It only gave them as decimals.
See screenshots and more here:
Equation Wizard
You can even get this software for free, with something called "TrialPay".
TrialPay allows you to purchase products by trying something else. Sign up with any one of our preselected partners and we will pay for your product.
Kamis, 23 Agustus 2007
Measure the circumference of the earth - contest
I got word of an interesting contest where school children will form teams and attempt to measure the circumference of the Earth using the same method as Eratosthenes used back in ancient times.
Any students from USA, Mexico, and Peru can form these teams, whether homeschooled, after-schooled, public schooled or whatever.
Whether you will participate or not, go see the animation that explains the method Eratosthenes used (in the left sidebar).
This sounds like an exciting opportunity to connect geometry, measuring, and math history in a project!
And here's some more information:
Any students from USA, Mexico, and Peru can form these teams, whether homeschooled, after-schooled, public schooled or whatever.
Whether you will participate or not, go see the animation that explains the method Eratosthenes used (in the left sidebar).
This sounds like an exciting opportunity to connect geometry, measuring, and math history in a project!
And here's some more information:
Please help us get the word out on this new, exciting student centered event!
Measure Your World!
Join us this fall as we pilot a new student-centered project where teams from the United States, Chile, and Mexico partner to replicate the technique introduced by Eratosthenes to determine the circumference of the Earth. Around 240 BC, Eratosthenes used trigonometry and knowledge of the angle of elevation of the Sun at noon in Alexandria and in Syene to calculate the size of the Earth. Windows to the Universe, Educared, and CREA are working together to offer school children in the U.S., Chile, and Mexico the opportunity to form partnerships, take local measurements, and collaborate using the Eratosthenes method to Measure Your World.
All of the information necessary to participate in this pilot student project can be found on the Measure Your World Web sites (www.measureyourworld.org and www.MideTuMundo.org). Student teams must have a parent or adult sponsor to participate. At least one of the team members or adult sponsors must be fluent in both English and Spanish. This event is open to all students in the three participating countries and does not have to be affiliated with a formal K-12 school. Home-schooled children and children participating in after-school programs (e.g. the Scouts, 4-H, etc.) are welcome to participate.
In addition to taking the measurements and calculating the circumference of the Earth, student teams will be encouraged to learn more about their partners in the other participating countries. Suggested activities to promote cultural exchange can be found on the Web site.
Registration for the Measure Your World event will be open from August 13 — September 14, 2007. Student teams will be notified of their partners by September 21, 2007. The time period for taking the measurements will be September 29 till October 7, 2007.
Changes in the blog appearance
I upgraded the blog template to the new one that Blogger provides, and then added the searchable "labels" in the side bar.
So now you can click on any of those "labels" (down on the right side), and find my past posts on that topic. I've written nearly 300 posts since I started the blog (in late 2005). Of course not all of those posts are of mathematical topics, but there is still quite a bit of material that is still as good as ever.
Hope this new feature improves the functionality of this blog. Enjoy reading!
So now you can click on any of those "labels" (down on the right side), and find my past posts on that topic. I've written nearly 300 posts since I started the blog (in late 2005). Of course not all of those posts are of mathematical topics, but there is still quite a bit of material that is still as good as ever.
Hope this new feature improves the functionality of this blog. Enjoy reading!
Selasa, 21 Agustus 2007
Some percent basics
The word "percent" means "per hundred", as if dividing by hundred — a hundredth part of something.
We treat some quantity (say 65 or $489 or 1.392 or anything) as "one whole". This "one whole" is then divided to hundred equal parts in our minds, and each such part is one percent of the whole.
If the "one whole" is 650 people, then 1% of it would be 6.5 people (if you have a practical application, you'd need to round such an answer to whole peoples of course).
If the "one whole" is $42, then 1% of it is $0.42. Also, 2% of it would be $0.84.
So to find 1% of something, divide by 100.
To find 24% or 8% or any other percentage, you can technically first find the 1%, then take that times 24 or 8 or whatever is your percentage.
For example:
To find 7% of $41.50, first go $41.50/100 to find 1% or 1/100 of $41.50, then multiply that by 7. But this is the same as (7/100) x $41.50, and 7/100 is 0.07 as a decimal. In most calculations, it is more practical to use decimals instead of this "divide by 100, then multiply" stuff.
So to find 7% of $41.50, I simply calculate 0.07 x $41.50 with a calculator.
To find 10% of something, you could first divide by 100 and then multiply by 10, but it's far quicker to simply divide by 10.
For example:
10% of 90 is 90/10 = 9.
10% of 250.6 is 25.06.
When you can find 10%, it's so easy to find 20%, 30%, 40%, etc., and 5% of anything just by using the 10% as a starting point.
For example:
20% of 52. First find 10% of 52 as 5.2, then double that: it is 10.4.
Example:
A gadget costs $48 and is discounted by 15%. What is the new price?
Imagine the price $48 is divided to 100 equal parts. Then you take a way 15 of those parts. That leaves 85 of those parts - what is the dollar amount that is left? Remember you're not taking away $15 but 15% of the total.
The student needs to realize that $48 is 100% - a "one whole", and 15 of those 100 parts will be taken away.
Solution:
10% of $48 would be $4.80.
5% of $48 would be $2.40 (half of 10%).
So 15% of $48 is $7.20. Subtract that from the orignal price to get the discount price of $40.80.
With a calculator, I'd go 0.85 x $48. (MAKE SURE YOU FIGURE OUT WHERE THE 0.85 COMES FROM!)
Of the class of 34 students, 12 are girls. How many percent of the class are girls?
Here, the "one whole" is 34, the whole class. The problem is, if that 34 "one whole" was divided to 100 parts, how many of those parts would we need to make 12 students?
Or, you could compare 34 people side-by-side with 100 "something". Imagine all those 34 people put head-to-toe so they form a long line, and those 12 girls are at the one end of that line. If you'd find 100 equal-size measuring units that would total exactly the same length as your people-line, how many of those measuring units would the 12 girls equal?
This easily leads to a percent proportion:
12/34 = x / 100
Solving x, you'd get
x = (12/34) × 100.
After you do this kind of proportion a few times, you notice that each time we just compare the part to the total using division, such as 12/34 in my example. So it's quite fast then to just write that directly, when solving "what percent" problems.
For example:
A $199 guitar was discounted by $40. How many percent discount was that?
Here, the "one whole" is the original (total) price, $199. It's simply asking how many percent is 40 of 199? Just calculate 40/199, and multiply the given decimal by 100 (which is easy to do mentally).
Hope this helps some. We'll tackle the percent of change next time.
We treat some quantity (say 65 or $489 or 1.392 or anything) as "one whole". This "one whole" is then divided to hundred equal parts in our minds, and each such part is one percent of the whole.
If the "one whole" is 650 people, then 1% of it would be 6.5 people (if you have a practical application, you'd need to round such an answer to whole peoples of course).
If the "one whole" is $42, then 1% of it is $0.42. Also, 2% of it would be $0.84.
So to find 1% of something, divide by 100.
To find 24% or 8% or any other percentage, you can technically first find the 1%, then take that times 24 or 8 or whatever is your percentage.
For example:
To find 7% of $41.50, first go $41.50/100 to find 1% or 1/100 of $41.50, then multiply that by 7. But this is the same as (7/100) x $41.50, and 7/100 is 0.07 as a decimal. In most calculations, it is more practical to use decimals instead of this "divide by 100, then multiply" stuff.
So to find 7% of $41.50, I simply calculate 0.07 x $41.50 with a calculator.
To find 10% of something, you could first divide by 100 and then multiply by 10, but it's far quicker to simply divide by 10.
For example:
10% of 90 is 90/10 = 9.
10% of 250.6 is 25.06.
When you can find 10%, it's so easy to find 20%, 30%, 40%, etc., and 5% of anything just by using the 10% as a starting point.
For example:
20% of 52. First find 10% of 52 as 5.2, then double that: it is 10.4.
Example:
A gadget costs $48 and is discounted by 15%. What is the new price?
Imagine the price $48 is divided to 100 equal parts. Then you take a way 15 of those parts. That leaves 85 of those parts - what is the dollar amount that is left? Remember you're not taking away $15 but 15% of the total.
The student needs to realize that $48 is 100% - a "one whole", and 15 of those 100 parts will be taken away.
Solution:
10% of $48 would be $4.80.
5% of $48 would be $2.40 (half of 10%).
So 15% of $48 is $7.20. Subtract that from the orignal price to get the discount price of $40.80.
With a calculator, I'd go 0.85 x $48. (MAKE SURE YOU FIGURE OUT WHERE THE 0.85 COMES FROM!)
How many percent is it?
Of the class of 34 students, 12 are girls. How many percent of the class are girls?
Here, the "one whole" is 34, the whole class. The problem is, if that 34 "one whole" was divided to 100 parts, how many of those parts would we need to make 12 students?
Or, you could compare 34 people side-by-side with 100 "something". Imagine all those 34 people put head-to-toe so they form a long line, and those 12 girls are at the one end of that line. If you'd find 100 equal-size measuring units that would total exactly the same length as your people-line, how many of those measuring units would the 12 girls equal?
This easily leads to a percent proportion:
12/34 = x / 100
Solving x, you'd get
x = (12/34) × 100.
After you do this kind of proportion a few times, you notice that each time we just compare the part to the total using division, such as 12/34 in my example. So it's quite fast then to just write that directly, when solving "what percent" problems.
For example:
A $199 guitar was discounted by $40. How many percent discount was that?
Here, the "one whole" is the original (total) price, $199. It's simply asking how many percent is 40 of 199? Just calculate 40/199, and multiply the given decimal by 100 (which is easy to do mentally).
Hope this helps some. We'll tackle the percent of change next time.
Minggu, 19 Agustus 2007
Master's degree in mathematics teaching and learning
First of all, I've updated my post about the percent problem — just scroll down to it.
Then, I thought maybe I have some math teachers in my readership that might be interested in a new Master's degree program offered by the university of Drexel, in collaboration with the Math Forum!
Knowing how much expertise the folks at Math Forum have this might be a unique opportunity for those math teachers who want to extend their education.
Online Master's in Mathematics Learning and Teaching - "Preparing teachers to incorporate creative, problem-based, student-centered instruction in their classroom."
The rest of you... can just continue reading my blog : )
I will post some more about the concept of percent soon.
Then, I thought maybe I have some math teachers in my readership that might be interested in a new Master's degree program offered by the university of Drexel, in collaboration with the Math Forum!
Knowing how much expertise the folks at Math Forum have this might be a unique opportunity for those math teachers who want to extend their education.
Online Master's in Mathematics Learning and Teaching - "Preparing teachers to incorporate creative, problem-based, student-centered instruction in their classroom."
The rest of you... can just continue reading my blog : )
I will post some more about the concept of percent soon.
Kamis, 16 Agustus 2007
Latex to images - online tool
Here's a handy math tool for those who know Latex (university folks and such). You type in a n mathematica expression using Latex language, and it makes an image. It even gives you a readily copyable code you can paste to a webpage.
Texify.com.
Here's an example of one such image; it's hotlinked from their server.
Texify.com.
Here's an example of one such image; it's hotlinked from their server.
Rabu, 15 Agustus 2007
So many percent more
Updated with an answer... see below
I'm continuing to catch up after vacation, and spotted a good discussion about problems with percent, at MathNotations. (via Let's Play Math blog).
Here's a problem to solve, first of all:
The answer is NOT that 40% are boys and 60% are girls...
You see, let's say there were 40 boys and 60 girls, 100 students total. If there are 40 boys, then 20% more than that would be 40 + 4 + 4 = 48 girls and not 60!
Try solve it. I'll let you think a little before answering it myself. Don't just rush over to the Mathnotations blog either! Use your thinking caps! I've already given you a big hint!
Update:
You can easily solve this problem by taking any example number for the number of boys. Like I did above, if you have 40 boys, you'd need 48 girls and there'd be 88 students total. What percent of the seniors are girls then? It'd be 48/88 = 0.545454... ≈ 54.55%. And 45.45% are boys.
But the same works even if you choose that there'd be 10 boys, which then means that there are 12 girls, and then the percent of girls in the class is 12/22 * 100% = 54.55%.
NOTE that this problem includes two DIFFERENT "wholes". First of all, it says "There are 20% more girls than boys in the senior class." This is a comparison, and the total number of boys is the "one whole" or the "100%". The number of girls is 20% more, or 120% of the boys.
In algebra terms, if there are p boys, then there will be 1.2p girls.
Then the final question involves a totally different "one whole" or 100%: it asks how many percent of the seniors are girls.
So the group of seniors becomes the 100% or the "whole", and all percent calculations are based on that. Therefore one will then compare the number of girls to the total number of seniors.
In algebra terms, the final answer as a decimal is 1.2p/(p + 1.2p) = 1.2/2.2
See also Denise's post about searching for the 100% in percent problems.
I'm continuing to catch up after vacation, and spotted a good discussion about problems with percent, at MathNotations. (via Let's Play Math blog).
Here's a problem to solve, first of all:
There are 20% more girls than boys in the senior class.
What percent of the seniors are girls?
The answer is NOT that 40% are boys and 60% are girls...
You see, let's say there were 40 boys and 60 girls, 100 students total. If there are 40 boys, then 20% more than that would be 40 + 4 + 4 = 48 girls and not 60!
Try solve it. I'll let you think a little before answering it myself. Don't just rush over to the Mathnotations blog either! Use your thinking caps! I've already given you a big hint!
Update:
You can easily solve this problem by taking any example number for the number of boys. Like I did above, if you have 40 boys, you'd need 48 girls and there'd be 88 students total. What percent of the seniors are girls then? It'd be 48/88 = 0.545454... ≈ 54.55%. And 45.45% are boys.
But the same works even if you choose that there'd be 10 boys, which then means that there are 12 girls, and then the percent of girls in the class is 12/22 * 100% = 54.55%.
NOTE that this problem includes two DIFFERENT "wholes". First of all, it says "There are 20% more girls than boys in the senior class." This is a comparison, and the total number of boys is the "one whole" or the "100%". The number of girls is 20% more, or 120% of the boys.
In algebra terms, if there are p boys, then there will be 1.2p girls.
Then the final question involves a totally different "one whole" or 100%: it asks how many percent of the seniors are girls.
So the group of seniors becomes the 100% or the "whole", and all percent calculations are based on that. Therefore one will then compare the number of girls to the total number of seniors.
In algebra terms, the final answer as a decimal is 1.2p/(p + 1.2p) = 1.2/2.2
See also Denise's post about searching for the 100% in percent problems.
Selasa, 14 Agustus 2007
Geometry fun with GeoMag
While on vacation, a friend of mine gave my older daughter a set of Geomag. You might already know about it, but it was new for us.
This has proved to be a fantastic learning toy! She's thoroughly enjoying building various shapes.
For example, she made a cube with sides 2 bars long and was proudly explaining to me how to do it: "First do a square, then put legs up from each corner, and then another square."
I made a tetrahedron that also had 2 bars on each side, according to the model. She thought it was neat and built that one several times herself last night.
I can see how the toy can help build geometric insight and beautifully demonstrate the common three-dimensional figures.
We've already ordered another set to accompany the small 42-piece set she got. You can find Geomag kits of various sizes and colors at Amazon
.
Senin, 13 Agustus 2007
A girly math book
I was traveling for two weeks so that's why this blog has been quiet... but now I'm back with tons of emails to go through and so.
One of them had an interesting link:
Math Doesn't Suck - a book whose cover looks like a girl magazine, but inside it should be solid fractions, percents, ratios and similar middle school math topics.
It's written in a "girly" style to middle school age girls, by the actress Danica McKellar (best known as Winnie Cooper in "The Wonder Years", inspiring them with the idea that math is not far from girls' life and activities. The book includes mentions of baking cookies, fashion, makeup, and such girly topics.
She wants to encourage girls to study math so that more of them would take it in college. But the problem is so many start dropping out and getting behind during the middle school years, and especially because of not understanding fractions and related topics.
You can read an interview of the author here and another good one here.
Disclaimer: I haven't read this myself. Some might not like it because of the style; talking about makeup, fashion, cheer leading and others. See some discussion about that here — the book sure has created controversy. I personally might have to be selective and careful, if I was using it with my daughters.
One of them had an interesting link:
Math Doesn't Suck - a book whose cover looks like a girl magazine, but inside it should be solid fractions, percents, ratios and similar middle school math topics.
It's written in a "girly" style to middle school age girls, by the actress Danica McKellar (best known as Winnie Cooper in "The Wonder Years", inspiring them with the idea that math is not far from girls' life and activities. The book includes mentions of baking cookies, fashion, makeup, and such girly topics.
She wants to encourage girls to study math so that more of them would take it in college. But the problem is so many start dropping out and getting behind during the middle school years, and especially because of not understanding fractions and related topics.
You can read an interview of the author here and another good one here.
Disclaimer: I haven't read this myself. Some might not like it because of the style; talking about makeup, fashion, cheer leading and others. See some discussion about that here — the book sure has created controversy. I personally might have to be selective and careful, if I was using it with my daughters.
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