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.
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