*Warning: Always figure the percentage of change relative to the Standardized Test Prep ACCUPLACER Math ACT Math ASVAB Math CBEST Math CHSPE Math CLEP Math COMPASS Math FTCE Math GED Math GMAT Math GRE Math MTEL Math NES Math PERT Math PRAXIS Math SAT Math TABE Math TEAS Math TSI Math more tests...*

*Warning: Always figure the percentage of change relative to the Standardized Test Prep ACCUPLACER Math ACT Math ASVAB Math CBEST Math CHSPE Math CLEP Math COMPASS Math FTCE Math GED Math GMAT Math GRE Math MTEL Math NES Math PERT Math PRAXIS Math SAT Math TABE Math TEAS Math TSI Math more tests...*

Consider using my prefered problem solving model here. Students are used to having at least three numbers and solving for an unknown fourth when solving proportion word problems.

This can be a bit of a slippery slope as they get older and the problems get more complex, or if there is a table involved.

The format displayed above, "(this number) is (some percent) of (that number)", always holds true for percents.

In any given problem, you plug your known values into this equation, and then you solve for whatever is left.

$$a=r\cdot b$$ $$47\%=0.47a$$ $$=0.47\cdot 34$$ $$a=15.98\approx 16$$ 16 of the students wear either glasses or contacts.

We often get reports about how much something has increased or decreased as a percent of change.

What I finally got correct this year is the importance of setting up the labels.

I am not talking about writing part/whole; I am referring to describing what the part is in relation to describing the whole. If students understand that the part is related to the percent by means of the label, then they will be so much more successful in setting up the percent proportion correctly and with more confidence.

We begin by subtracting the smaller number (the old value) from the greater number (the new value) to find the amount of change.

$0-150=90$$ Then we find out how many percent this change corresponds to when compared to the original number of students $$a=r\cdot b$$ $=r\cdot 150$$ $$\frac=r$$ $[[

We often get reports about how much something has increased or decreased as a percent of change.

What I finally got correct this year is the importance of setting up the labels.

I am not talking about writing part/whole; I am referring to describing what the part is in relation to describing the whole. If students understand that the part is related to the percent by means of the label, then they will be so much more successful in setting up the percent proportion correctly and with more confidence.

We begin by subtracting the smaller number (the old value) from the greater number (the new value) to find the amount of change.

$$240-150=90$$ Then we find out how many percent this change corresponds to when compared to the original number of students $$a=r\cdot b$$ $$90=r\cdot 150$$ $$\frac=r$$ $$0.6=r= 60\%$$ We begin by finding the ratio between the old value (the original value) and the new value $$percent\:of\:change=\frac=\frac=1.6$$ As you might remember 100% = 1.

||We often get reports about how much something has increased or decreased as a percent of change.What I finally got correct this year is the importance of setting up the labels.I am not talking about writing part/whole; I am referring to describing what the part is in relation to describing the whole. If students understand that the part is related to the percent by means of the label, then they will be so much more successful in setting up the percent proportion correctly and with more confidence.We begin by subtracting the smaller number (the old value) from the greater number (the new value) to find the amount of change.$$240-150=90$$ Then we find out how many percent this change corresponds to when compared to the original number of students $$a=r\cdot b$$ $$90=r\cdot 150$$ $$\frac=r$$ $$0.6=r= 60\%$$ We begin by finding the ratio between the old value (the original value) and the new value $$percent\:of\:change=\frac=\frac=1.6$$ As you might remember 100% = 1.When I begin to teach part, whole, and percent problems, I explain to my students that there is nothing that I teach in my class that I use more often in my real life.Students reflect on where they see percents: the grocery store, sales and discounts, and their grades.Since this is a TEKS readiness skill and a skill with heavy real-world application, I think that it is important to spend several days covering it.Here is an example of how I would sequence the skill over the course of a couple of days.Example 47% of the students in a class of 34 students has glasses or contacts.How many students in the class have either glasses or contacts?

]].6=r= 60\%$$ We begin by finding the ratio between the old value (the original value) and the new value $$percent\:of\:change=\frac=\frac=1.6$$ As you might remember 100% = 1.

## Comments Solving Percentage Word Problems

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