Maria's Math News, Vol. 49, January 2011
Hello again! We're all starting a new year. This time the newsletter a review of a new math facts software, algebra word problems, and some other websites you might try. 1. Review of Math Rider software 2. Fraction of a fraction word problem 3. Two algebra 1 word problems (systems of linear equations) 4. Work & workers word problem 5. Tidbits 1. Review of Math Rider softwareIf your children like horses, they might fall in love with Math Rider game!Math Rider is a "math facts" software game; that is, it specifically trains and practices the basic addition, subtraction, multiplication, and division facts. Some special features include:
Here is the home screen. Continue reading my review 2. Fraction of a fraction word problemA problem about fraction of a fraction...The sixthgraders have a fundraiser. They raise enough money to reach 7/8 of their goal. Nikki raises 3/4 of this money. What fraction of the goal does Nikki raise?The picture below shows first of all 7/8. Nikki raises 3/4 of this goal. We need to find 3/4 of 7/8. It's not easy to directly see what is 3/4 of 7/8. So to do that, I divide each 1/8 piece into four pieces, and then color three of the four. That way I color 3/4 of each of the seven eighths. Of course, those tiny pieces are now 1/32 parts. I have colored 3 x 7 = 21 of them. So, the colored part represents the fraction 21/32. This problem is also simple to solve without a picture, if you understand what is asked. To find 3/4 of 7/8, you simply multiply those two fractions. The word "of" translates into MULTIPLICATION in fraction math! 3 7 21  x  =  4 8 32 3. Two algebra 1 word problems (systems of linear equations)Here are two problems for you to solve... OR to learn from me when I solve them. Both problems are for algebra 1, and use a system of 2 linear equations.By the way, the comments have some wonderful ideas for solving these mentally, without using algebra. So please read them too! Problem: John bought red pens for $4 apiece and blue pens for $2.80 apiece. If John purchased a total of 24 pens for $84, how many red pens did he purchase? Solution: This is a typical problem that will have two variables and two equations. Continue reading! 4. Work & workers word problemHere's another one of those job / workers word problems (inverse or direct variation). Try and see if you can solve it using the "table" method instead of equations:A certain job can be done by 18 clerks in 26 days. How many clerks are needed to perform the job in 12 days? Again, we can set up a table and reason this out. Initially set it up like this: jobs  clerks  days  1  18  26  1    1  ?  12Then think of the "days" column. We want to "go" from 26 to 12. You could use a proportion here... or first figure out how many clerks are needed to do this job in 2 days, and then from that go to 12 days. If 18 clerks do it in 26 days, then how many clerks would do it in 2 days... which is 1/13 the amount of time.... so we need 13 times as many clerks. 13 x 18 = 234 clerks are needed. jobs  clerks  days  1  18  26  1  234  2  1  ?  12 Now, if 234 clerks do it in 2 days, how many clerks would do it in 12 days? Now, the time increases 6fold, so we need only 1/6 as many workers. 234 / 6 = 39. So 39 clerks are needed. 5. Tidbits
Till next time, Maria Miller Miss something from the earlier volumes? See newsletter archives. Feel free to forward this issue to a friend. Subscribe here.
