Tuesday, May 5, 2015

3.MD.3 Create a Scaled Picture Graph

3.MD.3 Draw a scaled picture graph and a scaled bar graph to represent data with several categories. Solve one-and two-step "how many more" and "how many less" problems using information presented in scaled bar graphs.  For example, draw a bar graph in which each square in the bar graph might represent 5 pets.

Evelyn and Jason surveyed two classrooms to see how many animal pets the children had at home. The data they collected showed that the students had 22 mammals, 14 birds, 4 reptiles, 8 fish, and 2 amphibians.







The class decided to make a picture graph to represent this data.

But how can we organize this information?






The graph would be too wide if each animal picture only represented one animal.

What if each animal picture represents two animals? Twenty-two mammals would need eleven dog pictures and that is still too much!









What if we make each animal picture represent four animals? We can count by fours: 4, 8, 12, 16, 20, 24.

But what do we do if there are twenty-two mammals?








Hector figured it out! We can cut one of the dog pictures in half.









Some of the children were not sure that would work, so Andrew showed everyone that if you cut one of the pictures in half then
4 = 2 + 2.

So half a picture represents two animals.








Maria explains to Jason why this works, and tells him how she made her graph.











Luis did not understand that a picture graph and a bar graph both have to start at the same point on the graph and either grow to the right or grow up.

His turtle correctly represents four reptiles, but is it glued in the right place? In addition, he is counting 11 whole animal squares as the number of total animals. Maybe his partner can help him understand what scale means.









Lupita shows the class how to find the difference of "how many more" mammals there are than fish.










William disagrees. He says, "You are right that there are 22 mammals and 8 fish. but 22 minus 8 equals 14--not 26."

Everyone struggled with #2.
Reptiles (4) + birds (14) = 18 reptiles and birds

And there are only 2 amphibians.



  
Hector shows the class how many pets the students in both classes have altogether. 
Wow, Hector!! We have a lot of pets.



Friday, December 6, 2013

4.NF.1 Performance Assessment: Leapfrog Fractions

 4.NF.1 

Explain why a fraction a/b is equivalent to a fraction  (× a)/(× b) by using fraction models, with attention to how the number and size of the parts differ even though the two fractions themselves are the same size. Use this principle to recognize and generate equivalent fractions.


See the Leapfrog problem.




  This summer I had the wonderful opportunity to tutor my friend, Dylan. He wanted to better understand fractions, so I challenged him with this MARS Task assessment. After he read the directions, I helped him think of some problem-solving strategies. In order to find the answers, Dylan made a number line graph that helped him visually compare the various fractional parts. These included: halves, fourths, thirds, sixths, eighths, and ninths. 




Using 1/2 inch graph paper, Dylan made number lines that were each 18 boxes long, so it was easy for him to make halves, thirds, sixths, and ninths. Then he figured out how to make fourths by cutting the halves in half.










Next, he was able to find the eighths by cutting the fourths in half.
In this way, he could easily compare some of the necessary equivalent fractions for this task. 







Dylan said that Frog 1 was easy to solve. He just had to determine the missing fraction that made the equation add to one, and help the frog hop from the lily pad to the island. 
So...
here is how Frog 1 hops to the island:




Dylan drew a double number line on his paper to compare halves and fourths for Frog 1.



With each step in the Leapfrog problem, Dylan was able to measure and compare the fractions on his number line graph with the fractions that were on the assessment.












Here is his solution for Frog 2:

This time Dylan's double number line compared thirds and sixths.  You can also see how he helped the frog hop across the number line to solve for the missing fraction.
 








This is Dylan's solution for
Frog 3:

His double number line compares fifths and tenths. Notice that this time these fractions were not on his number line graph sheet, so he had to decide how to compare them. He colored in the 3/5 and 1/10, but then wasn't sure what to do. I marked the missing part and wrote a question mark.





Immediately, he saw the answer! "It's 3/10," he beamed. "Too easy!"

I was so proud of Dylan when he figured out how to make equal parts with fifths and tenths and compare these with the same one whole.  Great job, Dylan!!




















Here are the Frog 3 and Frog 4 problems:

"That's terrific that you knew 6/8 was equal to 3/4," I smiled.

"Grandma and I are so impressed!!

But watch out for Frog 5," I continued.








Dylan saw that he had to add sixths and ninths, so he made another double number line and figured out how to make equal parts with each that represented the same whole.





Again, Dylan wasn't sure how to make the last hop to the island, so I showed him the missing portion on his number line and wrote another question mark.


Once more the eyes sparkled and he wrote "2/6 = 1/3."   

Fabulous!









Now for the "big enchilada"! Could Dylan figure out Frog 6 by himself?  I quietly watched as he read the problem out loud, started drawing a double number line as before, when suddenly he stopped and put down his pencil. He thought for a moment and then realized that this time he needed to draw a third number line. First, he added 1/4, then 10/20....but when he highlighted the 1/5 with the orange marker he realized that it was just a little bit short of the island.  

Wow!! He did it! I reminded him that he had to use pictures, numbers, and words to prove his answer so he wrote, "no he will not make it because it is short."   Way to go, Dylan!!  Good job!!



Here are Frog 5 and Frog 6:




And the studious man hard at work...









Happy Birthday this week, Dylan!!!   :)


Saturday, November 2, 2013

4.OA.3 "A New Coat for Anna" Technology Research Math Project

4.OA.3    Solve multistep word problems posed with whole numbers and having whole-number answers using the four operations, including problems in which remainders must be interpreted. Represent these problems using equations with a letter standing for the unknown quantity. Assess the reasonableness of answers using mental computation and estimation strategies including rounding. 


See our original movie of Anna's New Coat.

Thanks to the generosity of Tech4Learning, an educational software company, I was able to pilot their original Pixie software program in 2007. 

That year I taught 5th grade and was also involved as a technology coach at my school. While my own students were off track, I worked with a colleague's 3rd grade students for two weeks to help them create our math movie.  


Since I was familiar with the 3rd grade Open Court stories, I wanted to use "A New Coat for Anna" for a math investigation, using the Pixie draw program for the artwork.

The story of Anna takes place just after WWII in Eastern Europe. The stores are empty and few people have any money, so Mom needs to barter several items to get a new coat for her daughter, Anna. 



While reading the story, I wondered, "What would these items be worth on eBay today?"

In order to assist these 8-year old children with their research, I created a Word document of internet hotlinks. Then I brought them into the computer lab in order to help them determine the current day value of these antique items.




Some of the items Mom bartered were difficult to find exact matches for on eBay, so I found an expert jeweler to answer our questions.










The children were so engaged in the project, that they came back into the computer lab during recess, lunch, and after school. Fascinated by the project, Erick and Saul talked to their parents and had extra research time at home.







As the amounts of money grew larger and larger, the children were enthralled with the lesson!










They were certain they had won the lottery and their faces sparkled as we found similar items on eBay that matched the four items included in our search. 







Assumptions we made:

In the story, Mother barters Grandfather's gold pocket watch for bags of wool. That's expensive wool!! We decide that if Anna is 8-years old, Mom is about 32, and Grandfather is somewhere around 57 years old.

Therefore, Grandfather must have been born around 1888.



Here's how we determined that:  1945 - 57 = 1988. 



Louie told us that since the story took place in Eastern Europe, Grandfather's gold watch was probably a fine crafted Swiss watch. He also said that recently a Swiss gold pocket watch made by a famous watchmaker had sold for $2,000,000! The children were blown away! This was REAL math in the making!


It was also difficult to find a garnet necklace with a gold neck chain resembling the one our illustrator had drawn in the story. Since garnet is not a fancy gemstone, it is easier to find it fashioned in silver than in gold. We also discovered that garnet comes in different "grades" or qualities. Some cheaper grades of garnet are dyed, while others are natural. We found Grade C and Grade B, but remembered Louie's advice and decide that Mother must have had a natural blood red stone. Certainly this was Grade AAA!

Since this story takes place before modern day manufacturing, the gold garnet necklace was probably gemstone quality.  In fact, based on how the illustrator drew the picture, we were certain it was "AAA" quality grade.



Unfortunately, it was impossible to find this combination of jewelry on eBay. Still, the children learned a lot about the pricing of gemstones.   :)

Not only were they interested in the large amounts of money, now they wanted to reread the story for setting and scenery details to include in their computer drawings.



What Did It Cost 100 Years Ago?

As a child, I remember how pleased my father was when he finally made an annual salary of $10,000. He drooped out of high school in 1944 to join the air force, and later worked for IBM in 1952.

After I got married, Dad told me that his buying power in those earlier days was much greater than mine. He bought his first brand new house in Hawthorne, California, in 1947 for $7,000. I bought my first home in 1980 for $89.000. Later, my husband  and I bought a 50-year old home for $230,000. Now the children understand the meaning of inflation!

Older students can graph changes over time to the cost of common goods. 

Cost of common items years ago