## week 12

March 9, 2014 § Leave a comment

**What fraction of this square region is shaded? Stripes are equal in width, and the figure is drawn to scale.**

**Ss1**— I don’t have the exact answer, but I know it’s more than 1/2. I visually drew in grid lines.

**Ss2**— I mentally drew in grid lines like Ss1 did. It’s a 6×6 square, so 36 square units. Starting in the left corner, I saw 1 white square. Diagonally from there, I notice these are consecutive odd numbers. I just added the white parts to get 15. Subtract 15 from 36 gives 21. So, 21/36 is the answer.

**Ss 3**— I got 7/12. I split it up into smaller squares too and got 36 because it’s sides are 6 each. There are 21 shaded squares, 21/36 simplifies to 7/12.

What is the equation for the number of green dots in this pattern?

**Ss1**— I see the diagonal, then 3 groups of n on the outside, then the bottom group. My equation is

**D = (n+2) + 3n + (n+1)**

**Ss2**— I have the exact same equation as

**Ss1**, but I saw it differently, like this:

**Ss3**— Left and right, each is n. Top and bottom, each is (n+2). The leftover dot is 1 less than n. My equation:

**D = 2n + 2(n+2) + (n-1)**.

**Ss4**— I saw these 4 in the right corner in every step. My equation is

**D = 4 + 2n + 2(n-1) + (n+1)**.

**Can you figure out the rule: 2, 4, 8, …?**

*I had fun watching this video from Veritasium, so I posed the question to the kids. The quickest answer came from a 6th grader, and he even used the word “ascending” in his rule!*

What is the equation for the number of blue rods in this pattern?

**Ss1**— I see top and bottom as 2 groups of (n-1). The middle has (n+1) by (n). My equation is

**R = 2(n-1) + n(n+1)**.

**Ss2**— I move one rod from the bottom row to the top to make it a complete “rectangle” of (n+2) by (n). The leftover rods on the bottom row is (n-2). My equation is

**R = n(n+2) + (n-2)**.

**Three corn dogs cost $5.25. How much does one corn dog cost?**

*I asked this question because Justin Lanier shares how he and his friend had done this math problem mentally.*

**Ss1**— I just divided like I normally would on paper. Yea, I just set it up in my head because it’s not too bad.

**Ss2**— I knew that if the price had been $6, then each corn dog would cost $2. But it costs 75 cents less than that, so 75 divided by 3 is 25. I subtracted 25 cents from $2 to get $1.75 for each.

**Ss3**— I knew each must cost at least $1. Then I split up the leftover $2.25 into thirds: three 25-cents are not enough, three 50-cents are still not enough, so I tried three 75-cents, and that added up to $2.25.

**Ss4**— I used the divisibility rule to see if $5.25 is divisible by 3: 5+2+5 = 12, and 12 is divisible by 3, so I knew the amount is divisible by 3. I just tried some numbers and $1.75 works.

**Ss5**— I changed $5.25 to 21 quarters. 21 divided by 3 is 7 — 7 quarters is $1.75.

**Ss6**— I tried $1.50 first, that got me $4.50 for 3 corn dogs. I tried $2, and that’s $6 total. So I tried the middle number of $1.75, and it worked.

*When I shared with the students how Mr. Lanier had done the problem, they said they would never have guessed that! I said to them, “Math Munch people are just cool like that.” 🙂*

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