few_shots: 
Example 1:
Concept/skill: 
Recognize and represent proportional relationships between quantities.

Problem/activity:
You’re mixing ingredients for cookies. The recipe says to combine 6 tablespoons, or ¾ stick, of butter with 1 cup of sugar. You accidentally mix in a whole stick of butter (8 tablespoons) with the cup of sugar. How can you fix this?


Answer: 
No. Usually problems that teach recognizing and representing proportional relationships between quantities involve deciding whether two quantities are in a proportional relationship or identifying the constant of proportionality, and then representing proportional relationships by equations or explaining proportional relationship using coordinate graphs.

Example 2:
Concept/skill: 
Apply and extend previous understandings of multiplication and division and of fractions to multiply and divide rational numbers.

Problem/activity:
For each pair of numbers, determine if the values are equivalent. Explain your reasoning for each one.
a.
$$-0.25$$
and
$$\frac{1}{-4}$$
b.
$$-\left ( \frac{1}{4} \right )$$
and
$$\frac{-1}{-4}$$
c.
$$\frac{-1}{4}$$
and
$$-\frac{1}{4}$$


Answer: 
Yes. This problem involves understanding fractions and division to divide rational numbers, such as 1, -4, 4, and -1. This problem also involves an understanding of multiplication: if p and q are integers, then -(p/q) = (-p)/q = p/(-q).

Example 3:
Concept/skill: 
Prove that, given a system of two equations in two variables, replacing one equation by the sum of that equation and a multiple of the other produces a system with the same solutions.

Problem/activity:
Problem 1
The table below contains single equations in a single variable, single equations in two variables, and pairs of simultaneous equations in two variables. Check the appropriate box to show whether there is no solution, exactly 1 solution, or if there is an infinite number of solutions.
{"table": [{"_attributes": {"align": "center", "class": ["table", "table-bordered"], "style": "table-layout:fixed; width:70%"}, "tbody": [{"tr": [{"td": [{"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}, "strong": [{"_value": "No solution"}]}, {"_attributes": {"style": "text-align:center"}, "strong": [{"_value": "Exactly 1 solution"}]}, {"_attributes": {"style": "text-align:center"}, "strong": [{"_value": "Infinite number of solutions"}]}]}, {"td": [{"_attributes": {"style": "text-align:center"}, "span": [{"_attributes": {"class": ["math-tex"]}, "_value": "$${9=4x+7}$$"}]}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}]}, {"td": [{"_attributes": {"style": "text-align:center"}, "span": [{"_attributes": {"class": ["math-tex"]}, "_value": "$${y=3y+5}$$"}]}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}]}, {"td": [{"_attributes": {"style": "text-align:center"}, "span": [{"_attributes": {"class": ["math-tex"]}, "_value": "$${y={x\\over2}}$$"}]}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}]}, {"td": [{"p": [{"_attributes": {"style": "text-align:center"}, "span": [{"_attributes": {"class": ["math-tex"]}, "_value": "$${3x+6y-9=0}$$"}]}]}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}]}, {"td": [{"p": [{"_attributes": {"style": "text-align:center"}, "span": [{"_attributes": {"class": ["math-tex"]}, "_value": "$${x=y}$$"}]}, {"_attributes": {"style": "text-align:center"}, "span": [{"_attributes": {"class": ["math-tex"]}, "_value": "$${1.25x=1.25y}$$"}]}]}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}]}, {"td": [{"p": [{"_attributes": {"style": "text-align:center"}, "span": [{"_attributes": {"class": ["math-tex"]}, "_value": "$${9.9=6x+8y}$$"}]}, {"_attributes": {"style": "text-align:center"}, "span": [{"_attributes": {"class": ["math-tex"]}, "_value": "$${9x=2.5y-8.8}$$"}]}]}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}]}, {"td": [{"p": [{"_attributes": {"style": "text-align:center"}, "span": [{"_attributes": {"class": ["math-tex"]}, "_value": "$${11x-2y=1.5}$$"}]}, {"_attributes": {"style": "text-align:center"}, "span": [{"_attributes": {"class": ["math-tex"]}, "_value": "$${11x-2y=2.5}$$"}]}]}, {"_attributes": {"style": "text-align:center"}}, {"_attributes": {"style": "text-align:center"}}, {}]}]}]}]}
Problem 2
Kim has a small container and a large container as shown.
[IMG]
It takes 16 of the small containers to fill the large container. Three small containers leave 1.95 gallons of space in the big container.
What is the size of each of the two containers?


Answer: 
No. Completing neither of these problems requires replacing one equation by the sum of that equation and a multiple of the other equation.

Now, assess whether the following problem or activity is 'aligned' with the given Common Core standard in Mathematics.

shot_info: fl_problem_000972_neg_0

6

fl_problem_001097_pos_0

7

fl_problem_001392_neg_0

8
standard_description: Understand the concept of a unit rate a/b associated with a ratio a:b with b ≠0, and use rate language in the context of a ratio relationship. For example, “This recipe has a ratio of 3 cups of flour to 4 cups of sugar, so there is 3/4 cup of flour for each cup of sugar.” “We paid $75 for 15 hamburgers, which is a rate of $5 per hamburger.” (Expectations for unit rates in this grade are limited to non-complex fractions.)
problem_activity: Warm-up
This warm-up is the first time students are asked to find
\(A\%\)
of
\(B\)
when
\(B\)
is not 100 or 1.
Students may approach the problem in a few different ways, with or without filling in values on the provided double number line diagram. For example, they may understand from the fundraising context of the problem that \$40 is 100% because it is the amount of the goal. From there, they may simply find half of \$40 for the 50% value and add that value to \$40 to find the 150% value. Other students may use equivalent ratio reasoning to calculate the value at 50% and 150%. As students work, notice the different strategies used and any misconceptions so they can be addressed during discussion.
Launch
Remind students that in the previous lesson, we found percentages of 100 and of 1 using double number lines. Explain that in this lesson we will find percentages of other numbers. Give students 2 minutes of quiet work time, and follow with a whole-class discussion. Encourage students to create a double number line to help them answer the questions if needed.
Student Facing
Each of three friends—Lin, Jada, and Andre—had the goal of raising
$
40. How much money did each person raise? Be prepared to explain your reasoning.
Lin raised 100% of her goal.
Jada raised 50% of her goal.
Andre raised 150% of his goal.
Student Response
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for free access to Student Response.
Anticipated Misconceptions
Students may be surprised by a percentage greater than 100. If they are puzzled by this, explain that Andre raised more money than the goal.
Activity Synthesis
Consider displaying this double number line for all to see.
[IMG]
Invite a few students to share their solving strategies. One way to highlight the different techniques students use is to invite several students to explain how they calculated the amount of money raised by Andre. We can find 150% of 40 in several ways. For example, we can add the values of 50% of 40 and 100% of 40. We can also reason that since
\(100 \boldcdot (1.5) = 150\)
then
\(40 \boldcdot (1.5) = 60\)
, which means \$60 is 150% of \$40.
If not uncovered in students’ explanations, explain that we are finding percentages of \$40, since this number—not 100 cents or 1 dollar—is the fundraising goal for the three friends. Since 100% of a goal of \$40 is \$40, the 100% and \$40 are lined up on the double number line.
Students who relied on the visual similarity between, for example, \$0.25 is 25% of 1 dollar in the previous lesson find this strategy unworkable here (as \$50 is not 50% of \$40). To encourage students to use their understanding of equivalent ratios to reason about percent problems, ask the class to explain—either when the above misconception arises or as a closing question—why \$50 is not 50% of \$40, but 50% of 100 cents is 50 cents.


aligns: no