few_shots: 
Example 1:
Problem/activity:
Task
Materials
One 10-frame per pair of students (see example below)
[IMG]
Counters, 10 per pair of students
Decks of number cards 0-5 (with 4 of each number) or two 0-5 dice
Setup
This activity is done in pairs. The students sit side by side with 10-frame and deck of cards or dice. If they have cards, the cards are shuffled and placed face down in front of them.
Action
Player 1 flips a card or rolls the die. They place that many counters on the 10-frame. Player 2 flips a card or rolls the die. They place that many more counters on the 10-frame next to the counters already on the ten frame. The students count the total together.
A recording sheet can be added when the students have been introduced to equations and number sentences. For those students who can handle higher numbers, a deck of 0-10 cards and two 10-frames can be used.
This game can also be played alone.


Description: 
Represent addition and subtraction with objects, fingers, mental images, drawings (Drawings need not show details, but should show the mathematics in the problem. (This applies wherever drawings are mentioned in the Standards.)), sounds (e.g., claps), acting out situations, verbal explanations, expressions, or equations.

Answer: 
Yes. This task involves representing addition with objects such as number cards and dice.

Example 2:
Problem/activity:
Task
 Arrange the fractions in order from least to greatest. Explain your answer with a picture.
 $\frac{1}{5}, \frac{1}{7}, \frac{1}{3} $
 $\frac{2}{5}, \frac{2}{7}, \frac{2}{3} $
 $\frac{5}{6}, \frac{3}{6}, \frac{1}{6} $
 $\frac{5}{12}, \frac{8}{12}, \frac{4}{12} $

Description: 
Explain equivalence of fractions in special cases, and compare fractions by reasoning about their size.
 Grade 3 expectations in this domain are limited to fractions with denominators 2, 3, 4, 6, and 8.

Answer: 
Yes. By arranging the fractions in this problem, students learn to compare fractions by reasoning about their size.

Example 3:
Problem/activity:
Problem 1
Solve using the standard algorithm. If you get stuck, use an area model and/or the partial products algorithm to help. Then assess the reasonableness of your answer.
a.
$$5,803\times46$$
b.
$$63\times2,116$$
Problem 2
Dunkin' introduced a croissant doughnut in 2014. In the first three months of its debut, they sold 7,870 dozen of them per day. How many croissant doughnuts were sold in the first three months?


Description: 
Fluently multiply multi-digit whole numbers using the standard algorithm.

Answer: 
Yes. This set of problems involves multi-digit whole numbers because the numbers in each problem are four and two digit integers, and the mathematical calculations involved in both problems focus on multiplication. The first problem explicitly asks the student to solve its subproblems using the standard algorithm.

Now, assess whether the following problem or activity aligns with the provided description.

shot_info: im_task_001477_pos_0

K

im_task_000157_pos_0

3

fl_problem_000674_pos_0

5
standard_description: Find whole-number quotients of whole numbers with up to four-digit dividends and two-digit divisors, using strategies based on place value, the properties of operations, and/or the relationship between multiplication and division. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models.
problem_activity: Narrative
The goal of this activity is to use the distributive property to explain why multiplying a number by a fraction greater than one increases the size of the number while multiplying by a fraction less than one decreases the size of the number. Expressions are particularly useful here because they show explicitly how the size of the number relates to the product. For example writing
\(\frac{3}{5}\)
as
\(1 - \frac{2}{5}\)
and then multiplying by
\(\frac{4}{7}\)
gives:
\(\displaystyle \left(1 - \frac{2}{5}\right) \times \frac{4}{7} = \frac{4}{7} - \left(\frac{2}{5} \times \frac{4}{7}\right)\)
The revealing part of this calculation is that the structure of the right hand side shows that it is less than
\(\frac{4}{7}\)
without calculating the exact value (MP7). It must be less than
\(\frac{4}{7}\)
because it is
\(\frac{4}{7}\)
minus some other number.
Engagement: Internalize Self-Regulation.
Provide students an opportunity to self-assess and reflect on their own progress. For example, provide students with questions that relate to the size of the factors for them to reflect on once the activity is complete.
Supports accessibility for: Conceptual Processing, Attention, Memory
Launch
Groups of 2
Activity
1–2 minutes: quiet think time
8–10 minutes: partner work time
Monitor for students who use the expressions in the first problem to make the comparisons and then generalize about what happens when you multiply a number by any fraction greater than 1 or less than 1.
Student Facing
Rewrite each expression as a sum or difference of 2 products.
\(\left(1 - \frac{2}{5}\right) \times \frac{4}{7}\)
\(\left(1 + \frac{1}{5}\right)\times \frac{4}{7}\)
\(\left(1 - \frac{3}{8}\right)\times \frac{4}{7}\)
\(\left(1 + \frac{1}{8}\right)\times \frac{4}{7}\)
Fill in each blank with
\(<\)
or
\(>\)
to make the inequality true.
\(\left(1 - \frac{2}{5}\right)\times \frac{4}{7} \,\underline{\hspace{0.9cm}} \,\frac{4}{7}\)
\(\left(1 + \frac{1}{5}\right)\times \frac{4}{7} \,\underline{\hspace{0.9cm}}\, \frac{4}{7}\)
\(\left(1 - \frac{3}{8}\right)\times \frac{4}{7} \,\underline{\hspace{0.9cm}} \,\frac{4}{7}\)
\(\left(1 + \frac{1}{8}\right)\times \frac{4}{7} \,\underline{\hspace{0.9cm}} \,\frac{4}{7}\)
Describe the value of the product when
\(\frac{4}{7}\)
is multiplied by a fraction greater than 1. Explain your reasoning.
Describe the value of the product when
\(\frac{4}{7}\)
is multiplied by a fraction less than 1. Explain your reasoning.
Student Response
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Activity Synthesis
Invite students to share their expressions for the products in the first problem.
Display the equation:
\(\left(1 - \frac{2}{5}\right)\times \frac{4}{7} = \frac{4}{7} - \left(\frac{2}{5} \times \frac{4}{7}\right)\)
“How can you see that the value of the expression is less than
\(\frac{4}{7}\)
?” (It’s
\(\frac{4}{7}\)
minus something.)
“Does this reasoning also work for
\(\left(1 - \frac{3}{8}\right)\times \frac{4}{7}\)
?” (Yes, it’s again
\(\frac{4}{7}\)
minus some other number.)
“Will this reasoning work whenever you multiply a number less than 1 by
\(\frac{4}{7}\)
?” (Yes, I’ll always get
\(\frac{4}{7}\)
minus an amount so that’s less than
\(\frac{4}{7}\)
.)


aligns: no