# data[].attributes.attachment_link.readable_text
Alignments to Content Standards: 7.RP.A.2

Task

Coffee costs \$18.96 for 3 pounds.

1. What is the cost for one pound of coffee?

2. At this store, the price for a pound of coffee is the same no matter how many pounds you buy. Let $x$ be the number of pounds of coffee and $y$ be the total cost of $x$ pounds. Draw a graph of the relationship between the number of pounds of coffee and the total cost.

3. Where can you see the cost per pound of coffee in the graph? What is it?

IM Commentary

The purpose of this task is for students to find a unit rate in a context where two quantities are in a proportional relationship and to draw the graph of that proportional relationship.

This is a task where it would be appropriate for students to use technology such as a graphing calculator or GeoGebra, making it a good candidate for students to engage in MP5, Use appropriate tools strategically. A variant of this problem is appropriate for 8th grade; see 8.EE.5 Coffee by the Pound.

Solution

1. You can find the cost for one pound of coffee by dividing the total cost by 3. Coffee costs \$6.32 per pound.

2. We may graph the proportional relationship between the total cost and the number of pounds by plotting the line through the origin and (3, 18.96).

3. The cost of one pound, \$6.32, may be seen on the graph in two ways:

As the point (1, 6.32)

As the slope of the line: \$6.32 per pound.

Note: Students aren't explicitly required to see the connection between the unit rate and the slope until 8th grade (see 8.EE.5) but they may still see it in 7th grade.

# data[].attributes.attachment_link.description
Alignments to Content Standards: 7.RP.A.2 Coffee costs \$18.96 for 3 pounds. What is the cost for one pound of coffee? At this store, the price for a pound of coffee is the same no matter how many pounds you buy. Let $x$ be the number of pounds of coffee and $y$ be the total cost of $x$ pounds.

# data[].attributes.attachment_link.readable_text
Alignments to Content Standards: 6.RP.A.3

Task

A store has two different brands of laundry detergent. Brand A can do 80 loads of laundry and costs \$12.75. Brand B does 36 loads of laundry and costs \$6.75. Which laundry detergent costs less per load?

IM Commentary

This purpose of this task is to provide a context for comparing ratios. Many techniques can be used including making a ratio table, finding unit rates, and carefully selecting equivalent ratios which can be compared directly. The numbers selected for the task statement make it well suited for the 6th grade. If students make all of their calculations by hand then they will build skill related to the standard 6.NS.3, '' Fluently add, subtract, multiply, and divide multi-digit decimals using the standard algorithm for each operation.'' Alternatively, they can work with decimals and mental math as is shown in the third solution.

This task is based upon problem 5 from section 7.1 of Sybilla Beckmann's Mathematics for Elementary School Teachers, Pearson, 2014. Ideas from the commentary and solution have been gathered from a fall 2013 class of prospective teachers at the University of New Mexico.

Solutions

Solution: 1 Making a table

A table can be used to solve this problem in many different ways. For example we can use a table to find the cost to do 36 loads with Detergent A:

<

So detergent A is cheaper since it costs less than \$6 to do 36 loads while detergent B costs \$6.75 for 36 loads.

This first approach with a table has the downside of running into complicated decimals that do not carry any meaning within the context of money which is always rounded off to the nearest cent. A second method with a table which avoids this would be to look for a common multiple of 80 and 36. The least common mulitple of these is 720. Then we can make two tables to find the cost with each detergent brand for 720 loads of wash:

<

<

We see again that detergent A costs less money per load than detergent B. In these tables, we could skip the middle line and go directly to the cost for 720 loads. An extra step was included because, particularly with the second table, each step (multiplying by 2 and then multiplying by 10) can be done mentally. For the first table, multiplying by 3 twice is probably easier to do mentally than multiplying by 9.

If the ratio $$ \text{loads of wash} : \text{cost of detergent} $$ is viewed as a fraction $\frac{\text{loads of wash}}{\text{cost of detergent}}$, the two methods outlined above correspond to comparing fractions by finding a common numerator and finding a common demoninator respectively.

Solution: 2 Finding a unit rate

We can calculate the approximate cost per load of wash with each detergent and then compare these numbers directly. With detergent A we can wash 80 loads and the cost is \$12.75. This means that the cost per load of wash with detergent A is $$ \$ 12.75 \div 80 \approx \$ 0.16. $$ The exact value is slightly less than \$0.16 but closer to \$0.16 than to \$0.15. With detergent B we can wash 36 loads and the cost is \$6.75. So the coast per load with detergent B is $$ \$ 6.75 \div 36 = \$0.1875 $$ We can see that detergent A costs less per load of wash than detergent B, by almost 3 cents per load.

Solution: 3 Mental math

We can compare costs of the two detergents mentally. With detergent B we pay \$6.75 for 36 loads. With two detergent B's we will pay 2 $\times$ \$6.75 = \$13.50 for 2 $\times$ 36 = 72 loads. This is more money for fewer loads of wash compared to detergent A: \$0.75 cents more for 4 fewer loads of wash. Since detergent A does more loads of wash for less money it costs less per load of wash than detergent B.

# data[].attributes.attachment_link.description
Alignments to Content Standards: 6.RP.A.3 A store has two different brands of laundry detergent. Brand A can do 80 loads of laundry and costs \$12.75. Brand B does 36 loads of laundry and costs \$6.75. Which laundry detergent costs less per load? This purpose of this task is to provide a context for comparing ratios.

# data[].attributes.attachment_link.readable_text
Alignments to Content Standards: 3.OA.A.3

Task

Many problems can be solved in different ways. Decide if the following word problems can be solved using multiplication. Explain your thinking. Then solve each problem.

1. Liam is cooking potatoes. The recipe says you need 5 minutes for every pound of potatoes you are cooking. How many minutes will it take for Liam to cook 12 pounds of potatoes?

2. Mel is designing cards. She has 4 different colors of paper and 7 different pictures she can glue on the paper. How many different card designs can she make using one color of paper and one picture?

3. Nina can practice a song 6 times in an hour. If she wants to practice the song 30 times before the recital, how many hours does she need to practice?

4. Owen is building a rectangular tile patio that is 4 tiles wide and 6 tiles long. How many tiles does he need?

IM Commentary

The purpose of this task is for students to analyze different contexts in which multiplication is appropriate. Research indicates that students often do not make sense of word problems. Instead, to come up with an answer, they just apply the most recent algorithm taught, chose operations based on the types of numbers involved, or apply other approaches that do not require sense making. In this task, the students are asked to analyze the problems and explain their thinking before they find an answer. In the process, they are thinking more carefully about different contexts that can be represented by multiplication.

Note that none of the contexts involve multiplicative comparison, which is intentionally left until 4th grade (see 4.OA.A.1).

Solutions

Solution: Analysis

1. While you can solve this by repeated addition (5+5+5+....), this problem can also be solved by multiplication since adding 5 together 12 times is the same as $12\times5$.

2. This can be solved by multiplication, because for each of the four colors, there are seven different pictures from which to choose, so 7 pictures on the first color + seven pictures on the second color + 7 pictures on the third color + 7 pictures on the fourth color gives four groups of seven or $4\times 7$.

3. This is a division problem since the number of hours needed can be determined by counting the number of groups of six needed to make 30, or $$30 \div 6 = ?$$ On the other hand, we can think about this in terms of multiplication by asking what times six equals 30, which we can write symbolically as $$? \times 6 = 30$$

4. It is a good idea to draw a picture:

We can count the number of tiles one by one, or we can see that there are four rows, each with six tiles, so we can think of it as a multiplication problem $4 \times 6$.

Solution: Solving explicitly

1. $12\times5$ is 12 groups of 5, which we can see in the picture below:

12 groups of 5 is the same as 6 groups of 10:

which is 60. So Liam needs 60 minutes to cook the potatoes.

2. An array makes it easier to see:

There are 28 different possible cards.

3. If we multiply the number of hours she practices by 6, we know the answer must be 30: $$?\times 6 = 30$$ Since 5 times 6 is 30, she has to play for 5 hours if she wants to practice the song 30 times.

4. Looking at the picture above, we can see that Owen needs 24 tiles.

# data[].attributes.attachment_link.description
Alignments to Content Standards: 3.OA.A.3 Many problems can be solved in different ways. Decide if the following word problems can be solved using multiplication. Explain your thinking. Then solve each problem. Liam is cooking potatoes. The recipe says you need 5 minutes for every pound of potatoes you are cooking.

# data[].attributes.attachment_link.readable_text
Task

Each expression below gives the rate of change of temperature in degrees Celsius over a certain period of time. Interpret each expression below in terms of the temperature change.

1. $-\frac{2 \text{ degrees}}{3 \text{ hours}}$

2. $\frac{-2 \text{ degrees}}{3 \text{ hours}}$

3. $\frac{2 \text{ degrees}}{-3 \text{ hours}}$

IM Commentary

The goal of this task is to provide a context for interpreting the expressions $-\frac{p}{q}$, $\frac{-p}{q}$, and $\frac{p}{-q}$ (with $q \neq 0$): this matches the last part of the standard 7.NS.2.b, ''Interpret quotients of rational numbers by describing real-world contexts,'' though in this case the numerator and denominator are integers. Because of the context, students will also gain experience working with rates.

Each expression calculates the rate for the change in temperature over the three hour period and this means that the three expressions must be equal. The teacher may wish to discuss the fact that the task shows that $$-\frac{p}{q} = \frac{-p}{q} = \frac{p}{-q}$$ (when $p = 2$ and $q = 3$) but this is not formally requested. The teacher may wish to have students choose their own expression indicating that the temperature is falling 2 degrees every 3 hours and then have them share and consider other possible expressions: while it is unlikely any one would use $\frac{2}{-3}$, the other two expressions may both appear and would naturally lead to a good discussion.

Solution

1. In the first expression, $-\frac{2 \text{ degrees }}{3 \text{ hours}}$, the negative sign indicates that the temperature was decreasing. The expression $\frac{2 \text{ degrees}}{3 \text{ hours}}$ indicates that it was falling at a rate of 2 degrees for every 3 hours. This is the same as falling at a unit rate of $\frac{2}{3}$ of a degree per hour.

2. The expression $\frac{-2 \text{ degrees }}{3 \text{ hours}}$ means that the rate of change of temperature is -2 degrees in 3 hours. This means that the temperature is falling at a rate of 2 degrees every 3 hours.

3. The expression $\frac{2 \text{ degrees }}{-3 \text{ hours}}$ means that the temperature is increasing by 2 degrees every -3 hours. This is an odd way of talking about a rate of change but it makes sense. For every 3 hours we go back in time, the temperature goes up by 2 degrees. Again, the temperature is decreasing 2 degrees every 3 hours (moving forward in time).

This way of viewing the change of temperature does not feel natural because in our experience time flows forward and this expression requires looking at time in reverse.

# data[].attributes.attachment_link.description
Each expression below gives the rate of change of temperature in degrees Celsius over a certain period of time. Interpret each expression below in terms of the temperature change.

# data[].attributes.attachment_link.readable_text
Alignments to Content Standards: 5.NF.A

Task

Lucy has measuring cups of sizes 1 cup, $\frac{1}{2}$ cup, $\frac{1}{3}$ cup, and $\frac{1}{4}$ cup. She is trying to measure out $\frac{1}{6}$ of a cup of water and says, ''If I fill up the the $\frac{1}{2}$ cup and then pour that into the $\frac{1}{3}$ cup until it is full, there will be $\frac{1}{6}$ of a cup of water left.''

1. Is Lucy's method to measure $\frac{1}{6}$ of a cup of water correct? Explain.

2. Lucy wonders what other amounts she can measure. Is it possible for her to measure out $\frac{1}{12}$ of a cup? Explain.

3. What other amounts of water can Lucy measure?

IM Commentary

This task could form part of a classroom activity where students are encouraged to find as many different ways as possible to make different fractions such as $\frac{1}{12}$ and then share their methods. While measuring cups would be helpful to experiment with and explore the different possibilities of how to use the cups, students will need to use mathematics in order to calculate the precise amount of water left in any given cup.

Part (c) of the question has been deliberately left vague. On the one hand, students might provide one other amount that Lucy can measure, such as 7/12 of a cup which they can make by combining 1/3 of a cup and 1/4 of a cup in the one cup container. On the other hand, they could attempt to find all possible amounts that Lucy can measure, experimenting with different possible combinations, making lists, and sharing their ideas. Finally, for advanced students, they might be challenged to produce a complete list and explain why it is complete (some indications how to do this are provided in the solution).

The mathematics here is very closely related to work with Egyptian fractions (see for example www.illustrativemathematics.org/illustrations/839). Egyptian fractions, however, are sums of unit fractions while this work with measuring cups gives both sums and differences. Moreover, Lucy can use the same measuring cup multiple times. A different variant on this task, which would capture the spirit of Egyptian fractions, would allow Lucy to use each measuring cup at most a single time.

Work on this task can support many of the standards of mathematical practice including

MP2, Reason Abstractly and Quantitatively: students need to find a common denominator to calculate what happens as water is poured from one cup to another and the third part of the question requires analyzing the situation from an abstract viewpoint.

MP3, Construct viable arguments and critique the reasoning of others: part (c) of the question encourages students to create other fractions using the measuring cup model. They will need to explain why their method works and, if done as a group activity, can share and critically discuss their numbers and methods. The more deeply students go into part (c), the richer the opportunities will be to engage in MP3.

MP4, Model With Mathematics: students must describe with fractions what happens when water is poured from one measuring cup into another.

Solution

1. Lucy starts out with $\frac{1}{2}$ cup of water in one container and then uses that to fill up the $\frac{1}{3}$ cup container. This means that there will be $\frac{1}{2} - \frac{1}{3}$ cups left in the $\frac{1}{2}$ cup container. We can use 6 as a common denominator to calculate this fraction:

\begin{align} \frac{1}{2} - \frac{1}{3} &= \frac{1}{2} \times \frac{3}{3} - \frac{1}{3} \times \frac{2}{2} \\ &= \frac{3}{6} - \frac{2}{6} \\ &= \frac{1}{6}. \end{align}

So there is $\frac{1}{6}$ of a cup of water left in the $\frac{1}{2}$ cup container.

Below is a way to picture this

The different measuring cups are not pictured but we can see $\frac{1}{2}$ of a full cup shaded in both green and blue while $\frac{1}{3}$ of a cup is shaded in green. So after pouring $\frac{1}{3}$ of a cup out of the $\frac{1}{2}$ cup container what will remain is in blue. The full cup is divided into six equal pieces and one of them is blue. This means that $$ \frac{1}{2} - \frac{1}{3} = \frac{1}{6} $$

and $\frac{1}{6}$ of a cup of water remains in the $\frac{1}{2}$ cup container.

2. One way Lucy can measure out $\frac{1}{12}$ of a cup is to follow the same method as in part (a) except this time pour water from the $\frac{1}{3}$ cup container into the $\frac{1}{4}$ cup container. Once the smaller container is full, there will be $\frac{1}{3} - \frac{1}{4}$ cups of water left in the $\frac{1}{3}$ cup container. We calculate this amount as we did in part (a):

\begin{align} \frac{1}{3} - \frac{1}{4} &= \frac{1}{3} \times \frac{4}{4} - \frac{1}{4} \times \frac{3}{3} \\ &= \frac{4}{12} - \frac{3}{12} \\ &= \frac{1}{12}. \end{align}

So there will be $\frac{1}{12}$ of a cup of water left in the $\frac{1}{3}$ cup container. This is pictured in two steps below:

With pictures, in order to show both $\frac{1}{3}$ and $\frac{1}{4}$ of a cup we draw one cup, divided into 12 equal pieces:

This shows that after pouring $\frac{1}{4}$ of a cup out of the $\frac{1}{3}$ cup container, $\frac{1}{12}$ of a cup will remain.

3. We have seen in parts (a) and (b) that Lucy can measure $\frac{1}{6} = \frac{2}{12}$ of a cup and also $\frac{1}{12}$ of a cup. Her three containers measure $\frac{1}{4} = \frac{3}{12}$ of a cup, $\frac{1}{3} = \frac{4}{12}$ of a cup, and $\frac{1}{2} = \frac{6}{12}$ of a cup. For $\frac{5}{12}$ of a cup, she can measure $\frac{1}{12}$ of a cup and put this in her $\frac{1}{2}$ cup container: then she can fill up the $\frac{1}{3} = \frac{4}{12}$ cup container and add this giving $\frac{5}{12}$ of a cup. In order to get $\frac{7}{12}$, $\frac{8}{12}$, $\frac{9}{12}$, $\frac{10}{12}$, $\frac{11}{12}$, and $\frac{12}{12}$ of a cup, Lucy can add $\frac{1}{2} = \frac{6}{12}$ to the one cup container and then measure out the appropriate number of twelfths of a cup (1,2,3,4, 5, or 6) using the other containers and then add this to the one cup container.

Lucy can only measure a whole number of twelfths of a cup--for example, she could measure $\frac{17}{12} = 1\frac{5}{12}$--with the cups she has. To see why, notice that each cup measures a whole number of $\frac{1}{12}$ cups. Pouring water out of one container and into another allows her to add or subtract these amounts. Adding and subtracting a whole number of twelfths will always lead to a whole number of twelfths.

# data[].attributes.attachment_link.description
Alignments to Content Standards: 5.NF.A Lucy has measuring cups of sizes 1 cup, $\frac{1}{2}$ cup, $\frac{1}{3}$ cup, and $\frac{1}{4}$ cup. She is trying to measure out $\frac{1}{6}$ of a cup of water and says, ''If I fill up the the $\frac{1}{2}$ cup and then pour that into the $\frac{1}{3}$ cup until it is full, there will be $\frac{1}{6}$ of a cup of water left.'' Is Lucy's method to measure $\frac{1}{6}$ of a cup of water correct?

# data[].attributes.attachment_link.readable_text
Alignments to Content Standards: 3.OA.A.2

Task

Suppose there are 4 tanks and 3 fish in each tank. The total number of fish in this situation can be expressed as $4 \times 3 = 12$.

1. Describe what is meant in this situation by $12 \div 3 = 4$

2. Describe what is meant in this situation by $12 \div 4 = 3$

IM Commentary

The purpose of this task is to provide an opportunity for students to demonstrate that they can interpret within a given context the meaning of whole number quotients. Since this particular context is an asymmetric model of multiplication, there are two distinct interpretations of the quotients that can arise.

The given context is an asymmetric model of multiplication in the sense that 4 tanks with 3 fish per tank is different than 3 tanks with 4 fish per tank, even though in both cases there are a total of 12 fish. The convention used here is that the number of groups is listed first and the number of objects in each group is listed second, although in this case the context determines whether 3 or 4 is the group (respectively, number of objects in each group). Although these terms are not mentioned in the standards, the number of groups is sometimes called the multiplier and the number of objects in each group is called the mutiplicand.

Note: It is important that students can make sense of both of these interpretations of division arising from a single multiplication context with the factors and quotients all whole numbers. These interpretations will be extended to contexts with fractions and decimal in later grades.

Solution

The students' language may vary. What is important is the structure. The student needs to match what is known (either the number of groups or the amount in each group) and what is being determined by the quotient operation in this context.

1. Describe what is meant in this situation by 12 ÷ 3 = 4

Solution: There are 12 fish. If 3 fish are put in each tank there will be 4 tanks.

Comment: This is a measurement model of division, the number of objects in each group is known, and we are determining the number of groups.

2. Describe what is meant in this situation by 12 ÷ 4 = 3

Solution: There are 12 fish. If we want to partition (share) them equally among 4 tanks, we will end up with 3 fish in each tank.

Comment: This a sharing model of division, the number of groups is known and we are trying to determine the number in each group.

# data[].attributes.attachment_link.description
Alignments to Content Standards: 3.OA.A.2 Suppose there are 4 tanks and 3 fish in each tank. The total number of fish in this situation can be expressed as $4 \times 3 = 12$.

# data[].attributes.attachment_link.readable_text
Alignments to Content Standards: 5.NF.A

Task

Lucy has measuring cups of sizes 1 cup, $\frac{1}{2}$ cup, $\frac{1}{3}$ cup, and $\frac{1}{4}$ cup. She is trying to measure out $\frac{1}{6}$ of a cup of water and says, ''If I fill up the the $\frac{1}{2}$ cup and then pour that into the $\frac{1}{3}$ cup until it is full, there will be $\frac{1}{6}$ of a cup of water left.''

1. Is Lucy's method to measure $\frac{1}{6}$ of a cup of water correct? Explain.

2. Lucy wonders what other amounts she can measure. Is it possible for her to measure out $\frac{1}{12}$ of a cup? Explain.

3. What other amounts of water can Lucy measure?

IM Commentary

This task could form part of a classroom activity where students are encouraged to find as many different ways as possible to make different fractions such as $\frac{1}{12}$ and then share their methods. While measuring cups would be helpful to experiment with and explore the different possibilities of how to use the cups, students will need to use mathematics in order to calculate the precise amount of water left in any given cup.

Part (c) of the question has been deliberately left vague. On the one hand, students might provide one other amount that Lucy can measure, such as 7/12 of a cup which they can make by combining 1/3 of a cup and 1/4 of a cup in the one cup container. On the other hand, they could attempt to find all possible amounts that Lucy can measure, experimenting with different possible combinations, making lists, and sharing their ideas. Finally, for advanced students, they might be challenged to produce a complete list and explain why it is complete (some indications how to do this are provided in the solution).

The mathematics here is very closely related to work with Egyptian fractions (see for example www.illustrativemathematics.org/illustrations/839). Egyptian fractions, however, are sums of unit fractions while this work with measuring cups gives both sums and differences. Moreover, Lucy can use the same measuring cup multiple times. A different variant on this task, which would capture the spirit of Egyptian fractions, would allow Lucy to use each measuring cup at most a single time.

Work on this task can support many of the standards of mathematical practice including

MP2, Reason Abstractly and Quantitatively: students need to find a common denominator to calculate what happens as water is poured from one cup to another and the third part of the question requires analyzing the situation from an abstract viewpoint.

MP3, Construct viable arguments and critique the reasoning of others: part (c) of the question encourages students to create other fractions using the measuring cup model. They will need to explain why their method works and, if done as a group activity, can share and critically discuss their numbers and methods. The more deeply students go into part (c), the richer the opportunities will be to engage in MP3.

MP4, Model With Mathematics: students must describe with fractions what happens when water is poured from one measuring cup into another.

Solution

1. Lucy starts out with $\frac{1}{2}$ cup of water in one container and then uses that to fill up the $\frac{1}{3}$ cup container. This means that there will be $\frac{1}{2} - \frac{1}{3}$ cups left in the $\frac{1}{2}$ cup container. We can use 6 as a common denominator to calculate this fraction:

\begin{align} \frac{1}{2} - \frac{1}{3} &= \frac{1}{2} \times \frac{3}{3} - \frac{1}{3} \times \frac{2}{2} \\ &= \frac{3}{6} - \frac{2}{6} \\ &= \frac{1}{6}. \end{align}

So there is $\frac{1}{6}$ of a cup of water left in the $\frac{1}{2}$ cup container.

Below is a way to picture this

The different measuring cups are not pictured but we can see $\frac{1}{2}$ of a full cup shaded in both green and blue while $\frac{1}{3}$ of a cup is shaded in green. So after pouring $\frac{1}{3}$ of a cup out of the $\frac{1}{2}$ cup container what will remain is in blue. The full cup is divided into six equal pieces and one of them is blue. This means that $$ \frac{1}{2} - \frac{1}{3} = \frac{1}{6} $$

and $\frac{1}{6}$ of a cup of water remains in the $\frac{1}{2}$ cup container.

2. One way Lucy can measure out $\frac{1}{12}$ of a cup is to follow the same method as in part (a) except this time pour water from the $\frac{1}{3}$ cup container into the $\frac{1}{4}$ cup container. Once the smaller container is full, there will be $\frac{1}{3} - \frac{1}{4}$ cups of water left in the $\frac{1}{3}$ cup container. We calculate this amount as we did in part (a):

\begin{align} \frac{1}{3} - \frac{1}{4} &= \frac{1}{3} \times \frac{4}{4} - \frac{1}{4} \times \frac{3}{3} \\ &= \frac{4}{12} - \frac{3}{12} \\ &= \frac{1}{12}. \end{align}

So there will be $\frac{1}{12}$ of a cup of water left in the $\frac{1}{3}$ cup container. This is pictured in two steps below:

With pictures, in order to show both $\frac{1}{3}$ and $\frac{1}{4}$ of a cup we draw one cup, divided into 12 equal pieces:

This shows that after pouring $\frac{1}{4}$ of a cup out of the $\frac{1}{3}$ cup container, $\frac{1}{12}$ of a cup will remain.

3. We have seen in parts (a) and (b) that Lucy can measure $\frac{1}{6} = \frac{2}{12}$ of a cup and also $\frac{1}{12}$ of a cup. Her three containers measure $\frac{1}{4} = \frac{3}{12}$ of a cup, $\frac{1}{3} = \frac{4}{12}$ of a cup, and $\frac{1}{2} = \frac{6}{12}$ of a cup. For $\frac{5}{12}$ of a cup, she can measure $\frac{1}{12}$ of a cup and put this in her $\frac{1}{2}$ cup container: then she can fill up the $\frac{1}{3} = \frac{4}{12}$ cup container and add this giving $\frac{5}{12}$ of a cup. In order to get $\frac{7}{12}$, $\frac{8}{12}$, $\frac{9}{12}$, $\frac{10}{12}$, $\frac{11}{12}$, and $\frac{12}{12}$ of a cup, Lucy can add $\frac{1}{2} = \frac{6}{12}$ to the one cup container and then measure out the appropriate number of twelfths of a cup (1,2,3,4, 5, or 6) using the other containers and then add this to the one cup container.

Lucy can only measure a whole number of twelfths of a cup--for example, she could measure $\frac{17}{12} = 1\frac{5}{12}$--with the cups she has. To see why, notice that each cup measures a whole number of $\frac{1}{12}$ cups. Pouring water out of one container and into another allows her to add or subtract these amounts. Adding and subtracting a whole number of twelfths will always lead to a whole number of twelfths.

# data[].attributes.attachment_link.description
Alignments to Content Standards: 5.NF.A Lucy has measuring cups of sizes 1 cup, $\frac{1}{2}$ cup, $\frac{1}{3}$ cup, and $\frac{1}{4}$ cup. She is trying to measure out $\frac{1}{6}$ of a cup of water and says, ''If I fill up the the $\frac{1}{2}$ cup and then pour that into the $\frac{1}{3}$ cup until it is full, there will be $\frac{1}{6}$ of a cup of water left.'' Is Lucy's method to measure $\frac{1}{6}$ of a cup of water correct?

# data[].attributes.attachment_link.readable_text
Alignments to Content Standards: 3.OA.A.3

Task

1. Juanita spent \$9 on each of her 6 grandchildren at the fair. How much money did she spend?

2. Nita bought some games for her grandchildren for \$8 each. If she spent a total of \$48, how many games did Nita buy?

3. Helen spent an equal amount of money on each of her 7 grandchildren at the fair. If she spent a total of \$42, how much did each grandchild get?

IM Commentary

The first of these is a multiplication problem involving equal-sized groups. The next two reflect the two related division problems, namely, "How many groups?" and "How many in each group?"

Sometimes the second type of problem is referred to as a measurement division or repeated subtraction problem. The third type of problem is sometimes called a partitive division or sharing problem. It asks how large is each share when a whole is divided equally into a specified number of pieces. It specifies the size of each share and asks how many of that size are in the whole. The language used in the solution reflects the language in the common core, which also refers to them "Number of Groups Unknown" or "Group Size Unknown," respectively.

The Standards for Mathematical Practice focus on the nature of the learning experiences by attending to the thinking processes and habits of mind that students need to develop in order to attain a deep and flexible understanding of mathematics. Certain tasks lend themselves to the demonstration of specific practices by students. The practices that are observable during exploration of a task depend on how instruction unfolds in the classroom. While it is possible that tasks may be connected to several practices, only one practice connection will be discussed in depth. Possible secondary practice connections may be discussed but not in the same degree of detail.

This particular task helps illustrate Mathematical Practice Standard 4, Model with mathematics. Third graders apply the mathematics they know to solve problems arising in everyday life. They identify the mathematical elements of each situation and determine the solution pathway that is best for them to follow. For these problems, students may use drawings or tape diagrams to depict each situation. They may also write a multiplication equation to represent how many altogether in Part (a) or write multiplication or division equations to represent how may groups in Part (B) and how many in each group in Part (c). Depending upon their solution pathways, students will have opportunities to explain their solutions to each other and critique each other's reasoning (Mathematical Practice Standard 3). For example, the teacher could pair students who have used different representations to represent each situation (one with an equation and one with a tape diagram) and have them describe how both representations depict (or do not depict) the same situation. Additionally, students can comment on why ? × 8 = 48 and 48 ÷ 8 = ? are equivalent expressions and can both provide the solution to the question.

Solutions

Solution: Tape diagram

This task needs a tape diagram solution; one is under development.

Solution: Writing multiplication equations for division problems

1. Sandra spent 6 groups of \$9, which is $6 \times 9 = 54$ dollars all together.

2. Since the number of games represent the number of groups, but we don't know how many games she bought, this is a "How many groups?" division problem. We can represent it as $$? \times 8 = 48$$ or $$48 \div 8 = ?$$ So Nita must have bought 6 games.

3. Here we know how many grandchildren there are (so we know the number of groups), but we don't know how much money each one gets (the number of dollars in each group). So this is a "How many in each group?" division problem. We can represent it as $$7 \times ? = 42$$ or $$42 \div 7 = ?$$ So Helen must have given each grandchild \$6.

# data[].attributes.attachment_link.description
Alignments to Content Standards: 3.OA.A.3 Juanita spent \$9 on each of her 6 grandchildren at the fair. How much money did she spend? Nita bought some games for her grandchildren for \$8 each. If she spent a total of \$48, how many games did Nita buy?

# data[].attributes.attachment_link.readable_text
Task

There are almost 40 thousand fourth graders in Mississippi and almost 400 thousand fourth graders in Texas. There are almost 4 million fourth graders in the United States.

We write 4 million as 4,000,000. How many times more fourth graders are there in Texas than in Mississippi? How many times more fourth graders are there in the United States than in Texas? Use the approximate populations listed above to solve.

There are about 4 thousand fourth graders in Washington, D.C. How many times more fourth graders are there in the United States than in Washington, D.C.?

IM Commentary

The purpose of this task is to help students understand the multiplicative relationship between commonly used large numbers (thousands and millions) by using their understanding of place value. This task also connects to students' work on multiplicative comparison (see 4.OA.A.1). The task 4.NBT Threatened and Endangered is a good task to do before this one as it requires the same kind of reasoning but the numbers are smaller. The numbers in this task come from the National Center for Educational Statistics:

http://nces.ed.gov/programs/digest/2013menu_tables.asp

The Standards for Mathematical Practice focus on the nature of the learning experiences by attending to the thinking processes and habits of mind that students need to develop in order to attain a deep and flexible understanding of mathematics. Certain tasks lend themselves to the demonstration of specific practices by students. The practices that are observable during exploration of a task depend on how instruction unfolds in the classroom. While it is possible that tasks may be connected to several practices, only one practice connection will be discussed in depth. Possible secondary practice connections may be discussed but not in the same degree of detail.

This particular task supports the demonstration of Mathematical Practice Standard 6, Attend to precision. Fourth graders apply their understanding of place value and multiplicative reasoning to compare thousands and millions in this task. They need to pay close attention to precision to determine the value of each digit in each number and to compare the two numbers. They make use of the fact that the value of each place is ten times the value of the place immediately to the right and are able to explain this using precise language.

Solution

​We write 4 thousand as 4,000

We write 40 thousand as 40,000

We write 400 thousand as 400,000

The value of each place is ten times the value of the place immediately to the right.

So:

40,000 is 10 times 4,000

400,000 is 10 times 40,000.

4,000,000 is 10 times 400,000.

Thus, 400,000 = 10 $\times$ 40,000, and there are about 10 times as many fourth graders in Texas as there are in Mississippi.

Also, 4,000,000 = 10 $\times$ 400,000, and there are about 10 times as many fourth graders in the US as there are in Texas.

Finally, to go from 4,000 to 4,000,000, we have to multiply by 10 three times. We see that $$10\times10\times10=10\times100=1000$$ So there are about 1,000 times as many fourth graders in the US as there are in Washington DC.

# data[].attributes.attachment_link.description
There are almost 40 thousand fourth graders in Mississippi and almost 400 thousand fourth graders in Texas. There are almost 4 million fourth graders in the United States. We write 4 million as 4,000,000. How many times more fourth graders are there in Texas than in Mississippi?

# data[].attributes.attachment_link.readable_text
Alignments to Content Standards: 3.OA.A.2

Task

Presley has 18 markers. Her teacher gives her three boxes and asks her to put an equal number of markers in each box.

Anthony has 18 markers. His teacher wants him to put 3 markers in each box until he is out of markers.

1. Before you figure out what the students should do, answer these questions:

What is happening in these two situations? How are they similar? How are they different?

2. Figure out how many markers Presley should put in each box. Show your work. Then figure out how many boxes Anthony should fill with markers. Show your work.

IM Commentary

The purpose of this task is for students to compare two problems that draw on the same context but represent the two different interpretations of division, namely, the "How many groups?" interpretation and the "How many in each group?" interpretation. Before solving them, the students should analyze and understand the problem structures. Students may work in pairs, discussing and recording their ideas. After students have analyzed the problem structures they can work individually to solve the problems, showing their solution strategies. The attached PDF shows a classroom vignette that indicates how a teacher might wish to lead a classroom discussion for the first part of this task. As students work, the teacher should be monitoring students' progress, looking for students who solved the problem in different ways who will later share out. While there are many possibilities for the share out, one goal would be the idea of partitioning, or dealing out, one by one to a given number of groups. Another goal would be creating groups one at a time. Why were different actions used to solve the problems? Would each strategy work for the other problem? Why or why not?

The Standards for Mathematical Practice focus on the nature of the learning experiences by attending to the thinking processes and habits of mind that students need to develop in order to attain a deep and flexible understanding of mathematics. Certain tasks lend themselves to the demonstration of specific practices by students. The practices that are observable during exploration of a task depend on how instruction unfolds in the classroom. While it is possible that tasks may be connected to several practices, only one practice connection will be discussed in depth. Possible secondary practice connections may be discussed but not in the same degree of detail.

This particular task helps illustrate Mathematical Practice Standard 1, Make sense of problems and persevere in solving them. Problem solving is based upon students engaging in a task in which a solution pathway is not known in advance. As third graders approach these two problems, they will analyze the problems to make sense of what each is asking, working to understand the structures and the two interpretations of division. Through this analysis students will understand that the numbers and answers are the same, however, the actions or strategies students use for solving the two problems are quite different. A common misconception is that division can always be thought of as repeated subtraction. This works for a "How many groups?" problem, but not for a "How many in each group?" problem, and only for whole numbers. Students need experience with both kinds of division problems to address this misconception; this will support students later when they divide fractions. As an extension, students can design story problems to match both the "How many groups?" and "How many in each group?" problem types.

Solutions

Solution: Some similarities and differences for part (a)

Some things the students might say about comparing the two problems:

Both problems are about students putting 18 markers in boxes.

Both problems have a 3.

But the 3's are different in each problem. In the first problem there are 3 boxes and Presley is going to put an equal number of markers in each box. In the second problem, Anthony puts 3 markers in each box.

In the first problem we are trying to figure out how many are in each group, and in the second we are trying to figure out how many groups there are. So the first is a "How many in each group?" division problem and the second is a "How many groups?" division problem.

Solution: Level 1 solution to part (b): Direct Modeling

To solve Problem 1, students may draw three boxes and distribute the markers, one at a time from left to right until all markers have been distributed, similar to dealing out cards. Then the student might count the markers in each group to reach the answer of 6. There are 6 markers in each box.

A direct modeling solution to Problem 2 would look quite different. Students may count out 18 cubes. Next they may set out three cubes in a group and build additional groupings of 3 until all 18 cubes have been used.

Notice the differences in solution strategies, problem 1 requires more of a dealing or partitioning into groups that are already determined, while problem 2 requires forming the groups. Students must have experiences with both types of problems.

Solution: Level 2 solution to part (b): Counting on

Students may use a trial and error method of counting, such as those listed below, to solve problem 1.

3 markers in each...3, 6, 9. There will be more than 3 markers in each box.

5 markers in each box...5, 10, 15. 15 is getting closer...it's only 3 away from 18, so I think 6 markers are in each box.

6, 12, 18 ... There are 6 markers in each box

Counting strategies for problem 2 may appear more strategic as the "group of 3" may elicit students to count 3, 6, 9, 12, 15, 18. Then the student will track the number of 3's counted (6) by marking with fingers, a visual or physical representation.

Solution: Level 3 solution to part (b): Convert to an Easier Equivalent Problem

Students may solve problems 1 and 2 using what they know about the relationship of multiplication and division. For example, * Using Inverse Operation $18 \div 3 = n$. I know multiplication is the inverse of division and I know $3 \times 6 = 18$, so $n = 6$ * Students may also use their understanding of multiplication facts and the properties of operations.

$$ \begin{align} 3 \times 5 &= 15 \\ 3 \times 1 &= 3 \\ 15+3 &=18 \end{align} $$

So $3 \times 6 =18$ and there are 6 markers in each box. * Another strategy students may use to solve problem 2 is repeated subtraction. This strategy is likely to be used to solve this problem because the problem calls for the markers being in groups of 3. $$18-3 = 15$$ $$15-3=12$$ $$12-3=9$$ $$9-3 = 6$$ $$6-3=3$$ $$3-3=0$$ Because we subtracted three 6 times, there are 6 boxes. This is only a sampling of strategies students may use to solve the problem. For additional reading: Progressions for the Common Core Standards in Math K, Counting and Cardinality; K-5 Operations and Algebraic Thinking

# data[].attributes.attachment_link.description
Alignments to Content Standards: 3.OA.A.2 Presley has 18 markers. Her teacher gives her three boxes and asks her to put an equal number of markers in each box. Anthony has 18 markers. His teacher wants him to put 3 markers in each box until he is out of markers.

# data[].attributes.attachment_link.readable_text
Alignments to Content Standards: 4.OA.A

Task

There are two snakes at the zoo, Jewel and Clyde. Jewel was six feet and Clyde was eight feet. A year later Jewel was eight feet and Clyde was 10 feet. Which one grew more?

IM Commentary

The purpose of this task is to foster a classroom discussion that will highlight the difference between multiplicative and additive reasoning. Some students will argue that they grew the same amount (an example of "additive thinking"). Students who are studying multiplicative comparison problems might argue that Jewel grew more since it grew more with respect to its original length (an example of "multiplicative thinking"). This would set the stage for a comparison of the two perspectives. In the case were the students don't bring up both arguments, the teacher can introduce the missing perspective.

In later grades, students will learn that "which grows more" means "which has the greater absolute increase?" and "which has the greater growth rate?" means "which has the greater increase relative to the starting amount?" but students won't see this type of language for two or three years. Teachers need to be aware of this and work to ask questions as unambiguously as possible; for example, when asking for multiplicative comparisons, use language such as, "How many times greater is $x$ than $y$." They should also be prepared to address this potential for confusion along the way.

Solution

Viewing this additively, both snakes grew 2 feet and therefore grew the same amount. Viewing it multiplicatively, Jewel grew $\frac{2}{6}$ its length, while Clyde grew $\frac{2}{8}$ its length. From this perspective, Jewel grew more. Given the purposeful phrasing of the problem, both interpretations are reasonable, but the goal is to understand the two perspectives, thus the difference between additive and multiplicative reasoning.

# data[].attributes.attachment_link.description
Alignments to Content Standards: 4.OA.A There are two snakes at the zoo, Jewel and Clyde. Jewel was six feet and Clyde was eight feet. A year later Jewel was eight feet and Clyde was 10 feet. Which one grew more?

# data[].attributes.attachment_link.readable_text
Task

In January, Georgia signed up for a membership at Anytime Fitness. The plan she chose cost \$95 in start-up fees and then \$20 per month starting in February. Edwin also signed up at Anytime Fitness in January. His plan cost \$35 per month starting in February, and his start-up fees were waived.

1. Create tables for both Georgia and Edwin that compare the number of months since January to the total cost of their gym memberships. Continue this table for one year.

2. Plot the points from the two tables in part (a) on a coordinate plane.

3. Decide if either or both gym memberships are described by a proportional relationship, and write an equation representing any such relationship. Explain how parts (a) and (b) could be used to support your answer.

IM Commentary

In this task, students are presented with two situations in a single context and asked which one represents a proportional relationship. Students are asked to understand this proportional relationship from a variety of perspectives -- a table, a graph, a verbal context, and an equation. As such, this task might be used as a synthesis of these various perspectives that one learns about when studying proportional relationships. Alternatively, it could be used as an introduction to the various ways one might be presented with a proportional relationship. In this case, instructors should be prepared for students who may not be familiar with using one of the perspectives (in particular, tables of values).

Solution

1. The table for Georgia's gym membership cost for 12 months is below:

The table for Edwin's gym membership cost for 12 months is below:

2. We plot the points from the two tables in part (a) on the coordinate axes below, where number of months since January is on the horizontal axis and the total cost is on the vertical axis. The red dashed line contains Edwin's table of values and the blue dashed line contains the values from Georgia's table.

Note that we are connecting the plotted points with a dashed line only to better see the general trend. Since this is actually discrete data a solid line would not be a suitable representation.

3. Georgia's plan does not represent a proportional relationship, and Edwin's plan does represent a proportional relationship. That Edwins' plan is proportional can be seen from the table by observing that whenever we multiply the number of months by a constant, the total cost multiplies by that same constant -- for example, doubling the number of months from 3 to 6 has the effect of doubling the cost from \$105 to \$210. This does not hold true for Georgia's plan, as can be seen by similarly doubling.

We could also see this from our response to part (b). Proportional relationships can be visualized graphically as being described by lines that go through the origin. Since Edwin's line (in red above) does go through the origin, it describes a proportional relationship, and likewise, Georgia's does not.

Finally, we find an equation to describe Edwin's plan. Since his relationship is proportional, every one month that passes will cost him \$35. So after $n$ months, he will have paid \$35 dollars $n$ times, for a total cost of \$35$n$ dollars. Thus the total cost $c$ of Edwin's plan is related to the number of months passed by the equation $c=35n$.

# data[].attributes.attachment_link.description
In January, Georgia signed up for a membership at Anytime Fitness. The plan she chose cost \$95 in start-up fees and then \$20 per month starting in February. Edwin also signed up at Anytime Fitness in January. His plan cost \$35 per month starting in February, and his start-up fees were waived.

# data[].attributes.attachment_link.readable_text
Alignments to Content Standards: 4.OA.A.3

Task

Every year a carnival comes to Hallie's town. The price of tickets to ride the rides has gone up every year.

1. In 2008, Hallie's allowance was \$9.00 a month. How many carnival tickets could she buy with one month's allowance?

2. If her allowance had stayed the same, \$9.00 a month, how many carnival tickets could she buy in 2012?

3. In 2012, Hallie's allowance was \$14.00 per month. How much did her monthly allowance increase between 2008 and 2012?

4. How much more did a carnival ticket cost in 2012 than it did in 2008?

5. Was Hallie able to buy more carnival tickets in 2008 or in 2012 with one month's allowance?

6. What would Hallie's allowance need to be in 2012 in order for her to be able to buy as many carnival tickets as she could in 2008?

7. What happens to your ability to buy things if prices increase and your allowance doesn't increase?

IM Commentary

The purpose of this task is for students to solve multi-step problems in a context involving a concept that supports financial literacy, namely inflation. Inflation is a sustained increase in the average price level. In this task, students can see that if the price level increases and people's incomes do not increase, they aren't able to purchase as many goods and services; in other words, their purchasing power decreases. If the price level rises and people's incomes increase at a slower rate, their purchasing power increases but not as much as if their income increases at the same rate as the cost of goods and services. This task is a variation on another task 6.NS Movie tickets that also addresses inflation. From a mathematical perspective, students are asked to solve word problems involving operations only with whole numbers because students are not required in fourth grade to compute with decimal numbers. However, they are asked to understand decimal notation for fractions with denominators of 10 and 100 (see 4.NF.6) and so this task capitalizes on this by representing whole numbers with decimal notation and including dollar amounts that are not whole numbers in the table. Also, students in 4th grade should be comfortable with two-column tables (see e.g. 4.MD.1), so this task gives them some practice reading information in a table. Note that the numbers were chosen specifically so that there would be remainders for them to interpret (as described in 4.OA.3). This task is part of a set collaboratively developed with Money as You Learn, an initiative of the President's Advisory Council on Financial Capability. Integrating essential financial literacy concepts into the teaching of the Common Core State Standards can strengthen teaching of the Common Core and expose students to knowledge and skills they need to become financially capable young adults. A mapping of essential personal finance concepts and skills against the Common Core State Standards as well as additional tasks and texts will be available at http://www.moneyasyoulearn.org.

The Standards for Mathematical Practice focus on the nature of the learning experiences by attending to the thinking processes and habits of mind that students need to develop in order to attain a deep and flexible understanding of mathematics. Certain tasks lend themselves to the demonstration of specific practices by students. The practices that are observable during exploration of a task depend on how instruction unfolds in the classroom. While it is possible that tasks may be connected to several practices, only one practice connection will be discussed in depth. Possible secondary practice connections may be discussed but not in the same degree of detail.

This particular task helps illustrate Mathematical Practice Standard 2, Reason abstractly and quantitatively. Students make sense of quantities and how they are related in a problem situation. In the task at hand, students decontextualize each step of the problem and represent it with numbers and symbols. As they work through each step, they will pause to make sense of the quantities and operations that the symbols represent. The problem culminates in students making connections between their solutions in each step to the bigger idea of inflation. Problems that begin with a context and are represented with mathematical objects or symbols are also examples of modeling with mathematics (MP.4).

Solution

1. If she has \$9 and tickets cost \$2 each, we divide 9 by 2 to find out how many tickets she has (this is a "How many groups?" division problem). $$9 ÷ 2 = 4 \text{, with a remainder of }1$$ So she could buy 4 carnival tickets with one month's allowance in 2008 and would have \$1.00 left over.

2. As before, we divide to find how many tickets she can buy. 9 ÷ 4 = 2 with a remainder of 1. So if her allowance was \$9.00 in 2012, she could only buy 2 tickets with one month's allowance.

3. Her allowance was \$9.00 and now it is \$14.00, and $$14 - 9 = 5$$ so her allowance increased by \$5.00.

4. Since Tickets cost \$4 in 2012 and they cost \$2.00 in 2008, and $$4 - 2 = 2$$ a carnival ticket cost \$2.00 more in 2012 than it did in 2008.

5. We will divide as we did in parts (a) and (b) $$14 \div 4 = 3 \text{, with a remainder of }2$$ So she could buy 3 carnival tickets with one month's allowance in 2012 and would have \$2.00 left over. Even though her allowance went up, the number of tickets she could buy went down.

6. Since carnival tickets cost \$4.00 in 2012 and she could buy 4 tickets in 2008 , and 4 × 4 = 16, her allowance in 2012 would need to be at least \$16.

7. You can buy fewer things if prices increase but your allowance doesn't increase.

# data[].attributes.attachment_link.description
Alignments to Content Standards: 4.OA.A.3 Every year a carnival comes to Hallie's town. The price of tickets to ride the rides has gone up every year. In 2008, Hallie's allowance was \$9.00 a month. How many carnival tickets could she buy with one month's allowance?

# data[].attributes.attachment_link.readable_text
Alignments to Content Standards: 5.NF.A.1

Task

Ancient Egyptians used unit fractions, such as $\frac{1}{2}$ and $\frac{1}{3}$, to represent all fractions. For example, they might write the number $\frac{2}{3}$ as $\frac{1}{2} + \frac{1}{6}$.

We often think of $\frac{2}{3}$ as $\frac{1}{3}+\frac{1}{3}$, but the ancient Egyptians would not write it this way because they didn't use the same unit fraction twice.

1. Write each of the following Egyptian fractions as a single fraction:

1. $\frac{1}{2} + \frac{1}{3}$,

2. $\frac{1}{2}+ \frac{1}{3} + \frac{1}{5}$,

3. $\frac{1}{4} + \frac{1}{5} + \frac{1}{12}$.

2. How might the ancient Egyptians have writen the fraction we write as $\frac{3}{4}$?

IM Commentary

One goal of this task is to help students develop comfort and ease with adding fractions with unlike denominators. Another goal is to help them develop fraction number sense by having students decompose fractions. Because the Egyptians represented fractions differently than we do, it can also help students understand that there can be many ways of representing the same number. This helps prepare them for writing algebraic expressions in 6th grade. This task is an instructional task; the teacher may wish to supplement part (b) of the question in two directions:

1. As is indicated in the solution, there are many ways to write $\frac{3}{4}$ as a sum of unit fractions and student work may naturally lead to a discussion of this issue.

2. Students might be prompted to try to express some other fractions, such as $\frac{2}{5}$ or $\frac{3}{5}$ as sums of unit fractions.

For historical accuracy, it should be noted that the ancient Egyptians had special hieroglyphic symbols for the two fractions $\frac{2}{3}$ and $\frac{3}{4}$ considered in this problem. In general, however, they used tables in order to break down fractions into sums of unit fractions.

One issue which comes up in the solution is that there are apparently many ways to write an Egyptian fraction. This leads to an interesting question: how do you find the "best'' Egyptian fraction representing a given fraction? Here a reasonable interpretation of "best'' would be one which has the smallest number of unit fractions. This problem will be addressed in a second task which falls under the high school algebra standards.

Solution

1. For the Egyptian fraction $\frac{1}{2} + \frac{1}{3}$, a common denominator would be $6$ since $6$ is divisible by both $2$ and $3$. Converting to this common denominator we find

\begin{align} \frac{1}{2} + \frac{1}{3} &= \frac{3 \times 1}{3 \times 2} + \frac{2 \times 1}{2 \times 3} \\ &= \frac{3}{6} + \frac{2}{6} \\ &= \frac{5}{6}. \end{align}

For $\frac{1}{2} + \frac{1}{3} + \frac{1}{5}$ we could use what we have just found, namely that $\frac{1}{2} + \frac{1}{3} = \frac{5}{6}$. To add $\frac{5}{6}$ and $\frac{1}{5}$ we can use $5 \times 6 = 30$ as a common denominator:

\begin{align} \frac{1}{2} + \frac{1}{3} + \frac{1}{5} &= \frac{5}{6} + \frac{1}{5} \\ &= \frac{5 \times 5}{5 \times 6} + \frac{6 \times 1}{6 \times 5} \\ &= \frac{25 + 6}{30}\\ &= \frac{31}{30}. \end{align}

For $\frac{1}{4}$, $\frac{1}{5}$, and $\frac{1}{12}$ note that $12$ is divisible by $4$ so we can look for a common denominator of $\frac{1}{5}$ and $\frac{1}{12}$ and this will also work with $\frac{1}{4}$. For $\frac{1}{5}$ and $\frac{1}{12}$ we can use $5 \times 12$ as a common denominator:

\begin{align} \frac{1}{4} + \frac{1}{5} + \frac{1}{12} &= \frac{15 \times 1}{15 \times 4} + \frac{12 \times 1}{12 \times 5} + \frac{5 \times 1}{5 \times 12} \\ &= \frac{15}{60} + \frac{12}{60} + \frac{5}{60} \\ &= \frac{15+12+5}{60} \\ &= \frac{32}{60}. \end{align}

As a parenthetical note, this gives an example where even though we found the least common denominator to perform the addition, the resulting fraction is not in reduced form: the reduced form is $\frac{8}{15}$.

2. To write $\frac{3}{4}$ as an Egyptian fraction, we might notice that

\begin{align} \frac{3}{4} &= \frac{2+1}{4} \\ &= \frac{2}{4} + \frac{1}{4} \\ &= \frac{1}{2} + \frac{1}{4}. \end{align}

Alternatively, since $\frac{1}{2}$ is the largest of the unit fractions that is less than $\frac{3}{4}$ it would be reasonable to take $\frac{1}{2}$ as one of the unit fractions in the Egyptian fraction expression for $\frac{3}{4}$. Performing subtraction gives

\begin{align} \frac{3}{4} - \frac{1}{2} &= \frac{3}{4} - \frac{2 \times 1}{2 \times 2} \\ &= \frac{3}{4} - \frac{2}{4} \\ &= \frac{1}{4}. \end{align}

This gives us the same expression as above: $\frac{3}{4} = \frac{1}{2} + \frac{1}{4}$.
There are many other ways to write $\frac{3}{4}$ as an Egyptian fraction. Since

\begin{align} \frac{1}{2} &= \frac{3}{6} \\ &= \frac{2}{6} + \frac{1}{6} \\ &= \frac{1}{3} + \frac{1}{6} \end{align}

and since $\frac{3}{4} = \frac{1}{2} + \frac{1}{4}$ we have another expression of $\frac{3}{4}$ as an Egyptian fraction, namely $$ \frac{3}{4} = \frac{1}{3} + \frac{1}{6} + \frac{1}{4}. $$
All Egyptian fractions share this same property: there are always endless ways to write an Egyptian fraction.

# data[].attributes.attachment_link.description
Alignments to Content Standards: 5.NF.A.1 Ancient Egyptians used unit fractions, such as $\frac{1}{2}$ and $\frac{1}{3}$, to represent all fractions. For example, they might write the number $\frac{2}{3}$ as $\frac{1}{2} + \frac{1}{6}$. We often think of $\frac{2}{3}$ as $\frac{1}{3}+\frac{1}{3}$, but the ancient Egyptians would not write it this way because they didn't use the same unit fraction twice.

# data[].attributes.attachment_link.readable_text
Task

Chichén Itzá was a Mayan city in what is now Mexico. The picture below shows El Castillo, also known as the pyramid of Kukulcán, which is a pyramid located in the ruins of Chichén Itzá.

The temple at the top of the pyramid is approximately 24 meters above the ground, and there are 91 steps leading up to the temple. How high above the ground would you be if you were standing on the 50th step?

IM Commentary

The purpose of this task is for students to solve a straight-forward problem involving a proportional relationship in a context. In order to solve the problem, students must assume that the steps are of uniform height, which looks reasonable given the picture.

Two different and important approaches to the problem are presented: calculating a unit rate (height per step) and writing an equation (the height of the 50th step is $\frac{50}{91}$ the height of the pyramid). The first of these approaches falls under the 6th grade ratio and proportion standards while the second aligns better with the 7th grade. The main goal of the third approach, using Egyptian fractions, is to provide an example of calculating fluently. The methods of the first two solutions are vital for a deep understanding of proportional reasoning.

The numbers make this an advanced 6th grade task. It could be used as a transition from 6th to 7th grade expectations for work with ratios and proportional relationships, or it could be used as an assessment task in 7th grade (either formative or summative, depending on the time and way it is used). If used in an instructional setting, the teacher could ask the students to estimate before they compute. Since there are 91 steps and we want to know about the 50th step, we know that the height should be a little more than half of the total 24 meters in height.

This task could also be used to transition to the 8th grade standard on using similar triangles to prove that the slope of a line is well-defined. Looking at the cross-section of this pyramid and the similar triangles that we are essentially asking about would be a useful extension task:

The image is courtesy Wikimedia Commons:
http://en.wikipedia.org/wiki/File:Chichen_Itza_3.jpg
Solutions

There are 91 steps needed to get you to the top, which is 24 meters high. So each step is about $24\div91$ meters tall. Since

$$24\div91=\frac{24}{91}\approx 0.264$$

we estimate that each step is about 0.264 m or 26.4 cm tall on average. Since

$$50\times \frac{24}{91}\approx 13.2 $$

we'd expect to be about 13 meters above the ground if we stood on the 50th step.

Assuming the steps are all the same size, the ratio of the height, $h$, of the 50th step to the height of the 91st step will be 50:91. So we know that

$$\frac{50}{91}=\frac{h}{24}$$

If we multiply both sides of this equation by 24, we get that

$$h=24\times\frac{50}{91}\approx 13.2$$

So the 50th step is about 13 meters off of the ground.

Solution: The solution for people who like Egyptian Fractions

Half of 91 is 45.5. One tenth of one half (or one twentieth) of 91 is 4.55. These add up to 50.05 and tell us that the 50th step is about $\frac{1}{2} + \frac{1}{20}$ of the way up the structure. The top is 24 meters. So we can estimate the height of the 50th step by adding $\frac12$ of 24 which is 12 and $\frac{1}{20}$ of 24 which is 1.2. Since $$12+1.2 = 13.2$$ we know we are about 13 meters above the ground.

# data[].attributes.attachment_link.description
Chichén Itzá was a Mayan city in what is now Mexico. The picture below shows El Castillo, also known as the pyramid of Kukulcán, which is a pyramid located in the ruins of Chichén Itzá. The temple at the top of the pyramid is approximately 24 meters above the ground, and there are 91 steps leading up to the temple.

# data[].attributes.attachment_link.readable_text
Alignments to Content Standards: 7.RP.A.2

Task

Carli's class built some solar-powered robots. They raced the robots in the parking lot of the school. The graphs below are all line segments that show the distance $d$, in meters, that each of three robots traveled after $t$ seconds.

1. Each graph has a point labeled. What does the point tell you about how far that robot has traveled?

2. Carli said that the ratio between the number of seconds each robot travels and the number of meters it has traveled is constant. Is she correct? Explain.

3. How fast is each robot traveling? How did you compute this from the graph?

Solution

1. The point (1, 5) tells that robot A traveled 5 meters in 1 second.
The point (6, 9) tells that robot B traveled 9 meters in 6 seconds.
The point (5, 2) tells that robot C traveled 2 meters in 5 seconds.

2. Carli is correct. Whenever the ratio between two quantities is constant, the graph of the relationship between them is a straight line through (0,0). We can also say that for each robot, the relationship between the time and distance is a proportional relationship.

3. The speed can be seen as the $d$-coordinate of the graph when $t = 1$. This is the robot's unit rate:

Robot A traveled 5 meters per second, as shown by the point (1, 5) on its graph.
Robot B traveled 1.5 meters per second, as shown by the point (1, 1.5) on its graph.
Robot C traveled 0.4 meters per second, as shown by the point (1, 0.4) on its graph.

The speed of each robot can also be seen in the steepness of its graph, which is quantified as slope. But that perspective is not expected until grade 8.

# data[].attributes.attachment_link.description
Alignments to Content Standards: 7.RP.A.2 Carli's class built some solar-powered robots. They raced the robots in the parking lot of the school. The graphs below are all line segments that show the distance $d$, in meters, that each of three robots traveled after $t$ seconds. Each graph has a point labeled.

# data[].attributes.attachment_link.readable_text
Alignments to Content Standards: 3.OA.A.2

Task

Suppose there are 4 tanks and 3 fish in each tank. The total number of fish in this situation can be expressed as $4 \times 3 = 12$.

1. Describe what is meant in this situation by $12 \div 3 = 4$

2. Describe what is meant in this situation by $12 \div 4 = 3$

IM Commentary

The purpose of this task is to provide an opportunity for students to demonstrate that they can interpret within a given context the meaning of whole number quotients. Since this particular context is an asymmetric model of multiplication, there are two distinct interpretations of the quotients that can arise.

The given context is an asymmetric model of multiplication in the sense that 4 tanks with 3 fish per tank is different than 3 tanks with 4 fish per tank, even though in both cases there are a total of 12 fish. The convention used here is that the number of groups is listed first and the number of objects in each group is listed second, although in this case the context determines whether 3 or 4 is the group (respectively, number of objects in each group). Although these terms are not mentioned in the standards, the number of groups is sometimes called the multiplier and the number of objects in each group is called the mutiplicand.

Note: It is important that students can make sense of both of these interpretations of division arising from a single multiplication context with the factors and quotients all whole numbers. These interpretations will be extended to contexts with fractions and decimal in later grades.

Solution

The students' language may vary. What is important is the structure. The student needs to match what is known (either the number of groups or the amount in each group) and what is being determined by the quotient operation in this context.

1. Describe what is meant in this situation by 12 ÷ 3 = 4

Solution: There are 12 fish. If 3 fish are put in each tank there will be 4 tanks.

Comment: This is a measurement model of division, the number of objects in each group is known, and we are determining the number of groups.

2. Describe what is meant in this situation by 12 ÷ 4 = 3

Solution: There are 12 fish. If we want to partition (share) them equally among 4 tanks, we will end up with 3 fish in each tank.

Comment: This a sharing model of division, the number of groups is known and we are trying to determine the number in each group.

# data[].attributes.attachment_link.description
Alignments to Content Standards: 3.OA.A.2 Suppose there are 4 tanks and 3 fish in each tank. The total number of fish in this situation can be expressed as $4 \times 3 = 12$.

# data[].attributes.attachment_link.readable_text
Alignments to Content Standards: 5.G.A.1

Task

Materials

The students will need grid paper and colored pencils; some color for the ships and (for example) red for explosions on their ships and their enemy's ships. This is how they will keep track of what ordered pairs have been called.

Setup

Students begin by folding the grid paper in half. They need to draw coordinate axes on both the top half and the bottom half and label the $x$ and $y$ axes with the numbers 1-10 on each axis. The students will need to draw in 5 ships on ordered pairs and label the ordered pairs. They should draw:

Two ships that are sitting on 2 ordered pairs,

One ship that is sitting on 3 ordered pairs,

One ship that is sitting on 4 ordered pairs, and

One ship sitting on 5 ordered pairs.

Remind them the bottom half has their boats or (Navy) and the top half has their opponent's boats.

Actions

Students play in pairs sitting opposite each other and take turns calling out ordered pairs. Players should keep a list of the ordered pairs they call out written in $(x,y)$ form on a piece of paper that both players can see so there is no disagreement later on about what has been called (it is common for students to transpose the coordinates). Then they are to mark the ordered pair they call out on the top coordinate plane. They should mark in black if they missed and red if they hit their opponent's boat. On the bottom half of the grid paper they are to color black for the ordered pairs their opponent calls out and color red for the ordered pairs that hit their ship.

IM Commentary

The purpose of this task is to give students practice plotting points in the first quadrant of the coordinate plane and naming coordinates of points. It could be easily adapted to plotting points with negative coordinates. It also provides teachers with a good opportunity to assess how well their students understand how to plot ordered pairs and identify the coordinates of points.

Students can play this after the teacher demonstrates how to find the point on the coordinate plane that corresponds to an ordered pair. The teacher should help students set up their game boards the first time they play and visit with each group of students to listen and watch what they are doing to assess their understanding of the game and how to find ordered pairs. Also, it is good to listen to see if they are properly labeling their axes and generating, identifying, forming, and correctly graphing the ordered pairs on the coordinate plane.

Submitted by Ryan Glenn from McKissick Elementary to the second Illustrative Mathematics task writing contest.

Solution

The students will be able to demonstrate understanding of ordered pairs by playing this game. The teacher can assess the students' understanding through listening, watching, and checking the students' work as they play with a partner. The partners will also make sure the other student is not cheating because you can reward the winner with a prize.

# data[].attributes.attachment_link.description
Alignments to Content Standards: 5.G.A.1 The students will need grid paper and colored pencils; some color for the ships and (for example) red for explosions on their ships and their enemy's ships. This is how they will keep track of what ordered pairs have been called.

# data[].attributes.attachment_link.readable_text
Alignments to Content Standards: 6.G.A

Task

Christo and Jeanne Claude are artists who wrap large things - bridges, buildings, even islands. (See www.christojeanneclaude.net for images.)

Imagine that Jean Claude and Christo are preparing for their next large project - that of wrapping a large building. They are building a scale model in order to plan out the project.

The model building will measure $11 \frac12$ inches high, and $7 \frac12$ inches wide, and 6 inches long.

They are considering building the model with blocks that measure $\frac12 \text{ in}$ on each side.

1. What is the volume of the building in cubic inches?

2. How many of the blocks that measure $\frac12 \text{ in}$ on each side would Jeanne Claude and Christo need to make the model building?

3. Draw a picture of the model building and determine how much fabric Jean Claude and Christo will need to cover the model building.

4. The way Jean Claude and Christo have made their model, they know that they the number of square feet needed to cover the real building is 144 times the number of square inches needed to cover the model. How much fabric will Jean Claude and Christo will need to cover the real building?

IM Commentary

This task is primarily about volume and surface area, although it also gives students an early look at converting between measurements in scale models and the real objects they correspond to. Students are not required to understand the general principles for converting between measurements in scale drawings (or models) and measurements in their referents until 7th grade (see 7.G.1), but here they are given the conversion explicitly. This kind of task is more appropriate in an instructional setting rather than as an assessment item since it is laying the groundwork for work in seventh grade rather than representing a mature competency in 6th grade. (Thanks to Esther Koontz, a teacher who reviewed this task, for suggesting to include part (d).)

Submitted by Karen Gerow from the University of Georgia to the fourth Illustrative Mathematics contest.

Solution

1. $$ \begin{align} l \times w \times h &= \text{ Volume} \\ 11\frac12 \text{ in} \times 7\frac12 \text{ in} \times 6 \text{ in} &= \\ \frac{23}{2} \text{ in} \times \frac{15}{2} \text{ in} \times 6 \text{ in} &= \\ &= 517\frac12 \text{ in}^3 \end{align} $$

2. One way to do this is to figure out how many cubes that are $\frac12$ inch on each side it takes to make a cubic inch. The picture below shows that it takes $2\times2\times2=8$ of the blocks to make a cubic inch. Every cubic inch of volume requires 8 cubes, so we can think of a cubic inch as a group of 8 smaller cubes). Since the volume is given in cubic inches, we multiply the volume by 8: $$517.5\times8=4140$$ So they would need 4140 of the blocks to make the model.

Another way to do it: we can figure out how many blocks long, how many blocks wide, and how many blocks tall the model will be if we divide each of the dimensions by $\frac12$. $$ \begin{align} &6 \div \frac12 = 12 \text{ blocks long} \\ &7 \frac12 \div \frac12 = 15 \text{ blocks wide}\\ &11 \frac12 \div \frac12 = 23 \text{ blocks tall} \end{align} $$

So the base will be 12 blocks long and 15 blocks wide. $$12\times15=180$$ so 180 blocks will make up the base. The building will be 23 blocks tall, and each layer will contain 180 blocks.
$$ 23 \times 180 = 4140 \text{ blocks} $$ Again we see that they would need 4140 of the blocks to make the model, which might take a long time to glue together. Perhaps Jean Claude and Christo might want to take another approach to building their model.

3. The building needs to have the four sides and the top covered; here is a net for the faces of the building that will be covered by fabric:

It has:

two sides that are $11.5$ inches by $6$ inches,

two sides that are $11.5$ inches by $7.5$, and

the top of the building which is $6$ inches by $7.5$ inches.

We need to add the areas of all the faces: $$\begin{align} [2\times(11.5 \times 6) ] + [2\times(11.5 \times 7.5) ] + (6 \times 7.5 ) & = 138 + 172.5 + 45\\ &= 355.5 \end{align}$$ So the area of the fabric needed to cover the model building is $355.5 \text{ in}^2$.

4. The way Jean Claude and Christo have made their model, they know that they the number of square feet needed for the real building is 144 times the number of square inches needed for the model. In part (c) we calculated that number to be 355.5, so they need $144 \times 355.5 = 51,192$ square feet of fabric for the real building. That is a lot of fabric!

# data[].attributes.attachment_link.description
Alignments to Content Standards: 6.G.A Christo and Jeanne Claude are artists who wrap large things - bridges, buildings, even islands. (See www.christojeanneclaude.net for images.) Imagine that Jean Claude and Christo are preparing for their next large project - that of wrapping a large building.

# data[].attributes.attachment_link.readable_text
Alignments to Content Standards: 3.OA.C.7

Task

The game is played with one to four players.

To play, two decks of cards are needed: the "matching" cards and the "target" cards.

An array of matching cards, which have numbers on their faces, is dealt face down. There can be twelve to twenty-four cards in the array. A single target card is dealt face up.

A turn consists of a student flipping over matching cards one at a time, then trying to combine them to meet the criterion of the target card. The quantity of flipped cards and method of combination can be varied to utilize different skills and give variety to the game. Two possible versions of the game are described below. The student wins a point if they state a correct mathematical relationship between the matching cards and the target card.

In one version, a player reveals two cards each turn. The player wins a point if they can produce a product which matches the target card. Those who wish to keep track of points can do so by keeping target cards that are correctly "matched". Target cards can be single numbers but will more generally be descriptions of possible products. For example, the target card could be $24$, in which case the player would be looking for pairs $6$ and $4$, $8$ and $3$, or $12$ and $2$. But the target could also say "$30$ to $35$". In this case, if a player turns over a $4$, for example, they would then need to try to find an $8$, and to win the point they would need to say "$8$ times $4$ is $32$, which is between $30$ and $35$."

Another version of the game lets each player reveal three cards each turn. The player wins a point if they can produce an equation with their three cards and the target card that involves any of the operations of addition, subtraction, multiplication, or division. For example, suppose the target card is $7$ and the player turns $15$, $3$, and $2$. The player can win a point by stating that $15 \div 3 + 2 = 7$.

In each case, once a target card is "matched", a new target card is revealed. If a target card is not matched, the next player has a chance to match it.

Both of these versions of Kiri's Mathematical Match Game could be played as no-memory, light-memory, or memory games.

In the no-memory version, the chosen matching cards are left face-up after they are revealed.

In the light-memory version, the chosen matching cards are left face-up until some player is able to "match" the target card and wins a point. When a match is made, all of the matching cards that are face-up are turned back face-down, the board is replenished, and a new target card is drawn, so the next player sees the initial set-up.

In the memory version, the chosen matching cards are turned back face-down after every turn, so students would need to remember the locations of previously overturned cards in order to make use of them.

IM Commentary

This game is a fun way for the student to practice arithmetic skills to the point where the single-digit facts are committed to memory. It reinforces the relationship between multiplication and division, and depending on the target cards can also connect these recall skills with other skills such as estimation and understanding of properties. The only necessary materials, the cards, can be produced easily and can be re-used. After playing regularly, students could be engaged in making new target cards.

Here are some guidelines for constructing the "matching" and "target" card decks. The matching cards should typically be labeled between $1$ and $10$ or $12$. Larger numbers are possible, especially for variants that also involve addition. The target cards should be constructed to give a few possibilities, and promote reflection. For example, a target card might be just a single integer, but if so it should probably have many factors or else the game would proceed too slowly. A target card could be a small range of integers, such as "30 to 35". The targets can be even more interesting, too. For example, a target of "ones digit is $5$" invites students to notice that for this target, one of the two factors must be a $5$, and the other must be odd. The target card deck can be varied to encourage using different skills.

The second version of the game described in this task requires the student to process, if they are flipping three cards, up to $96$ possible arithmetic expressions (not counting commutativity) for matching the target card. Going through many of those possibilities fluidly would indicate a strong level of proficiency not only in multiplication facts but in use of properties as well.

Points could be ignored entirely, or used to determine "winners", or could be used to earn rewards collectively or individually.

This is a third-grade version of 1.OA Kiri's Mathematical Match Game.

Solution

Students meet the basic mathematical requirements of the game by evaluating products within $100$. To play the game fluidly and successfully, students will need most products committed to memory. If students are not already at this level, the versions of the game described in this task can help them commit these facts to memory while also allowing them to compute more quickly by recognizing patterns and structure in arithmetic.

# data[].attributes.attachment_link.description
Alignments to Content Standards: 3.OA.C.7 The game is played with one to four players. To play, two decks of cards are needed: the "matching" cards and the "target" cards. An array of matching cards, which have numbers on their faces, is dealt face down. There can be twelve to twenty-four cards in the array.

# data[].attributes.attachment_link.readable_text
Alignments to Content Standards: 4.OA.B.4

Task

The 20 students in Mr. Wolf's 4th grade class are playing a game in a hallway that is lined with 20 lockers in a row.

The first student starts with the first locker and goes down the hallway and opens all the lockers.

The second student starts with the second locker and goes down the hallway and shuts every other locker.

The third student stops at every third locker and opens the locker if it is closed or closes the locker if it is open.

The fourth student stops at every fourth locker and opens the locker if it is closed or closes the locker if it is open.

This process continues until all 20 students in the class have passed through the hallway.

1. Which lockers are still open at the end of the game? Explain your reasoning.

2. Which lockers were touched by only two students? Explain your reasoning.

3. Which lockers were touched by only three students? Explain your reasoning.

4. Which lockers were touched the most?

IM Commentary

The purpose of this instructional task is for students to deepen their understanding of factors and multiples of whole numbers. This is a classic mathematical puzzle; often it is stated in terms of 100 lockers, so students have to make the connection to factors and multiples to solve it. In this version, students can just go through all the rounds of opening and closing locker doors and observe after the fact that there is a relationship between the factors a locker number has and whether it is open or closed at the end.

This task provides students with an excellent opportunity to engage in MP7, Look for and make use of structure (if they see early on that there is a relationship with factors and multiples) or MP8, Look for and express regularity in repeated reasoning (if they start to see and describe the pattern as they imagine students opening and closing the lockers).

Because the total number of lockers is only 20, students might answer the questions without thinking about the underlying reasons for their answers. In the first question for example, a student might say "1, 4, 9, 16 are all still open because I tried it out and those were the ones that were left open." If a student goes this route, the teacher can steer the conversation back to factors by asking the students if they notice anything special about the numbers of the lockers which are still open. Asking students to try a larger number of lockers (say 50 or 100) and repeating the game can also help students look for a pattern since going through all the rounds of the game becomes less and less feasible as the number of lockers increases. Once students see the pattern, they should be pressed to explain why it works out this way.

Solution

1. The lockers that are still open are perfect squares, specifically $1\times 1$, $2\times 2$, $3\times 3$, and $4\times 4$. To see why this is true, first note that:

The 1st student changes all the lockers with a locker number that is a multiple of 1.

The 2nd student changes all the lockers with a locker number that is a multiple of 2.

The 3rd student changes all the lockers with a locker number that is a multiple of 3.

The 4th student changes all the lockers with a locker number that is a multiple of 4.

and so on. So a locker is touched by a student if that student's number is a factor of that locker's number.

Now consider the factors of 16 which are 1, 2, 4, 8, and 16. This means the 16th locker was opened by the 1st student, then it was closed by the 2nd student, then it was opened by the 4th student, then it was closed by the second 8th student, and then opened one last time by the 16th student. Now note that the square numbers are the only numbers with an odd number of factors (all other numbers have factors which come in pairs.) Since an odd number of factors means the locker will be changed an odd number of times, it must be open at the end.

With only 20 lockers, students can simply observe this fact about the number of factors to be true for the numbers 1 to 20, which gives practice with finding factors of whole numbers. The explanation for why this is always true goes a bit beyond 4th grade.

2. The lockers that were touched by only two students are the prime numbers (since prime numbers only have two factors.) In our case they are 2, 3, 5, 7, 11, 13, 17, and 19.

3. The lockers which are touched by only three students are the squares of prime numbers (because these are the only numbers which have three factors.) In our case they are 4 and 9. (For example, 4 has three factors: 1, 2, and 4)

4. The lockers which are touched the most are those lockers whose number has the most factors. In our case 12, 18, and 20 all have 6 factors, so they were each touched 6 times. $$ 12 = 1\times 12 = 2\times 6 = 3\times 4$$ $$ 18 = 1\times 18 = 2\times 9 = 3\times 6$$ $$ 20 = 1\times 20 = 2\times 10 = 4\times 5$$

# data[].attributes.attachment_link.description
Alignments to Content Standards: 4.OA.B.4 The 20 students in Mr. Wolf's 4th grade class are playing a game in a hallway that is lined with 20 lockers in a row. The first student starts with the first locker and goes down the hallway and opens all the lockers.

# data[].attributes.attachment_link.readable_text
Alignments to Content Standards: 7.NS.A.3

Task

The three seventh grade classes at Sunview Middle School collected the most boxtops for a school fundraiser, and so they won a $600 prize to share among them. Mr. Aceves' class collected 3,760 box tops, Mrs. Baca's class collected 2,301, and Mr. Canyon's class collected 1,855. How should they divide the money so that each class gets the same fraction of the prize money as the fraction of the box tops that they collected?

IM Commentary

This task requires students to be able to reason abstractly about fraction multiplication as it would not be realistic for them to solve it using a visual fraction model. Even though the numbers are too messy to draw out an exact picture, this task still provides opportunities for students to reason about their computations to see if they make sense. A teacher might start out by asking questions like, "Which class should get the most prize money? Should Mr. Aceves' class get more or less than half of the money? Mr. Aceves' class collected about twice as many box tops as Mr. Canyon's class - does that mean that Mr. Aceves' class will get about twice as much prize money as Mr. Canyon's class?"

This task also represents an opportunity for students to engage in Standard for Mathematical Practice 5 Use appropriate tools strategically. Fraction tasks in earlier grades (see 5.NF.6, for example) would be inappropriate to use a calculator with because the point of those tasks is to develop an understanding of the meaning of fraction multiplication and to practice some of those computations. Here, there is little benefit in students doing the computations by hand (few adults would), and so provides an opportunity to discuss the value of having a calculator and when it is (and is not) appropriate to use it.

Solution

All together, the students collected $3,760 + 2,301 + 1,855 = 7,916$ box tops.

Mr. Aceves' class collected $\displaystyle \frac{3760}{7916}$ of the box tops.

Mrs. Baca's class collected $\displaystyle \frac{2301}{7916}$ of the box tops.

Mr. Canyon's class collected $\displaystyle \frac{1855}{7916}$ of the box tops.

The amount for Mr. Aceves' class is $\displaystyle \frac{3760}{7916} \times 600 \approx 284.99$

The amount for Mrs. Baca's class is $\displaystyle \frac{2301}{7916} \times 600 \approx 174.41$

The amount for Mr. Canyon's class is $\displaystyle \frac{1855}{7916} \times 600 \approx 140.60$

So \$284.99 should go to Mr. Aceves' class, \$174.41 should go to Mrs. Baca's class, and \$140.60 should go to Mr. Canyon's class.

# data[].attributes.attachment_link.description
Alignments to Content Standards: 7.NS.A.3 The three seventh grade classes at Sunview Middle School collected the most boxtops for a school fundraiser, and so they won a $600 prize to share among them. Mr. Aceves' class collected 3,760 box tops, Mrs. Baca's class collected 2,301, and Mr. Canyon's class collected 1,855.

# data[].attributes.attachment_link.readable_text
Task

1. For each shape below, draw a scaled copy with a scale factor of 2. Explain why your drawing is accurate.

2. Below are two pairs of polygons. For each pair, explain whether or not one is a scaled version of the other.

IM Commentary

The goal of this task is to gather together knowledge and skills from the seventh grade in a context which prepares students for the important eighth grade notion of similarity. These important ideas include:

Scaled shapes have the same angle measures: concretely, when we scale two line segments meeting at a point, the lengths of the segments change but they still are part of the same pair of rays making the same angle.

Corresponding segments of scaled shapes have proportional lengths. If $a$ and $b$ are side lengths of shape S and $c$ and $d$ are the lengths of the corresponding sides of shape $T$, then $$c:a :: d:b$$ and the common ratio tells us the scale factor going from shape $S$ to shape $T$.

These ideas are sufficient to identify similar polygons and also to produce a scaled version of a given polygon.

The teacher will want to make sure that students have access to rulers, protractors, and possibly patty paper as well to help with their reasoning. In many of the answers, the original angle or polygon is used as part of the scaled version. This is a good technique which saves some time but students may be more comfortable or more inclined to produce a separate scaled angle or polygon. The teacher will need to make sure that students understand that scaling only changes size, not shape: in particular, it does not impact angles. If students try to double the measure of the angle on the first part of (a), the teacher can ask them if the two images are the same shape. The teacher may also encourage students to measure the angles of the shapes in part (a) to verify that they are the same.

Scaled Polygons are one of the most important examples of similar shapes, introduced in 8.G.A. The ideas developed in 7.G.1 and 7.RP.2 are not sufficient, however, to motivate the abstract notion of similarity transformation introduced in the eighth grade. A more abstract, challenging example involving scaled shapes is found in https://www.illustrativemathematics.org/tasks/1818. This task has been aligned with 8.G.A because, while the ideas needed to do this work are available in the seventh grade, the motivation for this example and the synthesis of these important ideas is also appropriate for the eighth grade.

Solution

1. We can double the length of the two segments as pictured below

(We can physically do this using a ruler, or overlaying some grid paper, or drawing a circle of radius $\overline{CB}$ centered at $C$, etc.) Regardless of the construction, we then have that segment $\overline{BC'}$ is a scaled version of $\overline{BC}$ with a scale factor of 2 and, similarly, $\overline{BA'}$ is the scaled version of $\overline{BA}$ with scale factor two. Scaling does not change angles and we can see this in the picture as $\angle A'BC'$ is the same as $\angle ABC$.

2. Scaling $\triangle DEF$ can be accomplished as in (i), thinking of $D$, $E$, and $F$ as three points lying on an angle and then drawing in the remaining side after scaling the angle:

We can check by measuring that the length of $\overline{E'F'}$ is twice the length $\overline{EF}$ so all three sides of $\triangle DEF$ have been scaled by a factor of 2.

3. The grid can help draw our scaled quadrilateral as it allows us to measure each side length. Below is a picture of the quadrilateral along with a scaled copy:

Starting at vertex G, we make the blue quadrilateral by moving right two units and up 4 units to H, then we move right one unit and down 5 units to I, then moving left one unit and up 3 units to J, finally left two units and down 2 units back to G. If we double all of these numbers, this makes the red quadrilateral starting at its bottom left vertex G$^\prime$. So the red quadrilateral is a scaled version of the blue quadrilateral with scale factor 2.

1. Triangle $NOP$ is not a scaled version of triangle $KLM$. To see why, note that the shortest side of $\triangle KLM$, $\overline{KL}$, is about half the length (perhaps a little less) of the shortest side of $\triangle NOP$, $\overline{PO}$. But the longest side of $\triangle KLM$, $\overline{LM}$, is more than half the length of the longest side of $\triangle NOP$, $\overline{NO}$. This means that there is no possible common scale factor for the sides of these triangles.

Alternatively, we could estimate the angle measures with a protractor and find that they are different for the two triangles. Because of what we found in part (i) of (a), this means that the two triangles are not scaled versions of one another.

2. Starting with vertex Q of the blue quadrilateral we move right two units and up one to vertex R, then right three units and down three units to vertex S, then left three units and up two units to vertex T, and finally left two units back to vertex Q. Starting with vertex U of the red quadrilateral and we go right four units and down two to vertex X, then right six units and up six units to vertex W, then left six units and down four units to vertex V, and finally left four units back to vertex U. If we compare these directions, they are the same except that all lengths have been doubled and up has been replaced with down. So the red shape is a scaled version of the blue shape with a scale factor of two.

# data[].attributes.attachment_link.description
For each shape below, draw a scaled copy with a scale factor of 2. Explain why your drawing is accurate. Below are two pairs of polygons. For each pair, explain whether or not one is a scaled version of the other.

# data[].attributes.attachment_link.readable_text
Alignments to Content Standards: 6.RP.A

Task

In a marching band there are 24 trombones and 15 snare drums.

1. Heather says, "The ratio of trombones to snare drums is 24:15." Audrey says, "No, the ratio of trombones to snare drums is 8:5." Who is right, and why?

2. A different marching band has 30 snare drums, but its trombone to snare drum ratio is the same as the first marching band. How many trombones does it have?

3. Explain what these ratios all have in common: 8:5, 16:10, 24:15. Create a diagram to support your answer.

4. Mathematicians call these "equivalent ratios." Come up with a new set of ratios that are equivalent to 6:21.

5. Explain in your own words the meaning of "equivalent ratios."

IM Commentary

The purpose of this task is to make explicit the meaning of equivalent ratios. Students create, analyze, and draw diagrams of two different sets of equivalent ratios, and then they write their own definition of "equivalent ratios" in their own words.

This task should not be students' first encounter with equivalent ratios, but rather, should cap off lots of work with sets of equivalent ratios in many contexts using different representations.

We suggest showing a few minutes of a performance by a large university marching band (there's an example embedded at the bottom of this commentary) before starting this task (or maybe during a short break). Students who participate in band might think some of the quantities given are unrealistic, but some marching bands are quite large. A teacher could take a few minutes to see how many instruments students can identify.

Part (a) is meant to clarify that even though 24:15 and 8:5 are written with different numbers, that they both correctly describe the situation. Using language like "there are 24 trombones for every 15 snare drums" and "there are 8 trombones for every 5 snare drums" can help reinforce this, especially with an appropriately constructed diagram. When experienced practitioners talk about "the ratio," they often automatically assume a reduced form written with the smallest possible whole numbers; but there's nothing about the mathematics that requires this.

Students who have worked with unit rates, especially if they have come to see unit rates as useful, may generate something like $\frac{8}{5}:1$ in part (c) and $\frac{2}{7}:1$ in part (d). This could present an opportunity for a nice conversation about keeping a context in mind when representing quantities (involving aspects of MP2). There's nothing mathematically wrong with saying that $\frac{8}{5}:1$ is a ratio equivalent to the others in part (c), and it may be a good use of time to point out how we can see $\frac{8}{5}:1$ in a diagram. At the same time, we can recognize that it would be impossible for a marching band to contain $\frac{8}{5}$ (a little more than one and a half) trombones.

The solution to part (e) is given in a form that a student might produce. More formally, we might say "Given a ratio $a : b$, an equivalent ratio is any ratio of the form $sa : sb$ where $s$ is a positive number." Students are likely to say "multiply or divide" instead of just "multiply." A teacher might decide to let this go or to take the time to talk about how in math we like to write definitions as simply as possible. Since, for example, dividing by 4 is the same as multiplying by $\frac{1}{4}$, we can get away with just saying "multiply."

Solution

1. They are both right! These are different ways of correctly describing the situation.

2. Here is the given information and what we want to find organized in a ratio table:

In the snare drum row, I notice that 30 is 6 groups of 5. So the number of trombones indicated by the ? must be 6 groups of 8, or 48 trombones.

Each ratio is a multiple of 8 and 5. Here is a diagram that shows this:

But drawing all those squares is cumbersome! We could choose a different representation; for example, here is a double number line diagram:

1. Answers may vary. Some ratios that are equivalent to 6:21 are 2:7, 4:14, and 12:42.

2. You can get an equivalent ratio if you start with the two numbers that make the ratio and multiply them by the same number.

# data[].attributes.attachment_link.description
Alignments to Content Standards: 6.RP.A In a marching band there are 24 trombones and 15 snare drums. Heather says, "The ratio of trombones to snare drums is 24:15." Audrey says, "No, the ratio of trombones to snare drums is 8:5." Who is right, and why?

# data[].attributes.attachment_link.readable_text
Task

Alex, Bryan, and Cynthia are about to eat lunch, and they have two sandwiches to share.

1. Draw a picture to show how they could equally share the sandwiches. How much of a sandwich does each person get?

2. Write an equation involving addition to show how together these parts make up the 2 sandwiches. Explain how the equation you wrote represents this situation.

3. Write an equation involving multiplication to show how all the parts make up the 2 sandwiches. Explain how the equation you wrote represents this situation.

4. Write an equation using multiplication to show the fraction of a sandwich each student gets. Explain how the equation you wrote represents this situation.

5. Write an equation using division to show the fraction of a sandwich each student gets. Explain how the equation you wrote represents this situation.

IM Commentary

This task requires students to think about how a single situation involving fractions can be accurately represented using addition or multiplication.

Students are first asked to draw a picture in part a showing this situation to reinforce the fact that the same situation can be modeled using three different operations. Students are asked to divide two sandwiches into three equal pieces, and must reason what that will look like. Students should be able to realize fairly easily that each student will get $\frac{2}{3}$ of the sandwich, and to model this using in addition in part b. As students' facility with operations on fractions matures, they should not be expected to always draw a picture.

In part c, students are asked to represent the situation using multiplication, and to explain their representation. Teachers may want to show multiple correct explanations, including interpretations of products as partitions of equal parts, visual displays (such as a number line or fraction diagram), or a mathematical statement such as $\frac23 + \frac23 + \frac23 = 3 \times \frac23 = 2$ If students struggle, the teacher can tell the students to think of the problem as $2$ sandwiches $= 3 \times$ a fraction of a sandwich. A teacher might also want to ask students to demonstrate multiple correct explanations, particularly if they notice students relying only on one method and being confused by others.

Parts d and e shift the focus away from using operations to represent all of the sandwiches, instead using operations to focus on each student's share of the sandwiches. In part d, students may notice that each person will get $\frac13$ of the $2$ sandwiches, since there are $3$ people sharing them equally. Students may then need help realizing that since they will get $\frac13$ of each of the two sandwiches, they then will get $\frac13 \times 2$ sandwiches, which equals $\frac23$ The second diagram in the solution to part a may help, as each of the two sandwiches is divided into three equal parts (therefore,
$\frac13 \times 2$).

Finally, students are asked to consider the meaning of a fraction in part e - that is, the division of a numerator by a denominator. This task reinforces the concept that any fraction can be described as dividing a whole number into the number of pieces specified in the denominator. The explanations that the students produce for each of the three tasks may be slightly different, but will result in the same conclusion (that each student gets $\frac23$ of a sandwich). If students struggled with this concept, the teacher might note that if there were six sandwiches and each person got the same amount, we would use division to determine that each person gets two sandwiches $(6 \div 3 = 2)$. The situation here is the same, except instead of $6$ wholes, the unit is $6$ thirds.

This task was written as part of a collaborative project between Illustrative Mathematics, the Smarter Balanced Digital Library and the Teaching Channel.

Solutions

Solution: Solution to part A

A. Students should say that each person gets $\frac23$ of one sandwich. The picture could be anything that shows the sandwiches being divided equally three ways. For example:

Or

The second solution has a few advantages. First, this type of picture prepares students for the tape diagrams they will see in the sixth grade when they study ratios. Secondly, this method of division into equal parts would apply if, for example, one of the sandwiches were tuna fish and the other was grilled cheese. Each person would receive an equal part of each type of sandwich. Third, for a more complex fraction such as $\frac{3}{5}$ the equal portions can be seen very readily (without counting) with this type of model.

Solution: Solution to part B

B. $\frac23 + \frac23 + \frac23 = \frac63 = 2$. Other reasonable equations, such as $\frac23 + \frac23 + \frac13 + \frac13 = \frac63 = 2$, could be acceptable. Each person gets $\frac23$ of a sandwich and there are three people so if you add up these three pieces of size $\frac23$ you get a total of 2.

Solution: Solution to part C

C. $3 \times \frac23 = 2$. Acceptable explanations include accurate descriptions (e.g. "You add two-thirds three times, which is the same as multiplying it by $3$"), a number line model (showing $\frac23$, $\frac43$, and $2$) or a visual fraction diagram.

Solution: Solution for part D

D. $2 \times \frac13 = \frac23$. Each person gets $\frac13$ of each sandwich and there are two sandwiches so two groups of $\frac13$ is equal to $\frac23$.

Solution: Solution for part E

E. $2 \div 3 = \frac23$. Acceptable explanations include accurate descriptions (e.g. "You divide two wholes into three equal parts, and each part will be $\frac23$"), a number line model (focusing on the dividing two wholes into three parts), or a visual fraction diagram.

# data[].attributes.attachment_link.description
Alex, Bryan, and Cynthia are about to eat lunch, and they have two sandwiches to share. Draw a picture to show how they could equally share the sandwiches. How much of a sandwich does each person get? Write an equation involving addition to show how together these parts make up the 2 sandwiches.

# data[].attributes.attachment_link.readable_text
Alignments to Content Standards: 4.NF.A

Task

Cruz and Erica were both getting ready for soccer.

Erica said,

I ran more laps, so I ran farther.

Cruz said,

4 laps around the school is 1 mile, but it takes 12 laps around the playground to go 1 mile. My laps are much longer, so I ran farther.

Who is right? Draw a picture to help you explain your answer.

IM Commentary

The purpose of this task is for students to compare two fractions that arise in a context. Because the fractions are equal, students need to be able to explain how they know that. Some students might stop at the second-to-last picture and note that it looks like they ran the same distance, but the explanation is not yet complete at that point. Just because to pictures look the same doesn't mean they are. Think of a picture representing $\frac{49}{100}$ and $\frac{50}{99}$--these will look very similar, but they are not, in fact, equal. To explain why Erica and Cruz ran the same distance, we have to explain why $\frac14 = \frac{3}{12}$.

Note that this task illustrates the cluster "4.NF.A Extend understanding of fraction equivalence and ordering" because it addresses elements of both the standards within this cluster.

Solution

Let's begin with a picture representing $1$ mile.

With this picture in mind, we can see that one lap around the school is $\frac14$ mile:

and we can see that one lap around the playground is $\frac{1}{12}$ mile:

Since Erica ran 3 laps around the playground, she ran $\frac{3}{12}$ mile. Since Cruz ran 1 lap around the school, he ran $\frac14$ mile:

Now looking at the pictures above, it looks like they actually ran the same distance. We can show this is true if we divide each of the laps around the school into three equal pieces:

When we subdivide each $\frac14$ mile lap into three equal pieces, our mile is divided into 12 equal pieces. The piece that represents Cruz's run is now divided into 3 pieces, and so $\frac14 = \frac{3}{12}$ he ran the same distance that Erica ran.

# data[].attributes.attachment_link.description
Alignments to Content Standards: 4.NF.A Cruz and Erica were both getting ready for soccer. Erica said, I ran more laps, so I ran farther. Cruz said, 4 laps around the school is 1 mile, but it takes 12 laps around the playground to go 1 mile.

# data[].attributes.attachment_link.readable_text
Alignments to Content Standards: 5.OA.A

Task

1. State the meaning of each of the following expressions and draw a picture that represents it. $$(3\times5)+4$$ $$3\times(5+4)$$

2. State the meaning of this expression: $$3\times5+4$$ How do you know?

3. State the meaning of each of the following expressions and draw a picture that represents it. $$(3+5)\times4$$ $$3+(5\times4)$$

4. State the meaning of this expression: $$3+5\times4$$ How do you know?

IM Commentary

The purpose of this task is to help students think about the reason for the mathematical convention known as the "order of operations." The task could be used in two ways: to present an apparently unresolvable mathematical situation that can lead to a classroom discussion about the need for an order of operations, or to help students who know already know the order of operations to pause and think more deeply about why they are necessary. For students to get the full benefit out of this task, they will need to hear a bit more from the teacher about the idea of mathematical conventions and why we need to have them when a mathematical statement would otherwise be ambiguous.

Addition, subtraction, multiplication, and division are all operations that take two numbers (named the addends, the minuend and subtrahend, the factors, or the divisor and the dividend, respectively) and produce a single number (called the sum, the difference, the product, or the quotient, respectively)--this is why they are called binary operations. In principle, grouping symbols are needed whenever there are more than 2 numbers that will be combined using one or more of these binary operations. This is because it is impossible to know a priori whether, for example,

$$8 + 3 \times 2$$

means we should add the 8 and the 3 first, which seems reasonable since in many languages, including English, we read from left to right, or multiply the 3 and the 2 first. In mathematics, we resolve the meaning of numeric expressions that include more than two numbers but do not contain grouping symbols in one of two ways:

We convince ourselves that, in some instances, the result is the same regardless of which pair of numbers we choose to combine first.

We choose conventions for the order in which we evaluate a numeric expression because we will get different results if we simplify the expressions in different ways.

The associative property of addition:

$$a + (b + c) = (a + b) + c$$

and the associative property of multiplication:

$$a \times (b \times c) = (a \times b) \times c$$

are examples of the first of these. The so-called "order of operations" are how we have settled the second of these. To go back to our original example, we know that

$$8 + 3 \times 2 = 8 + (3 \times 2)$$

because we have all agreed that if we have an expression with at least three numbers that are combined using addition and multiplication, we will multiply before we add. There are actually some good mathematical reasons we have structured the order of operations as we have; to read more about it, see "Order of operations" and other oddities in school mathematics by H. Wu.

Solution

1. The expression $(3\times5)+4$ means we should multiply 3 and 5 and then add 4 to the result. Here is a picture:

The expression $3\times(5+4)$ means we should multiply 3 by the sum of 5 and 4. Here is a picture:

2. The expression $3\times5+4$ means the same thing as the expression $(3\times5)+4$. If we haven't already learned about the order of operations, then it is possible for $$3\times5+4$$ to mean either $$(3\times5)+4$$ or $$3\times(5+4)$$ The only way to know what $3\times5+4$ means is to understand that we have all agreed that when we don't have grouping symbols, we should multiply before we add.

3. The expression $(3+5)\times4$ means we should add 3 and 5 and then multiply the result by 4. Here is a picture:

The expression $3+(5\times4)$ means we should add 3 to the product of 5 and 4. Here is a picture:

4. The expression $3+5\times4$ means the same thing as the expression $3+(5\times4)$. If we haven't already learned about the order of operations, then it is possible for $$3+5\times4$$ to mean either $$(3+5)\times4$$ or $$3+(5\times4)$$ The only way to know what $3+5\times4$ means is to understand that we have all agreed that when we don't have grouping symbols, we should multiply before we add.

# data[].attributes.attachment_link.description
Alignments to Content Standards: 5.OA.A State the meaning of each of the following expressions and draw a picture that represents it. $$(3\times5)+4$$ $$3\times(5+4)$$ State the meaning of this expression: $$3\times5+4$$ How do you know? State the meaning of each of the following expressions and draw a picture that represents it.

# data[].attributes.attachment_link.readable_text
Alignments to Content Standards: 8.EE.C.7

Task

You have a coupon worth \$18 off the purchase of a scientific calculator. At the same time the calculator is offered with a discount of 15%, but no further discounts may be applied. For what tag price on the calculator do you pay the same amount for each discount?

IM Commentary

This task involves solving equations with rational coefficients, and requires students to use the distributive law ("combine like terms"). The equation also provides opportunities for students to observe structure in the equation to find a quicker solution, as in the second solution presented.

A somewhat less directed version of this question might simply ask students which discount they'd prefer. After their intuition or calculations show that the answer depends on the price, a further probe asking them to delineate how it depends on the price would lead them to set up and solve the same equation.

Solutions

Solution: Coupon versus discount

If $p$ is the tag price in dollars then $p-18$ is the price using the coupon while $0.85p$ is the price using the 15% discount. Then when the discounts are the same,

\begin{eqnarray*} p-18 &=& 0.85p\\ p - 0.85p &=& 18\\ 0.15p &=& 18\\ p &=& 120. \end{eqnarray*}

Thus, the tag price is \$120.

Solution: Coupon versus discount: solution by comparing the price reduction

The cost is the same when the reduction in price is the same. The coupon always gives a \$18 reduction in price; if $p$ is the tag price in dollars then the discount gives a reduction in price of $\$0.15p$. So the cost is the same when

\begin{eqnarray*} 18 &=& 0.15p\\ p &=& 120. \end{eqnarray*}

Thus, the tag price is \$120.

# data[].attributes.attachment_link.description
Alignments to Content Standards: 8.EE.C.7 You have a coupon worth \$18 off the purchase of a scientific calculator. At the same time the calculator is offered with a discount of 15%, but no further discounts may be applied. For what tag price on the calculator do you pay the same amount for each discount?

# data[].attributes.attachment_link.readable_text
Task

The 150 students at Skokie School were asked if they prefer seeing the movie Hunger Games or Divergent. The data showed that 100 preferred Hunger Games and 50 preferred Divergent.

1. Look at the following statements and decide if each accurately reports the results of the survey and explain how you know.

1. At Skokie School, 1/3 of the students prefer Hunger Games.

2. Students prefer Hunger Games to Divergent in a ratio of 2 to 1.

3. The ratio of students who prefer Divergent to students who prefer Hunger Games is 1 to 2.

4. The number of students who prefer Hunger Games is 50 more than the number of students who prefer Divergent.

5. The number of students who prefer Hunger Games is two times the number of students who prefer Divergent.

2. Compare statements (iv) and (v) above. In what ways is the information given by these statements similar? In what ways is it different? Explain.

IM Commentary

The goal of this task is to introduce ratio language and have students apply reasoning about ratios in a context. In order to answer the questions, students need to understand the difference between a fraction and a ratio and need to analyze the part to whole and the part to part relationships in a ratio.

This is an engaging introductory lesson for a unit on ratio and proportional relationships. Students can relate to the context and the numbers are "friendly" enough that students can focus on the concepts without getting bogged down in calculations.

This task was written as part of a collaborative project between Illustrative Mathematics, the Smarter Balanced Digital Library, the Teaching Channel, and Desmos.

Solution

1. There are 150 students who took the survey so 1/3 of the students would be 50 students. But 100 of the students surveyed preferred Hunger Games to Divergent so this is not true. It is an easy mistake to make, however, because it is true that 1/3 more of the 150 students prefer Hunger Gamescompared to those who prefer Divergent.

2. The ratio of students who prefer Hunger Games to students who prefer Divergent is 100:50. This is equivalent to the ratio 2:1 as we can see by multiplying 2 and 1 by 50. This can also be shown in steps with a ratio table:

Students who prefer Hunger Games

Students who prefer Divergent

100

50

10

5

2

1

The second line comes from the first by multiplying both entries by $\frac{1}{10}$ and then the third row is the second row multiplied by $\frac{1}{5}$.

3. The ratio of students who prefer Divergent to students who prefer Hunger Games is 50:100. This is equivalent to 1:2 as we can see by multiplying 1 and 2 by 50. This can be shown with the ratio table of part (b) or with a double number line as below:

The given information is furthest to the right and then the equivalent ratios, calculated by multiplying by $\frac{1}{10}$ and then $\frac{1}{5}$, are to the left closer to 0.

4. Since 100 students prefer Hunger Games and 50 prefer Divergent and 100 = 50 + 50 it is true that 50 more students prefer Hunger Games.

5. Since 100 students prefer Hunger Games and 50 prefer Divergent and 100 = 2 $\times$ 50 it is true that the number of students who Hunger Games is twice the number who prefer Divergent.

2. The two statements give the same information if we know how many students took the survey. If we know that 150 students took the survey then both (d) and (e) tell us that 100 students preferred Hunger Games and 50 preferred Divergent. On the other hand, without this information only (e) tells us the ratio of students who preferred Hunger Games to Divergent

# data[].attributes.attachment_link.description
The 150 students at Skokie School were asked if they prefer seeing the movie Hunger Games or Divergent. The data showed that 100 preferred Hunger Games and 50 preferred Divergent. Look at the following statements and decide if each accurately reports the results of the survey and explain how you know.

# data[].attributes.attachment_link.readable_text
Alignments to Content Standards: 3.NF.A.3

Task

Alec and Felix are brothers who go to different schools. The school day is just as long at Felix' school as at Alec's school. At Felix' school, there are 6 class periods of the same length each day. Alec's day is broken into 3 class periods of equal length. One day, it snowed a lot so both of their schools started late. Felix only had four classes and Alec only had two. Alec claims his school day was shorter than Felix' was because he had only two classes on that day. Is he right?

IM Commentary

The purpose of this task is for students to investigate a claim about a comparison of two fractions in a context. Many fraction problems are set in food contexts or a situation where a physical thing is being shared. It is important for students to work on more abstract quantities like time as well.

The Standards for Mathematical Practice focus on the nature of the learning experiences by attending to the thinking processes and habits of mind that students need to develop in order to attain a deep and flexible understanding of mathematics. Certain tasks lend themselves to the demonstration of specific practices by students. The practices that are observable during exploration of a task depend on how instruction unfolds in the classroom. While it is possible that tasks may be connected to several practices, only one practice connection will be discussed in depth. Possible secondary practice connections may be discussed but not in the same degree of detail.

This particular task helps to illustrate Mathematical Practice Standard 3, Construct viable arguments and critique the reasoning of others. Students are asked to critique the reasoning of Alec's claim that his school day was shorter than his brother's. This type of task provides students with an opportunity to distinguish a reasonable explanation from that which is flawed. If there is a flaw in the argument they can further explain why it is flawed. Students will need to determine what information is necessary to calculate in order to support or dispute the claim. To distinguish if Alec's claim is true or false, students will have to uncover the fractional part of the day Alec and Felix were at school and not just rely on the number of classes they each attended. Learning how to argue whether a claim is true or false concisely and precisely becomes a routine part of a student's mathematical work.

Solution

Felix has six equal class periods each day.

So each class period lasts for $\frac16$ of the day.

Alec has three equal class periods each day.

So each class period lasts for $\frac13$ of the day.

Felix only had 4 class periods, so he went to school for $\frac46$ of a full day. Alec only had 2 class periods, so he went to school for $\frac23$ of a full day.

But a full day is equal for the two brothers, so two of Felix' class periods are the same length as one of Alec's. The brothers actually went to school for the same amount of time on the snow day.

# data[].attributes.attachment_link.description
Alignments to Content Standards: 3.NF.A.3 Alec and Felix are brothers who go to different schools. The school day is just as long at Felix' school as at Alec's school. At Felix' school, there are 6 class periods of the same length each day. Alec's day is broken into 3 class periods of equal length.

# data[].attributes.attachment_link.readable_text
Alignments to Content Standards: 4.OA.B

Task

The table below shows all the products of pairs of numbers between 1 and 9.

1. Which numbers appear only once in the table of products? Where are these numbers located in the table? Why?

2. Which numbers appear the most frequently in the table? Why?

3. Which numbers between 10 and 20 do not appear in the table? What do these numbers have in common?

IM Commentary

The goal of this task is to encourage students to study the multiplication table, a familiar object, from a novel point of view. The table shows some, but not necessarily all, factorizations of different numbers. For example, 24 appears 4 times in the table: $$ 24 = 3 \times 8 = 8 \times 3 = 6 \times 4 = 4 \times 6. $$ For the 9 by 9 multiplication table shown, only the numbers 1 through 9 appear with all of their factorizations. Working through the table to see where different numbers appear, the students will have a good opportunity to observe the symmetry of the table which comes from the commutative property of multiplication: $a \times b = b \times a$.

The last part of this question gives an opportunity for the teacher to discuss prime numbers since the list of numbers the students produce will all be primes. This makes sense in the context because a prime number cannot be written as a product of two smaller whole numbers. The first non-prime number not appearing in the table is $22 = 2 \times 11$, the product of the smallest prime number with the smallest prime number bigger than 9.

The main standard for mathematical practice which aligns with this task is MP7, Look For and Make Use of Structure, since the goal of the task is to help students understand the multiplication table from the point of view of factorizations. If students work on this task in groups and share their insights, then it is also a good opportunity to engage in MP3, Construct Viable Arguments and Critique the Reasoning of Others.

Solution

1. The numbers which only appear once in the table are 1, 25, 49, 64, and 81. These are all along the diagonal of the table. These squares are each shaded in different colors in the picture below:

This makes sense because a number such as 48 = 6 $\times$ 8, not on the diagonal, also appears as 48 = 8 $\times$ 6. This is the commutative property of multiplication and several of these squares are shaded below:

For a number on the diagonal like $49 = 7 \times 7$, the two factors are the same and so when we switch their order, we don't get a different entry in the table. Some products on the diagonal, such as 2 $\times$ 2 = 4, do also appear elsewhere because they have a different factorization (4 $\times$ 1 and 1 $\times$ 4 in this case). If we extended the table to include products of all numbers, then every number $n$ (including all perfect squares) would appear at least twice as $n\times1$ and $1\times n$.

2. Several numbers appear 4 times in the table: 6, 8, 12, 18, and 24. These are shown below:

There are some numbers, such as 9, 16, and 36, which appear 3 times. The numbers which appear 4 times have two distinct factorizations into single digit numbers (up to the order of the factors): for example, $24 = 4 \times 6 = 3 \times 8$. If the table were continued further, other larger numbers such as 72 and 48 would appear many more than 4 times. The number 72 will eventually appear 12 times and the number 48 will appear 10 times.

3. We see that 10, 12, 14, 15, 16, 18, and 20 all appear in the table so the missing numbers between 10 and 20 are $$ 11, 13, 17, 19. $$ The multiplication table contains all products of single digit numbers. So a number not in the table can not be factored as a product of two single digit numbers. In the case of 11, 13, 17, and 19, this means that they do not factor at all, except as a product of 1 and themselves. So this list is the list of prime numbers between 10 and 20.

# data[].attributes.attachment_link.description
Alignments to Content Standards: 4.OA.B The table below shows all the products of pairs of numbers between 1 and 9. Which numbers appear only once in the table of products? Where are these numbers located in the table? Why? Which numbers appear the most frequently in the table? Why?

# data[].attributes.attachment_link.readable_text
Alignments to Content Standards: 6.RP.A

Task

1. If you have a cup of tea, will it be sweeter if you add one teaspoon of sugar or two?

2. If you have two identical cups of tea and add one teaspoon of sugar to each, they will taste the same.

1. If you pour both cups into a larger container and mixed them thoroughly, will the result taste the same or different?

2. How many cups of tea are in the mixture? How many teaspoons of sugar are in the mixture?

3. If you mix two cups of tea with two teaspoons of sugar, will it taste the same or different as one cup of tea with one teaspoon of sugar?

3. A small pitcher contains 4 cups of tea and 2 teaspoons of sugar. A large pitcher contains 6 cups of tea and 4 teaspoons of sugar. Will the tea in the large pitcher be more, less, or have the same sweetness as the tea in the small pitcher? Explain and draw a picture to illustrate.

IM Commentary

The purpose of this task is to help students understand why equivalent ratios of food ingredients result in mixtures that taste the same even if the individual quantities are different, and to use this understanding to determine whether two different mixtures will taste the same or not. The key move in reasoning is to compare ratios where one quantity is the same to see whether the other quantity is the same or different.

Parts (a) and (b) might seem trivial to an adult, but this task was written because the conclusions might not be obvious to students with no experience with mixture tasks. The hope is that asking these explicit questions will make it more likely that students will find success working with equivalent mixtures in future work.

Students should have seen and used diagrams to represent ratios before this task. Ideally, they will have drawn either discrete or tape diagrams, and can use one of these to represent their thinking for part (c). Part (c) may result in many different, interesting student responses that a teacher could use to assess where students are in their understanding.

Solution

1. A cup of tea would be sweeter with two teaspoons of sugar added than it would be with only one teaspoon of sugar added.

2. No response necessary.

1. Because they tasted the same to start, mixing them together will result in tea that will taste the same as each cup tasted separately.

2. There are two cups of tea and two teaspoons of sugar in the mixture.

3. Two cups of tea with two teaspoons of sugar will taste the same as one cup of tea with one teaspoon of sugar.

3. The tea in the large pitcher will be sweeter than the tea in the small pitcher.

There are many ways to draw diagrams of this situation, and many ways to reason about any given diagram. The key is to compare some like quantity between the two pitchers. In this example, we compare the amount of sugar in 2 cups of tea for each pitcher.

In the small pitcher (the top diagram), you can see in the diagram that for every 2 cups of tea, there is 1 teaspoon of sugar. But in the large pitcher (the bottom diagram), you can see that for every 2 cups of tea, there is more than 1 teaspoon of sugar.

# data[].attributes.attachment_link.description
Alignments to Content Standards: 6.RP.A If you have a cup of tea, will it be sweeter if you add one teaspoon of sugar or two? If you have two identical cups of tea and add one teaspoon of sugar to each, they will taste the same.

# data[].attributes.attachment_link.readable_text
Alignments to Content Standards: 4.OA.A

Task

There are two snakes at the zoo, Jewel and Clyde. Jewel was six feet and Clyde was eight feet. A year later Jewel was eight feet and Clyde was 10 feet. When asked which one grew more, students gave varying answers.

Mia said, "Since the two snakes both grew two feet ($8-6=2$ and $10-8=2$) then I would say that they grew the same amount."

Raul said, "They both grew 2 feet, but Jewel was only 6 feet to start with while Clyde was 8 feet to start with. That means Jewel grew more compared to her original length (2 is a larger part of six than it is of eight)."

Compare the two arguments. Describe the difference in the way the two students are thinking about the problem. Suppose a one-foot snake grew two feet and a 20-foot snake grew two feet. Could the two students still make the same type of argument?

IM Commentary

The purpose of this task is to assess students' understanding of multiplicative and additive reasoning. We would hope that students would be able to see identify that Student A is just looking at how many feet are being added on, while the Student B is comparing how much the snakes grew in comparison to how long they were to begin with. The follow up question is meant to highlight the difference in the two ways of thinking about growth. Some students will be more likely to agree with Student B when the difference in relative growth is accentuated as it is in the follow-up problem.

In later grades, students will learn that "which grows more" means "which has the greater absolute increase?" and "which has the greater growth rate?" means "which has the greater increase relative to the starting amount?" but students won't see this type of language for two or three years. Teachers need to be aware of this and work to ask questions as unambiguously as possible; for example, when asking for multiplicative comparisons, use language such as, "How many times greater is $x$ than $y$." They should also be prepared to address this potential for confusion along the way.

Solution

Viewing this additively, both snakes grew 2 feet and therefore grew the same amount. Viewing it multiplicatively, Jewel grew $\frac{2}{6}$ its length, while Clyde grew $\frac{2}{8}$ its length. From this perspective, Jewel grew more. Given the purposeful phrasing of the problem, both answers are correct, but the goal is to see if the student understand the two perspectives, and thus the difference between additive and multiplicative reasoning.

# data[].attributes.attachment_link.description
Alignments to Content Standards: 4.OA.A There are two snakes at the zoo, Jewel and Clyde. Jewel was six feet and Clyde was eight feet. A year later Jewel was eight feet and Clyde was 10 feet. When asked which one grew more, students gave varying answers.

# data[].attributes.attachment_link.readable_text
Alignments to Content Standards: 7.G.B.4

Task

The figure below is composed of eight circles, seven small circles and one large circle containing them all. Neighboring circles only share one point, and two regions between the smaller circles have been shaded. Each small circle has a radius of $5 \text{ cm}$.

Calculate:

1. The area of the large circle.

2. The area of the shaded part of the figure.

IM Commentary

The purpose of this task is to strengthen students' understanding of area. It could be assigned in class to individuals or small groups or given as a homework exercise to generate interesting discussions the following day. The relatively high levels of complexity and technical demand enhance its instructional value. Seventh grade student may require some teacher guidance to be successful with this task.

Solution

The radius of each of the seven small circles is $5$ cm. This makes the radius of the large circle $3\cdot5 = 15$ cm.

$$ \text{Area of a Circle} = \pi r^2 $$

The area of the large circle is $\pi(15\text{ cm})^2=225 \pi \text{ cm}^2$.

The area of each small circle is $\pi(5\text{ cm})^2=25 \pi \text{ cm}^2$.

There are seven small circles in all, so the area of all the small circles together is $$7\times25\pi \text{ cm}^2=175\pi \text{ cm}^2.$$

If we take the area of the large circle and subtract the area of the seven small circles, we will be left with all of the area contained in the large circle that is not contained in a small circle, that is, the area around the small circles. This area is $225 \pi\text{ cm}^2 - 175 \pi\text{ cm}^2 = 50 \pi\text{ cm}^2$ . Notice that the exact shape of the shaded region is repeated six times in the large circle. This makes the shaded region $\frac16$ of the area that is contained in the large circle that is not contained in a small circle. Thus the shaded region has an area of $$\frac16 \times 50 \pi \text{ cm}^2 = \frac{50\pi}{6}\text{ cm}^2 = \frac{25 \pi}{3}\text{ cm}^2.$$

# data[].attributes.attachment_link.description
Alignments to Content Standards: 7.G.B.4 The figure below is composed of eight circles, seven small circles and one large circle containing them all. Neighboring circles only share one point, and two regions between the smaller circles have been shaded. Each small circle has a radius of $5 \text{ cm}$.

# data[].attributes.attachment_link.readable_text
Alignments to Content Standards: 6.RP.A

Task

John and Will ran for 6th grade class president. There were 36 students voting. John got two votes for every vote Will got. How many votes did each get?

IM Commentary

This is the first and most basic problem in a series of seven problems, all set in the context of a classroom election. Every problem requires students to understand what ratios are and apply them in a context. The problems build in complexity and can be used to highlight the multiple ways that one can reason about a context involving ratios.

This first problem can be used to solidify students' understanding of ratio tables or can be used to highlight how one can use unit rates to reason in a ratio context as described in the solution below.

There are several ways students can solve this problem, and encouraging multiple solutions can lead to good connections between different mathematical ideas, as noted in the comment by sbingham67.

Solutions

Solution: Using a ratio table

One approach is to use a ratio table that has a column for votes for Will, a column for votes for John , and a column for the total number of voters.

Solution: Computing votes

There are multiple ways to approach this problem; here is one that uses the fact that the 2 to 1 ratio means there are three equal parts in this case.

One way to consider the votes in three equal parts is to determine the number of votes in each part. Since it is given that there are 36 votes, each part must then be $\frac13$ × 36 = 12 votes. It is also give that John has 2 parts and Will has 1 part. Therefore:

2 × 12 = 24 votes for John

1 × 12 = 12 votes for Will

Solution: Applying fractions

An alternate approach is to recognize that John gets 2/3 of the votes, Will gets 1/3 of the votes, and to work with the fractions rather than the number of votes.

$\frac23$ × 36 = 24 votes for John

$\frac13$ × 36 = 12 votes for Will

Solution: With connections to 6.EE.7

Students might also approach this algebraically, as noted in the comment by sbingham67:

The 2:1 ratio means for every 2 votes for John, Will gets 1. But we don't know how many groups of "2 votes to 1 vote" we have, so the ratio could also be represented as $$2x : 1x$$

Where $x$ is how many groups of each quantity in the ratio we have. If you put John's and Will's votes together, there are $3x$ votes all together. Since 36 students voted, $$3x=36$$ which is equivalent to the equation $$x=\frac{36}{3}$$ so $x=12$. Multiply $x$ by 2 to get John's votes: $2\times12=24$.

# data[].attributes.attachment_link.description
Alignments to Content Standards: 6.RP.A John and Will ran for 6th grade class president. There were 36 students voting. John got two votes for every vote Will got. How many votes did each get? This is the first and most basic problem in a series of seven problems, all set in the context of a classroom election.

# data[].attributes.attachment_link.readable_text
Task

Maned wolves are a threatened species that live in South America. People estimate that there are about 24,000 of them living in the wild.

The dhole is an endangered species that lives in Asia. People estimate there are ten times as many maned wolves as dholes living in the wild.

About how many dholes are there living in the wild?

IM Commentary

This task addresses 4.NBT.1 in that in requires students to understand multiplying and dividing whole numbers by powers of 10. It also addresses 4.OA.1 since it helps students push their understanding of multiplication from a 3rd grade understanding of repeated addition to a 4th grade understanding that we can also use multiplication to understand "times as many" comparison problems.

A tape-diagram may be helpful in this problem. Though this task seems straightforward, it actually requires students to work backwards from the known value that is 'ten times as many.' Students may impulsively want to multiply the number of maned wolves by a factor of 10. The tape diagram will help students organize the different numbers and animals in this problem. A challenging follow-up task to this one is 4.NBT Millions and Billions of People.

Population facts from http://wwf.panda.org/

Images from http://imgur.com/a/Jexvo and http://www.outdoorconservation.eu/

Solutions

There are about 24,000=20,000+4,000 maned wolves. 20,000 is ten times as many as 2,000, and 4,000 is ten times as many as 400. So there are about 2,000+400 or 2,400 dholes estimated to live in the wild.

We can check this:

$$10\times(2000+400)=10\times2,000+10\times400=20,000+4,000=24,000$$

We can also use a tape diagram to show that the number of maned wolves (24,000) is ten times the number of dholes. The number of dholes must be 2,400 (which a student could then check with multiplication or repeated addition.)

For a student to use this strategy effectively, s/he would have to have some understanding of how to divide 24,000 by 10.

For students struggling to figure out the division, it may help to model the following thought process: "Could one unit be worth 1,000 animals...2,000 animals...3,000 animals?" Understanding that each unit must represent somewhere between 2,000 and 3,000 animals would be a likely first step.

From there, the thought process might be: "If we know each unit represents 2,000 animals, then we still need to account for 4,000 more animals. Could one unit contain an additional 100 animals...200 animals...300 animals...400 animals...500 animals?"

Through this reasoning, students could come to the conclusion that each unit must represent 2,400 animals.

We can also use a place value organizer to think about dividing 24,000 by 10.

We know that no matter the number in our base ten system, if we divide by 10 then every digit will shift to the right in place value. This place value organizer shows how we can represent dividing 24,000 by 10 using place value.

Using this organizer, we can see that 24,000 divided by 10 is 2,400. Therefore, there are about 2,400 dholes.

# data[].attributes.attachment_link.description
Maned wolves are a threatened species that live in South America. People estimate that there are about 24,000 of them living in the wild. The dhole is an endangered species that lives in Asia. People estimate there are ten times as many maned wolves as dholes living in the wild.

# data[].attributes.attachment_link.readable_text
Alignments to Content Standards: 6.RP.A

Task

John and Will ran for 6th grade class president. There were 36 students voting. John got two votes for every vote Will got. How many more votes did John get than Will?

IM Commentary

This is the second in a series of tasks that are set in the context of a classroom election. It requires students to understand what ratios are and apply them in a context. The simple version of this question just asked how many votes each gets. This has the extra step of asking for the difference between the votes.

The problem is useful as a means of highlighting the multiple ways that one can reason about a situation involving ratios. This problem can be used to solidify students' understanding of ratio tables or can be used to highlight how one can use unit rates to reason in a ratio context, which is explained in the solutions. Each task has some commentary or solutions that clarify some of the opportunities made available by the particular task.

Solutions

Solution: Computing votes

A straightforward approach is to simply compute the number of votes each person receives and then subtract the two to find the difference. The votes are in three equal parts, two for John and one for Will. Since it is given that there are 36 votes, each part must then be $\frac13 × 36 = 12$ votes. Therefore:

$2 × 12 = 24$ votes for John

$1 × 12 = 12$ votes for Will

$24 - 12 = 12$ votes difference

Solution: Applying fractions

A faster and perhaps more sophisticated approach would be to realize that with John getting $2$ of the equal shares of votes and Will getting only $1$, then John gets one whole share more than Will, or $1/3$ of the votes.

$1/3 × 36 = 12$ more votes for John.

This approach, which can probably be done without paper and pencil, avoids having to compute the votes for both candidates and subtracting.

Solution: Ratio tables

As with variation 1, a more basic approach could be taken using a ratio table. Ultimately students should be able to reason as in solutions 1 and 2, but for those students not ready for that level of abstraction, ratio tables provide an important bridge.

# data[].attributes.attachment_link.description
Alignments to Content Standards: 6.RP.A John and Will ran for 6th grade class president. There were 36 students voting. John got two votes for every vote Will got. How many more votes did John get than Will? This is the second in a series of tasks that are set in the context of a classroom election.

# data[].attributes.attachment_link.readable_text
Alignments to Content Standards: 4.OA.A.2

Task

1. Helen raised \$12 for the food bank last year and she raised 6 times as much money this year. How much money did she raise this year?

2. Sandra raised \$15 for the PTA and Nita raised \$45. How many times as much money did Nita raise as compared to Sandra?

3. Luis raised \$45 for the animal shelter, which was 3 times as much money as Anthony raised. How much money did Anthony raise?

IM Commentary

The purpose of this task is for students to solve three comparisons problems that are related by their context but are structurally different. Multiplicative comparison is purposefully excluded from third grade (see 3.OA.3 and 3.MD.2), making this task appropriate for fourth but not third grade. In these multiplicative comparison problems, one factor and the product are amounts of money and the other factor represents the number of times bigger one amount is than the other. Sometimes this second factor is called a "scale factor." In part (a), the larger amount (which is the product) is unknown, while in part (b) the scale factor is unknown and in part (c) the smaller amount of money is unknown. Students will study multiplicative comparison problems involving scale factors that are fractions in 5th grade; see 5.NF.B.5. Note that in fourth grade, scale factors must always be bigger than 1, so students often think that "multiplying makes bigger"; however in 5th grade they will see that when the scale factor is less than 1, the product will actually be smaller than the initial quantity. Note that the numbers in parts (b) and (c) are related by the fact family $3\times15=45$. This allows for a classroom discussion about the different interpretations of the factors in a multiplicative comparison context. To see an annotated version of this and other Illustrative Mathematics tasks as well as other Common Core aligned resources, visit Achieve the Core.

The Standards for Mathematical Practice focus on the nature of the learning experiences by attending to the thinking processes and habits of mind that students need to develop in order to attain a deep and flexible understanding of mathematics. Certain tasks lend themselves to the demonstration of specific practices by students. The practices that are observable during exploration of a task depend on how instruction unfolds in the classroom. While it is possible that tasks may be connected to several practices, only one practice connection will be discussed in depth. Possible secondary practice connections may be discussed but not in the same degree of detail.

This particular task helps illustrate Mathematical Practice Standard 1, Make sense of problems and persevere in solving them. Problem solving is based upon students engaging in a task in which a solution pathway is not known in advance. As fourth graders approach these three problems, they will analyze the problems to make sense of what each is asking, working to understand the structures and the unknowns. Through this analysis, students will understand that the numbers in parts (b) and (c) are both factors related by the same fact family but each serves a different role in their respective problems. Students need experience with unknowns that play different roles in multiplicative comparison problems in order to support a deeper understanding of this type of problem situation.

Solutions

Solution: Tape diagram

1. She raised six times as much money (as shown in the diagram) so she raised $6 \times 12 = 72$.

Helen raised \$72 this year.

2. $? \times 15 = 45$ is equivalent to $45 \div 15 = ?$

Nita raised 3 times as much as Sandra.

3. $3 \times ? = 45$ is equivalent to $45 \div 3 = ?$

Anthony raised \$15.

Solution: Writing multiplication equations for division problems

1. Helen raised $6 \times \$12$ this year, so she raised \$72 this year.

2. This is a "Number of Groups Unknown" problem. We can represent the question as $$? \times 15 = 45$$ or $$45 \div 15 = ?$$ So Nita raised 3 times as much money as Sandra.

3. This is a "Group Size Unknown" problem. We can represent the question as $$3 \times ? = 45$$ or $$45 \div 3 = ?$$ So Anthony raised \$15.

# data[].attributes.attachment_link.description
Alignments to Content Standards: 4.OA.A.2 Helen raised \$12 for the food bank last year and she raised 6 times as much money this year. How much money did she raise this year? Sandra raised \$15 for the PTA and Nita raised \$45. How many times as much money did Nita raise as compared to Sandra?

# data[].attributes.attachment_link.readable_text
Alignments to Content Standards: 6.RP.A

Task

John and Will ran for 6th grade class president. There were 36 students voting. John got two votes for every vote Will got. It was necessary to get more than half the votes to get elected. How many more votes than half did John get?

IM Commentary

This problem, the third in a series of tasks set in the context of a class election, is more than just a problem of computing the number of votes each person receives. In fact, that isn't enough information to solve the problem. One must know how many votes it takes to make one half of the total number of votes. Although the numbers are easy to work with, there are enough steps and enough things to keep track of to lift the problem above routine.

It is worth noting that these numbers have units: votes. Helping students build the habit of carefully tracking the units in the context of simple problems will help them prepare for future situations that involve more complex and multiple types of units.

One final note: as with variations 1 and 2, this task can be solved with a ratio table and a modest amount of additional reasoning. However, the next task in the series requires students to go beyond ratio tables. This would be a good task type to help students see the connection between ratio tables and more abstract approaches to solving the problem.

Solutions

Solution: Computing votes

The step-by-step way to solve this problem is to compute the number of votes John receives and compare it to the one half of the total votes. To do this, the student must determine 1) the number of votes for John, 2) the number equal to $1/2$ of the votes, and 3) the difference between these two numbers.

Since the votes are in three equal parts, two for John and one for Will:

$36 ÷ 3 = 12$ votes in each equal part of the ratio

$2 × 12 = 24$ total votes for John

$36 ÷ 2 = 18$ votes in each half of the total votes

$24 - 18 = 6$ more votes than half for John.

Solution: Applying fractions

There is another way to solve the problem that may be more difficult for some students to see because it is more abstract.

Given that John received $2/3$ of the votes and wanting to know how much more that is than $1/2$ of the votes, one can write and solve the following equation:

$2/3 - 1/2 = 1/6$ of the votes.

$1/6 × 36 = 6$ votes, which answers the question.

# data[].attributes.attachment_link.description
Alignments to Content Standards: 6.RP.A John and Will ran for 6th grade class president. There were 36 students voting. John got two votes for every vote Will got. It was necessary to get more than half the votes to get elected. How many more votes than half did John get?

# data[].attributes.attachment_link.readable_text
IM Commentary

This task can be made more hands-on by asking the students to determine about how many pennies are needed to make a stack one inch high. In addition, for part (b) students could be invited to research this question. Useful information which would need to be compiled to get total mintage figures can be found at Annual mintage of Lincoln pennies. From 1909 through 2009 the total number of pennies minted is 455,627,740,918 according to Total mintage of Lincoln pennies.

The goal of this task is to give students a context to investigate large numbers and measurements. Students need to fluently convert units with very large numbers in order to successfully complete this task. The total number of pennies minted either in a single year or for the last century is phenomenally large and difficult to grasp. One way to assess how large this number is would be to consider how far all of these pennies would reach if we were able to stack them one on top of another: this is another phenomenally large number but just how large may well come as a surprise. In particular, before they start working on the problem, the teacher may wish to ask the students whether or not they think it would be possible to reach the moon with this giant stack of pennies.

Four solutions are offered to the problem stressing different mathematical ideas:

1. A division approach taking the distance from the earth to the moon and dividing by the height of a single penny.

2. An algebraic approach where $x$ denotes the number of pennies necessary to reach the moon.

3. A ratio approach which could be solved algebraically, linking to solution (a), or through a ratio table.

4. Arithmetic with scientific notation.

The first of these solutions uses multiplication as well as division because there is conversion of units involved. Note that the second of these solutions leads naturally to the introduction of a function (representing the height of a stack of $x$ pennies) and so, although it also strongly stresses ratio language, it would need to be adapted in order to be appropriate at the sixth grade level. As written, the answer to part (c) meets the 6-RP.3 standard. Students should eventually be comfortable with all three approaches. Note too that throughout the teacher needs to provide some guidance in terms of the level of accuracy with which results should be recorded: because the number of pennies minted is an estimate and the thickness of the pennies is an estimate, no more than one or two significant digits should be recorded in the answers; this is important because it makes each successive calculation easier if numbers are rounded.

The number of pennies made in the last 100 years is about 100 times the number of pennies made in 2011. The vast majority of these pennies, however, have been made in the last 40 years. In 1922, for example, only a little over 7 million pennies were made while in 1982 the total exceeded a staggering 17 billion.

The Standards for Mathematical Practice focus on the nature of the learning experiences by attending to the thinking processes and habits of mind that students need to develop in order to attain a deep and flexible understanding of mathematics. Certain tasks lend themselves to the demonstration of specific practices by students. The practices that are observable during exploration of a task depend on how instruction unfolds in the classroom. While it is possible that tasks may be connected to several practices, the commentary will spotlight one practice connection in depth. Possible secondary practice connections may be discussed but not in the same degree of detail.

This task helps illustrate Mathematical Practice Standard 4, Model with mathematics. There are 3 parts to this task and each part provides scaffolding opportunities for students as they apply the mathematics they know to find a solution pathway. Because this task has multiple entry points, students may apply several different approaches depending upon their own understanding, such as, arithmetical, algebraic, ratio reasoning, or scientific notation. Students may make assumptions and approximations while solving this task as the number of pennies minted and the thickness of each penny are estimates. As a result, the level of accuracy of results must be resolved which connects to Mathematical Practice 6, Attend to precision.

Solutions

Solution: 1 Arithmetic (6-NS.2 and 6-NS.3)

1. Five billion pennies, each $\frac{1}{16}$ of an inch thick, will make a stack $$ 5,000,000,000 \times \frac{1}{16} \,\,\text{inches} $$ high. Dividing $5,000,000,000$ by $16$ gives a little more than $300,000,000$ inches. We need to convert this to miles so we can begin by finding how many feet there are in $300,000,000$ inches. This will be $$ 300,000,000 \div 12 =25,000,000. $$ Finally, to find out how many miles there are in $25,000,000$ feet we need to divide by $5280$ because there are $5280$ feet in each mile: $$ 25,000,000 \div 5280 \approx 5000. $$ So the stack of pennies would be about $5000$ miles high.

2. A stack of $500$ billion pennies is the same as $100$ stacks of $5$ billion pennies so using the answer from part (a), the stack of $500$ billion pennies would be about $100 \times 5000 = 500,000$ miles high.

3. Here we need to perform the division $239,000 \,\,miles \div \frac{1}{16} inches$. Because the units, miles and inches, are different we need to find out how many inches are in each mile (or equivalently, convert the $239,000$ miles to inches). We have

\begin{align} 239,000\,\,\, \text{miles} & = 239,000 \,\, \text{miles} \times \frac{5280 \,\,\text{feet}}{\text{mile}} \times \frac{12 \,\, \text{inches}}{\text{foot}} \\ &={239,000 \times 5280 \times 12} \,\, \text{inches} \end{align}

Multiplying this out gives $15,143,040,000$ or about $15$ billion inches. Now we can calculate $$ 15,000,000,000 \div \frac{1}{16} = 240,000,000,000. $$ So it takes about $240$ billion pennies to make a stack high enough to reach the moon.

Solution: 2 Algebra (6-EE.6)

1. Since it takes $16$ pennies to make a stack one inch high, with $5,000,000,000$ pennies we could make a stack $$ \frac{5,000,000,000}{16} \text{inches} $$ high. The question asks how many miles this is so we need to convert inches to miles. For this, we use the facts that there are $12$ inches in a foot and $5280$ feet in a mile:

\begin{align} \frac{5,000,000,000}{16} \text{inches} & = \frac{5,000,000,000}{16 \times 12} \text{inches} \times \frac{\text{foot}}{\text{inches}} \\ &= \frac{5,000,000,000}{16 \times 12 \times 5280} \text{inches} \times \frac{\text{foot}}{\text{inches}} \times \frac{\text{mile}}{\text{feet}} \\ &\approx 5000 \,\, \text{miles}. \end{align}

Note that the exact value of the quotient above is a little more than $4932$ miles but because $5,000,000,000$ is an estimate for the number the number of pennies minted in $2011$ it is apprpriate to list the height of the stack as about $5000$ miles.

2. A stack of $500,000,000,000$ pennies would reach $100$ times as far as a stack of $5,000,000,000$ pennies. So this would be about $100 \times 5000$ miles or about $500,000$ miles.

3. If we write $x$ for the number of pennies in a stack which reaches the moon then we have the equation $$ x \,\times \frac{1}{16} \text{inch} = 239,000 \,\, \text{miles}. $$ We can solve this equation as follows: $$ x \,\,\, \text{inches} = 16 \times 239,000 \,\, \text{miles} \times \frac{5280 \,\,\,\text{feet}}{\text{mile}} \times \frac{12 \,\,\,\text{inches}}{\text{foot}} $$ Simplifying we find

\begin{align} x \,\,\, \text{inches} &= 16 \times 239,000 \times 5280 \times 12 \,\, \text{inches} \end{align}

So $x$ is $16 \times 239,000 \times 5280 \times 12 \approx 240,000,000,000$. So it takes about $240$ billion pennies (or almost a quarter of a trillion!) to reach the moon.

Solution: 3 Ratio (6-RP.3, 8-F.1)

Since the pennies being stacked all have the same thickness, the ratio of the number of pennies in a stack to the height of the stack does not depend on how many pennies are in the stack. In other words if $a$ denotes the number of pennies in a stack and $h$ is the function so that $h(a)$ denotes the height of that stack then the ratios $(a:h(a))$ are equivalent for any positive number $a$ of pennies.

1. Here we have $a = 5,000,000,000$. According to the previous paragraph we need to find $h(5,000,000)$ we need to solve the equation $$ \left(1:\frac{1}{16} \text{inch}\right) = \left( 5,000,000,000: x \,\, \text{miles} \right). $$ We have $$ \left(1:\frac{1}{16} \text{inch}\right) = \left( 5,000,000,000: 5,000,000,000 \times \frac{1}{16}\,\, \text{inches} \right). $$ So to solve for $x$ we need to convert $\frac{5,000,000,000}{16} \text{inches}$ to miles which can be done as in the previous solution:

\begin{align} \frac{5,000,000,000}{16} \text{inches} & = \frac{5,000,000,000}{16 \times 12} \text{inches} \times \frac{\text{foot}}{\text{inches}} \\ &= \frac{5,000,000,000}{16 \times 12 \times 5280} \text{inches} \times \frac{\text{foot}}{\text{inches}} \times \frac{\text{mile}}{\text{feet}} \\ &\approx 5000 \,\, \text{miles}. \end{align}

So $x$, the height of the stack of $5,000,000,000$ pennies, is about $5000$ miles.

2. If we multiply the number of pennies in a stack by $10$, the height of the stack is also multiplied by $10$ (so that the ratio is the same). Since $h(500,000,000,000) = 10h(50,000,000,000)$ and $h(50,000,000,000)$ is about $5000$ miles, we have that $h(500,000,000,000) \approx 500,000$ miles.

3. We need to find $x$ to solve the following equation of ratios: $$ \left(1:\frac{1}{16} \text{inch}\right) = \left( x: 239,000 \,\, \text{miles} \right). $$ We can first convert miles to inches as in part (a): $$ 239,000 \,\,\text{miles} = 12 \times 5280 \times 239,000 \,\,\frac{\text{inches}}{\text{foot}} \times \frac{\text{feet}}{\text{mile}} \times \text{miles}. $$ Putting this into the previous equation gives $$ \left(1:\frac{1}{16} \text{inch}\right) = \left( x: 12 \times 5280 \times 239,000 \,\, \text{inches} \right). $$ Solving this gives $$ x = 16 \times 12 \times 5280 \times 239,000 \approx 240,000,000,000. $$

Solution: 4 Scientific notation (8-EE.3 and 8-EE.4)

Scientific notation is appropriate for this problem as the number of pennies involved is very large and nicely represented using exponential notation. Either of the above solutions can be adapted for the use of exponential notation and we have chosen here to use the first method.

1. One billion is one with $9$ zeroes or $1 \times 10^9$. So $5$ billion pennies is $5 \times 10^9$ pennies. Since it takes $16$ pennies to make a stack one inch high, with $5 \times 10^9$ pennies we could make a stack $$ \frac{5 \times 10^9}{16} \text{inches} $$ high. The question asks how many miles this is so we need to convert inches to miles. For this, we use the facts that there are $12$ inches in a foot and $5280$ feet in a mile:

\begin{align} \frac{5 \times 10^9}{16} \text{inches} & = \frac{5 \times 10^9}{16 \times 12} \text{inches} \times \frac{\text{foot}}{\text{inches}} \\ &= \frac{5 \times 10^9}{16 \times 12 \times 5280} \text{inches} \times \frac{\text{foot}}{\text{inches}} \times \frac{\text{mile}}{\text{feet}} \\ &\approx \frac{5 \times 10^9}{1 \times 10^6}\,\, \text{miles}\\ &= 5 \times 10^3 \,\,\text{miles}. \end{align}

In the last step, we use the law of exponents $$ \frac{10^9}{10^6} = 10^{9-6}. $$

2. A stack of $500,000,000,000$ pennies is $5 \times 10^{11}$ pennies or $100$ times as many as the stack of $5 \times 10^9$ considered in part (a). So this stack would reach $100$ times as far as a stack of $5 \times 10^9$ pennies or about $$ 100 \times 5 \times 10^3 = 5 \times 10^5 \,\,\text{miles}. $$

3. If we write $x$ for the number of pennies in a stack which reaches the moon then we have the equation $$ x \,\times \frac{1}{16} \text{inch} = 239,000 \,\, \text{miles}. $$ We can solve this equation as follows: $$ x \,\,\, \text{inches} = 16 \times 239,000 \,\, \text{miles} \times \frac{5280 \,\,\,\text{feet}}{\text{mile}} \times \frac{12 \,\,\,\text{inches}}{\text{foot}} $$ Simplifying we find

\begin{align} x \,\,\, \text{inches} &= 16 \times 239,000 \times 5280 \times 12 \,\, \text{inches} \end{align}

So $x$ is $16 \times 239,000 \times 5280 \times 12 \approx 2.4 \times 10^{11}$. So it takes about $2.4 \times 10^{11}$ pennies (or almost a quarter of a trillion!) to reach the moon.

# data[].attributes.attachment_link.description
This task can be made more hands-on by asking the students to determine about how many pennies are needed to make a stack one inch high. In addition, for part (b) students could be invited to research this question. Useful information which would need to be compiled to get total mintage figures can be found at Annual mintage of Lincoln pennies.