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Alignments to Content Standards: 7.RP.A

Task

Jessica gets her favorite shade of purple paint by mixing 1/3 cup of blue paint with 1/2 cup of red paint. How many cups of blue and red paint does Jessica need to make 20 cups of her favorite purple paint?

IM Commentary

The goal of this task is to provide a context for students to develop their ratio and proportional reasoning skills. Many of the techniques developed in the sixth grade can be readily applied here: for example, solution 2 below uses a ratio table. Other techniques, such as double number lines or tape diagrams, can also be applied (see solution 6) although the fractions make these more challenging. If the fractions in the problem (1/3 and 1/2) were more complex (1/3 and 1/5 for example) then there is genuine motivation to examine more abstract techniques. The numbers chosen make this a good bridge problem, allowing students to practice the techniques learned in the sixth grade while also working with the more abstract seventh grade ideas.

The sixth grade version of this problem, https://www.illustrativemathematics.org/tasks/2049, replaces the fractions 1/3 and 1/2 in the prompt with whole numbers 2 and 3. If students are successful applying double number lines, tape diagrams, and ratio tables with fractions then they have demonstrated mastery of the sixth grade techniques and are ready to move into the more abstract ratio and proportion methods exemplified by solutions 3 and 4.

If this task is being used to motivate proportional reasoning techniques, the teacher may wish to give students more complex fractions than 1/2 and 1/3. As the numbers become more complex, physical representations such as tape diagrams and double number lines become more difficult to manipulate. Methods such as scaling (solution 1), setting up a proportion (solution 3), or setting up an equation (solution 4) work for any fractions, provided students are fluent doing fraction arithmetic.

In addition to being used to bridge sixth and seventh grade ratio and proportional thinking, this task can also be used after students have experience with the seventh grade language and techniques. In this case, a large variety of responses, as shown, should be encouraged and expected and the teacher may wish to have students share their different approaches. The vast array of techniques available to solve this problem are indicative of the central role of ratio and proportion in the middle school curriculum. Building upon arithmetic with fractions, it prepares students for using expressions and equations, graphing lines, and understanding the meaning of functions.

This task was developed with the assistance of a group of teachers from Washington and Illinois in connection with an SBAC digital library project. In the lesson, the statement of the problem was ''If Perfect Purple Paint is made by mixing 1/3 cups blue paint to 1/2 cup red paint, how much of each is needed for 20 cups?''

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

Solutions

Solution: 1 Arithmetic and Scaling

One batch of purple paint contains 1/3 cup of blue paint and 1/2 cup of red paint. This will make a total of 1/2 + 1/3 = 5/6 of a cup of purple paint. In order to make 20 cups of purple paint, we need 20 $\div$ 5/6 = 24 batches. Each batch has 1/3 of a cup of blue paint so 24 batches will contain 24 $\times$ 1/3 = 8 cups of blue paint. Each batch has 1/2 cup of red paint so 24 batches will contain 24 $\times$ 1/2 = 12 cups of red paint.

Solution: 2 Ratio table

We can use a ratio table to find how much blue and red paint will be in 20 cups of Jessica's perfect purple paint:

For the second row, we take 6 batches of the purple paint mixture, in order to get a whole number of cups of purple paint. From here, doubling this mixture twice shows that there are 8 cups of blue paint and 12 cups of red paint in 20 cups of purple paint.

Solution: 3 Proportions

Combining 1/3 cup of blue paint and 1/2 cup of red paint makes 1/3 + 1/2 = 5/6 cups of blue paint. To find how much blue paint is in 20 cups of purple paint we can use a proportion:$$\frac{1}{3}:\frac{5}{6} :: \,\,?:20.$$ Note that $\frac{1}{3} = \frac{2}{6}$ so $\frac{1}{3} \div \frac{5}{6}= \frac{2}{6} \div \frac{5}{6} = \frac{2}{5}$. So ? satisfies the equation $$\frac{?}{20} = \frac{2}{5}.$$ We can solve this equation to find ? = 8. There are 8 cups of blue paint in 20 cups of Jessica's purple paint.

Since the rest of the paint in the 20 cups of purple paint is red, this means that there are 12 cups of red paint in the purple paint mixture.

Solution: 4 Equations

One batch of purple paint is 1/3 + 1/2 = 5/6 cups. The amount of blue paint is 1/3 = 2/6 cups. This means that there is 5/2 times as much purple paint as blue paint. If $p$ is the amount of purple paint and $b$ the amount of blue paint, this means $$p = \frac{5}{2}b.$$ So if we have 20 cups of purple paint then to find out how much blue paint there is we can solve $$20 = \frac{5}{2}b$$ to find that there are 8 cups of blue paint.

Similarly, if $r$ denotes the red paint in the mixture then $$p = \frac{5}{3}r$$ and if $p = 20$ then we find $r = 12$ so there are 12 cups of red paint in 20 cups of Jessica's purple paint.

Solution: 5 Using Percent

There are 1/3 + 1/2 = 5/6 cups of purple paint in a single batch of Jessica's favorite purple. Since 1/3 = 2/6 this means that 2 out of 5 equal parts, or 40%, of the purple paint comes from blue paint. Since 40% of 20 cups is 8 cups, 8 cups of blue paint are needed to make 20 cups of purple paint.

Similarly, the remaining 60% of the purple paint comes from the added red paint. Since 60% of 20 cups is 12 cups, 12 cups of red paint are needed to make 20 cups of purple paint.

Solution: 6 Tape diagrams

We put together batches of the purple paint until we find whole number of cups of both red paint and blue paint:

Here we have taken 6 batches, each consisting of 1/3 cup of blue paint and 1/2 cup of red paint. Together, this makes 5 cups of purple paint. So if we combine four of these that makes 20 cups of perfect purple paint. Four groups of 2 cups of blue paint make 8 cups of blue paint and four groups of 3 cups of red paint make 12 cups. So 20 cups of perfect purple paint contains 8 cups of blue paint and 12 cups of red paint.

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Alignments to Content Standards: 7.RP.A Jessica gets her favorite shade of purple paint by mixing 1/3 cup of blue paint with 1/2 cup of red paint. How many cups of blue and red paint does Jessica need to make 20 cups of her favorite purple paint?

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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.

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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.

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Alignments to Content Standards: 5.NF.A.1

Task

Find two different ways to add these two numbers:

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

IM Commentary

The purpose of this task is to help students realize there are different ways to add mixed numbers and is most appropriate for use in an instructional setting. The two primary ways one can expect students to add are converting the mixed numbers to fractions greater than 1 or adding the whole numbers and fractional parts separately. It is good for students to develop a sense of which approach would be better in a particular context.

Note that the solutions show the sum as a single fraction and as a mixed number. While both are perfectly acceptable ways to write the sum, students should be encouraged to check to see that the two methods yield the same answer. Some students will convert the fraction to a mixed number, and some will write the mixed number as an equivalent fraction. Teachers should encourage both approaches.

Solution

1. One way to add these numbers is to first convert the mixed numbers into equivalent fractions:

$$ \begin{align} 1\frac{1}{3}&=\frac{1}{1}+\frac{1}{3}\\ &=\frac{1\times3}{1\times3}+\frac{1}{3}\\ &=\frac{3+1}{3}\\ &=\frac{4}{3} \end{align} $$

and

$$ \begin{align} 2\frac{3}{5} &= \frac{2}{1}+\frac{3}{5}\\ &= \frac{2\times5}{1\times5}+\frac{3}{5} \\ &= \frac{10+3}{5} \\ &= \frac{13}{5} \end{align} $$

The picture below illustrates these equivalences:

Next, we need to find a common denominator in order to add these fractions with unlike denominators. The denominators are 3 and 5 and $3\times5 = 15$ is one possible common denominator (any other common multiple of 3 and 5 could also be used). Then:

$$ \begin{align} 1\frac{1}{3}+2\frac{3}{5}&=\frac{4}{3}+\frac{13}{5}\\ &=\frac{4\times5}{3\times5}+\frac{13\times3}{5\times3}\\ &=\frac{20}{15}+\frac{39}{15}\\ &=\frac{59}{15} \end{align} $$

The picture below illustrates the sum:

Thus, $1\frac{1}{3}+2\frac{3}{5} = \frac{59}{15}$.

2. Another way is to add the two whole numbers and the the fractions separately. So:

$$ \begin{align} 1\frac{1}{3}+2\frac{3}{5} &= (1 + \frac13) +(2+ \frac35)\\ &= (1+2) + (\frac13 + \frac35)\\ &= 3 + \frac{5}{15} + \frac{9}{15}\\ &= 3 + \frac{14}{15} \\ &= 3 \frac{14}{15} \end{align} $$

The picture below illustrates the sum:

As written, the two methods look like the result in different sums. A quick check shows that $$3 \frac{14}{15} = \frac{(3\times 15 + 14)}{15} = \frac{59}{15}$$ and so the answers are equivalent.

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Alignments to Content Standards: 5.NF.A.1 Find two different ways to add these two numbers: $$1\frac{1}{3} + 2\frac{3}{5}$$ The purpose of this task is to help students realize there are different ways to add mixed numbers and is most appropriate for use in an instructional setting.

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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.

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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?

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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.

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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?

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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.

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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.

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would be great to do a simulation with

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Alignments to Content Standards: 8.EE.C.7

Task

For a science project, Sammy observed a chipmunk and a squirrel stashing acorns in holes. The chipmunk hid 3 acorns in each of the holes it dug. The squirrel hid 4 acorns in each of the holes it dug. They each hid the same number of acorns, although the squirrel needed 4 fewer holes. How many acorns did the chipmunk hide?

IM Commentary

This task provides a context for setting up a linear equation whose solution requires some algebraic manipulation. Because the numbers involved are not too large, students can also experiment with some small values and eventually find the solution this way; a first solution with a table is provided showing this method. On the other hand, the reasoning required without using an equation is complex enough that the simplicity and elegance of the algebraic approach can be highlighted.

For the algebraic solution, since a variable is not provided, students might choose different variables providing an opportunity for an interesting class discussion. For example, they could let $x$ be the number of holes the squirrel has made and this would lead to the equation $$ 4x = 3(x + 4). $$ This is slightly less natural because the question is about the number of acorns buried by the chipmunk, but this is the same as the number of acorns buried by the squirrel and so this approach also works.

This task was adapted from problem #7 on the 2012 American Mathematics Competition (AMC) 10B Test. For the 2012 AMC 10B, which was taken by 35,086 students, the multiple choice answers for the problem had the following distribution:

Of the 35,086 students, 17,169 (49%) were in 10th grade, 9,928 (28%) were in 9th grade, and the remainder were below than 9th grade.

Solutions

Solution: 1 Table

We start by making a table with the number of holes dug by the chipmunk and squirrel and the number of acorns they have buried. We are looking for a common number of acorns and then need to study the number of holes.

Notice that the first common number of acorns we find is 12. The chipmunk hides 12 acorns in 4 holes while the squirrel hides 12 acorns in 3 holes. This is a difference of only one hole so if we want a difference of 4 holes we can repeat this scenario four times to get:

So the chipmunk and squirrel each buried 48 acorns. This is the only answer that works because with more acorns the difference in the number of holes dug goes up and with fewer acorns this difference becomes smaller.

Solution: 2 Using equations

We can introduce a variable $h$ for the number of holes that the chipmunk has dug. Since the chipmunk hides 3 acorns in each hole this make a total of $3h$ acorns that the chipmunk hides. The squirrel, on the other hand, has dug 4 fewer holes than the chipmunk: this is represented by the expression $h - 4$. The squirrel hides 4 acorns in each hole so this means thas squirrel has hidden $4\times (h - 4)$ acorns in total. We are given that the chipmunk and squirrel have hidden the same number of acorns so $$ 3h = 4 \times (h - 4). $$ Using the distributive property on the right hand side gives $$ 3h = 4h - 16. $$ Subtracting 3 from both sides and adding 16 to both sides gives $$ 16 = h. $$ So the squirrel and chipmunk have each made 16 holes. Since the chipmunk hides 3 acorns in each hole this means that the chipmunk (and squirrel) have each hidden 48 acorns.

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Alignments to Content Standards: 8.EE.C.7 For a science project, Sammy observed a chipmunk and a squirrel stashing acorns in holes. The chipmunk hid 3 acorns in each of the holes it dug. The squirrel hid 4 acorns in each of the holes it dug.

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Alignments to Content Standards: 6.RP.A

Task

John and Will ran for 6th grade class president. John got two votes for every vote Will got. It was necessary to get more than half the votes to get elected. What fraction of the votes more than half did John get?

IM Commentary

This is the fourth in a series of tasks about ratios set in the context of a classroom election. What makes this problem interesting is that the number of voters is not given. This information isn't necessary, but at first glance some students may believe it is. On the other hand, students that solved Voting for Two, Variation #3 by recognizing it can be solved as a difference of fractions will be able to apply the same strategy here. Not knowing the number of voters makes the problem more abstract.

Note that for the first three variations in this series, students could use ratio tables instead of the more abstract approaches illustrated for each one. Here is the first time that students will have to go beyond a ratio table to solve it. Students who have had a chance to see the connection between ratio tables and other solution methods will have an easier time making this transition.

Solution

One way to think about the problem is to realize that John gets $2$ votes for every $1$ of Will's votes. Thus, John gets $2$ out of every $3$ votes or $\frac23$ of the votes. The difference between $2/3$ of the votes and $1/2$ of the votes is:

$\frac23 - \frac12 =$

$\frac46 - \frac36 = \frac16$

So, the answer is $\frac16$ of the votes.

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Alignments to Content Standards: 6.RP.A John and Will ran for 6th grade class president. John got two votes for every vote Will got. It was necessary to get more than half the votes to get elected. What fraction of the votes more than half did John get?

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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. $$ 12 = 1\times 12 = 2\times 6 = 3\times 4$$ $$ 18 = 1\times 18 = 2\times 9 = 3\times 6$$ $$ 20 = 1\times 20 = 2\times 10 = 4\times 5$$

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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.

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Alignments to Content Standards: 5.NF.A.1

Task

Nick and Tasha are buying supplies for a camping trip. They need to buy chocolate bars to make s'mores, their favorite campfire dessert. Each of them has a different recipe for their perfect s'more. Nick likes to use $\frac{1}{2}$ of a chocolate bar to make a s'more. Tasha will only eat a s'more that is made with exactly $\frac{2}{5}$ of a chocolate bar.

1. What fraction of a chocolate bar will Nick and Tasha use in total if they each eat one s'more?

2. Nick wants to cut one chocolate bar into pieces of equal size so that he and Tasha can make their s'mores. How many pieces should he cut the chocolate bar into so that each person will get the right amount of chocolate to make their perfect s'more?

3. After Nick cuts the chocolate bar into pieces of equal size, how many pieces of the chocolate bar should he get? How many pieces of the chocolate bar should he give to Tasha?

IM Commentary

The purpose of this instructional task is to motivate a discussion about adding fractions and the meaning of the common denominator. The different parts of the task have students moving back and forth between the abstract representation of the fractions and the meaning of the fractions in the context.

Note that any common multiple of the denominators of the fractions given in this problem could be used in order to find a solution to the addition problem in part (a). It is also important to understand that the common denominator that is used to add the two given fractions corresponds to the number of pieces that Nick will cut the chocolate into in part (b). A classroom discussion of this problem might include the pros and cons for different common denominators when we consider the context of this problem; for example, a smaller denominator will be easier for Nick because he won't have to cut the chocolate bar up into as many pieces.

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. In the task at hand, fifth graders make sense of fractions and how they are related in a problem situation. They decontextualize the problem and represent it with fractions and symbols. As they work through each step, they pause to make sense of the quantities and operations. For instance, students must find a common denominator which really is the number of pieces the chocolate bar is cut into so that both Nick and Tasha can make their s'mores. In part (c) the number of pieces each person receives becomes the numerator of their fractional part of the chocolate bar. Problems such as this one 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 Nick uses $\frac{1}{2}$ of a chocolate bar and Tasha uses $\frac{2}{5}$ of a chocolate bar, they will use $\frac{1}{2}+\frac{2}{5}$ all together. If we first find a common denominator, we can easily add the fractions. 10 is a common multiple of 2 and 5, so 10 could be used as a common denominator. Since $$\frac{1}{2} = \frac{1\times5}{2\times5} =\frac{5}{10}$$ and $$\frac{2}{5} = \frac{2\times2}{5\times2} = \frac{4}{10}$$ We have that

$$ \begin{align} \frac{1}{2}+\frac{2}{5} &= \frac{5}{10}+\frac{4}{10}\\ &= \frac{9}{10} \end{align} $$

The picture below illustrates the amount of chocolate they will use:

2. In part (a) we used the fact that 10 is a common denominator to write both fractions in terms of tenths. So if Nick cuts the bar into 10 equal sized pieces, each person can use some of those pieces to make their perfect s'more. Note that any common denominator would work, but since all common denominators are multiples of 10, it would be more work for Nick if we used a larger denominator.

3. Since $\frac{1}{2}=\frac{5}{10}$, Nick should take 5 of the ten pieces. Since $\frac{2}{5}=\frac{4}{10}$, Tasha should take 4 of the ten pieces. This is easy to see in the picture shown above.

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Alignments to Content Standards: 5.NF.A.1 Nick and Tasha are buying supplies for a camping trip. They need to buy chocolate bars to make s'mores, their favorite campfire dessert. Each of them has a different recipe for their perfect s'more. Nick likes to use $\frac{1}{2}$ of a chocolate bar to make a s'more.

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Task

Your teacher was just awarded \$1,000 to spend on materials for your classroom. She asked all 20 of her students in the class to help her decide how to spend the money. Think about which supplies will benefit the class the most.

1. Write down the different items and how many of each you would choose. Find the total for each category.

Supplies

Books and maps

Puzzles and games

Special items

2. Create a bar graph to represent how you would spend the money. Scale the vertical axis by \$100. Write all of the labels.

3. What was the total cost of all your choices? Did you have any money left over? If so, how much?

4. Compare your choices with a partner. How much more or less did you choose to spend on each category than your partner? How much more or less did you choose to spend in total than your partner?

IM Commentary

The purpose of this task is for students to "Solve problems involving the four operations" (3.OA.A) and "Draw a scaled picture graph and a scaled bar graph to represent a data set with several categories" (3.MD.3). Additionally, students will engage in MP3, Model with mathematics. In this task students are asked to decide how to spend \$1,000 on supplies and materials for their classroom; students will have to make choices and be careful not to exceed the budget. Students are asked to decide which supplies will benefit the class the most and will compare their choices with other students' choices. Because the budget does not allow students to buy one of everything, this task provides an opportunity for teachers to discuss costs and benefits. A benefit is something that satisfies your wants. A cost is what you give up when you decide to do something.

In third grade students are asked to fluently add and subtract within 1,000 (3.NBT.3) which is why the total budget is \$1,000. Students are also multiplying and dividing within 100 (3.OA.7), so they might choose, for example, to buy 20 boxes of markers at \$5 each so that there is a box of markers for every student in the class. It is possible that students will choose to purchase a number of one of the items where the total is greater than \$100; while students are not expected to be fluent in multiplication above 100, they should be able to use their multiplication strategies to find such products. This task provides students with a natural opportunity to use addition, subtraction, and multiplication, and they might also use division depending on how they approach the task; thus it is well aligned to 3.OA.8.

Bar graphs make it easy for students to compare their allocations. If all of the students in the class include all categories on their graphs (whether they allotted any spending to them or not), list the categories in the same order that they are listed in the data table, and use the same colors for each category on a final draft, the teacher can put all of the final graphs up for display and the class can see whether there is a general consensus for how to spend the \$1000 or not.

As an extension, the teacher might consider asking students to represent their total purchases with an equation; for example, if a student chooses 15 boxes of markers, 3 boxes of crayons, and 2 beanbag chairs, she could write:

$$15\times 5 + 3 \times 8 + 2 \times 65 = 75 + 24 + 65 + 65 = 229$$

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.

Solution

1. Solutions will vary. Here is one possible set of choices.

Supplies

Cost per item

Number of items

Total cost

Books and maps

Puzzles and games

Special Items

8 boxes of markers will cost $8\times5 = 4\times2\times5=4\times10=40$ dollars.

4 boxes of crayons will cost $4\times 8 = 4\times4\times2=16\times2 =10\times2+6\times2=20+12=32$ dollars.

2 boxes of pencils will cost $2\times5=10$ dollars.

1 box of printer paper costs 40 dollars.

2 packages of lined paper cost $2\times15 = 2 \times 10 + 2 \times 5 = 20 + 10 = 30$ dollars.

3 boxes of construction paper cost $3\times32 = 3\times30+3\times2=90+6=96$ dollars.

The total for the supplies is $40 + 32+10+40+30+96=248$ dollars.

12 books cost $12\times8 = 10\times8+2\times8=80+16=96$ dollars.

The total cost for the books and maps is $250+96+45 = 391$ dollars.

The total cost for the puzzles and games is $10\times12+6\times15 = 120 + 3\times30=120+90=210$ dollars.

The total for the special items is 130 dollars.

2. Here is a bar graph showing these totals:

3. The total from all the purchases would be $248 + 391+ 210 + 130 = 979$. So these purchases would total \$979 and \$21 would be left over.

4. Comparisons will vary.

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Your teacher was just awarded \$1,000 to spend on materials for your classroom. She asked all 20 of her students in the class to help her decide how to spend the money. Think about which supplies will benefit the class the most. Write down the different items and how many of each you would choose.

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http://tasks.illustrativemathematics.org/content-standards/3/NBT/A/2/tasks/1315