Classroom Case Studies, 6-8 Part C: Problems That Illustrate Reasoning About Number and Operations (55 minutes)

The points are all equally spaced on the number line. Identify the four points that represent the following:

1.3/4 – 1/6

2.d • f

3.e – a

4.b  d

a. Solution:

1. The answer is e. First you need to convert the denominators so that 3/4 equals 9/12, and 1/6 equals 2/12. The difference between 9/12 and 2/12 is 7/12. Since 3/4 is seven points to the right of 1/6, you know that the distance between adjacent points is 1/12. Now you can label all the points on the line: a = 3/12 or 1/4, b = 4/12 or 1/3 . . . and h = 11/12. The answer to your subtraction problem, 7/12, is point e.
2. The answer is b (d • f = 1/2 • 2/3 = 1/3 = 4/12).
3. The answer is b (e – a = 7/12 – 1/4 = 4/12).
4. The answer is f (b  d = 1/3  1/2 = 2/3 = 8/12).

b.This problem deals with the concept of fractions and computation with fractions. Students need to identify common denominators — in this case, 12. The problem also looks at the placement of rational numbers on the number line. At this level, students need to be able to think about mathematical problems in more abstract terms. There is also some inference and proof, since they need to build conjectures on previous conjectures and justify their answers. They also need to compare the fractions. Notice that at this grade level the complexity of the problem is slightly higher than simply ordering fractions.

c.This problem requires students to understand the concept of fractions. They must be able to place fractions on a number line; add, subtract, multiply, and divide fractions, and estimate the magnitude and reasonableness of such operations; and order fractions.

d. Here are some questions that may help students who are struggling:

• What is the difference between 3/4 and 1/6? (7/12) How do you know? (By converting the fractions to the same denominator, we get 3/4 – 1/6 = 9/12 – 2/12 = 7/12. At this level, students need to be able to do this type of computation fluently.)
• Three-fourths is how many points to the right of 1/6? (The answer is seven. Again, after converting the fractions to the same denominator, students need to be able to make such comparisons and understand how they play out on the number line.)
• How does this relate to the difference between the two numbers? (The difference is 7/12, which corresponds to the seven equal lengths between 3/4 and 1/6. Therefore, the difference between adjacent points must be 1/12. Again, this type of question strengthens students’ understanding of fractions and the representation of fractions on the number line. It also connects those understandings with their understanding of operations with fractions.)

e. A question you could ask to extend students’ thinking is, “Where would you place 3/8 on this number line?” In order to place it, they need to change 3/8 to 9/24. They would then realize that 9/24 is between 8/24 and 10/24, or 4/12 and 5/12, and use benchmarks to place 24ths halfway between 12ths.

The yearly changes in the enrollment in your school for the last four years were, respectively, a 20% increase, a 20% increase, a 20% decrease, and a 20% decrease.

1.What is the net change over the four years, to the nearest percent?

2.Would the answer change if the decreases came before the increases?

a. Solution:

1. Enrollment has decreased by 8%. Here’s how you know:

• The enrollment at the end of the first year is 120% of the enrollment of the year before; 120% written as a decimal is 1.2.
• The enrollment at the end of the second year is 120% of the enrollment of the year before; this is (120%) • (120%), or (1.2) • (1.2) times the original enrollment.
• The enrollment at the end of the third year is 80% of the enrollment of the year before; 80% written as a decimal is 0.8. This is (120%) • (120%) • (80%), or (1.2) • (1.2) • (0.8) times the original enrollment.
• The enrollment at the end of the fourth year is 80% of the enrollment of the year before; this is (120%) • (120%) • (80%) • (80%), or (1.2) • (1.2) • (0.8) • (0.8) times the original enrollment.

Now do the math: (1.2) • (1.2) • (0.8) • (0.8) = 0.9216. The enrollment is 92% of what it was four years ago, which means that it has decreased by 8%.

2. The order of the changes does not matter, since the same four numbers will be multiplied, just in a different order. Multiplication is commutative and associative, so the order of the factors does not change the product.

b. This problem deals with percents, computation with percents, and the notion that percent problems can be represented as a proportion of data values compared to a ratio of the percentage to 100. When working with multiple percents, you need to determine the value of the whole, which changes after each computation.

c. Students need to understand the concept of percent, how to represent percent increases and percent decreases, and how to compute with percents.

d. Here are some questions that may help students who are struggling:

• If there were 100 students enrolled in your school to start and there was a 20% increase, how many students would there be at the end of the first year? (120)
• How do you know? (100 • 120% = 100 • 1.2 = 120)
• If there were 120 students enrolled in your school to start and there was a 20% increase, how many students would there be at the end of that year? (144)
• How do you know? (120 • 120% = 120 • 1.2 = 144)
• If there were 144 students enrolled in your school to start and there was a 20% decrease, how many students would there be at the end of that year? (115)
• How do you know? (144 • 80% = 144 • 0.8 = 115.2, which rounds down to 115)

e. Here are some questions you could ask to extend students’ thinking:

• Is the change in enrollment of 40% up and then 15% down the same as 15% up and then 40% down? (The students will need to understand that this is not the same.)
• If the first year was down by 20%, what increase would bring it back to the original enrollment? (The answer is 25%.)

The sum of nine consecutive integers is 63. What is the smallest positive integer that could be part of the sum?

a. Solution: The smallest positive integer that could be part of the sum is 3. Students need to know that with nine consecutive numbers, the center number will be 63 9. Any time you are adding an odd number of integers, the middle is the average of all of them. Since 63 is the sum, the average is 63  9, or 7, meaning that the string of consecutive integers starts with 3 (i.e., 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 = 63).

b. This problem deals with number theory, consecutive integers, divisibility, odd and even numbers, patterns, and logical reasoning.

c. Students would need to understand odd and even numbers and some divisibility tests. Logical and systematic thinking are also necessary skills. Students would need to think about the behavior of sums of consecutive integers. They also need to reason through the pattern by trying out a couple of cases and thinking about similarities in those situations.

d. If students are struggling, you could review the behavior of sums of consecutive integers, as shown above in (a). Students should experiment, using specific examples to test these conjectures. For example, to prove that the sum of three consecutive integers is divisible by 3, they might try the following: 1 + 2 + 3 = 6; 2 + 3 + 4 = 9; 3 + 4 + 5 = 12. This should convince them that the sum of three consecutive integers is always divisible by 3. It is important that they can reason out what the pattern has to be by trying a couple of cases and thinking about the symmetry of the situation. For example, you can show seven consecutive intergers in the following form:

Ask what happens if they add up these numbers. Then ask what happens if they have 9, 11, or another odd number of numbers. Ask why they think this trick doesn’t work for summing an even number of numbers.

e. Here are some questions you could ask to extend students’ thinking:

• Could 63 be the sum of two consecutive integers? (Yes, 31 and 32. Any odd number can be obtained by adding two consecutive integers.)
• Could 63 be the sum of three consecutive integers? (Yes, 20, 21, and 22. Any number that is divisible by 3 can be obtained by adding three consecutive integers. The middle number of the consecutive integers can be found by dividing the original number by three.)
• Could 63 be the sum of four consecutive integers? (No, 63 is not divisible by 4. Any four consecutive integers, when added together, will yield an even number.)
• If 63 is the sum of six consecutive integers, what is the largest number that could be part of the sum? (It’s 13.)
• The sum of five consecutive integers is 5. What is the smallest of the five integers? (Recognizing that consecutive integers also “go the other way” is important; … -3, -2, -1, 0, … are all part of the sequence we care about. You might begin by dividing the sum, 5, by 5 (the number of consecutive integers). This yields 1 as the middle number. So the integers will be -1, 0, 1, 2, 3. The -1 and 1 cancel each other out, so the sum is 5.)

Lucky Edgar has been hired by a movie star to be his assistant on the set of a new film. For his salary, Edgar has been given a choice: He could get paid either $1,000 per day for 20 days or $1 on the first day, $2 on the second day, $4 on the third day, and so on, doubling each day for 20 days. Which salary should Edgar choose?

a. Solution:

			Option 1				Option 2
	Day		This Day’s Pay		Total		This Day’s Pay		Total
	1 $1,000 $1,000 $1 $1 2 $1,000 $2,000 $2 $3 3 $1,000 $3,000 $4 $7 4 $1,000 $4,000 $8 $15 n $1,000 $1,000n 2n-1 (2 • 2n-1) – 1 20 $1,000 $20,000 219 = $524,288 $1,048,575

The second option is clearly better!

b. The content addressed by this problem is exponential growth, computation with large numbers, reasonable estimation, and the graphing of linear and exponential equations. (To learn more about exponential growth, go to Patterns, Functions, and Algebra, Session 7, Part B.)

c. Students need to understand powers, as well as how to compute with exponents, how to organize data, and how to graph data. However, they will likely find that their estimation skills do not extend to situations dealing with exponential growth. We do not often think about problems of exponential growth, so our intuition can be faulty.

d. For students who are struggling, you might ask, “At what point in this process do you catch up? In other words, when do the two options cross each other?” It might be helpful for the students to fill out the table with more values before they jump to Day 20. They may also need assistance in recognizing the patterns, particularly in the exponential function.

e. For Option 2, students will likely write each day’s pay by doubling the previous day’s pay. After they have done five days, ask students to determine the pay on Day 10 without doing the numbers in between. You could try to have the students notice that each day’s pay is 2 to the exponent that is one less than the day number.

If they are filling in a chart, you can try to have students notice that the totals in the “Total” column are always one fewer than the next day’s pay.

You can also have students graph the functions on one coordinate system to compare the two different types of growth: linear and exponential.