(7th Grade Level)
Proficiency 1: Numbers and Number Relationships| ADVANCED | PROFICIENT | NOVICE | BEGINNER |
Three students order chicken McNuggets. They all fall asleep. One awakes and eats 1/3. The second awakes and eats 1/3 of what is left. The third awakes and eats 1/3 of what is left. Eight nuggets remain.
1. How many nuggets are left after they all have eaten their share?
2. If a student eats 1/3 of the nuggets, how many thirds are left?
3. How many nuggets were there in the beginning?
4. Now assume that there are 'X' students. When each student awakes, he or she will eat just 1/x of what is left, leaving (x-1)/x nuggets. Generalize how many nuggets remain after each of the students eats their share if there are initially a total of 'Y' nuggets.
Write a letter that explains your reasoning and shows your process used to find solutions to the above questions. Include diagrams, tables, formulas, numbers or mathematical symbols to help express your thinking. Explain why you think your conclusions are accurate.
See Stella's solutions to these problems
Question 1: I know that there are eight nuggets left at the end because this information is given in the problem.
| Question 2: If a student eats 1/3 from a whole (which is 3/3), then 2/3 are left. | 3 | - | 1 | = | 2 |
| 3 | 3 | 3 |
Question 3: Let's work backwards. If a student eats 1/3, then we know that there are 2/3 left. The 8 nuggets represents what is left over or rather 2/3 of what was there after the 3rd student ate.
So the real question is: 8 is 2/3 of what number?
If we can answer this, then we will know how many nuggets there were before the 3rd student ate.
If 8 is 2/3, then.....4 must be 1/3. And if 4 is 1/3, then...
4 (which is 1/3) + 4 (which is 1/3) + 4 (which is 1/3) = 12 (which is 3/3 or a whole)
So, before the 3rd student ate any nuggets, there were 12 nuggets left.
Now these 12 nuggets represent what is left over or rather, 2/3 of what was there after the 2nd student ate.
So the real question is: 12 is 2/3 of what number?
If we can answer this, then we will know how many nuggets there were before the 2nd student ate.
If 12 is 2/3, then....6 must be 1/3. And if 6 is 1/3, then...
6 (which is 1/3) + 6 (which is 1/3) + 6 (which is 1/3) = 18 (which is 3/3 or a whole)
So, before the 2nd student ate any nuggets there were 18 nuggets left.
We must do this process one more time because the first student ate 1/3 of the nuggets. So the 18 represents 2/3 of the original number of nuggets. So.....
If 18 represents 2/3 of the nuggets, then...9 must be 1/3. And if 9 is 1/3, then.....
9 (which is 1/3) + 9 (which is 1/3) + 9 ( which is 1/3) = 27 (which is 3/3 or a whole)
So, originally, there were 27 nuggets.
I actually could have solved question 3 in a quicker fashion using algebra. I do not know how many nuggets there were originally, but I do know how many were left. So I can set up an algebraic equation. Let Y represent the number of nuggets that were originally there. If I eat 1/3, then 2/3 of Y are left or 2/3 * Y ('of' means multiply). Then another student eats 2/3 of (multiply) that or rather 2/3 * 2/3 * Y. Then another student eats 2/3 of that or rather 2/3 * 2/3* 2/3* Y. This product equals what is left, which is 8. So the equation would be 2/3 * 2/3 * 2/3 * Y = 8 or (2/3)3 * Y = 8. Simplify to 8/27 * Y = 8. Solve for Y. To undo the multiplication, I divide. But I have a fraction, so when you divide fractions, multiply by the reciprocal. So:
| 27 | * | 8 | Y | = | 8 | * | 27 |
| 8 | 27 | 1 | 8 |
Question 4: Notice in our algebraic equation in question C we had 2/3 left because 1/3 was eaten. If 1/X nuggets are eaten, then 1 less than X out of X nuggets are left over, resulting in X - 1/X. . We had raised 2/3 to the 3rd power because we repeated the process 3 times for the 3 students. So, if X students repeat this process, then we raise the fraction (X-1)/X to the Xth power: [(X-1)/X]x. If there were originally Y nuggets and [(X-1)/X]x of Y nuggets are leftover then the expression to represent this amount would be:
Well thanks for your letter. I hope this all made sense to you. If you are still confused you know that you can always ask your teacher for further explanations. Take care and have a good one!
Stella
Back to Proficiency 1 Questions
| ADVANCED | PROFICIENT | NOVICE | BEGINNER |
A rectangular yard has an area of 120 cm². Its length and width are whole numbers.
1. What is the area of the rectangular garden?
2. What are all the possibilities for the length and the width?
3. Which set of dimensions give the smallest perimeter?
4. Prove that there is a better shape for the yard that maintains the same area but decreases the overall perimeter of the yard. This time, there are no restrictions on the types of numbers that you choose to use.
Write a letter that explains your reasoning and shows your process used to find solutions to the above questions. Include diagrams, tables, formulas, numbers or mathematical symbols to help express your thinking. Explain why you think your conclusions are accurate.
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Dear [Name],
Well, we're on to Geometry. I hope you are still having as much fun as me. I've been traveling a lot lately. So, if it has been taking a little longer for me to get back to you - blame it on the US mail. Sorry. Well here we go with the math.
Question 1: The area of the rectangular yard is 120 cm². I know this because this answer is given.
Question 2: In order to determine all of the lengths and widths of the garden, I must first recall the formula for the area of a rectangle.
FORMULA: AREA = LENGTH * WIDTH
With this in mind, I need to know what numbers divide into 120 evenly, or rather, what are the factors of 120. Here is my list:
| width | x | length | = | area | 2(width) | + | 2(length) | = | perimeter |
|---|---|---|---|---|---|---|---|---|---|
| 1 cm | x | 120 cm | = | 120 cm² | 2*1 cm | + | 2*120 cm | = | 242 cm |
| 2 cm | x | 60 cm | = | 120 cm² | 2*2 cm | + | 2*60 cm | = | 124 cm |
| 3 cm | x | 40 cm | = | 120 cm² | 2*3 cm | + | 2*40 cm | = | 86 cm |
| 4 cm | x | 30 cm | = | 120 cm² | 2*4 cm | + | 2*30 cm | = | 68 cm |
| 5 cm | x | 24 cm | = | 120 cm² | 2*5 cm | + | 2*24 cm | = | 58 cm |
| 6 cm | x | 20 cm | = | 120 cm² | 2*6 cm | + | 2*20 cm | = | 52 cm |
| 8 cm | x | 15 cm | = | 120 cm² | 2*8 cm | + | 2*15 cm | = | 46 cm |
| 10 cm | x | 12 cm | = | 120 cm² | 2*10 cm | + | 2*12 cm | = | 44 cm |
Question 3: In order to determine the smallest perimeter for the gardens, I need to first find all of the perimeters for each possible garden by adding up all of their sides . FORMULA: PERIMETER = 2 LENGTHS + 2 WIDTHS. Using a chart again, I generated the perimeters and found my answer. Obviously, from my chart above, I can see that the 10 cm x 12 cm yields the smallest perimeter.
Question 4: In order to decrease the perimeter, but maintain the same area, I noticed the perimeter of the garden decreased as its shape approached a square. So, I believe a square would be the best type of rectangle to minimize my perimeter.
AREA of Square 120 = side * side or s² square root of 120 = s about 10.95cm = s
Therefore, the perimeter is 4 * 10.95 or about 43.8 cm which is smaller than 44 cm.
I can even find a smaller perimeter, if I could only get rid of those wasteful corners. I actually, can get rid of the corners by creating a circle! Watch.
You may not already know the formulas for Area and Perimeter (Circumference) of a circle....but if you don't I'll teach you. Area = pi*r² and Circumference (perimeter for a circle) = 2pi*r. where r = radius of the circle.
To maintain the same area, we plug 120 cm into the equation:
A = pi*r² 120 = pi*r² pi = 3.14 (approximation) 120 = (3.14)r² To solve for r, divide both sides by 3.14 3.14 3.14 38.26 = r² To solve for r square root each side 6.18 cm = r (radius)
Now, I can use the radius information, to substitute into my Circumference equation.
So, you can see that a circle gives the smallest perimeter of 38.81 cm.
C = 2pi*r C = 2(3.14)(6.18) C = 38.81 cm
I know that this last question was quite a challenge, but that's what life is all about. Take care!
Stella
Back to Proficiency 2 Questions
| ADVANCED | PROFICIENT | NOVICE | BEGINNER |
Suppose you roll two six-sided dice.
1. How many sides are on one die?
2. How many ways can two dice, added together give a sum of 3?
3. What is the probability that when two dice are rolled their sum will be divisible by 3?
4. What is the probability that when you roll an X-sided pair of dice, you will roll doubles?
Write a letter that explains your reasoning and shows your process used to find solutions to the above questions. Include diagrams, tables, formulas, numbers or mathematical symbols to help express your thinking. Explain why you think your conclusions are accurate.
See Stella's solutions to these problems
Dear [Name],
Stella here again. I cannot believe the week has gone by so fast!! I hope the letter I sent you on your last letter was helpful. If it wasn't, please let me know what I can do in the future. Also, it never hurts to ask your teacher for help. We're good friends and I'm sure if you need extra help, your teacher would be happy to help. Well, on to the fun stuff.
Question 1: As stated in the information, one die has six sides.
Question 2: There are two ways to roll two dice in order to get a sum of 3. Suppose one die is red and the other is white. I made a chart to illustrate the two rolls.
red white sum 1 2 3 2 1 3
Question 3: For this question I first made a chart to find all of the combinations that you can get with two dice.
Next, I need to define what it means for 'the sum to be divisible by 3'. To find the sum, the two dice must be added together. Then, check each sum to see if 3 goes into it evenly. The only sums that are divisible by 3 are: 3, 6, 9, and 12. I then highlighted the above combinations that added up to these totals. There are 12 different sums that are divisible by 3. However, 12 is not the probability, because a probability is expressed as a fraction. This fraction is created by placing the 12 over the number of total possible combinations (36). So, the probability is 12/36 or when reduced, 1/3.
red die white
die
1 2 3 4 5 6 1 1,1 1,2 1,3 1,4 1,5 1,6 2 2,1 2,2 2,3 2,4 2,5 2,6 3 3,1 3,2 3,3 3,4 3,5 3,6 4 4,1 4,2 4,3 4,4 4,5 4,6 5 5,1 5,2 5,3 5,4 5,5 5,6 6 6,1 6,2 6,3 6,4 6,5 6,6
Question 4: Using the above chart, I underlined all of the doubles. Since there are 6 out of 36, the probability for doubles when rolling two 6-sided dice is 6/36 or when reduced 1/6. A quicker way to determine the total number of possible combinations is to simply multiply the number of sides of each die. (6 * 6 = 36 total combinations) . So, for a pair of 4-sided dice, there will be 4 * 4 = 16 total combinations. If you were to create the chart you would see that there are 4 sets of doubles. This creates a probability of 4/16 which reduces to 1/4 for a pair of 4-sided dice. Consequently, for a pair of X-sided dice, you will have X*X = X2 total combinations. X of the combinations will be doubles. Therefore, the probability that you will roll doubles with a pair of X-sided dice will be formally written as:
P(doubles with an X-sided pair of dice) = X or 1 X² X
Ciao for Now!
Stella
Back to Proficiency 3 Questions
| ADVANCED | PROFICIENT | NOVICE | BEGINNER |
In weighing fruit on an equal arm balance, Joe found that 3 bananas and 1 grapefruit just balanced 10 apples. In another weighing, he found that 1 banana and 6 apples balanced 1 grapefruit.
1. What balances with 1 grapefruit?
2. Draw a picture of the 2 balances and then create 2 equations using variables that would represent the two balances.
3. Use algebra to help Joe determine how many apples would be required to balance with 1 grapefruit.
4. There is more than one algebraic procedure that generates the correct solution. Provide a second algebraic method that supports your solution.
Write a letter that explains your reasoning and shows your process used to find solutions to the above questions. Include diagrams, tables, formulas, numbers or mathematical symbols to help express your thinking. Explain why you think your conclusions are accurate.
See Stella's solutions to these problems
Question 1: One grapefruit balances with 1 banana and 6 apples. This information is given in the problem.
Question 2:
! = banana m = grapefruit S = apple Balance 1: !!!m = SSSSSSSSSS Balance 2: !SSSSSS = m b = banana g = grapefruit a = apple Equation 1: 3b + 1g = 10a Equation 2: 1b + 6a = 1g
Question 3: According to equation 2: 1 grapefruit is the same as 1 banana(1b) and 6 apples (6a). So, every time I see a grapefruit in equation 1, I can substitute it with 1b and 6a.
I know from Equation 2 that: 1g = 1b + 6a.
So, when I see 1g in Equation 1, I will substitute it with (1b + 6a) because they weigh the same.
Originally Equation 1 looked like this: 3b + 1g = 10a Now Equation 1 looks like this: 3b + (1b + 6a) = 10a or 4b + 6a = 10a
Now let's solve to see what 1b equals. If I remove one apple off of the scale on the left side of the equation, I must also remove one apple off the right side of the equation. Actually, there are six apples I can remove on the left side, so I can remove six apples from the right side leaving just four apples on the right side.
4b + 6a = 10a - 6a -6a Subtract both sides by 6a. 4b = 4a 4 4 Divide both sides by 4. 1b = 1a
If 4 bananas weighs the same as 4 apples, then, 1 banana must equal 1 apple. So, every time I see a banana in Equation 2, I can replace it with an apple.
Originally Equation 2 looked like this: 1g = 1b + 6a Now Equation 2 looks like this: 1g = 1a + 6a or 1g = 7a
So, 1 grapefruit equals 7 apples.
Question 4: Another way to solve this algebraically is to solve the equations by using the Multiplication-Addition Algorithm or as others refer to it the Cancellation Method.
Equation 1: 3b + 1g = 10a Rewrite this equation so that each variable lines up Equation 2: 1b + 6a = 1g with the variables in Equation 1. So, subtract 6a - 6a -6a from both sides 1b = 1g - 6a Now, subtract 1g from both sides -1g -1g Equation 2: (Rewritten) 1b - 1g = -6a Now the equation looks like this Equation 1: 3b + 1g = 10a Next, we want to get the bananas to cancel each other out, so that we are just left with Equation 2: -3(1b - 1g = -6a) grapefruits and apples. So, multiply each term in Equation 2 by -3 Equation 2: (New) -3b + 3g = 18a (Remember: negative * negative = positive)
Now add together Equation 2 (New) and Equation 1.
Equation 2: (New) -3b + 3g = 18a + Equation 1: 3b + 1g = 10a Notice that -3b and 3b cancel 4g 28a Get g by itself by dividing both sides by 4. 4 4 1g = 7a Thus, 1 grapefruit equals 7 apples.
I hear some of you saying, "You've got to be kidding me!!! How was I supposed to know this method?" The answer: You're not. Not yet anyway. But, it does not hurt to expose you to these methods because you will be learning them soon and you can say that you have already heard of them before. Also, you will be given another Stella where you can try this method out. Good luck and adios!
Stella
Back to Proficiency 4 Questions
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