Alignment 1: 7.RP.1Cooking with the Whole Cup
Grade 7
Domain Ratios and Proportional Relationships
Cluster Analyze proportional relationships and use them to solve real-world and mathematical
problems.
Standard Compute unit rates associated with ratios of fractions, including ratios of lengths, areas and
other quantities measured in like or different units. For example, if a person walks mile in each hour,
compute the unit rate as the complex fraction
miles per hour, equivalently 2 miles per hour.
Travis was attempting to make muffins to take to a neighbor that had just moved in down the
street. The recipe that he was working with required 3/4 cup of sugar and 1/8 cup of butter.
a. Travis accidentally put a whole cup of butter in the mix.
i.
What is the ratio of sugar to butter in the original recipe? What amount of
sugar does Travis need to put into the mix to have the same ratio of sugar to
butter that the original recipe calls for?
ii.
If Travis wants to keep the ratios the same as they are in the original recipe,
how will the amounts of all the other ingredients for this new mixture
compare to the amounts for a single batch of muffins?
iii.
The original recipe called for 38 cup of blueberries. What is the ratio of
blueberries to butter in the recipe? How many cups of blueberries are needed
in the new enlarged mixture?
b. This got Travis wondering how he could remedy similar mistakes if he were to dump in a
single cup of some of the other ingredients. Assume he wants to keep the ratios the same.
i.
How many cups of sugar are needed if a single cup of blueberries is used in
the mix?
ii.
How many cups of butter are needed if a single cup of sugar is used in the
mix?
iii.
How many cups of blueberries are needed for each cup of sugar?
Commentary:
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While the task as written does not explicitly use the term "unit rate," most of the work students
will do amounts to finding unit rates. A recipe context works especially well since there are so
many different pair-wise ratios to consider.
This task can be modified as needed; depending on the choice of numbers, students are likely to
use different strategies which the teacher can then use to help students understand the connection
between, for example, making a table and strategically scaling a ratio.
The choice of numbers in this task is already somewhat strategic: in part (a), the scale factor is a
whole number and in part (b), the scale factors are fractions. Because of this difference, students
will likely approach the parts of the task in different ways. The teacher can select and sequence a
discussion of the different approaches to highlight the structure of the mathematics and allow for
connections to proportional relationships.
This task was submitted by Travis Lemon for the first IMP task writing contest 2011/12/122011/12/18.
Solution: Solution
a.
i. The ratio of cups of sugar to cups of butter is 3/4:1/8. If we multiply both numbers in the
ratio by 8, we get an equivalent ratio that involves 1 cup of butter.
8×3/4=6
And
8×1/8=1
In other words, 3/4:3/8 is equivalent to 6:1, and so six cups of sugar is needed if there is one
cup of butter.
ii. In the previous part we saw that we have 8 times as much butter, so all the ingredients need
to be increased by a factor of 8. That is, the quantity of each ingredient in the original recipe
needs to be multiplied by 8 in order for all the ratios to be the same in the new mixture.
iii. The ratio of cups of blueberries to cups of butter is 38:18 in the original recipe, so Travis will
need to add 8×3/8=3 cups of blueberries to his new mixture.
b.
i. The ratio of cups of sugar to cups of blueberries is 3/4:3/8. If we multiply both numbers in
the ratio by 8/3, we get an equivalent ratio.
8/3×3/4=2 and 8/3×3/8=1.
Since 3/4:3/8 is equivalent to 2:1, two cups of sugar is needed if there is one cup of
blueberries.
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ii. The ratio of cups of butter to cups of sugar is 1/8:3/4. If we multiply both numbers in the
ratio by 4/3, we get an equivalent ratio.
4/3×1/8=1/6 and 4/3×3/4=1.
In other words, 1/8:3/4 is equivalent to 1/6:1, and 16 cup of butter is needed if there is one
cup of sugar.
iii. The ratio of cups of blueberries to cups of sugar is 3/8:3/4. If we multiply both numbers in
the ratio by 4/3, we get an equivalent ratio.
4/3×3/8=1/2 and 4/3×3/4=1.
Since 3/8:3/4 is equivalent to 1/2:1, Travis would need 1/2 cup of blueberries if there is one
cup of sugar.
Instructional Note: For part (b), I have encouraged students to think about unit fractions as an
intermediate step to developing an understanding of how to multiply by fractions. With the
emphasis on unit fractions in the CCSSM, I decided to use this approach this year and have
found success. Students see the value of scaling to a unit fraction and then going from there.
For example, if a student realizes that 3/8 needs to become 1 to answer part (b.i), she can first
take 1/3 of the amount to create a unit fraction of 1/8 and then multiply this by 8 to create 1.
The composite result of these calculations is equivalent to multiplying by 8/3. Students often find
that the two calculations (taking 1/3 of the amount to create a unit fraction of 1/8 and then
multiply this by 8) made independently are more mentally accessible, which makes them a nice
intermediate step in understanding the composite calculation of multiplying by the reciprocal.
Solution: Using tables
a.
i. The ratio of cups of sugar to cups of butter is 3/4:1/8. If we set up a table, we can
successively double the amounts:
cups of sugar
3/4
6/4
12/4
24/4=6
cups of butter
1/8
2/8
4/8
8/8=1
So six cups of sugar is needed if there is one cup of butter.
ii. In the previous part, we had to double the quantities three times: 2⋅2⋅2=8. So Travis needs 8
times as much butter as the original recipe required. If we want to keep all the ingredients in
the same ratio, Travis needs to multiply the amount of each ingredient by 8.
iii. The ratio of cups of blueberries to cups of butter is 3/8:1/8 in the original recipe, so Travis
will need to add 8⋅3/8=3cups of blueberries.
b.
i. It is much harder to solve this problem using a table because the scale factor is no longer a
whole number. Students who solved the first part using a table may need guidance from their
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classmates or the teacher to see that multiplying both numbers in the ratio by the reciprocal
of the amount of blueberries will give an equivalent ratio with 1 cup of blueberries. Here is
where the teacher should highlight the importance of being able to find a unit rate.
The ratio of cups of sugar to cups of blueberries is 3/4:3/8. If we multiply both numbers in
the ratio by 8/3, we get an equivalent ratio.
3/4×8/3=2 and 3/8×8/3=1.
So two cups of sugar is needed if there is one cup of blueberries.
ii. The ratio of cups of butter to cups of sugar is 18:34. If we multiply both numbers in the ratio
by 4/3, we get an equivalent ratio.
1/8×4/3=1/6 and 3/4×4/3=1.
So 1/6 cup of butter is needed if there is one cup of sugar.
iii. The ratio of cups of blueberries to cups of sugar is 3/8:3/4. If we multiply both numbers in
the ratio by 4/3, we get an equivalent ratio.
3/8×4/3=1/2 and 3/4×4/3=1.
So 1/2 cup of blueberries is needed if there is one cup of sugar.
Page 4 of 26
Alignment 1: 7.RP.2 - Robot Races
Grade7
DomainRP: Ratios and Proportional Relationships
ClusterAnalyze proportional relationships and use them to solve real-world and mathematical
problems.
StandardRecognize and represent proportional relationships between quantities.
Carli’s class built some solar-powered robots. They raced the robots in the parking lot of the
school. The graphs below show the distance d, in meters, that each of three robots traveled
after t seconds.
a. Each graph has a point labeled. What does the point tell you about how far that robot has
traveled?
b. 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.
c. How fast is each robot traveling? How can you see this in the graph?
Solution: Answers
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.
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a. 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.
b. 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.
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Alignment 1: 7.RP.2
Buying Coffee
Grade 7
Domain RP: Ratios and Proportional Relationships
Cluster Analyze proportional relationships and use them to solve real-world and mathematical
problems.
Standard Recognize and represent proportional relationships between quantities.
Coffee costs $18.96 for 3 pounds.
a. What is the cost per pound of coffee?
b. Let x be the number of pounds of coffee and y be the total cost of x pounds. Draw a graph of
the proportional relationship between the number of pounds of coffee and the total cost.
c. How can you see the cost per pound of coffee in the graph?
Commentary:
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 Standard for
Mathematical Practice 5 Use appropriate tools strategically. A variant of this problem is
appropriate for 8th grade; see 8.EE.5 Coffee by the Pound.
Solution: Drawing the graph
a. You can find the cost for one pound of coffee by dividing the total cost by 3. Coffee
costs $6.32 per pound.
b. 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).
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c. 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.
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Alignment 1: 7.RP.2
Art Class- Variation 1
Grade7
DomainRP: Ratios and Proportional Relationship
ClusterAnalyze proportional relationships and use them to solve real-world and mathematical
problems.
StandardRecognize and represent proportional relationships between quantities.
Art Class- Variation 1
The students in Ms. Baca’s art class were mixing yellow and blue paint. She told them that two
mixtures will be the same shade of green if the blue and yellow paint are in the same ratio.
The table below shows the different mixtures of paint that the students made.
A
B
C
D
E
Yellow
1 part
2 parts
3 parts
4 parts
6 parts
Blue
2 part
3 parts
6 parts
6 parts
9 parts
a. How many different shades of paint did the students make?
b. Some of the shades of paint were bluer than others. Which mixture(s) were the bluest? Show
work or explain how you know.
c. Carefully plot a point for each mixture on a coordinate plane like the one that is shown in the
figure. (Graph paper might help.)
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d. Draw a line connecting each point to (0,0). What do the mixtures that are the same shade of
green have in common?
Commentary:
Giving the amount of paint in "parts" instead of a specific standardized unit like cups might be
confusing to students who do not understand what this means. Because this is standard language
in ratio problems, students need to be exposed to it, but teachers might need to explain the
meaning if their students are encountering it for the first time.
Solution: Equations
a. The students made two different shades: mixtures A and C are the same, and mixtures B, D,
and E are the same.
b. To make A and C, you add 2 parts blue to 1 part yellow. To make mixtures B, D, and E, you
add 3/2 parts blue to 1 part yellow. Mixtures A and C are the bluest because you add more
blue paint to the same amount of yellow paint.
c. See the figure.
d. If two mixture are the same shade, they lie on the same line through the point (0,0).
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Alignment 1: 7.RP.2 Art Class- Variation 2
The students in Ms. Baca’s art class were mixing yellow and blue paint. She told them that two
mixtures will be the same shade of green if the blue and yellow paint are in the same ratio.
The table below shows the different mixtures of paint that the students made.
A
B
C
D
E
F
Yellow
1 part
2 parts
3 parts
4 parts
5 parts
6 parts
Blue
2 part
3 parts
6 parts
6 parts
8 parts
9 parts
a. How many different shades of paint did the students make?
b. Write an equation that relates y, the number of parts of yellow paint, and b, the number of
parts of blue paint for each of the different shades of paint the students made.
Commentary:
Giving the amount of paint in "parts" instead of a specific standardized unit like cups might be
confusing to students who do not understand what this means. Because this is standard language
in ratio problems, students need to be exposed to it, but teachers might need to explain the
meaning if their students are encountering it for the first time.
Solution: Equations
a. The students made three different shades: mixtures A and C are the same, mixtures B, D, and
F are the same, and mixture E is different than the other two.
b. An equation for mixtures A and C is b=2y. An equivalent equation is y=1/2b.
The equation for mixtures B, D, and F is b=3/2y. An equivalent equation is y=2/3b.
The equation for mixture E is b=8/5y. An equivalent equation is y=5/8b.
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Alignment 1: 7.RP.2
Music Companies- Variation 1
Grade 7
Domain RP: Ratios and Proportional Relationships
Cluster Analyze proportional relationships and use them to solve real-world and mathematical
problems.
StandardRecognize and represent proportional relationships between quantities.
Music Companies- Variation 1
BeatStreet, TunesTown, and MusicMind are music companies. BeatStreet offers to buy 1.5
million shares of TunesTown for$561 million. At the same time, MusicMind offers to buy 1.5
million shares of TunesTown at $373 per share. Who would get the better deal, BeatStreet or
MusicMind? What is the total price difference?
Commentary:
This problem requires a comparison of rates where one is given in terms of unit rates, and the
other is not. See "7.RP Music Companies, Variation 2" for a task with a very similar setup but is
much more involved and so illustrates 7.RP.3.
Teachers should be aware that the context of stock purchase may not be familiar to 7th graders.
The context should be explained to students if needed.
Solution: Compute the unit rate
One simple way to attack the problem is to compute the unit rate for the BeatStreet offering.
The unit rate that BeatStreet is offering, i.e. the price per share, is
561,000,000 dollars /1,500,000 shares =374 dollars/ share
The BeatStreet offer unit rate is thus 374 dollars per share. MusicMind is offering to buy shares
at the unit rate of 373 dollars per share and so has the better deal. MusicMind gets the shares
at $1 per share less than BeatStreet. The total difference is
1.5 million shares ×1 dollar /share=1.5 million dollars
In summary, MusicMind gets the better deal and would save $1.5 million over what BeatStreet
would pay.
Solution: Compute total cost
Since we have to figure out the difference in cost anyway, an alternative way to solve this
problem is to just compute how much MusicMind would have to pay altogether.
1,500,000 shares ×373 dollars/share =559,500,000 dollars .
This is less than BeatStreet offered, so MusicMind gets the better deal. Using this approach not
only answers the first question, but the answer to the second is now a simple subtraction
problem:
561,000,000 dollars −559,500,000 dollars =1,500,000 dollars .
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Alignment 1: 7.RP.2-3
Music Companies- Variation 2
Grade 7
Domain RP: Ratios and Proportional Relationships
Cluster Analyze proportional relationships and use them to solve real-world and mathematical
problems.
StandardUse proportional relationships to solve multistep ratio and percent problems.
Examples: simple interest, tax, markups and markdowns, gratuities and commissions, fees,
percent increase and decrease, percent error.
BeatStreet, TunesTown, and MusicMind are music companies. BeatStreet and MusicMind are
teaming up together to make an offer to acquire 1.5 million shares of TunesTown worth $374 per
share. They will offer TunesTown 20 million shares of BeatStreet worth $25 per share. To make
the swap even, they will offer another 2 million shares of MusicMind.
What price per share (in dollars) must each of these additional shares be worth?
Commentary:
This problem has multiple steps. In order to solve the problem it is necessary to compute:
the value of the TunesTown shares;
the total value of the BeatStreet offer of 20 million shares at $25 per share;
the difference between these two amounts; and
the cost per share of each of the extra 2 million shares MusicMind offers to equal to the
difference.
See "7.RP Music Companies, Variation 1" for a task with a very similar setup that focuses on
comparing unit rates to illustrate 7.RP.2.
Teachers should be aware that the context of stock purchase may not be familiar to 7th graders.
The context should be explained to students if needed.
Solution: Step-by-step computing
The value of the TunesTown shares is 1.5 million times $374, or $561 million, i.e.
374 dollars/ 1 share ×1.5 million shares =561 million dollars .
BeatStreet offers 20 million shares at $25 per share. This is worth
20,000,000 shares ×25 dollars/ 1 share =500,000,000 dollars
MusicMind still needs to offer $561 million minus $500 million, i.e. $61 million. They want this
to be 2 million shares, and the unit cost must be:
61,000,000 dollars /2,000,000 shares =30.5 dollars/ 1 share
So each of the additional shares must be worth $30.50 per share in order to make an even swap.
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Solution: Set up an equation
Set up an equation:
374 dollars 1 share ⋅1,500,000 shares =20,000,000 shares ⋅25 dollars 1 share +2,000,000 shares ⋅
x dollars 1 share .
Cancel the share units and divide by the dollar units to get:
374⋅1,500,000=20,000,000⋅25+2,000,000⋅x.
Solving for x:
x=374⋅1,500,000−20,000,000⋅25 / 2,000,000=30.5.
There are no units for x because it is the NUMBER of dollars per share.
Page 14 of 26
Alignment 1: 7.RP.3
Friends Meeting on Bikes
Grade 7
Domain RP: Ratios and Proportional Relationships
Cluster Analyze proportional relationships and use them to solve real-world and mathematical
problems.
Standard Use proportional relationships to solve multistep ratio and percent problems.
Examples: simple interest, tax, markups and markdowns, gratuities and commissions, fees,
percent increase and decrease, percent error.
Taylor and Anya are friends who live 63 miles apart. Sometimes on a Saturday, they ride toward
each other's houses on their bikes and meet in between. One day they left their houses at 8 am
and met at 11 am. Taylor rode at 12.5 miles per hour. How fast did Anya ride?
Commentary:
There is a more scaffolded version of this same problem; see 6.RP.3 Friends Meeting on
Bicycles.
Additional questions for a student who doesn't know where to start: "How long did the bike ride
take? How far did Taylor ride? How far did Anya ride?"
Comparing the solutions below using distance and using speed, there is an opportunity to point
out the distributive property. If we take 8.5 mph + 12.5 mph = 21 mph and multiply by 3 hours,
we get 63 miles = 25.5 miles + 37.5 miles
Solution: Finding distances first
Since the bike ride took 3 hours, Taylor must have ridden 37.5 miles. Therefore Anya must have
ridden 63 - 37.5 - 25.5 miles. If we divide the distance Anya rode by the amount of time it took
her to ride, we will get her speed (assuming she rode at a constant speed).
25.5 miles ÷3 hours =8.5 miles per hour.
So if Anya rode at a constant speed, she was traveling 8.5 miles per hour.
Solution: Subtracting Taylor's speed from total speed
The friends are moving toward each other at a total speed of
63 miles ÷3 hours =21 miles per hour.
Since Taylor rides at 12.5 miles per hour, Anya's speed must be 21 - 12.5 = 8.5 miles per hour.
Assuming Anya rode at a constant speed, she was traveling 8.5 miles per hour.
Page 15 of 26
Alignment 1: 7.RP.3
Buying Protein Bars & Magazines
Grade 7
Domain RP: Ratios and Proportional Relationships
Cluster Analyze proportional relationships and use them to solve real-world and mathematical
problems.
Standard Use proportional relationships to solve multistep ratio and percent problems.
Examples: simple interest, tax, markups and markdowns, gratuities and commissions, fees,
percent increase and decrease, percent error.
Tom wants to buy some protein bars and magazines for a trip. He has decided to buy three times
as many protein bars as magazines. Each protein bar costs $0.70 and each magazine costs $2.50.
The sales tax rate on both types of items is 6½%. How many of each item can he buy if he has
$20.00 to spend?
Solution: Using a ratio table
The table below shows the cost for the protein bars and magazines in a 3 : 1 ratio.
Number of magazines
1
2
3
4
Number of protein bars
3
6
9
12
Value of the magazines
$2.50
$5.00
$7.50
$10.00
Value of the protein bars
$2.10
$4.20
$6.30
$8.40
Value of both magazines
and candy bars
$4.60
$9.20
$13.80
$17.40
Cost with tax
$4.90
$9.80
$14.70
$19.60
Looking at the last column of the table, we can see that Tom can buy 4 magazines and 12 protein
bars for $20 and that he cannot afford 5 magazines and 15 protein bars.
Solution: 1 magazine and 3 protein bars as a single unit
Tom’s decision to buy three times as many protein bars as magazines can be thought of as
deciding to buy in a unit consisting of 1 magazine AND 3 protein bars.
The cost of a unit then is $2.50 + 3×($0.70), which is $4.60.
With sales tax, this would be $4.60 × 1.065, which when rounded to the nearest cent would
be $4.90, or just under $5.00.
There are four groups of five in 20 and 4 × 4.899 = 19.596. This leaves $0.40 in change. So,
with $20, he can buy 4 magazines and 12 protein bars, with $0.40 in change.
Page 16 of 26
Alignment 1: 7.RP.3
Tax and Tip
Grade 7 Domain RP: Ratios and Proportional Relationships
Cluster Analyze proportional relationships and use them to solve real-world and mathematical
problems.
Standard Use proportional relationships to solve multistep ratio and percent problems.
Examples: simple interest, tax, markups and markdowns, gratuities and commissions, fees,
percent increase and decrease, percent error.
After eating at your favorite restaurant, you know that the bill before tax is $52.60 and that the
sales tax rate is 8%. You decide to leave a 20% tip for the waiter based on the pre-tax amount.
How much should you leave for the waiter? How much will the total bill be, including tax and
tip? Show work to support your answers.
Solution: An exact answer
To figure out the tip, you need to find 20% of $52.60.
0.2×52.6=10.52
You should leave $10.52 for the waiter if you want to leave him exactly 20%.
To figure the tax, you need to find 8% of $52.60.
0.08×52.6=4.208
Next, add them up:
52.6+10.52+4.208=67.328
The total bill, including tax and tip, will be $67.33.
Solution: Estimating an answer
If you are not concerned whether you give the waiter exactly 20%, you can simply estimate the total bill.
Tax and tip together are a little less than 30% and the bill is a little more than $50. Since 30% of 50 is 15,
the tax and tip together are approximately $15. (Note that the exact calculation is $14.73.)
We can also estimate the total bill as $52 + $15 = $67, which is very close to the exact calculation
of $67.33. This kind of estimating is a good way to check the answer to an exact calculation, and more
like the way you would probably compute tax and a tip in a real restaurant.
Solution: Exact tax, approximate tip
In a real-world context, restaurant patrons generally estimate the tip, but pay the exact tax based on the
calculation of their bill. Therefore, a third solution possibility is for students to estimate the tip portion of
the calculations and to find the exact tax when calculating the total bill.
If you wish to estimate the tip, you can round $52.60 to $50. Twenty percent of $50 is $10. So,
you would leave a $10 tip.
To figure tax, you need to find 8% of $52.60:
0.08×52.6=4.208.
Next, add the estimated tip, the calculated tax, and the pre-tax bill:
52.60+10+4.208=66.808.
The total bill will be $66.81.
Page 17 of 26
Alignment 1: 7.RP.3and7.G.6
Sand Under the Swing Set
Grade 7
Domain RP: Ratios and Proportional Relationships
Cluster Analyze proportional relationships and use them to solve real-world and mathematical
problems.
Standard Use proportional relationships to solve multistep ratio and percent problems.
Examples: simple interest, tax, markups and markdowns, gratuities and commissions, fees,
percent increase and decrease, percent error.
Domain G: Geometry
Cluster Solve real-life and mathematical problems involving angle measure, area, surface area,
and volume.
Standard Solve real-world and mathematical problems involving area, volume and surface area
of two- and three-dimensional objects composed of triangles, quadrilaterals, polygons, cubes,
and right prisms.
The 7th graders at Sunview Middle School were helping to renovate a playground for the
kindergartners at a nearby elementary school. City regulations require that the sand underneath
the swings be at least 15 inches deep. The sand under both swing sets was only 12 inches deep
when they started.
The rectangular area under the small swing set measures 9 feet by 12 feet and required 40 bags
of sand to increase the depth by 3 inches. How many bags of sand will the students need to cover
the rectangular area under the large swing set if it is 1.5 times as long and 1.5 times as wide as
the area under the small swing set?
Solution: Finding the scale factor the hard way
3 inches is 14=0.25 foot, so the volume of sand that was used is
0.25×9×12=27
cubic feet. The amount of sand needed for an area that is 1.5 times as long and 1.5 times as wide
would be
0.25×(1.5⋅9)×(1.5⋅×12)=60.75
cubic feet.
We know that 40 bags covers 27 cubic feet. Since the amount of sand for the large swing set is
60.75÷27=2.25
times as large, they will need 2.25 times as many bags. Since 2.25×40=90, they will need 90
bags of sand for the large swing set.
Solution: Finding the scale factor the easy way
Since we have to multiply both the length and the width by 1.5, the area that needs to be covered
is
1.52=2.25
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times as large. Since the depth of sand is the same, the amount of sand needed for the large
swing set is 2.25 times as much as is needed for the small swing set, and they will need 2.25
times as many bags.
Since 2.25×40=90, they will need 90 bags of sand for the large swing set.
Solution: Using a unit rate
The area they cover under the small swing set is 9×12=108 square feet. Since the depth is the
same everywhere, and we know that 40 bags covers 108 square feet, they can
cover 108÷40=2.7 square feet per bag.
The area they need to cover under the large swing set is
1.52=2.25
times as big as the area under the small swing set, which is
2.25×108=243
square feet. If we divide the number of square feet we need to cover by the area covered per bag,
we will get the total number of bags we need:
243÷2.7=90
So they will need 90 bags of sand for the large swing set.
Solution: The other unit rate
The area they cover under the small swing set is 9×12=108 square feet. Since the depth is the
same everywhere, and we know that 40 bags covers 108 square feet, they can
cover 40÷108=1027 bags per square foot.
The area they need to cover under the large swing set is
(3/2)2=9/4
times as big as the area under the small swing set, which is
9/4×108=243
square feet. If we multiply the number of square feet they need to cover by the number of bags
needed per square foot, we will get the total number of bags we need:
243×10/27=90
So they will need 90 bags of sand for the large swing set.
Page 19 of 26
Alignment 1: 7.RP.3
Comparing Years
Grade 7
Domain RP: Ratios and Proportional Relationships
Cluster Analyze proportional relationships and use them to solve real-world and mathematical
problems.
Standard Use proportional relationships to solve multistep ratio and percent problems.
Examples: simple interest, tax, markups and markdowns, gratuities and commissions, fees,
percent increase and decrease, percent error.
Historically, different people have defined a year in different ways. For example, an Egyptian
year is 365 days long, a Julian year is 36514 days long, and a Gregorian year is 365.2425 days
long.
a. What is the difference, in seconds, between a Gregorian year and a Julian year?
b. What is the percent decrease, to the nearest thousandth of a percent, from a Julian year to a
Gregorian year?
c. How many fewer days are there in 400 years of the Gregorian calendar than there are in 400
years of the Julian calendar?
Commentary:
Many students will not know that when comparing two quantities, the percent decrease between
the larger and smaller value is not equal to the percent increase between the smaller and larger
value. Students would benefit from exploring this phenomenon with a problem that uses smaller
values before working on this one.
Solution: Find the number of seconds in a year first and then subtract
a. To find how many seconds in a day, multiply the number of seconds in a minute by the
number of minutes in an hour by the number of hours in a day:
60 seconds/1 minute×60 minutes/1 hour×24 hours1 day=86,400 seconds/1 day
To find the number of seconds in a year, multiply the length of the year in days by the
number of seconds in a day.
For a Julian year:
365.25 days/1 year×86,400 seconds/1 day=31,557,600 seconds/1 year
For a Gregorian year:
365.2425 days/1 year×86,400 seconds/1 day=31,556,952 seconds/1 year
Now subtract the length of a Gregorian year from the length of a Julian year:
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31,557,600−31,556,952=648
So a Julian year is 648 seconds longer than a Gregorian year.
b. To find how much shorter a Gregorian year is than a Julian year (as a percentage of a Julian
year), divide the difference by the length of a Julian year:
648÷31557600=0.00002053...
which is approximately 2/100,000 or 0.002%.
So a Gregorian year is about 0.002% shorter than a Julian year.
c. Multiplying 365.25 by 400, we get 146,100 days for the Julian calendar. Multiplying
365.2425 by 400, we get 146,097 days for the Gregorian calendar. We see a decrease of 3
days in 400 years. If we round 146,100 days to 150,000 days, we see that 3 units less in
150,000 units is consistent with our earlier finding of approximately 1 unit less per 50,000
units.
Solution: Find the difference in the number of days first and then convert to seconds
a. We can subtract the length of a Gregorian year from a Julian year:
365.25−365.2425=0.0075
So there are 0.0075 more days in the Julian year. To convert the number of days to the
number of seconds, we can multiply number of seconds in one day with the number of days:
86,400×0.0075=648
So there are 648 more seconds in the Julian year than in the Gregorian year.
b. To find how much shorter a Gregorian year is than a Julian year (as a percentage of a Julian
year), divide the difference by the length of a Julian year:
648÷31557600=0.00002053...
which is 0.002053... %.
Rounding, we see that a Gregorian year is about 0.002% shorter than a Julian year.
c. Multiplying 365.25 by 400, we get 146,100 days for the Julian calendar. Multiplying
365.2425 by 400, we get 146,097 days for the Gregorian calendar. We see a decrease of 3
days in 400 years.
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Alignment 1: 7.RP.3
Chess Clubs
Grade 7
Domain RP: Ratios and Proportional Relationships
Cluster Analyze proportional relationships and use them to solve real-world and mathematical
problems.
Standard Use proportional relationships to solve multistep ratio and percent problems.
Examples: simple interest, tax, markups and markdowns, gratuities and commissions, fees,
percent increase and decrease, percent error.
There were 24 boys and 20 girls in a chess club last year. This year the number of boys increased
by 25% but the number of girls decreased by 10%. Was there an increase or decrease in overall
membership? Find the overall percent change in membership of the club.
Commentary:
This problem includes a percent increase in one part with a percent decrease in the remaining and
asks students to find the overall percent change. The problem may be solved using proportions or
by reasoning through the computations or writing a set of equations.
When using equations to solve the problem, the task of finding the number of club members this
year can be accomplished in two separate steps by finding the appropriate percent of last year’s
members and then adjusting the number of members by this amount. Alternatively, the number
can be determined in one step by finding the appropriate percent that will remain after the
change. The second approach requires a deeper understanding of the concept of percent change.
As with equations, when solving this problem using proportions, the number of new club
members can be found in one or two steps. Again, the second approach requires a deeper
understanding.
Solution: Using an equation to find the number of new club members in two steps
Last year there were 24 boys in chess club. This year the number increased by 25%:
24×0.25=6 more boys this year
24+6=30 total boys this year
Last year there were 20 girls in chess club. This year the number decreased by 10%:
20×0.1=2 fewer girls this year
20–2=18 total girls this year
Combining the number of girls and boys this year:
30+18=48 total members this year
Combining the number of girls and boys last year:
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24+20=44 total members last year
Finding the difference in the number of members this year and last year:
48–44=4 more members this year
To find the percent change, divide the change in the number of members by the number of
members last year:
4÷44=0.09 (repeating)
Change the decimal amount into a percent amount and round to the nearest whole percent:
0.09(repeating) ×100≈9% change in the chess club membership
Since there are 4 more members in the club this year when compared to last year, the change is
a 9% increase.
Solution: Using an equation to find the number of new club members in one step.
A 25% increase in the number of boys is equivalent to 125% of the number of boys last year:
24×1.25=30 total boys this year
A 10% decrease in the number of girls is equivalent to 90% of the number of girls last year:
20×0.9=18 total girls this year
The remaining steps to the solution are the same as shown in the solution above.
Solution: Using proportions and two steps to find the number of new members
Write a proportion where b represents the increase in the number of boys:
b/24=25/100
Multiplying both sides by 24, we get b=6 more boys this year
24+6=30 total boys this year
Write a proportion where g represents the decrease in the number of girls:
g/20=10/100
Multiplying both sides by 20, we get g=2 fewer girls this year
20–2=18 total girls this year
The remaining steps are the same as those shown in the first solution above.
Solution: Using proportions and one step to find the number of new members
Write a proportion where b represents the number of boys in the club this year:
b/24=125/100
Multiplying both sides by 24, we get b=30 total boys this year
Write a proportion where g represents the number of girls in the club this year:
g/20=90/100
Multiplying both sides by 20, we get g=18 total girls this year
The remaining steps are the same as those shown in the first solution above.
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Alignment 1: 7.RP.3
Selling Computers
Grade 7
Domain RP: Ratios and Proportional Relationships
Cluster Analyze proportional relationships and use them to solve real-world and mathematical
problems.
Standard Use proportional relationships to solve multistep ratio and percent problems.
Examples: simple interest, tax, markups and markdowns, gratuities and commissions, fees,
percent increase and decrease, percent error.
The sales team at an electronics store sold 48 computers last month. The manager at the store
wants to encourage the sales team to sell more computers and is going to give all the sales team
members a bonus if the number of computers sold increases by 30% in the next month. How
many computers must the sales team sell to receive the bonus? Explain your reasoning.
y Rounding Up
If the sales team is going to sell 30% more computers next month, they will have to sell
0.3×48=14.4
more computers. Of course, you can't sell four-tenths of a computer, so that means they will have
to sell 15 more computers. Since 48+15=63, they will need to sell 63 computers next month to
receive the bonus.
Solution: Making a table
If a student sees initially that 12 is 25% of 48, it would be quick to check the percent increase for
12 or more computers.
Number of additional computers
12
13
14
15
Percent increase
25 %
≈ 27%
≈ 29%
≈31%
Thus, they need to sell 15 more computers, or 63 computers, to get the bonus.
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Alignment 1: 7.NS.3
Sharing Prize Money
Grade7
DomainNS: The Number System
ClusterApply and extend previous understandings of operations with fractions to add, subtract,
multiply, and divide rational numbers.
StandardSolve real-world and mathematical problems involving the four operations with
rational numbers.
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 between 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?
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: Finding the fractions
All together, the students collected 3,760+2,301+1,855=7,916 box tops.
Mr. Aceves’ class collected3760/7916 of the box tops.
Mrs. Baca’s class collected 2301/7916 of the box tops.
Mr. Canyon’s class collected 1855/7916 of the box tops.
The amount for Mr. Aceves’ class is 3760/7916×600≈284.99
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The amount for Mrs. Baca’s class is 2301/7916×600≈174.41
The amount for Mr. Canyon’s class is 1855/7916×600≈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.
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