Everyday Mathematics
This curriculum is a single, unified project that puts students in the role of a Structural Engineer. They will manage a complex building project, tackling problems involving multi-step conversions, complex ratios, and density calculations. The activities are hands-on, practical, and directly tied to the skills measured on the test. Use your workbook to source additional practice problems for each section.
The Engineer's Toolkit: The Complete Supply List
For all lessons: Whiteboard or chalkboard, markers or chalk, student worksheets, one dry-erase board with a marker per student or group, and a simple calculator.
For Multi-Step Conversions: A large pitcher of water, a measuring cup with both metric (milliliters) and imperial (cups/ounces) markings, and a stopwatch.
For Force, Pressure, and Density: A simple kitchen scale, a small wooden block or object, and a ruler.
For Advanced Ratios and Proportions: A bag of two different-colored beads or small objects, and a container.
Lesson 1: Foundation and Materials 📐
Estimated Time: 50-60 minutes
Concepts: Multi-step conversions of area, volume, and weight; problems involving density, pressure, and force.
Objective: Teach students to solve multi-step conversion problems with area, volume, and weight.
Activity: "The Concrete Pour." Present a scenario: "Your crew is pouring concrete for a foundation. You need to fill a mold that is 3 meters long, 2 meters wide, and 0.5 meters deep. The concrete supplier sells concrete by the cubic yard. How many cubic yards of concrete do you need? (Note: 1 meter = 3.28 feet)."
Hands-On Application: "Let's model the conversion." Use the pitcher of water and the measuring cup.
The Math: First, have students calculate the volume in cubic meters: 3m x 2m x 0.5m = 3 cubic meters. Then, have them convert this to cubic feet: 3 m³ x (3.28 ft/m)³ = 105.9 cubic feet. Finally, convert to cubic yards: 105.9 cubic feet / 27 cubic feet per cubic yard = 3.92 cubic yards.
The "How-To": The physical model will demonstrate the concept of converting between units of volume.
a. Tell students: "The metric system and the imperial system are like two different languages for measuring. We're going to practice a simple translation."
b. Have a student pour exactly 200 milliliters of water from the pitcher into the measuring cup.
c. Tell them: "Now, look at the other side of the measuring cup and read the volume in fluid ounces. It should be approximately 6.76 fluid ounces. The abstract conversion (1 milliliter = 0.0338 fluid ounces) is happening right in front of you. This shows that we can convert a volume from one unit to another, just as we did with our concrete problem."
Practice Problems:
Problem A: A water tank has a volume of 4,000 liters. How many gallons is this? (Note: 1 gallon = 3.785 liters).
Problem B: A material weighs 50 kilograms. How many pounds is this? (Note: 1 kg = 2.2 lbs).
Solutions:
Problem A Solution:
Gallons = 4,000 liters / 3.785 liters/gallon = 1,056.8 gallons.
Problem B Solution:
Pounds = 50 kg x 2.2 lbs/kg = 110 lbs.
Objective: Teach students to solve problems involving density, pressure, and force.
Activity: "The Support Beam." Present a scenario: "A support beam with a base area of 20 square inches is holding a weight of 100 pounds. What is the pressure on the beam in pounds per square inch (PSI)?"
Hands-On Application: "Let's feel the pressure." Use the kitchen scale and the wooden block.
The Math: First, have students solve the problem: Pressure = Force / Area = 100 lbs / 20 sq. inches = 5 PSI.
The "How-To": The physical model will demonstrate the components of the formula.
a. Tell students: "We are going to find the pressure that this block exerts on our scale."
b. Have a student place the wooden block on the scale and measure its weight. This is the Force. Let's say it's 2 pounds.
c. Have the student use a ruler to measure the length and width of the base of the block. This allows them to calculate the Area. For example, a base of 4 inches x 2 inches = 8 square inches.
d. Have the student use their measured Force (2 lbs) and their calculated Area (8 sq. inches) to find the pressure: 2 lbs / 8 sq. inches = 0.25 PSI.
e. Tell them: "The formula works! The pressure is the result of the weight (force) being distributed over a specific area."
Practice Problems:
Problem A: A box weighs 25 lbs and has a volume of 5 cubic feet. What is the density of the box?
Problem B: A hydraulic piston exerts a force of 500 lbs on a surface with an area of 10 square inches. What is the pressure in PSI?
Solutions:
Problem A Solution:
Density = Weight / Volume = 25 lbs / 5 cubic feet = 5 lbs/cubic foot.
Problem B Solution:
Pressure = Force / Area = 500 lbs / 10 sq. inches = 50 PSI.
Lesson 2: Quality Control and Logistics 📈
Estimated Time: 40-50 minutes
Concepts: Problems involving a combination of geometric concepts and other math skills; advanced ratios and proportions.
Objective: Teach students to solve problems involving advanced ratios and proportions.
Activity: "The Concrete Mix." Present a scenario: "For every 2 parts cement, you need 3 parts sand. For every 5 parts sand, you need 4 parts gravel. If you have 40 bags of cement, how many bags of gravel do you need?"
Hands-On Application: "Let's model the mix." Use a container and two different-colored beads (e.g., black for cement and yellow for sand).
The Math:
First, find how many bags of sand are needed: (40 bags cement / 2 parts cement) x 3 parts sand = 60 bags of sand.
Now, use the second ratio to find the bags of gravel: (60 bags sand / 5 parts sand) x 4 parts gravel = 48 bags of gravel.
The "How-To": The physical model will demonstrate the multi-step nature of the ratio problem.
a. Tell students: "We're going to model this with beads."
b. To represent the first ratio, have a student place 2 black beads in a container and add 3 yellow beads.
c. Explain the first ratio: 2 parts cement to 3 parts sand.
d. To show the second ratio, have them remove the yellow beads and add 5 yellow beads for every 4 gray beads (gravel). This reinforces the concept of a series of proportions—the amount of sand from the first ratio becomes the input for the second.
Practice Problems:
Problem A: A recipe calls for 3 cups of flour for every 2 cups of sugar. For every 4 cups of sugar, you need 1 cup of butter. If you use 9 cups of flour, how much butter do you need?
Problem B: The ratio of supervisors to employees is 1:10. The ratio of team leaders to supervisors is 1:2. If a company has 120 employees, how many team leaders are there?
Solutions:
Problem A Solution:
Flour to sugar ratio: 3:2. (9 cups flour / 3 parts flour) = 3. So, we need 3 x 2 = 6 cups of sugar.
Sugar to butter ratio: 4:1. (6 cups sugar / 4 parts sugar) = 1.5. So, we need 1.5 x 1 = 1.5 cups of butter.
Problem B Solution:
(120 employees / 10 employees per supervisor) = 12 supervisors.
(12 supervisors / 2 supervisors per team leader) = 6 team leaders.
Objective: Teach students to solve complex problems involving geometric concepts.
Activity: "The Foundation Cost." Present a scenario: "You are pouring a circular concrete foundation with a radius of 10 feet and a depth of 1 foot. The concrete costs $200 per cubic yard. What is the total cost of the concrete? (Note: 1 cubic yard = 27 cubic feet)."
Hands-On Application: "Let's break down the geometry and the math."
The Math:
Volume of a cylinder is V = πr²h.
V = π x (10 ft)² x 1 ft = 314.16 cubic feet.
Convert volume to cubic yards: 314.16 ft³ / 27 ft³/yd³ = 11.64 cubic yards.
Calculate the cost: 11.64 yds³ x $200/yd³ = $2,328.
The "How-To": The physical model will demonstrate the components of the formula.
a. Tell students: "This can of tuna is a cylinder, just like our foundation."
b. Have a student use a ruler to measure the radius (distance from center to edge) and the height of the can.
c. Have them use their measurements to calculate the volume of the can using the same formula: V = πr²h.
d. Tell them: "Just like with our can, the volume of the foundation is found by multiplying the area of the circular base by its height. Once we have the volume, we can figure out the cost."
Practice Problems:
Problem A: You are painting a rectangular wall that is 20 feet long and 8 feet high. A single can of paint covers 40 square meters. How many cans of paint do you need? (Note: 1 meter = 3.28 feet).
Problem B: A cylindrical grain silo has a radius of 10 meters and a height of 20 meters. The grain weighs 50 lbs per cubic foot. What is the total weight of the grain in the silo in pounds? (Note: 1 meter = 3.28 feet).
Solutions:
Problem A Solution:
Wall Area = 20 ft x 8 ft = 160 sq. feet.
Convert area to square meters: 160 sq. ft / (3.28 ft/m)² = 160 / 10.76 = 14.87 sq. meters.
Cans needed = 14.87 sq. meters / 40 sq. meters per can = 0.37 cans, so you need 1 can.
Problem B Solution:
Silo Volume (in cubic meters) = V = πr²h = π x (10m)² x 20m = 6,283.19 cubic meters.
Convert volume to cubic feet: 6,283.19 m³ x (3.28 ft/m)³ = 6,283.19 x 35.31 = 221,800 cubic feet.
Total Weight = 221,800 cu. ft x 50 lbs/cu. ft = 11,090,000 lbs.