Problem Solving with Units - Ninth Grade Math
1. Scale Drawings: Word Problems
Scale Drawing: A drawing that represents an actual object but is either larger or smaller than the actual size
Scale: The ratio between the measurements on a drawing and the actual measurements
Scale Factor: The multiplier used to convert between drawing and actual measurements
Scale: The ratio between the measurements on a drawing and the actual measurements
Scale Factor: The multiplier used to convert between drawing and actual measurements
Scale Formula:
$$\text{Scale} = \frac{\text{Drawing Measurement}}{\text{Actual Measurement}}$$
Common Scale Notations:
• Ratio form: $1:100$ (means 1 unit on drawing = 100 units in real life)
• Equation form: $1 \text{ in} = 5 \text{ ft}$ (1 inch represents 5 feet)
• Fraction form: $\frac{1}{100}$ (drawing is $\frac{1}{100}$ the actual size)
$$\text{Scale} = \frac{\text{Drawing Measurement}}{\text{Actual Measurement}}$$
Common Scale Notations:
• Ratio form: $1:100$ (means 1 unit on drawing = 100 units in real life)
• Equation form: $1 \text{ in} = 5 \text{ ft}$ (1 inch represents 5 feet)
• Fraction form: $\frac{1}{100}$ (drawing is $\frac{1}{100}$ the actual size)
Proportion Method for Scale Drawings:
$$\frac{\text{Drawing Length 1}}{\text{Actual Length 1}} = \frac{\text{Drawing Length 2}}{\text{Actual Length 2}}$$
Finding Actual from Drawing:
$$\text{Actual Length} = \text{Drawing Length} \times \text{Scale Factor}$$
Finding Drawing from Actual:
$$\text{Drawing Length} = \frac{\text{Actual Length}}{\text{Scale Factor}}$$
$$\frac{\text{Drawing Length 1}}{\text{Actual Length 1}} = \frac{\text{Drawing Length 2}}{\text{Actual Length 2}}$$
Finding Actual from Drawing:
$$\text{Actual Length} = \text{Drawing Length} \times \text{Scale Factor}$$
Finding Drawing from Actual:
$$\text{Drawing Length} = \frac{\text{Actual Length}}{\text{Scale Factor}}$$
Steps to Solve Scale Drawing Problems:
Step 1: Identify the scale (ratio)
Step 2: Identify what you're looking for (drawing or actual measurement)
Step 3: Set up a proportion or use multiplication/division
Step 4: Solve for the unknown
Step 5: Check that units match and answer makes sense
Step 1: Identify the scale (ratio)
Step 2: Identify what you're looking for (drawing or actual measurement)
Step 3: Set up a proportion or use multiplication/division
Step 4: Solve for the unknown
Step 5: Check that units match and answer makes sense
Example 1: A scale drawing of a house uses a scale of 1 inch = 4 feet. If the actual width of the house is 32 feet, what is the width on the drawing?
Set up proportion:
$\frac{1 \text{ in}}{4 \text{ ft}} = \frac{x \text{ in}}{32 \text{ ft}}$
Cross-multiply:
$4x = 32$
$x = 8$ inches
Answer: The width on the drawing is 8 inches.
Set up proportion:
$\frac{1 \text{ in}}{4 \text{ ft}} = \frac{x \text{ in}}{32 \text{ ft}}$
Cross-multiply:
$4x = 32$
$x = 8$ inches
Answer: The width on the drawing is 8 inches.
Example 2: On a map with scale $1:50,000$, two cities are 6 cm apart. What is the actual distance in kilometers?
Scale means: 1 cm on map = 50,000 cm in reality
Actual distance: $6 \times 50,000 = 300,000$ cm
Convert to km: $300,000 \text{ cm} = 3,000 \text{ m} = 3$ km
Answer: The actual distance is 3 kilometers.
Scale means: 1 cm on map = 50,000 cm in reality
Actual distance: $6 \times 50,000 = 300,000$ cm
Convert to km: $300,000 \text{ cm} = 3,000 \text{ m} = 3$ km
Answer: The actual distance is 3 kilometers.
Example 3: A blueprint of a room has dimensions 3 inches by 4 inches. The scale is 1 inch = 8 feet. Find the actual dimensions and area of the room.
Actual length: $3 \times 8 = 24$ feet
Actual width: $4 \times 8 = 32$ feet
Actual area: $24 \times 32 = 768$ square feet
Answer: Actual dimensions are 24 ft × 32 ft with area 768 ft²
Actual length: $3 \times 8 = 24$ feet
Actual width: $4 \times 8 = 32$ feet
Actual area: $24 \times 32 = 768$ square feet
Answer: Actual dimensions are 24 ft × 32 ft with area 768 ft²
Important Notes:
• Keep units consistent when setting up proportions
• Scale applies to length, not area or volume
• If scale is 1:n, then area scale is 1:n² and volume scale is 1:n³
• Always check if your answer is reasonable
• Keep units consistent when setting up proportions
• Scale applies to length, not area or volume
• If scale is 1:n, then area scale is 1:n² and volume scale is 1:n³
• Always check if your answer is reasonable
2. Convert Rates and Measurements: Customary Units
Customary System: The measurement system commonly used in the United States (also called Imperial or US Standard)
Conversion Factor: A ratio equal to 1 used to convert between units
Conversion Factor: A ratio equal to 1 used to convert between units
Length Conversions (Customary)
| Conversion | Relationship |
|---|---|
| Inches (in) ↔ Feet (ft) | 1 ft = 12 in |
| Feet (ft) ↔ Yards (yd) | 1 yd = 3 ft |
| Yards (yd) ↔ Miles (mi) | 1 mi = 1,760 yd |
| Feet (ft) ↔ Miles (mi) | 1 mi = 5,280 ft |
| Inches (in) ↔ Yards (yd) | 1 yd = 36 in |
Weight Conversions (Customary)
| Conversion | Relationship |
|---|---|
| Ounces (oz) ↔ Pounds (lb) | 1 lb = 16 oz |
| Pounds (lb) ↔ Tons (T) | 1 T = 2,000 lb |
Capacity/Volume Conversions (Customary)
| Conversion | Relationship |
|---|---|
| Fluid Ounces (fl oz) ↔ Cups (c) | 1 c = 8 fl oz |
| Cups (c) ↔ Pints (pt) | 1 pt = 2 c |
| Pints (pt) ↔ Quarts (qt) | 1 qt = 2 pt |
| Quarts (qt) ↔ Gallons (gal) | 1 gal = 4 qt |
| Fluid Ounces (fl oz) ↔ Pints (pt) | 1 pt = 16 fl oz |
| Cups (c) ↔ Gallons (gal) | 1 gal = 16 c |
Time Conversions
| Conversion | Relationship |
|---|---|
| Seconds (sec) ↔ Minutes (min) | 1 min = 60 sec |
| Minutes (min) ↔ Hours (hr) | 1 hr = 60 min |
| Hours (hr) ↔ Days (d) | 1 d = 24 hr |
| Days (d) ↔ Weeks (wk) | 1 wk = 7 d |
Conversion Method:
Multiply by conversion factor: When converting to a smaller unit
Divide by conversion factor: When converting to a larger unit
Alternative: Use dimensional analysis (multiply by conversion fractions)
Multiply by conversion factor: When converting to a smaller unit
Divide by conversion factor: When converting to a larger unit
Alternative: Use dimensional analysis (multiply by conversion fractions)
Example 1: Convert 3.5 feet to inches.
$3.5 \text{ ft} \times \frac{12 \text{ in}}{1 \text{ ft}} = 42$ inches
Answer: 42 inches
$3.5 \text{ ft} \times \frac{12 \text{ in}}{1 \text{ ft}} = 42$ inches
Answer: 42 inches
Example 2: Convert 80 ounces to pounds.
$80 \text{ oz} \times \frac{1 \text{ lb}}{16 \text{ oz}} = 5$ pounds
Answer: 5 pounds
$80 \text{ oz} \times \frac{1 \text{ lb}}{16 \text{ oz}} = 5$ pounds
Answer: 5 pounds
Example 3: Convert a speed of 60 miles per hour to feet per second.
$60 \frac{\text{mi}}{\text{hr}} \times \frac{5,280 \text{ ft}}{1 \text{ mi}} \times \frac{1 \text{ hr}}{3,600 \text{ sec}}$
$= \frac{60 \times 5,280}{3,600} = \frac{316,800}{3,600} = 88$ ft/sec
Answer: 88 feet per second
$60 \frac{\text{mi}}{\text{hr}} \times \frac{5,280 \text{ ft}}{1 \text{ mi}} \times \frac{1 \text{ hr}}{3,600 \text{ sec}}$
$= \frac{60 \times 5,280}{3,600} = \frac{316,800}{3,600} = 88$ ft/sec
Answer: 88 feet per second
3. Convert Rates and Measurements: Metric Units
Metric System: A decimal system of measurement used worldwide (SI units)
Prefixes: Used to indicate powers of 10 (kilo-, hecto-, deka-, deci-, centi-, milli-)
Prefixes: Used to indicate powers of 10 (kilo-, hecto-, deka-, deci-, centi-, milli-)
Metric Prefixes
| Prefix | Symbol | Meaning | Multiplier |
|---|---|---|---|
| kilo- | k | thousand | $\times 1,000$ or $10^3$ |
| hecto- | h | hundred | $\times 100$ or $10^2$ |
| deka- | da | ten | $\times 10$ or $10^1$ |
| BASE UNIT | - | one | $\times 1$ or $10^0$ |
| deci- | d | tenth | $\times 0.1$ or $10^{-1}$ |
| centi- | c | hundredth | $\times 0.01$ or $10^{-2}$ |
| milli- | m | thousandth | $\times 0.001$ or $10^{-3}$ |
Metric Length Conversions
| Unit | Abbreviation | Relationship to Meter |
|---|---|---|
| Kilometer | km | 1 km = 1,000 m |
| Hectometer | hm | 1 hm = 100 m |
| Dekameter | dam | 1 dam = 10 m |
| Meter (BASE) | m | 1 m |
| Decimeter | dm | 1 m = 10 dm |
| Centimeter | cm | 1 m = 100 cm |
| Millimeter | mm | 1 m = 1,000 mm |
Metric Mass/Weight Conversions
| Unit | Abbreviation | Relationship to Gram |
|---|---|---|
| Kilogram | kg | 1 kg = 1,000 g |
| Gram (BASE) | g | 1 g |
| Centigram | cg | 1 g = 100 cg |
| Milligram | mg | 1 g = 1,000 mg |
| Metric Ton | t | 1 t = 1,000 kg |
Metric Capacity/Volume Conversions
| Unit | Abbreviation | Relationship to Liter |
|---|---|---|
| Kiloliter | kL | 1 kL = 1,000 L |
| Liter (BASE) | L | 1 L |
| Centiliter | cL | 1 L = 100 cL |
| Milliliter | mL | 1 L = 1,000 mL |
Metric Conversion Trick (King Henry Died By Drinking Chocolate Milk):
Kilo → Hecto → Deka → Base → Deci → Centi → Milli
• Move right: multiply by 10 for each step
• Move left: divide by 10 for each step
Kilo → Hecto → Deka → Base → Deci → Centi → Milli
• Move right: multiply by 10 for each step
• Move left: divide by 10 for each step
Example 1: Convert 3.5 kilometers to meters.
Move 3 places right (km → hm → dam → m)
$3.5 \text{ km} = 3.5 \times 1,000 = 3,500$ m
Answer: 3,500 meters
Move 3 places right (km → hm → dam → m)
$3.5 \text{ km} = 3.5 \times 1,000 = 3,500$ m
Answer: 3,500 meters
Example 2: Convert 450 milliliters to liters.
Move 3 places left (mL → cL → dL → L)
$450 \text{ mL} = 450 \div 1,000 = 0.45$ L
Answer: 0.45 liters
Move 3 places left (mL → cL → dL → L)
$450 \text{ mL} = 450 \div 1,000 = 0.45$ L
Answer: 0.45 liters
Example 3: Convert 7.5 grams to milligrams.
Move 3 places right (g → dg → cg → mg)
$7.5 \text{ g} = 7.5 \times 1,000 = 7,500$ mg
Answer: 7,500 milligrams
Move 3 places right (g → dg → cg → mg)
$7.5 \text{ g} = 7.5 \times 1,000 = 7,500$ mg
Answer: 7,500 milligrams
Helpful Relationships:
• 1 mL = 1 cm³ (cubic centimeter)
• 1 L = 1,000 cm³
• 1 kg ≈ 2.2 lb (customary)
• 1 inch ≈ 2.54 cm
• 1 mile ≈ 1.61 km
• 1 mL = 1 cm³ (cubic centimeter)
• 1 L = 1,000 cm³
• 1 kg ≈ 2.2 lb (customary)
• 1 inch ≈ 2.54 cm
• 1 mile ≈ 1.61 km
4. Unit Prices with Unit Conversions
Unit Price: The cost per one unit of measure (per pound, per liter, per item, etc.)
Also called: Unit rate, price per unit
Also called: Unit rate, price per unit
Unit Price Formula:
$$\text{Unit Price} = \frac{\text{Total Cost}}{\text{Total Quantity}}$$
With Unit Conversion:
1. Convert quantity to desired unit
2. Calculate unit price using converted quantity
$$\text{Unit Price} = \frac{\text{Total Cost}}{\text{Total Quantity}}$$
With Unit Conversion:
1. Convert quantity to desired unit
2. Calculate unit price using converted quantity
Steps for Unit Price with Conversion:
Step 1: Identify the total cost and quantity
Step 2: Convert the quantity to the desired unit
Step 3: Divide total cost by converted quantity
Step 4: Write answer with appropriate units ($/unit)
Step 1: Identify the total cost and quantity
Step 2: Convert the quantity to the desired unit
Step 3: Divide total cost by converted quantity
Step 4: Write answer with appropriate units ($/unit)
Example 1: A 32-ounce bottle of juice costs $4.80. Find the unit price per pound.
Step 1: Cost = $4.80, Quantity = 32 oz
Step 2: Convert: $32 \text{ oz} = 32 \div 16 = 2$ lb
Step 3: Unit price = $\frac{\$4.80}{2 \text{ lb}} = \$2.40$ per lb
Answer: $2.40 per pound
Step 1: Cost = $4.80, Quantity = 32 oz
Step 2: Convert: $32 \text{ oz} = 32 \div 16 = 2$ lb
Step 3: Unit price = $\frac{\$4.80}{2 \text{ lb}} = \$2.40$ per lb
Answer: $2.40 per pound
Example 2: A 500-gram package of coffee costs $12.50. Find the unit price per kilogram.
Convert: $500 \text{ g} = 500 \div 1,000 = 0.5$ kg
Unit price: $\frac{\$12.50}{0.5 \text{ kg}} = \$25$ per kg
Answer: $25 per kilogram
Convert: $500 \text{ g} = 500 \div 1,000 = 0.5$ kg
Unit price: $\frac{\$12.50}{0.5 \text{ kg}} = \$25$ per kg
Answer: $25 per kilogram
Example 3: Which is the better buy?
• Option A: 24 fl oz for $3.60
• Option B: 1 gallon for $18.00
Option A: $\frac{\$3.60}{24 \text{ fl oz}} = \$0.15$ per fl oz
Option B: Convert 1 gal = 128 fl oz
$\frac{\$18.00}{128 \text{ fl oz}} = \$0.1406$ per fl oz ≈ $0.14 per fl oz
Answer: Option B is the better buy ($0.14 vs $0.15 per fl oz)
• Option A: 24 fl oz for $3.60
• Option B: 1 gallon for $18.00
Option A: $\frac{\$3.60}{24 \text{ fl oz}} = \$0.15$ per fl oz
Option B: Convert 1 gal = 128 fl oz
$\frac{\$18.00}{128 \text{ fl oz}} = \$0.1406$ per fl oz ≈ $0.14 per fl oz
Answer: Option B is the better buy ($0.14 vs $0.15 per fl oz)
Example 4: Gasoline costs $3.50 per gallon. What is the cost per quart?
Given: 1 gallon = 4 quarts
Cost per quart: $\frac{\$3.50}{4} = \$0.875$ per quart
Answer: $0.875 or 87.5¢ per quart
Given: 1 gallon = 4 quarts
Cost per quart: $\frac{\$3.50}{4} = \$0.875$ per quart
Answer: $0.875 or 87.5¢ per quart
5. Multi-Step Problems with Unit Conversions
Multi-Step Problems: Problems that require multiple operations and/or conversions to solve
Dimensional Analysis: A method of converting units by multiplying by conversion factors
Dimensional Analysis: A method of converting units by multiplying by conversion factors
Strategy for Multi-Step Problems:
Step 1: Read carefully and identify what you need to find
Step 2: List all given information and needed conversions
Step 3: Break the problem into smaller steps
Step 4: Perform calculations in order
Step 5: Check units and reasonableness
Step 1: Read carefully and identify what you need to find
Step 2: List all given information and needed conversions
Step 3: Break the problem into smaller steps
Step 4: Perform calculations in order
Step 5: Check units and reasonableness
Dimensional Analysis Method:
Multiply by conversion fractions so unwanted units cancel:
$$\text{Start Value} \times \frac{\text{Want}}{\text{Have}} \times \frac{\text{Want}}{\text{Have}} = \text{Final Value}$$
Multiply by conversion fractions so unwanted units cancel:
$$\text{Start Value} \times \frac{\text{Want}}{\text{Have}} \times \frac{\text{Want}}{\text{Have}} = \text{Final Value}$$
Example 1: A recipe calls for 2.5 cups of flour per batch. If you want to make 6 batches and flour costs $0.25 per pound, how much will the flour cost? (1 cup flour ≈ 0.25 lb)
Step 1: Total cups needed = $2.5 \times 6 = 15$ cups
Step 2: Convert to pounds: $15 \text{ cups} \times 0.25 \frac{\text{lb}}{\text{cup}} = 3.75$ lb
Step 3: Calculate cost: $3.75 \text{ lb} \times \$0.25/\text{lb} = \$0.9375 \approx \$0.94$
Answer: The flour will cost approximately $0.94
Step 1: Total cups needed = $2.5 \times 6 = 15$ cups
Step 2: Convert to pounds: $15 \text{ cups} \times 0.25 \frac{\text{lb}}{\text{cup}} = 3.75$ lb
Step 3: Calculate cost: $3.75 \text{ lb} \times \$0.25/\text{lb} = \$0.9375 \approx \$0.94$
Answer: The flour will cost approximately $0.94
Example 2: A garden hose fills a pool at a rate of 8 gallons per minute. How many liters will it fill in 45 minutes? (1 gallon ≈ 3.785 L)
Step 1: Gallons filled = $8 \frac{\text{gal}}{\text{min}} \times 45 \text{ min} = 360$ gallons
Step 2: Convert to liters: $360 \text{ gal} \times 3.785 \frac{\text{L}}{\text{gal}} = 1,362.6$ L
Answer: 1,362.6 liters
Step 1: Gallons filled = $8 \frac{\text{gal}}{\text{min}} \times 45 \text{ min} = 360$ gallons
Step 2: Convert to liters: $360 \text{ gal} \times 3.785 \frac{\text{L}}{\text{gal}} = 1,362.6$ L
Answer: 1,362.6 liters
Example 3: A car travels 420 miles using 15 gallons of gas. What is the fuel efficiency in kilometers per liter? (1 mile ≈ 1.61 km, 1 gal ≈ 3.785 L)
Method: Dimensional Analysis
$\frac{420 \text{ mi}}{15 \text{ gal}} \times \frac{1.61 \text{ km}}{1 \text{ mi}} \times \frac{1 \text{ gal}}{3.785 \text{ L}}$
$= \frac{420 \times 1.61}{15 \times 3.785} = \frac{676.2}{56.775} \approx 11.91$ km/L
Answer: Approximately 11.91 kilometers per liter
Method: Dimensional Analysis
$\frac{420 \text{ mi}}{15 \text{ gal}} \times \frac{1.61 \text{ km}}{1 \text{ mi}} \times \frac{1 \text{ gal}}{3.785 \text{ L}}$
$= \frac{420 \times 1.61}{15 \times 3.785} = \frac{676.2}{56.775} \approx 11.91$ km/L
Answer: Approximately 11.91 kilometers per liter
Example 4: A rectangular room is 18 feet by 12 feet. Carpeting costs $25 per square yard. What will it cost to carpet the room?
Step 1: Area in sq ft = $18 \times 12 = 216$ sq ft
Step 2: Convert: 1 yd = 3 ft, so 1 sq yd = 9 sq ft
$216 \text{ sq ft} \div 9 = 24$ sq yd
Step 3: Cost = $24 \text{ sq yd} \times \$25/\text{sq yd} = \$600$
Answer: $600
Step 1: Area in sq ft = $18 \times 12 = 216$ sq ft
Step 2: Convert: 1 yd = 3 ft, so 1 sq yd = 9 sq ft
$216 \text{ sq ft} \div 9 = 24$ sq yd
Step 3: Cost = $24 \text{ sq yd} \times \$25/\text{sq yd} = \$600$
Answer: $600
6. Rate of Travel: Word Problems
Rate: A ratio that compares two quantities with different units
Speed/Rate of Travel: Distance traveled per unit of time
Average Speed: Total distance divided by total time
Speed/Rate of Travel: Distance traveled per unit of time
Average Speed: Total distance divided by total time
Distance-Rate-Time Formula (D = RT):
$$\text{Distance} = \text{Rate} \times \text{Time}$$
$$d = rt$$
Rearranged Forms:
$$\text{Rate} = \frac{\text{Distance}}{\text{Time}} \quad (r = \frac{d}{t})$$
$$\text{Time} = \frac{\text{Distance}}{\text{Rate}} \quad (t = \frac{d}{r})$$
$$\text{Distance} = \text{Rate} \times \text{Time}$$
$$d = rt$$
Rearranged Forms:
$$\text{Rate} = \frac{\text{Distance}}{\text{Time}} \quad (r = \frac{d}{t})$$
$$\text{Time} = \frac{\text{Distance}}{\text{Rate}} \quad (t = \frac{d}{r})$$
Remember D-R-T Triangle:
Draw a triangle with D on top, R and T on bottom
• To find D: multiply R × T
• To find R: divide D ÷ T
• To find T: divide D ÷ R
Draw a triangle with D on top, R and T on bottom
• To find D: multiply R × T
• To find R: divide D ÷ T
• To find T: divide D ÷ R
Steps for Rate of Travel Problems:
Step 1: Identify what you're looking for (d, r, or t)
Step 2: Write down given information
Step 3: Choose the appropriate formula
Step 4: Substitute and solve
Step 5: Check units and reasonableness
Step 1: Identify what you're looking for (d, r, or t)
Step 2: Write down given information
Step 3: Choose the appropriate formula
Step 4: Substitute and solve
Step 5: Check units and reasonableness
Example 1: A car travels at 65 miles per hour for 3.5 hours. How far does it travel?
Given: $r = 65$ mph, $t = 3.5$ hours
Find: Distance
Formula: $d = rt$
$d = 65 \times 3.5 = 227.5$ miles
Answer: 227.5 miles
Given: $r = 65$ mph, $t = 3.5$ hours
Find: Distance
Formula: $d = rt$
$d = 65 \times 3.5 = 227.5$ miles
Answer: 227.5 miles
Example 2: A train travels 240 kilometers in 3 hours. What is its average speed?
Given: $d = 240$ km, $t = 3$ hours
Find: Rate
Formula: $r = \frac{d}{t}$
$r = \frac{240}{3} = 80$ km/h
Answer: 80 kilometers per hour
Given: $d = 240$ km, $t = 3$ hours
Find: Rate
Formula: $r = \frac{d}{t}$
$r = \frac{240}{3} = 80$ km/h
Answer: 80 kilometers per hour
Example 3: How long will it take to travel 450 miles at an average speed of 60 mph?
Given: $d = 450$ miles, $r = 60$ mph
Find: Time
Formula: $t = \frac{d}{r}$
$t = \frac{450}{60} = 7.5$ hours
Answer: 7.5 hours (or 7 hours 30 minutes)
Given: $d = 450$ miles, $r = 60$ mph
Find: Time
Formula: $t = \frac{d}{r}$
$t = \frac{450}{60} = 7.5$ hours
Answer: 7.5 hours (or 7 hours 30 minutes)
Example 4 (Two travelers): Two cars leave from the same point, traveling in opposite directions. One travels at 50 mph and the other at 60 mph. How far apart will they be after 2 hours?
Car 1 distance: $d_1 = 50 \times 2 = 100$ miles
Car 2 distance: $d_2 = 60 \times 2 = 120$ miles
Total distance apart: $100 + 120 = 220$ miles
Answer: 220 miles apart
Car 1 distance: $d_1 = 50 \times 2 = 100$ miles
Car 2 distance: $d_2 = 60 \times 2 = 120$ miles
Total distance apart: $100 + 120 = 220$ miles
Answer: 220 miles apart
Example 5 (Round trip): Sarah drives to work at 40 mph and returns home at 60 mph. The total driving time is 2 hours. How far is her work from home?
Let: $d$ = distance to work
Time to work: $t_1 = \frac{d}{40}$
Time returning: $t_2 = \frac{d}{60}$
Total time: $\frac{d}{40} + \frac{d}{60} = 2$
Solve: Multiply by LCD (120):
$3d + 2d = 240$
$5d = 240$
$d = 48$ miles
Answer: 48 miles from home to work
Let: $d$ = distance to work
Time to work: $t_1 = \frac{d}{40}$
Time returning: $t_2 = \frac{d}{60}$
Total time: $\frac{d}{40} + \frac{d}{60} = 2$
Solve: Multiply by LCD (120):
$3d + 2d = 240$
$5d = 240$
$d = 48$ miles
Answer: 48 miles from home to work
7. Weighted Averages: Word Problems
Weighted Average: An average where some values contribute more than others based on their "weight" or importance
Weight: A factor that indicates the relative importance or frequency of a value
Weight: A factor that indicates the relative importance or frequency of a value
Weighted Average Formula:
$$\text{Weighted Average} = \frac{\sum (\text{Value} \times \text{Weight})}{\sum \text{Weights}}$$
Expanded Form:
$$\overline{x}_w = \frac{x_1w_1 + x_2w_2 + x_3w_3 + \cdots + x_nw_n}{w_1 + w_2 + w_3 + \cdots + w_n}$$
where $x_i$ = values and $w_i$ = weights
$$\text{Weighted Average} = \frac{\sum (\text{Value} \times \text{Weight})}{\sum \text{Weights}}$$
Expanded Form:
$$\overline{x}_w = \frac{x_1w_1 + x_2w_2 + x_3w_3 + \cdots + x_nw_n}{w_1 + w_2 + w_3 + \cdots + w_n}$$
where $x_i$ = values and $w_i$ = weights
Steps for Weighted Average Problems:
Step 1: Identify the values and their weights
Step 2: Multiply each value by its weight
Step 3: Add all the weighted values
Step 4: Add all the weights
Step 5: Divide total weighted values by total weights
Step 1: Identify the values and their weights
Step 2: Multiply each value by its weight
Step 3: Add all the weighted values
Step 4: Add all the weights
Step 5: Divide total weighted values by total weights
Example 1: In a class, there are 15 girls with an average age of 14 years and 10 boys with an average age of 15 years. What is the average age of the entire class?
Weighted Average:
$= \frac{(15 \times 14) + (10 \times 15)}{15 + 10}$
$= \frac{210 + 150}{25}$
$= \frac{360}{25} = 14.4$ years
Answer: 14.4 years
Weighted Average:
$= \frac{(15 \times 14) + (10 \times 15)}{15 + 10}$
$= \frac{210 + 150}{25}$
$= \frac{360}{25} = 14.4$ years
Answer: 14.4 years
Example 2: A student's grade is calculated as follows: homework (20%), quizzes (30%), and final exam (50%). If the student scored 85 on homework, 78 on quizzes, and 92 on the final exam, what is the overall grade?
Convert percentages to decimals:
$= (85 \times 0.20) + (78 \times 0.30) + (92 \times 0.50)$
$= 17 + 23.4 + 46$
$= 86.4$
Answer: 86.4% overall grade
Convert percentages to decimals:
$= (85 \times 0.20) + (78 \times 0.30) + (92 \times 0.50)$
$= 17 + 23.4 + 46$
$= 86.4$
Answer: 86.4% overall grade
Example 3: A store sells apples at $2 per pound and oranges at $3 per pound. If 40 pounds of apples and 30 pounds of oranges are sold, what is the weighted average price per pound?
Weighted Average:
$= \frac{(40 \times 2) + (30 \times 3)}{40 + 30}$
$= \frac{80 + 90}{70}$
$= \frac{170}{70} \approx 2.43$ dollars per pound
Answer: $2.43 per pound
Weighted Average:
$= \frac{(40 \times 2) + (30 \times 3)}{40 + 30}$
$= \frac{80 + 90}{70}$
$= \frac{170}{70} \approx 2.43$ dollars per pound
Answer: $2.43 per pound
Example 4: An investor puts $5,000 in stock A earning 8% annually and $8,000 in stock B earning 6% annually. What is the weighted average annual return rate?
Weighted Average:
$= \frac{(5,000 \times 8\%) + (8,000 \times 6\%)}{5,000 + 8,000}$
$= \frac{(5,000 \times 0.08) + (8,000 \times 0.06)}{13,000}$
$= \frac{400 + 480}{13,000}$
$= \frac{880}{13,000} \approx 0.0677 = 6.77\%$
Answer: 6.77% annual return
Weighted Average:
$= \frac{(5,000 \times 8\%) + (8,000 \times 6\%)}{5,000 + 8,000}$
$= \frac{(5,000 \times 0.08) + (8,000 \times 0.06)}{13,000}$
$= \frac{400 + 480}{13,000}$
$= \frac{880}{13,000} \approx 0.0677 = 6.77\%$
Answer: 6.77% annual return
Example 5 (Mixture Problem): How many liters of 20% acid solution must be mixed with 5 liters of 50% acid solution to get a 30% acid solution?
Let: $x$ = liters of 20% solution
Equation:
$(0.20)(x) + (0.50)(5) = (0.30)(x + 5)$
$0.20x + 2.5 = 0.30x + 1.5$
$2.5 - 1.5 = 0.30x - 0.20x$
$1 = 0.10x$
$x = 10$ liters
Answer: 10 liters of 20% solution
Let: $x$ = liters of 20% solution
Equation:
$(0.20)(x) + (0.50)(5) = (0.30)(x + 5)$
$0.20x + 2.5 = 0.30x + 1.5$
$2.5 - 1.5 = 0.30x - 0.20x$
$1 = 0.10x$
$x = 10$ liters
Answer: 10 liters of 20% solution
Common Weighted Average Applications:
• Grade calculations (homework, tests, exams with different weights)
• Average prices with different quantities
• Combined averages from different groups
• Mixture problems (concentrations, percentages)
• Investment returns with different amounts
• Grade calculations (homework, tests, exams with different weights)
• Average prices with different quantities
• Combined averages from different groups
• Mixture problems (concentrations, percentages)
• Investment returns with different amounts
Quick Reference Guide
Scale Drawings:
• Scale = $\frac{\text{Drawing}}{\text{Actual}}$
• Set up proportions to solve
• Area scale = (Linear scale)²
• Volume scale = (Linear scale)³
• Scale = $\frac{\text{Drawing}}{\text{Actual}}$
• Set up proportions to solve
• Area scale = (Linear scale)²
• Volume scale = (Linear scale)³
Key Customary Conversions:
• Length: 1 ft = 12 in, 1 yd = 3 ft, 1 mi = 5,280 ft
• Weight: 1 lb = 16 oz, 1 T = 2,000 lb
• Capacity: 1 gal = 4 qt = 8 pt = 16 c = 128 fl oz
• Time: 1 hr = 60 min, 1 min = 60 sec
• Length: 1 ft = 12 in, 1 yd = 3 ft, 1 mi = 5,280 ft
• Weight: 1 lb = 16 oz, 1 T = 2,000 lb
• Capacity: 1 gal = 4 qt = 8 pt = 16 c = 128 fl oz
• Time: 1 hr = 60 min, 1 min = 60 sec
Key Metric Conversions:
• Kilo: ×1,000 (km, kg, kL)
• Hecto: ×100 (hm, hg, hL)
• Deka: ×10 (dam, dag, daL)
• BASE: m, g, L
• Deci: ÷10 (dm, dg, dL)
• Centi: ÷100 (cm, cg, cL)
• Milli: ÷1,000 (mm, mg, mL)
• Kilo: ×1,000 (km, kg, kL)
• Hecto: ×100 (hm, hg, hL)
• Deka: ×10 (dam, dag, daL)
• BASE: m, g, L
• Deci: ÷10 (dm, dg, dL)
• Centi: ÷100 (cm, cg, cL)
• Milli: ÷1,000 (mm, mg, mL)
Unit Price:
$$\text{Unit Price} = \frac{\text{Total Cost}}{\text{Total Quantity}}$$
Remember to convert units first if needed!
$$\text{Unit Price} = \frac{\text{Total Cost}}{\text{Total Quantity}}$$
Remember to convert units first if needed!
Distance-Rate-Time:
• $d = rt$ (Distance = Rate × Time)
• $r = \frac{d}{t}$ (Rate = Distance ÷ Time)
• $t = \frac{d}{r}$ (Time = Distance ÷ Rate)
• $d = rt$ (Distance = Rate × Time)
• $r = \frac{d}{t}$ (Rate = Distance ÷ Time)
• $t = \frac{d}{r}$ (Time = Distance ÷ Rate)
Weighted Average:
$$\text{Weighted Avg} = \frac{\sum (\text{Value} \times \text{Weight})}{\sum \text{Weights}}$$
Multiply each value by its weight, add all products, divide by sum of weights
$$\text{Weighted Avg} = \frac{\sum (\text{Value} \times \text{Weight})}{\sum \text{Weights}}$$
Multiply each value by its weight, add all products, divide by sum of weights
Dimensional Analysis Steps:
1. Start with given value
2. Multiply by conversion fractions
3. Cancel units that appear in numerator and denominator
4. Simplify to get answer in desired units
1. Start with given value
2. Multiply by conversion fractions
3. Cancel units that appear in numerator and denominator
4. Simplify to get answer in desired units
Problem-Solving Success Tips:
✓ Always write down what you know and what you need to find
✓ Keep track of units throughout your calculations
✓ Check that your answer makes sense in context
✓ Use dimensional analysis for complex conversions
✓ Draw diagrams for scale and distance problems
✓ Set up proportions carefully with matching units
✓ Double-check conversion factors
✓ For weighted averages, ensure weights add up correctly
✓ Always write down what you know and what you need to find
✓ Keep track of units throughout your calculations
✓ Check that your answer makes sense in context
✓ Use dimensional analysis for complex conversions
✓ Draw diagrams for scale and distance problems
✓ Set up proportions carefully with matching units
✓ Double-check conversion factors
✓ For weighted averages, ensure weights add up correctly