Pre-Algebra Measurement and Unit Conversion Review

Homework Solutions and Conversion Fundamentals (Page 75, Section 2.4)

• Metric System Mnemonic & Unit Identification Rules:

  • The standard mnemonic for metric prefixes is "King Henry Doesn't Usually Drink Chocolate Milk" ($K-H-D-U-D-C-M$):
    • $K$: Kilo-
    • $H$: Hecto-
    • $D$: Deca-
    • $U$: Base Unit ($g$ for grams, $m$ for meters, $L$ for liters)
    • $D$: Deci-
    • $C$: Centi-
    • $M$: Milli-
  • When a single letter stands alone (e.g., $g$, $m$, or $l$), it represents the base unit ($U$) in the conversion table.
  • Prefix differentiation: A metric symbol without an $a$ (e.g., $dL$) indicates deci- (deciliters), whereas a symbol with an $a$ (e.g., $daL$ or $deca$) indicates deca- (decaliters).

• Page 75, Problems 1–24 Solutions & Explanations:

  • Problem 1: $20\,000$
  • Problem 2: $957\,g = 957\,000\,mg$. To convert grams to milligrams, move the decimal point $3$ spaces to the right and add three zeros.
  • Problem 3: Unit concept identification (single letter represents the base unit).
  • Problem 4: $4.1$
  • Problem 5: $5\,193\,mL = 0.5193\,daL$. Converting from milliliters ($mL$) to decaliters ($daL$) requires shifting the decimal point $4$ positions to the left.
  • Problem 6: $1020.7$
  • Problem 7: $160\,dL$ (deciliters).
  • Problem 8: $0.073$
  • Problem 9: $78\,600$. Converting hectometers to decimeters requires moving the decimal point $3$ places to the right ($786.00 \rightarrow 78\,600$).
  • Problem 10: $62\,dg = 0.0062\,kg$
  • Problem 11: $9\,mm$ (millimeters)
  • Problem 12: $1.24$ / $0.04$
  • Problem 13: $23\,hg = 2\,300\,g$
  • Problem 14: $0.004\,cm = 0.00004\,m$. Moving from centimeters to meters requires shifting the decimal point $2$ places to the left, resulting in exactly $4$ zeros between the decimal point and the digit $4$ ($0.00004$).
  • Problem 15: $2.725\,km = 2\,725\,000\,mm$
  • Problem 16: $68\,lb = 1\,088\,oz$
    • Conversion Factor: $16\,oz = 1\,lb$
    • Calculation: $68 \times 16 = 1\,088\,oz$
  • Problem 17: $9\,ft = 108\,in$
  • Problem 18: $93\,qt = 23\frac{1}{4}\,gal$
  • Problem 19: $5\,\text{weeks} = 35\,\text{days}$
  • Problem 20: $372\,fl\,oz = 46\frac{1}{2}\,cups$
  • Problem 21: $7\,\text{years} = 84\,\text{months}$
  • Problem 22: $20\,yd\,18\,in$
  • Problem 23: Mixed Time Subtraction: $23\,hr\,13\,min - 19\,hr\,42\,min$
    • Step 1: Subtracting $42\,min$ from $13\,min$ is not possible directly. Borrow $1\,hr$ ($60\,min$) from $23\,hr$, reducing it to $22\,hr$.
    • Step 2: Add $60\,min$ to $13\,min$ to get $73\,min$.
    • Step 3: Subtract the minutes: $73\,min - 42\,min = 31\,min$
    • Step 4: Subtract the hours: $22\,hr - 19\,hr = 3\,hr$
    • Final Answer: $3\,hr\,31\,min$
  • Problem 24: $3\,\text{weeks}\,4\,\text{days}$

Classwork Review: Key Terminology and Conversion Rules (Page 78)

• Page 78, Problems 1–10 Concepts & Answers:
  • Problem 1: When converting from a larger unit to a smaller unit, the required arithmetic operation is multiplication.
  • Problem 2: The conversion factor for seconds to minutes is written as the ratio $\frac{60\,\text{sec}}{1\,\text{min}}$.
  • Problem 3: Leap Year Rules:
    • A standard leap year must be evenly divisible by $4$.
    • Centurial Exception: If a year ends in $00$ (a century year), it must be evenly divisible by $400$ to be classified as a leap year.
  • Problem 4: The conversion factor relating fluid ounces to cups is $\frac{8\,fl\,oz}{1\,cup}$.
  • Problem 5: The basic unit for liquid capacity/volume in the metric system is the liter.
  • Problem 6: Standard capacity equivalencies: $1\,gal = 4\,qt$ and $1\,qt = 2\,pt$.
  • Problem 7: A fraction equal to $1$ that is used to convert units is called a conversion factor.
  • Problem 8: A measurement containing two or more different units of measure (e.g., $3\,hr\,31\,min$) is called a mixed measure.
  • Problem 9: When converting from grams ($g$) to centigrams ($cg$), move the decimal point to the right.
  • Problem 10: When converting from deciliters ($dL$) to kiloliters ($kL$), move the decimal point to the left.

Systematic Dimensional Analysis Principles and Worked Examples

• Definition & Core Objective:

  • Dimensional Analysis is a structured algebraic method of converting units by setting up a series of conversion factors as ratios such that unwanted units cancel diagonally out of the numerator and denominator, leaving only the target unit.

• Step-by-Step Dimensional Analysis Procedure:

  • Step 1: Identify all necessary conversion factors bridging the given unit to the target unit.
  • Step 2: Express the given quantity as a fraction over $1$.
  • Step 3: Multiply by the conversion factor ratio(s), arranging the units such that the unit to be eliminated appears in the opposite position (denominator vs. numerator).
  • Step 4: Cancel common units diagonally.
  • Step 5: Simplify numerical values through cross-cancellation and multiply remaining values across numerators and denominators.

• Detailed Exemplar Problem: Convert $32\,000\,oz$ to Tons ($tn$):

  • Step 1: Identify Conversion Factors
    • $\text{Factor 1}: 16\,oz = 1\,lb$
    • $\text{Factor 2}: 2\,000\,lb = 1\,tn$
  • Step 2 & 3: Set up Expression$\frac{32\,000\,oz}{1} \times \frac{1\,lb}{16\,oz} \times \frac{1\,tn}{2\,000\,lb}$
  • Step 4: Unit Cancellation
    • Ounces ($oz$) in numerator and denominator cancel out.
    • Pounds ($lb$) in numerator and denominator cancel out.
    • Remaining unit: Tons ($tn$).
  • Step 5: Numerical Simplification
    • Cancel three zeros from $32\,000$ and $2\,000$ (dividing both by $1\,000$):             $\frac{32}{1} \times \frac{1}{16} \times \frac{1\,tn}{2}$
    • Simplify $32$ in numerator with $2$ in denominator ($32 \div 2 = 16$):             $\frac{16}{1} \times \frac{1}{16} \times \frac{1\,tn}{1}$
    • Simplify $16$ in numerator with $16$ in denominator ($16 \div 16 = 1$):             $\frac{1}{1} \times \frac{1}{1} \times \frac{1\,tn}{1} = 1\,tn$
  • Final Result: $32\,000\,oz = 1\,tn$

Page 78 Guided Practice Problems (Numbers 21–23)

• Problem 21: Convert $64\,pt$ to Pecks ($pk$):

  • Conversion Factors: $2\,pt = 1\,qt$, $8\,qt = 1\,pk$
  • Setup: $\frac{64\,pt}{1} \times \frac{1\,qt}{2\,pt} \times \frac{1\,pk}{8\,qt}$
  • Execution: Cancel $pt$ and $qt$. Multiply denominators: $2 \times 8 = 16$. Divide numerator by denominator: $\frac{64}{16} = 4\,pk$
  • Answer: $4\,pk$

• Problem 22: Convert $16\,qt$ to Cups ($c$):

  • Conversion Factors: $1\,qt = 2\,pt$, $1\,pt = 2\,c$
  • Setup: $\frac{16\,qt}{1} \times \frac{2\,pt}{1\,qt} \times \frac{2\,c}{1\,pt}$
  • Execution: Cancel $qt$ and $pt$. Multiply numerators: $16 \times 2 \times 2 = 64\,c$
  • Answer: $64\,c$

• Problem 23: Convert $1\,mi$ to Inches ($in$):

  • Conversion Factors: $1\,mi = 5\,280\,ft$, $1\,ft = 12\,in$
  • Setup: $\frac{1\,mi}{1} \times \frac{5\,280\,ft}{1\,mi} \times \frac{12\,in}{1\,ft}$
  • Execution: Cancel $mi$ and $ft$. Multiply $5\,280 \times 12$:
    • $5\,280 \times 10 = 52\,800$
    • $5\,280 \times 2 = 10\,560$
    • $52\,800 + 10\,560 = 63\,360\,in$
    • Common Error Caveat: Ensure correct carrying during addition ($5 + 8 = 13$, resulting in $63\,360$, not $63\,460$).
  • Answer: $63\,360\,in$

Interactive Unit Conversion Board Drills (Page 78, Problems 11–19)

• Board Contest Rules & Requirements:

  • All work must be explicitly written out step-by-step; mental math solutions are prohibited.
  • Students must write using the SMART Board marker, avoiding direct hand or finger contact with the board.
  • Students raise hands immediately upon finishing to register completion.

• Drill Problems & Solutions:

  • Problem 11: Convert $6\,lb$ to ounces ($oz$).
    • Calculation: $6 \times 16 = 96$
    • Answer: $96\,oz$
  • Problem 13: Convert $12\,ft$ to yards ($yd$).
    • Calculation: $12 \div 3 = 4$
    • Answer: $4\,yd$
  • Problem 14: Convert $72\,in$ to yards ($yd$).
    • Calculation: $72 \div 36 = 2$
    • Answer: $2\,yd$
  • Problem 15: Convert $2\,gal$ to quarts ($qt$).
    • Calculation: $2 \times 4 = 8$
    • Answer: $8\,qt$
  • Problem 16: Convert $18\,tsp$ to tablespoons ($tbsp$).
    • Calculation: $18 \div 3 = 6$
    • Answer: $6\,tbsp$
  • Problem 18: Convert $3\,bu$ (bushels) to pecks ($pk$).
    • Calculation: $3 \times 4 = 12$
    • Answer: $12\,pk$
  • Problem 19: Convert $8\,\text{weeks}$ to days.
    • Calculation: $8 \times 7 = 56$
    • Answer: $56\,\text{days}$

Mixed Measures Independent Speed Drill (Page 79, Problems 26–31)

• Problem 26: $10\,lb\,10\,oz$
• Problem 27: $2\,qt\,1\,pt$
• Problem 28: $5\,min\,25\,sec$
• Problem 29: $78\,gal\,3\,qt$
• Problem 30: $4\,ft\,3\,in$
• Problem 31: $1\,\text{week}\,4\,\text{days}$

Homework Assignment & Study Directives

• Assigned Tasks:
  • Complete Chapter 2, Review A on page 79.
• Review Instructions:
  • Thoroughly review all metric measures and prefixes ($K-H-D-U-D-C-M$) in preparation for upcoming quizzes and evaluations.