Grade 5 Math Ready, Prep, Go! Comprehensive Notes


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Educational Methodology and Implementation

  • Mixed Standards Approach: Standards are intentionally mixed to mimic the STAAR assessment environment. This requires students to actively discern what a problem is asking and determine the appropriate solution strategy.

  • ELAR Integration: Story problems are written at a 5th-grade reading level. They are designed to practice math skills while simultaneously developing ELAR skills such as making inferences.

  • Usage Strategies:

    • Utilize for STAAR preparation as an alternative to simple multiple-choice questions.

    • Implement as a spiral review for taught skills.

    • Use in small group rotations or summer school to increase engagement.

    • Pair activities with released STAAR problems to assess mastery.


Detailed Readiness Standards (Grade 5)

  • Standard 5.2B: The student applies mathematical process standards to represent, compare, and order positive rational numbers and understand relationships related to place value. Specifically, the student is expected to compare and order two decimals to thousandths and represent comparisons using the symbols >>, <<, or ==.

  • Standard 5.3E: Solve for products of decimals to the hundredths, including money situations, using strategies based on place-value understandings and properties of operations.

  • Standard 5.3G: Solve for quotients of decimals to the hundredths, with up to four-digit dividends and two-digit whole number divisors, including the standard algorithm.

  • Standard 5.3K: Add and subtract positive rational numbers fluently.

  • Standard 5.3L: Divide whole numbers by unit fractions and unit fractions by whole numbers.

  • Standard 5.4B: Represent and solve multi-step problems involving the four operations with whole numbers using equations with a letter standing for the unknown quantity (xx).

  • Standard 5.4C: Generate a numerical pattern when given a rule in the form y=axy = ax or y=x+ay = x + a and graph.

  • Standard 5.4F: Simplify numerical expressions that do not involve exponents, including up to two levels of grouping.

  • Standard 5.4H: Represent and solve problems related to perimeter, area, and volume.

  • Standard 5.5A: Classify two-dimensional figures in a hierarchy of sets and subsets using graphic organizers based on attributes and properties.

  • Standard 5.8C: Graph in the first quadrant of the coordinate plane ordered pairs generated by number patterns or input-output tables.

  • Standard 5.9C: Solve one- and two-step problems using data from frequency tables, dot plots, bar graphs, stem-and-leaf plots, or scatterplots.


Detailed Supporting Standards (Grade 5)

  • Standard 5.3A: Estimate solutions to mathematical and real-world problems involving addition, subtraction, multiplication, or division.

  • Standard 5.3B: Multiply with fluency a three-digit number by a two-digit number using the standard algorithm.

  • Standard 5.3C: Solve for quotients of up to a four-digit dividend by a two-digit divisor.

  • Standard 5.3D: Represent multiplication of decimals to the hundredths using objects and pictorial models, including area models.

  • Standard 5.3H: Represent and solve addition and subtraction of fractions with unequal denominators using models.

  • Standard 5.3I: Represent and solve multiplication of a whole number and a fraction using models.

  • Standard 5.3J: Represent division of unit fractions and whole numbers (e.g., 13÷7\frac{1}{3} \div 7 and 7÷137 \div \frac{1}{3}) using models.

  • Standard 5.4E: Describe the meaning of parentheses and brackets in a numeric expression.

  • Standard 5.6B: Determine the volume of a rectangular prism based on the number of layers times the number of unit cubes in the area of the base.

  • Standard 5.7A: Solve problems by calculating conversions within a measurement system (customary or metric).

  • Standard 5.9B: Represent discrete paired data on a scatterplot.


Activity Guide: Big Turtle Chocolate Company

  • Core Concepts: Multiplication and division of decimals, money math, estimation, and modeling.

  • Problem Scenarios:

    • Unit Cost: Maria sells 20 Turtle Drops for $6.00\$6.00. The cost per drop is calculated as \6.00 \div 20 = \0.300.30.

    • Budgeting: A customer has $68.50\$68.50. At $6.00\$6.00 per box, she can buy 11 boxes (11×$6.00=$66.0011 \times \$6.00 = \$66.00).

    • Total Revenue: If 500 drops are sold in boxes of 20, the number of boxes is 500÷20=25500 \div 20 = 25. Revenue is 25×$6=$12525 \times \$6 = \$125.

    • Monthly Average: Selling $936\$936 over 3 months results in $312\$312 per month.

    • Tiny Tortoise Competitor: Carlos sells Baby Chocolate Shells for $0.50\$0.50. To make $73\$73, he must sell 73÷0.50=14673 \div 0.50 = 146 shells.

    • Decimal Ordering: Comparing Monday ($0.10\$0.10), Tuesday ($0.01\$0.01), Thursday ($0.11\$0.11), and Friday ($1.01\$1.01). Ordered greatest to least: $1.01,$0.11,$0.10,$0.01\$1.01, \$0.11, \$0.10, \$0.01.

    • Multi-Item Sales: 68 shells at $0.50\$0.50 ($34\$34), 100 penny shells at $0.01\$0.01 ($1\$1), and 7 giant shells at $2\$2 ($14\$14) totals $34+$1+$14=$49\$34 + \$1 + \$14 = \$49.

    • Quantity Comparison: Selling $26.50\$26.50 of penny shells (2,6502,650 units) vs. $1,325\$1,325 of $0.50\$0.50 shells (1,325÷0.5=2,6501,325 \div 0.5 = 2,650 units). He sold an equal amount of each.


Activity Guide: Spicky Spider

  • Core Concepts: Creative problem solving with money and decimal operations.

  • Key Problems:

    • Coin Counts: Spicky found $1.87\$1.87. He has 10 dimes ($1.00\$1.00) and 6 nickels ($0.30\$0.30). The remaining $0.57\$0.57 consists of 57 pennies.

    • Stack Management: 2 stacks each of nickels (4 per stack), dimes (4 per stack), quarters (3 per stack), and pennies (10 per stack). Total: 8 \text{ nickels } (\0.40) + 8 \text{ dimes } (\0.80) + 6 \text{ quarters } (\1.50) + 20 \text{ pennies } (\0.20)=$2.900.20) = \$2.90.

    • Averaging: Gathering $2.02\$2.02 per week for 15 weeks equals 15×$2.02=$30.3015 \times \$2.02 = \$30.30.

    • Nickel Conversion: $8.00\$8.00 in only nickels is calculated as 8.00÷0.05=1608.00 \div 0.05 = 160 nickels.

    • Splitting Earnings: Mickey ends the day with 12 dimes ($1.20\$1.20) and 12 nickels ($0.60\$0.60), totaling $1.80\$1.80. Since they split found money equally, the total found was \1.80 \times 2 = \3.603.60.

    • Daily Consistency: $29.15\$29.15 found over 5 days at the same rate per day is \29.15 \div 5 = \5.835.83 per day.

    • Quarter Count: 35 quarters are worth 35×0.25=$8.7535 \times 0.25 = \$8.75.

    • Wishes: Replacing 1 lucky penny with 25 nickels ($1.25\$1.25) results in a net gain of $1.24\$1.24.


Activity Guide: Don’t be Punny!

  • Core Concepts: Mastery of decimal comparison (<<, >>, and ==).

  • Comparison Solutions:

    1. 0.999>0.990.999 > 0.99 (Letter: S)

    2. 2.02=2.0202.02 = 2.020 (Letter: O)

    3. 8.80>8.088.80 > 8.08 (Letter: H)

    4. 2.2>2.022.2 > 2.02 (Letter: I)

    5. 5.001<5.015.001 < 5.01 (Letter: C)

    6. 1.11>1.1011.11 > 1.101 (Letter: L)

    7. 3.14<3.173.14 < 3.17 (Letter: E)

    8. 3.1=3.1003.1 = 3.100 (Letter: T)

    9. 7.36<7.3667.36 < 7.366 (Letter: G)

    10. 4.202<4.224.202 < 4.22 (Letter: A)

    11. 2.41>2.0412.41 > 2.041 (Letter: E)

  • Puzzle Answer: Why did the golfer bring two pairs of pants? "In case she got a hole in one!"


Activity Guide: Pattern Play (3 & 7 Family and 8 & 9 Family)

  • Core Concepts: Multiplication and division of whole numbers with unit fractions to build algebraic fluency.

  • The "3 & 7 Family" Patterns:

    • 3÷17=213 \div \frac{1}{7} = 21

    • 7÷13=217 \div \frac{1}{3} = 21

    • 3÷13=93 \div \frac{1}{3} = 9

    • 3×13=13 \times \frac{1}{3} = 1

    • 17÷3=121\frac{1}{7} \div 3 = \frac{1}{21}

    • 13÷7=121\frac{1}{3} \div 7 = \frac{1}{21}

    • 17×7=1\frac{1}{7} \times 7 = 1

  • The "8 & 9 Family" Patterns:

    • 8÷19=728 \div \frac{1}{9} = 72

    • 9÷18=729 \div \frac{1}{8} = 72

    • 8÷18=648 \div \frac{1}{8} = 64

    • 9×19=19 \times \frac{1}{9} = 1

    • 19÷8=172\frac{1}{9} \div 8 = \frac{1}{72}

    • 18÷9=172\frac{1}{8} \div 9 = \frac{1}{72}

  • Journal Questions:

    • Why do 8×188 \times \frac{1}{8}, 9×199 \times \frac{1}{9}, 8÷88 \div 8, and 9÷99 \div 9 all equal 1?

    • Why do 18÷18\frac{1}{8} \div \frac{1}{8} and 19÷19\frac{1}{9} \div \frac{1}{9} also equal 1?


Activity Guide: Bad Billy’s Button Boxes

  • Core Concepts: Perimeter, Area, and Volume involving whole numbers and fractions.

  • Geometric Formulas:

    • Area: L×W=AL \times W = A

    • Perimeter: L+L+W+W=PL + L + W + W = P or 2L+2W=P2L + 2W = P

    • Volume: L×W×H=VL \times W \times H = V

  • Problem Set 1 (Correct Calculations):

    • Box 1: 14 in wide, 6 in long. Area = 84 sq in84 \text{ sq in}. Perimeter = 40 in40 \text{ in}.

    • Box 2 (Rectangular Button): 12 in\frac{1}{2} \text{ in} long, 2 in wide. Area = 1 sq in1 \text{ sq in}. Perimeter = 5 in5 \text{ in}.

    • Blue Button Box: 5 in long, 2 in wide, 14 in\frac{1}{4} \text{ in} tall. Volume = 5×2×14=104=212 cubic inches5 \times 2 \times \frac{1}{4} = \frac{10}{4} = 2 \frac{1}{2} \text{ cubic inches}.

    • Orange Button Box: 5 in long, 7 in wide, 12 in\frac{1}{2} \text{ in} tall. Volume = 5×7×12=352=1712 cubic inches5 \times 7 \times \frac{1}{2} = \frac{35}{2} = 17 \frac{1}{2} \text{ cubic inches}.

  • Problem Set 2 (Error Analysis):

    • Warehouse: 31 ft long, 25 ft wide. Billy calculated 775 ft. Error: Calculated Area (31×2531 \times 25) instead of Perimeter (31+31+25+2531+31+25+25). Correct Perimeter: 112 feet112 \text{ feet}.

    • Beta Blue Box: 8 in long, 5 in wide, 13 in\frac{1}{3} \text{ in} tall. Billy calculated 13. Error: Improper fraction to mixed number conversion. Correct Volume: 8×5×13=403=1313 cubic inches8 \times 5 \times \frac{1}{3} = \frac{40}{3} = 13 \frac{1}{3} \text{ cubic inches}.

    • Toolbox: 3 ft wide, 13 ft long, 2 ft tall. Billy calculated 6 sq ft for the base. Error: Multiplied width by height (3×23 \times 2). Correct Area of Base: 3×13=39 square feet3 \times 13 = 39 \text{ square feet}.

    • Dumpster: 7 ft long, 7 ft wide, 6 ft tall. Billy calculated 26 cu ft. Error: Calculated perimeter (7+7+6+67+7+6+6) instead of volume. Correct Volume: 7×7×6=294 cubic feet7 \times 7 \times 6 = 294 \text{ cubic feet}.


Activity Guide: Garden Match Up

  • Core Concepts: Matching situation, equation, and solution.

  • Scenarios:

    • Tomato Garden: 13 m long, 16.5 m wide. Fence needed (PP). Equation: P=(13×2)+(16.5×2)P = (13 \times 2) + (16.5 \times 2). Solution: 59 m.

    • Dump Truck: 4.8 m long, 2.6 m wide, 3 m layer. Volume: V=4.8×2.6×3V = 4.8 \times 2.6 \times 3. Solution: 37.44 m3m^3.

    • Rosebushes: 13 m long, 16.5 m wide, 1 per m2m^2. Area: A=13×16.5A = 13 \times 16.5. Solution: 214.

    • Mud Patch: 4.8 m wide, 3.2 m long, grass pads (1 m21 \text{ m}^2 each). Area: A=4.8×3.2=15.36A = 4.8 \times 3.2 = 15.36. Must buy whole pads. Solution: 16.

  • Journal Challenges:

    • Revised Tomato Garden: Length is now 12\frac{1}{2} of original (6.5 m). New Area: 6.5×16.5=107.25 sq m6.5 \times 16.5 = 107.25 \text{ sq m}.

    • Combined Dump Trucks: Emma's truck stays same; brother's is 3 m longer (7.8 m). Combined: (7.8×3×2.6)+(4.8×3×2.6)=98.28 cubic meters(7.8 \times 3 \times 2.6) + (4.8 \times 3 \times 2.6) = 98.28 \text{ cubic meters}.

    • Riley’s Expanded Garden: 0.5 m longer and 0.5 m wider. New dimensions: 13.5 m by 17 m. Area: 13.5×17=229.5 sq m13.5 \times 17 = 229.5 \text{ sq m}.

    • Cole’s Picket Fence: Perimeter divided by picket width. (4.8×2)+(3.2×2)width\frac{(4.8 \times 2) + (3.2 \times 2)}{width}. If picket is 4 inches, student must convert or calculate based on the problem's provided width (note: journal text lists 2 meters and 4 inches in different spots; answer key assumes Pickett is specific to units used).


Activity Guide: Andre’s Animal Enclosure

  • Core Concepts: Multi-step problems using numerical expressions with variables.

  • Matches:

    • Otters (Weight): 3 otters at 14 lbs, 2 at 20 lbs. Equation: (3×14)+(2×20)=x(3 \times 14) + (2 \times 20) = x. Solution: 82.

    • Otters (Attendance): 36 people/day for 6 days, then Sunday is (36 - 6). Equation: (36×6)+(36−6)=x(36 \times 6) + (36 - 6) = x. Solution: 246.

    • Sea Otters (Population): 36 + 6 more, then divide by 2. Equation: (36+6)÷2=x(36 + 6) \div 2 = x. Solution: 21.

    • Black Bears (Weight Difference): 3 females (370 lbs) vs 2 males (550 lbs). Equation: (3×370)−(2×550)=x(3 \times 370) - (2 \times 550) = x. Solution: 10.

    • Lemurs (Tail Length): 14 lemurs, tails are 23 inches, bodies are 20 inches. Equation: (23×14)−(20×14)=x(23 \times 14) - (20 \times 14) = x. Solution: 42.

    • Teeth in Exhibit: 3 capybaras (20 teeth), 6 muskrats (36 teeth), 2 rabbits (28 teeth). Equation: (3×20)+(2×28)+(6×36)=x(3 \times 20) + (2 \times 28) + (6 \times 36) = x. Solution: 332.

    • Goat Transfers: 36 goats, 6 leave, then 2 leave, then divide remaining by 2. Equation: (36−6−2)÷2=x(36 - 6 - 2) \div 2 = x. Solution: 14.

    • Horse Colors: 28 horses, 14\frac{1}{4} brown (7), 12\frac{1}{2} black (14). Others = 28−14−7=728 - 14 - 7 = 7. Comparison: 14−7=714 - 7 = 7. Equation: 28−(28÷2)−(28÷4)=x28 - (28 \div 2) - (28 \div 4) = x. Solution: 7.


Activity Guide: Representing Linear Data

  • Additive vs. Multiplicative Rules:

    • Paul’s Run: Multiplicative. Rule: y=3xy = 3x. 7 days = 21 miles. 24 miles = Day 8.

    • Charlotte’s Blocks: Additive (Subtraction). Paul runs 1 block less than Charlotte. If Charlotte runs 6, Paul runs 5 (y=x−1y = x - 1).

    • Paul’s Hourly Rate: Multiplicative. Rule: y=4xy = 4x. 6 hours = 24 miles. Marathon (26.2 miles) takes approx 6.5 hours.

    • Charlotte’s Daily Rate: Multiplicative. Rule: y=6xy = 6x. 7 days = 42 miles. 33 miles takes approx 5.5 days.

    • Candy Spending: Multiplicative. Rule: y=5xy = 5x (xx is Saturdays). $102\$102 bike takes approx 21 weeks of saving.

    • Paul’s Savings Logic: Paul saves $3\$3. Equation: y=x−3y = x - 3. To spend $5\$5 on candy, he must earn $8\$8. To buy $10\$10 game and $3\$3 candy, he needs $16\$16 ($13+$3\$13 + \$3 savings).

    • Piggy Bank Contributions: Parents give extra $2\$2. Equation: y=x+2y = x + 2. To save $20\$20, she needs to earn $18\$18 total, but since her parents always give $2\$2, if she has already earned $10\$10, she needs $8\$8 more.

    • Bank Comparison: Charlotte has $4\$4 more than Paul. Equation: y=x+4y = x + 4. If P has $5\$5, C has $9\$9. If combined total is $20\$20, Charlotte has $12\$12 and Paul has $8\$8.


Activity Guide: Right or Wrong? Order of Operations

  • Problem 1: 5(2+3)−75(2 + 3) - 7. Wrong solution: 10+3−710+3-7. Correct Process: 5(5)−7=25−7=185(5) - 7 = 25 - 7 = 18.

  • Problem 2: 11(1+1)×(5−4)=11(2)×(1)=2211(1 + 1) \times (5 - 4) = 11(2) \times (1) = 22. RIGHT.

  • Problem 3: 4(17+7−2)+84(17 + 7 - 2) + 8. Wrong solution: 4(30)4(30). Correct Process: 4(22)+8=88+8=964(22) + 8 = 88 + 8 = 96.

  • Problem 4: 21−(3+1)÷(2+0)21 - (3 + 1) \div (2 + 0). Wrong solution: 17÷217 \div 2. Correct Process: 21−4÷2=21−2=1921 - 4 \div 2 = 21 - 2 = 19.

  • Problem 5: 14(9−1)+2(10−4)\frac{1}{4}(9 - 1) + 2(10 - 4). Wrong solution: 4×64 \times 6. Correct Process: 14(8)+2(6)=2+12=14\frac{1}{4}(8) + 2(6) = 2 + 12 = 14.

  • Problem 6: 2+(7−4)+3(2÷1)2 + (7 - 4) + 3(2 \div 1). Wrong solution: 8(2÷1)8(2 \div 1). Correct Process: 2+3+3(2)=5+6=112 + 3 + 3(2) = 5 + 6 = 11.

  • Problem 7: 25−[7+4(1+1)]=25−[7+4(2)]=25−15=1025 - [7 + 4(1 + 1)] = 25 - [7 + 4(2)] = 25 - 15 = 10. RIGHT.

  • Problem 8: 2+4×13−12 + 4 \times 13 - 1. Wrong solution: 6×13−16 \times 13 - 1. Correct Process: 2+52−1=532 + 52 - 1 = 53.


Activity Guide: Shoop Shoes Data

  • Data Types: Stem-and-leaf, bar graph, dot plot, frequency table, scatterplot.

  • Key Findings:

    • Comparison: Carrie Shop owns more shoes than the 5 students in her class with the combined fewest shoes.

    • White Shoes: Combined 39 pairs of white shoes. Timmy has about the same number of shoes.

    • Long-term Ownership: At the end of 2017, Timmy owned more pairs than Tom because he bought many in 2016 and 2017 while Tom did not.

    • Information Gap: Problem 4 (Total spending of Carrie and Sean) cannot be answered with provided charts.

    • Minimum Expense Equation: (3×0)+(3×31)+(3×61)+(4×91)+(3×121)+(4×151)=x(3 \times 0) + (3 \times 31) + (3 \times 61) + (4 \times 91) + (3 \times 121) + (4 \times 151) = x.

    • Range: The person with the most shoes owns 18 pairs more than the person with the least.

    • Gender Comparison: Shoop men vs. Shoop women; approximately 56 more shoes for men.

    • Age Calculation: The difference between the combined ages of the Shoop adults and children is 45 years.


Activity Guide: Emilio Honeysuckle Story Math

  • Part 1 Traps:

    • Food: 10 rations, eat 12\frac{1}{2} per day. Correct: 20 days. (Incorrect: Adventure ends).

    • Steps: Room area 50 ft by 15\frac{1}{5} of length (10 ft). Area = 500+68=568 sq ft500 + 68 = 568 \text{ sq ft}. Correct: 568 steps. (Incorrect: Pinned in rock).

    • Precise Stops: 64 steps total. Stop 1: 58\frac{5}{8} of way (4040 steps). Steps left: 2424. Stop 2: 14\frac{1}{4} of distance left (66 steps). Remaining: 1818 steps… wait, key specifies 20 steps for Part 1, Problem 3.

    • Lever Wait Time: Simplification of 55−14[12+(29−21)]55 - \frac{1}{4}[12 + (29 - 21)]. Process: 55−14[12+8]=55−14[20]=55−5=5055 - \frac{1}{4}[12 + 8] = 55 - \frac{1}{4}[20] = 55 - 5 = 50. Correct: 50 seconds.

  • Part 2 Traps:

    • The Fork: Right path (5×0.18=0.905 \times 0.18 = 0.90) vs. Left path (4×0.23=0.924 \times 0.23 = 0.92). Correct: Left (furthest). (Incorrect: Fallen into a hole).

    • Poison Room: 0.85 m paces ×15.20.85 \text{ m paces } \times 15.2. Correct: 12.92 meters. (Incorrect: Poison darts hitting fuzzily).

    • Torches: 96 torches ×0.2596 \text{ torches } \times 0.25. Correct: Count to 24. (Incorrect: Torch blown out by wind).

    • Treasure Path: 6 ft forward +(500 bricks ×0.3 yards per brick×3 feet per yard)6 \text{ ft forward } + (500 \text{ bricks } \times 0.3 \text{ yards per brick} \times 3 \text{ feet per yard}). Calculation: 6+(500×0.9)=6+450=4566 + (500 \times 0.9) = 6 + 450 = 456. Correct: 456 feet.


Activity Guide: Find a Figure

  • Core Concepts: Geometry attributes including sides, angles, parallel lines, and congruence.

  • Attributes Key:

    • 3 sides: Includes different triangles (acute, right, obtuse).

    • 4 sides: Rectangles, squares, rhombuses, trapezoids, parallelograms.

    • Parallelism: Exactly 1 pair (Trapezoid); 2 pairs (Rectangle, Square, Rhombus, Parallelogram).

    • Angles: 4 right angles (Rectangle, Square); all acute angles (some triangles).

    • Congruence: All sides congruent (Square, Rhombus, Equilateral Triangle).