Solving Worded Problems Using the Variable Strategy

Fundamentals of the "Use a Variable" Strategy

• The strategy use a variable serves as a primary method for translating real-world, verbal descriptions into formal mathematical expressions and workable algebraic equations.

• It is systematically applied across various categories of word problems, including:

  • Number relation problems

  • Consecutive integer problems

  • Digit problems

  • Age problems

  • Financial, cost, and revenue problems

  • Multi-item weight and quantity allocation problems

• Guiding Motto:

  • "Problems, problems go away / Mathematics is on the way"

• Primary Learning Objective:

  • Solve complex worded problems systematically by formulating algebraic equations using the "use a variable" strategy.

George Pólya’s Four-Step Problem-Solving Framework

• When applying variables to worded problems, George Pólya’s structured four-step methodology provides a rigorous procedural framework:

  • Step 1: Understand the Problem

  • Read the problem carefully and analyze all given information.

  • Identify explicitly what quantity is being asked for.

  • Assign an algebraic variable (such as $x$) to represent the primary unknown quantity.

  • Step 2: Devise a Plan (Translate)

  • Determine the underlying mathematical relationships between the known and unknown quantities.

  • Translate the English verbal phrases directly into equivalent mathematical operations and algebraic expressions to formulate an equation.

  • Step 3: Carry Out the Plan (Solve)

  • Apply standard algebraic properties (e.g., distributive property, addition and multiplication properties of equality) to manipulate and isolate the variable.

  • Solve for all unknown quantities requested in the problem statement.

  • Step 4: Look Back (Check)

  • Substitute the computed solution back into the original verbal conditions of the problem to verify correctness.

  • Confirm that the numeric result satisfies every constraint and is logically sound.

Number and Consecutive Integer Problems

• Example 1: Single Number Relationship

  • Problem Statement: Three times the difference of a number and two is the same as the number increased by $8$. Find the number.

  • Step 1 (Understand the Problem):

  • Let $x$ represent the unknown number.

  • Step 2 (Devise a Plan / Translate):

  • "Three times the difference of a number and two" translates to $3(x - 2)$.

  • "is the same as" translates to $=$.

  • "the number increased by 8" translates to $x + 8$.

  • Formulated equation:       $3(x - 2) = x + 8$

  • Step 3 (Carry Out the Plan / Solve):

  • Apply the distributive property on the left side:       $3x - 6 = x + 8$

  • Transpose terms involving $x$ to the left and constants to the right:       $3x - x = 8 + 6$

  • Combine like terms:       $2x = 14$

  • Divide both sides by the coefficient of $x$:       $\frac{2x}{2} = \frac{14}{2}$

  • Result:       $x = 7$

  • Step 4 (Look Back / Check):

  • Substitute $x = 7$ back into the original translated equation:       $3(x - 2) = x + 8$       $3(7 - 2) = 7 + 8$       $3(5) = 15$       $15 = 15$

  • Conclusion: The number is $7$.

• Example 2: Consecutive Integer Relation

  • Problem Statement: Find three consecutive integers if five times the third is equal to eighteen more than the sum of the first and the second.

  • Step 1 (Understand the Problem):

  • Let $x$ represent the first integer.

  • Let $x + 1$ represent the second consecutive integer.

  • Let $x + 2$ represent the third consecutive integer.

  • Step 2 (Devise a Plan / Translate):

  • "five times the third" translates to $5(x + 2)$.

  • "is equal to" translates to $=$.

  • "eighteen more than the sum of the first and the second" translates to $x + (x + 1) + 18$.

  • Formulated equation:       $5(x + 2) = x + (x + 1) + 18$

  • Step 3 (Carry Out the Plan / Solve):

  • Expand and collect terms on both sides:       $5x + 10 = 2x + 19$

  • Transpose variable terms to the left and constants to the right:       $5x - 2x = 19 - 10$

  • Simplify:       $3x = 9$

  • Divide by $3$:       $\frac{3x}{3} = \frac{9}{3}$       $x = 3$

  • Evaluate the remaining integers:

    • First integer: $x = 3$

    • Second integer: $x + 1 = 3 + 1 = 4$

    • Third integer: $x + 2 = 3 + 2 = 5$

  • Step 4 (Look Back / Check):

  • Substitute the values back into the equation:       $5(x + 2) = x + (x + 1) + 18$       $5(5) = 3 + 4 + 18$       $25 = 25$

  • Conclusion: The three consecutive integers are $3$, $4$, and $5$.

Digit Value and Number Relation Problems

• In base-10 positional numeral systems, a two-digit number having tens digit $t$ and unit digit $u$ has a total value equal to $10t + u$.

• Example 3: Two-Digit Number with Digit Operations

  • Problem Statement: The sum of the digits of a two-digit number is $10$. If the unit digit is divided by the tens digit, the quotient is four. Find the number.

  • Variable Definitions:

  • Let $u$ be the unit digit.

  • Let $t$ be the tens digit.

  • Translation into a System of Equations:

  • "The sum of the digits of a two-digit number is 10":       $t + u = 10$       (Designated as Equation 1)

  • "If the unit digit is divided by the tens digit, the quotient is four":       $\frac{u}{t} = 4$       (Designated as Equation 2)

  • Algebraic Solution:

  • Solve for $u$ in Equation 1:       $u = 10 - t$

  • Substitute this expression for $u$ into Equation 2:       $\frac{10 - t}{t} = 4$

  • Clear the denominator by multiplying both sides by $t$:       $10 - t = 4t$

  • Group like terms:       $4t + t = 10$       $5t = 10$

  • Divide by $5$:       $\frac{5t}{5} = \frac{10}{5}$       $t = 2$

  • Solve for the unit digit $u$:       $u = 10 - t$       $u = 10 - 2$       $u = 8$

  • Combine digits to form the number:

    • Tens digit: $2$

    • Units digit: $8$

    • Resulting two-digit number: $28$

  • Checking the Solution:

  • Check sum condition: $t + u = 2 + 8 = 10$ (True)

  • Check quotient condition: $u \div t = 8 \div 2 = 4$ (True)

  • Conclusion: The number is $28$.

Principles and Applications of Age Problems

• Core Temporal Rules for Age Problems:

  • Every individual in an age problem possesses three temporal ages: the past age, the present age, and the future age.

  • To calculate a past age, subtract the elapsed number of years from the present age ($\text{Present Age} - \text{Years Ago}$).

  • To calculate a future age, add the elapsed number of years to the present age ($\text{Present Age} + \text{Years Hence}$).

  • Constructing a structured data table is the standard procedure for mapping these quantities before equation setup.

• Example 4: Fractional Age Relations Across Time (Jubilant)

  • Problem Statement: One-fourth of Jubilant’s age twelve years ago is equal to one-sixth of her age eight years hence. How old is she now?

  • Tabulation of Data:

  • Present age: $x$

  • Past age ($12\,\text{years}$ ago): $x - 12$

  • Future age ($8\,\text{years}$ hence): $x + 8$

• Equation Formulation:

  • "One-fourth of Jubilant’s age twelve years ago is equal to one-sixth of her age eight years hence":       $\frac{1}{4}(x - 12) = \frac{1}{6}(x + 8)$

• Algebraic Solution:

  • Cross-multiply to eliminate denominators:       $6(x - 12) = 4(x + 8)$

  • Distribute both sides:       $6x - 72 = 4x + 32$

  • Rearrange terms:       $6x - 4x = 32 + 72$       $2x = 104$

  • Divide by $2$:       $\frac{2x}{2} = \frac{104}{2}$       $x = 52$

• Checking the Solution:

  • Substitute $x = 52$ back into the original condition:       $\frac{1}{4}(52 - 12) = \frac{1}{6}(52 + 8)$       $\frac{1}{4}(40) = \frac{1}{6}(60)$       $10 = 10$

  • Conclusion: Jubilant is currently $52\,\text{years old}$.

  • Example 5: Finding Unknown Years Into the Future (Patrick and Son)

• Problem Statement: Patrick is $28\,\text{years old}$, and his son is $2\,\text{years old}$. After how many years will Patrick’s age be twice his son’s age?

• Tabulation of Data:

  • Let $x$ represent the number of elapsed years from now.

  • Son: Present age = $2$; Future age ($x$ years from now) = $2 + x$

  • Patrick: Present age = $28$; Future age ($x$ years from now) = $28 + x$

• Equation Formulation:

  • "Patrick’s age will be twice his son’s age":       $28 + x = 2(2 + x)$

• Algebraic Solution:

  • Distribute the right-hand side:       $28 + x = 4 + 2x$

  • Transpose variable terms to the left and constants to the right:       $x - 2x = 4 - 28$       $-x = -24$

  • Multiply the entire equation by $-1$:       $x = 24$

• Checking the Solution:

  • Evaluate ages at $x = 24$:

    • Patrick’s future age: $28 + 24 = 52$

    • Son’s future age: $2 + 24 = 26$

    • Relationship: $52 = 2(26)$, giving $52 = 52$

  • Conclusion: In $24\,\text{years}$, Patrick will be $52\,\text{years old}$, which is twice his son's age ($26\,\text{years old}$).

  • Example 6: Relative Multi-Person Age Problem (Edward and Emer John)

• Problem Statement: Emer John is five years less than twice Edward’s age. Three years ago, Emer John’s age was $8\,\text{years}$ more than Edward’s age. How old are they now?

• Tabulation of Data:

  • Let Edward’s present age be $x$.

  • Emer John’s present age: $2x - 5$

  • Edward’s past age ($3\,\text{years}$ ago): $x - 3$

  • Emer John’s past age ($3\,\text{years}$ ago): $(2x - 5) - 3 = 2x - 8$

• Equation Formulation:

  • "Three years ago, Emer John’s age was 8 years more than Edward’s age":       $2x - 8 = x - 3 + 8$

• Algebraic Solution:

  • Rearrange and isolate $x$:       $2x - x = -3 + 8 + 8$       $x = 13$

  • Solve for each individual’s current age:

    • Edward’s age: $x = 13$

    • Emer John’s age: $2x - 5 = 2(13) - 5 = 26 - 5 = 21$

• Checking the Solution:

  • Emer John's age ($21\,\text{years}$) is $5\,\text{years}$ less than twice Edward's age ($2 \times 13 = 26$).

  • Three years ago, Emer John was $21 - 3 = 18\,\text{years old}$.

  • Three years ago, Edward was $13 - 3 = 10\,\text{years old}$.

  • Verify relationship: $18 = 10 + 8$, which satisfies $18 = 18$.

  • Conclusion: Edward is currently $13\,\text{years old}$, and Emer John is $21\,\text{years old}$.

Financial and Cost Allocation Problems

• In financial and purchasing scenarios, multiple item costs are typically expressed relative to a single common baseline variable, yielding a linear cost equation.

• Example 7: Combined Clothing Purchases

  • Problem Statement: A shirt and a pair of pants together cost $\text{Php}\,1,750.00$. The pair of pants costs $\text{Php}\,650.00$ more than the shirt. How much is the shirt?

  • Given Quantities and Variable Setup:

  • Let $x$ = cost of the shirt

  • Let $x + 650$ = cost of the pair of pants

  • Total combined cost = $\text{Php}\,1,750.00$

  • Equation Formulation:     $x + (x + 650) = 1,750$

  • Algebraic Solution:

  • Combine like terms:       $2x + 650 = 1,750$

  • Subtract $650$ from both sides:       $2x = 1,750 - 650$       $2x = 1,100$

  • Divide both sides by $2$:       $\frac{2x}{2} = \frac{1,100}{2}$       $x = \text{Php}\,550.00$

  • Compute cost of pants:       $\text{Cost of pants} = x + 650 = 550 + 650 = \text{Php}\,1,200.00$

  • Checking the Solution:

  • Sum of items: $550 + (550 + 650) = 550 + 1,200 = 1,750$

  • Difference: $\text{Php}\,1,200.00 - \text{Php}\,550.00 = \text{Php}\,650.00$

  • Conclusion: The shirt costs $\text{Php}\,550.00$, and the pair of pants costs $\text{Php}\,1,200.00$.

Multi-Item Quantity and Weight Problems

• When tracking chained relational quantities among multiple items, express each successive entity in terms of the initial baseline item.

• Example 8: Fruit Sales Weight Distribution

  • Problem Statement: A fruit vendor sells grapes, apples, and oranges. At the end of the day, she was able to sell three more kilos of oranges than grapes and six more kilos of apples than oranges. If $27\,\text{kilos}$ of fruits were sold for that day, how many kilos of each kind were sold?

  • Given Quantities and Setup:

  • Let $x$ = number of kilos of grapes sold

  • Let $x + 3$ = number of kilos of oranges sold

  • Let $(x + 3) + 6$ = number of kilos of apples sold

  • Total fruit weight sold = $27\,\text{kilos}$

  • Equation Formulation:     $x + (x + 3) + (x + 3) + 6 = 27$

  • Algebraic Solution:

  • Group like terms:       $3x + 12 = 27$

  • Subtract $12$ from both sides:       $3x = 27 - 12$       $3x = 15$

  • Divide both sides by $3$:       $\frac{3x}{3} = \frac{15}{3}$       $x = 5$

  • Determine the quantity sold for each fruit category:

    • Grapes: $x = 5\,\text{kilos}$

    • Oranges: $x + 3 = 5 + 3 = 8\,\text{kilos}$

    • Apples: $x + 3 + 6 = 5 + 3 + 6 = 14\,\text{kilos}$

  • Checking the Solution:

  • Total sum: $5 + 8 + 14 = 27\,\text{kilos}$

  • Oranges relative to grapes: $8 - 5 = 3\,\text{kilos more}$

  • Apples relative to oranges: $14 - 8 = 6\,\text{kilos more}$

  • $27 = 27$

  • Conclusion: The vendor sold $5\,\text{kilos}$ of grapes, $8\,\text{kilos}$ of oranges, and $14\,\text{kilos}$ of apples.

Group Discount and Ticket Revenue Problems

• Revenue problems with tiered pricing require multiplying the unit price of each demographic tier by its respective head count and summing to equal the total revenue.

• Example 9: Resort Entrance Fees and Senior Citizen Discounts

  • Problem Statement: A birthday celebration was held at a resort. The resort charges entrance fees of $\text{Php}\,350.00$ for adults and $\text{Php}\,200.00$ for children. Senior citizens were given a discount of $20\%$. The number of adults is four more than the number of children while the number of seniors is four less than the number of children. If $30\,\text{persons}$ attended the celebration and the entrance fees paid amounted to $\text{Php}\,8,580.00$, how many senior citizens received discounts?

  • Given Price and Discount Parameters:

  • Adult ticket fee = $\text{Php}\,350.00$

  • Child ticket fee = $\text{Php}\,200.00$

  • Senior citizen discount calculation:       $\text{Discount Amount} = \text{Php}\,350.00 \times 0.20 = \text{Php}\,70.00$       $\text{Senior Ticket Fee} = \text{Php}\,350.00 - \text{Php}\,70.00 = \text{Php}\,280.00$

  • Demographics Setup:

  • Let $x$ = number of children

  • Let $x + 4$ = number of adults

  • Let $x - 4$ = number of senior citizens

  • Total attendance validation:       $x + (x + 4) + (x - 4) = 3x = 30 \implies x = 10$

  • Revenue Equation Formulation:     $x(200) + (x + 4)(350) + (x - 4)(280) = 8,580$

  • Algebraic Solution:

  • Expand all terms:       $200x + 350x + 1,400 + 280x - 1,120 = 8,580$

  • Group like terms:       $(200 + 350 + 280)x + (1,400 - 1,120) = 8,580$       $830x + 280 = 8,580$

  • Subtract $280$ from both sides:       $830x = 8,580 - 280$       $830x = 8,300$

  • Divide by $830$:       $\frac{830x}{830} = \frac{8,300}{830}$       $x = 10$

  • Calculate head count across all categories:

    • Children: $x = 10$

    • Adults: $x + 4 = 10 + 4 = 14$

    • Senior citizens: $x - 4 = 10 - 4 = 6$

  • Checking the Solution:

  • Evaluate total fees:       $10(200) + 14(350) + 6(280) = 2,000 + 4,900 + 1,680 = 8,580$       $8,580 = 8,580$

  • Evaluate total attendees:       $10 + 14 + 6 = 30\,\text{persons}$

  • Conclusion: Exactly $6$ senior citizens received discounts.