Section 2.1: Algebraic Word Problems

Section 2.1: Algebraic Word Problems

The EDPT Arithmetic Reasoning section heavily tests your ability to interpret written descriptions of mathematical relationships and convert them into solvable equations. Algebraic word problems are not just about calculating; they are about translation. Your primary task is to decode the English language into the language of mathematics.

1. The Art of Translation

Before you can solve an equation, you must build it correctly. The most common mistake candidates make is rushing into calculations without a solid algebraic foundation.

Key Translation Rules

Here is a definitive guide to translating English words into mathematical symbols:

• Addition (+): sum, increased by, more than, combined, total, added to, older than, heavier than.
• Subtraction (-): difference, decreased by, less than, fewer than, reduced by, younger than. Trap: "5 less than x" translates to x - 5, not 5 - x.
• Multiplication (×): product, of, times, twice (2x), triple (3x), fraction of.
• Division (÷): quotient, ratio, divided by, per, out of.
• Equals (=): is, are, was, will be, gives, yields, amounts to.

Defining Variables

Always start by defining your unknown. If the problem asks for the number of apples, write down Let a = number of apples. If there are multiple unknowns, try to express them in terms of a single variable. For instance, if Bob has 5 more apples than Alice, let Alice's apples = x and Bob's apples = x + 5. Do not use x and y unless absolutely necessary, as this requires a system of equations.

2. Age Problems

Age problems are a classic category. They often involve comparing the ages of two or more people at different points in time (past, present, and future).

Strategic Tip: Use a Timeline Table

Whenever you encounter an age problem, immediately draw a table. Label the columns with the time periods (e.g., "5 Years Ago", "Present", "In 3 Years") and the rows with the people involved.

Worked Example 1: A father is currently three times as old as his son. In 12 years, the father will be twice as old as his son. What is the son's current age?

Step 1: Set up the variables using the present time. Let the son's present age = x. Since the father is three times as old, the father's present age = 3x.

Step 2: Express their ages in the future. In 12 years, the son's age will be x + 12. In 12 years, the father's age will be 3x + 12.

Step 3: Translate the future relationship into an equation. The problem states: In 12 years, the father will be twice as old as the son. Equation: Father's Future Age = 2 × (Son's Future Age) 3x + 12 = 2(x + 12)

Step 4: Solve for x. 3x + 12 = 2x + 24 Subtract 2x from both sides: x + 12 = 24 Subtract 12 from both sides: x = 12

The son's current age is 12. (And the father is 36).

Common Traps in Age Problems

• Forgetting to add/subtract the years to BOTH people. If 5 years pass, both the father and the son age by 5 years.
• Misplacing the multiplier. "Father is twice as old as the son" means you multiply the son's age by 2 to equal the father's age.

3. Mixture Problems

Mixture problems involve combining two or more substances of different concentrations to create a final mixture of a new concentration.

The Mixture Formula

The core concept is that the total amount of the "active ingredient" remains constant. Amount 1 + Amount 2 = Final Amount (Concentration 1 × Volume 1) + (Concentration 2 × Volume 2) = (Final Concentration × Final Volume)

Worked Example 2: A chemist has 20 liters of a 10% acid solution. How many liters of a 40% acid solution must be added to create a mixture that is 20% acid?

Step 1: Define the unknown. Let v = liters of the 40% solution to be added.

Step 2: Set up the equation using the mixture formula.

• Solution 1: 10% concentration, 20 liters. Amount of acid = 0.10(20).
• Solution 2: 40% concentration, v liters. Amount of acid = 0.40(v).
• Final Mixture: 20% concentration, total volume is (20 + v) liters. Amount of acid = 0.20(20 + v).

Equation: 0.10(20) + 0.40(v) = 0.20(20 + v)

Step 3: Solve the equation. 2 + 0.4v = 4 + 0.2v Subtract 0.2v from both sides: 2 + 0.2v = 4 Subtract 2 from both sides: 0.2v = 2 Divide by 0.2: v = 10

The chemist must add 10 liters of the 40% solution.

4. Consecutive Integer Problems

These problems ask you to find numbers that follow each other in order.

Strategic Tip: Expressing Consecutive Integers

• Consecutive integers: Let the numbers be n, n+1, n+2, n+3, etc.
• Consecutive even/odd integers: Let the numbers be n, n+2, n+4, n+6, etc. (Notice that whether you are looking for even or odd, the difference between consecutive numbers of the same parity is always 2).

Worked Example 3: The sum of three consecutive odd integers is 87. What is the largest integer?

Let the integers be n, n+2, and n+4. Equation: n + (n+2) + (n+4) = 87 3n + 6 = 87 3n = 81 n = 27

The question asks for the largest integer, which is n+4. 27 + 4 = 31. The largest integer is 31.

5. Coin and Value Problems

Another frequent category on the EDPT is the coin or value problem. These problems involve items that have an inherent monetary value (like coins, tickets, or stamps). The key to solving these is distinguishing between the number of items and the total value of those items.

Strategic Tip: The Value Equation

Just like mixture problems, you must account for the value of the items. Total Value = (Number of Item A × Value of Item A) + (Number of Item B × Value of Item B)

Worked Example 4: A vending machine contains only dimes and quarters. There are 45 coins in total, and their combined value is $7.50. How many dimes are in the machine?

Step 1: Define the variables. Since the problem asks for dimes, let d = number of dimes. Because there are 45 coins in total, the number of quarters must be 45 - d.

Step 2: Set up the value equation.

• The value of the dimes is 0.10 × d (or 10 cents).
• The value of the quarters is 0.25 × (45 - d) (or 25 cents).
• Total value = $7.50 (or 750 cents). To avoid decimals, it is often easier to work entirely in cents. Let's use cents:

Equation: 10d + 25(45 - d) = 750

Step 3: Solve the equation. Distribute the 25: 10d + 1125 - 25d = 750 Combine like terms: -15d + 1125 = 750 Subtract 1125: -15d = -375 Divide by -15: d = 25

There are 25 dimes in the machine. (And to verify, 45 - 25 = 20 quarters. 20 quarters = $5.00, 25 dimes = $2.50. Total = $7.50).

Common Traps in Coin Problems

• Mixing units: Do not write 10d + 25q = 7.50. You must either write 0.10d + 0.25q = 7.50 (all dollars) or 10d + 25q = 750 (all cents). Mixing them will guarantee a wrong answer.
• Forgetting the quantity relationship: Candidates often introduce two variables (d for dimes, q for quarters) and then get stuck. Always express one quantity in terms of the other using the total count given in the problem.

Final Review of Strategic Tips

1. Always double-check what the question is asking for. If you find x, make sure x is the actual answer and not just an intermediate step.
2. Plug your final answer back into the original word problem to ensure it makes logical sense. If you calculate an age to be negative, you have made a setup error.
3. Be rigorous with your signs when translating "difference" and "less than".