3.5 Algebraic Word Problems

3.5 Algebraic Word Problems

Algebraic word problems translate real-world scenarios into formal equations. Mastering systematic translation strategies for rate-time-distance, work rate, mixture, age, and consecutive integer problems is essential for achieving a top Quantitative Reasoning score.

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1. Translating English into Algebraic Equations

Success in word problems depends on precise translation of verbal phrases into algebraic expressions:

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2. Rate, Time, and Distance Problems

The fundamental relationship governing uniform motion is: $\text{Distance} = \text{Rate} \times \text{Time} \quad (d = r \cdot t)$

Average Speed Trap

WARNING: Average speed for a trip is NEVER the simple arithmetic mean of the speeds $\frac{r_1 + r_2}{2}$ (unless the travel times at each speed are equal). Average speed is strictly defined as: $\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}$

Worked Example 3.5.1: Round-Trip Average Speed

A car travels from Town X to Town Y at an average speed of 60 miles per hour, and returns along the exact same route from Town Y to Town X at an average speed of 40 miles per hour. Calculate the car's average speed for the entire round trip.

Solution:

1. Let the one-way distance between Town X and Town Y be $d = 120$ miles (smart number choice: LCM of 60 and 40).
2. Compute travel times:
   • Time outbound ($t_1$): $\frac{120 \text{ miles}}{60 \text{ mph}} = 2 \text{ hours}$.
   • Time return ($t_2$): $\frac{120 \text{ miles}}{40 \text{ mph}} = 3 \text{ hours}$.
3. Calculate Total Distance and Total Time:
   • Total Distance $= 120 + 120 = 240 \text{ miles}$.
   • Total Time $= 2 + 3 = 5 \text{ hours}$.
4. Compute Average Speed: $\text{Average Speed} = \frac{240 \text{ miles}}{5 \text{ hours}} = 48 \text{ mph}$ (Notice that 48 mph is lower than the simple average of 50 mph because the car spends more time traveling at the slower speed!)

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3. Work Rate Problems

Work rate problems measure job completion over time. If a person or machine completes 1 whole job in $T$ hours, their work rate $R$ is: $R = \frac{1}{T} \text{ (job per hour)}$

Combined Work Formula

When two workers work together simultaneously at constant rates $R_1 = \frac{1}{T_1}$ and $R_2 = \frac{1}{T_2}$, their combined rate $R_{\text{total}}$ is: $R_{\text{total}} = R_1 + R_2 = \frac{1}{T_1} + \frac{1}{T_2}$ The combined time $T_{\text{combined}}$ required to complete 1 job together is: $T_{\text{combined}} = \frac{T_1 \cdot T_2}{T_1 + T_2}$

Worked Example 3.5.2: Combined Work Time

Worker A can complete a project alone in 15 hours, while Worker B can complete the same project alone in 10 hours. How many hours will it take them working together at constant rates?

Solution:

1. Worker A's rate: $R_A = \frac{1}{15}$ job/hr.
2. Worker B's rate: $R_B = \frac{1}{10}$ job/hr.
3. Combined rate: $R_{\text{total}} = \frac{1}{15} + \frac{1}{10} = \frac{2}{30} + \frac{3}{30} = \frac{5}{30} = \frac{1}{6}$ job/hr.
4. Combined time: $T_{\text{combined}} = \frac{1}{1/6} = 6 \text{ hours}$.

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4. Concentration and Mixture Problems

Mixture problems involve combining solutions of different concentration percentages. The core principle is conservation of solute mass (or volume): $\text{Amount of Solute} = \text{Concentration (\%)} \times \text{Total Volume}$

Mixture Equation

When mixing Volume $V_1$ of concentration $C_1$ with Volume $V_2$ of concentration $C_2$: $C_1 V_1 + C_2 V_2 = C_{\text{final}} (V_1 + V_2)$

Worked Example 3.5.3: Solution Mixture

How many liters of a 20% saline solution must be added to 30 liters of a 60% saline solution to obtain a mixture that is 50% saline?

Solution:

1. Let $x$ be the volume (in liters) of 20% saline solution added.
2. Express total salt content: $0.20(x) + 0.60(30) = 0.50(x + 30)$
3. Expand and simplify: $0.20x + 18 = 0.50x + 15$
4. Subtract $0.20x$ and 15 from both sides: $3 = 0.30x \implies x = \frac{3}{0.30} = 10 \text{ liters}$ 10 liters of 20% saline solution must be added.

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5. Age Problems & Consecutive Integers

Age Problems

Always establish a table representing past, present, and future ages for each person:

• If a person is currently $x$ years old:
  • $N$ years ago, their age was $x - N$.
  • In $M$ years, their age will be $x + M$.

Consecutive Integer Representation

• Consecutive Integers: $n, n+1, n+2, n+3, \dots$
• Consecutive Even Integers: $n, n+2, n+4, n+6, \dots$ (where $n$ is even)
• Consecutive Odd Integers: $n, n+2, n+4, n+6, \dots$ (where $n$ is odd)

Worked Example 3.5.4: Consecutive Even Integers

The sum of four consecutive even integers is 140. Find the largest of these integers.

Solution:

1. Let the four consecutive even integers be $n, n+2, n+4, n+6$.
2. Set up sum equation: $n + (n+2) + (n+4) + (n+6) = 140$.
3. Combine like terms: $4n + 12 = 140 \implies 4n = 128 \implies n = 32$.
4. The largest integer is $n + 6 = 32 + 6 = 38$.

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6. GRE Practical Word Problem Tactics

• Backsolving (Plugging in Answers): When answer choices are discrete numbers, test Choice C first. If Choice C is too large, eliminate C, D, and E (for ascending choices).
• Picking Smart Numbers: In work and rate problems without given quantities, pick convenient numbers for distance or total job size (such as the LCM of rates/times).