1.4 Mathematical Word Problems

1.4 Mathematical Word Problems

Word problems translate real-world scenarios into algebraic equations. On the PMAEE, these questions test your logical modeling and speed. This section details formulas and strategies for rate-work, mixture, age, ratio-proportion, and speed-distance-time problems.

Rate and Work Problems

Work problems involve individuals or machines completing a task at a specific rate. The fundamental work equation is:

$\text{Work} = \text{Rate} \times \text{Time}$

When the total work is not specified, it is assumed to be $1$ (representing one complete job). Thus, the rate of an individual is $R = \frac{1}{t}$, where $t$ is the time it takes them to complete the job alone.

Combined Rates

When multiple entities work together, their individual rates of work are added:

$\text{Combined Rate} = R_1 + R_2 + \dots + R_n$

For two individuals with solo times $t_1$ and $t_2$, the combined time $T$ to complete $1$ job satisfies:

$\frac{1}{t_1} + \frac{1}{t_2} = \frac{1}{T} \Rightarrow T = \frac{t_1 t_2}{t_1 + t_2}$

Pipes and Pumps (Inflow vs. Outflow)

If a pipe fills a tank (inflow) while a drain empties it (outflow), the rates are subtracted:

$\text{Net Rate} = \frac{1}{t_{\text{in}}} - \frac{1}{t_{\text{out}}}$

Worked Example: Combined Rates with Inflow and Outflow

Pipe A fills a pool in 4 hours, and Pipe B fills it in 6 hours. Drain C empties a full pool in 12 hours. If all three are opened together on an empty pool, how long will it take to fill?

Solution: Write down the rate of each pipe per hour:

• Rate A = $+\frac{1}{4}$ pool/hour
• Rate B = $+\frac{1}{6}$ pool/hour
• Rate C = $-\frac{1}{12}$ pool/hour (negative because it removes water)

Find the net rate per hour by summing the rates:

$\text{Net Rate} = \frac{1}{4} + \frac{1}{6} - \frac{1}{12}$

Find the least common denominator (LCD = 12):

$\text{Net Rate} = \frac{3}{12} + \frac{2}{12} - \frac{1}{12} = \frac{4}{12} = \frac{1}{3} \text{ pool/hour}$

Since the net rate is $\frac{1}{3}$ pool per hour, the time to fill 1 pool is:

$\text{Time} = \frac{1}{\text{Net Rate}} = 3 \text{ hours}$

Mixture Problems

Mixture problems involve blending two or more substances with different concentrations to achieve a final target concentration. The key is to track the amount of the pure substance (e.g., pure acid, salt, or alcohol) before and after mixing:

$\text{Volume} \times \text{Concentration} = \text{Amount of Pure Substance}$

The standard equation for mixing two volumes $V_1$ and $V_2$ with concentrations $C_1$ and $C_2$ to get a total volume $V_f$ with concentration $C_f$ is:

$C_1 V_1 + C_2 V_2 = C_f V_f$

where $V_f = V_1 + V_2$.

Worked Example: Dilution

How many liters of a 20% acid solution must be mixed with 10 liters of a 50% acid solution to obtain a 30% acid solution?

Solution: Let $x$ be the volume of the 20% solution needed.

• Pure acid in first solution: $0.20x$
• Pure acid in second solution: $0.50(10) = 5$ liters
• Total volume of mixture: $x + 10$ liters
• Pure acid in final mixture: $0.30(x + 10)$

Set up the equation:

$0.20x + 5 = 0.30(x + 10)$ $0.20x + 5 = 0.30x + 3$

Subtract $0.20x$ and $3$ from both sides:

$2 = 0.10x \Rightarrow x = 20 \text{ liters}$

Thus, 20 liters of the 20% solution must be added.

Age Problems

Age problems compare the ages of people at different points in time (past, present, and future). The best approach is to:

1. Define a variable (usually representing the current age of one person).
2. Construct a table to organize the expressions for their ages 'now', '$N$ years ago', and 'in $M$ years'.
3. Write an equation using the relationships specified in the problem statement.

Worked Example: Age Comparison

A father is currently 3 times as old as his son. In 12 years, the father will be twice as old as his son. How old are they now?

Solution: Let $x$ be the son's age now. Then the father's age now is $3x$.

• In 12 years:
  • Son's age = $x + 12$
  • Father's age = $3x + 12$

According to the problem, the father's age in 12 years will be twice the son's age in 12 years:

$3x + 12 = 2(x + 12)$ $3x + 12 = 2x + 24$

Subtract $2x$ and $12$ from both sides:

$x = 12$

So, the son is 12 years old now, and the father is $3(12) = 36$ years old.

Ratio and Proportion

A ratio is a comparison of two quantities, written as $a:b$ or $\frac{a}{b}$. A proportion is an equation stating that two ratios are equal: $\frac{a}{b} = \frac{c}{d}$.

Types of Proportion

1. Direct proportion: As one variable increases, the other increases proportionally ($y = kx$).

$\frac{x_1}{y_1} = \frac{x_2}{y_2}$

1. Inverse proportion: As one variable increases, the other decreases proportionally ($y = \frac{k}{x}$).

$x_1 y_1 = x_2 y_2$

1. Partitive proportion: A whole is divided into parts based on a given ratio. For example, dividing $100$ in the ratio $2:3$ means the parts are $2x$ and $3x$, where $2x + 3x = 100 \Rightarrow 5x = 100 \Rightarrow x = 20$. The parts are 40 and 60.

Speed, Distance, and Time

The fundamental formula for motion is:

$\text{Distance} (d) = \text{Speed} (r) \times \text{Time} (t)$

Relative Speed

• Approaching / Opposite Directions: When two objects move towards each other, their relative speed is the sum of their individual speeds:

$r_{\text{rel}} = r_1 + r_2$

• Overtaking / Same Direction: When one object chases another, the relative speed is the difference:

$r_{\text{rel}} = r_1 - r_2 \quad (\text{where } r_1 > r_2)$

River Current Problems

• Downstream (moving with the current): The current increases the boat's speed.

$\text{Effective Speed} = v_{\text{still}} + v_{\text{current}}$

• Upstream (moving against the current): The current decreases the boat's speed.

$\text{Effective Speed} = v_{\text{still}} - v_{\text{current}}$

graph LR
    subgraph Downstream["Downstream Motion"]
        B1["Boat Speed (v)"] -->|"+"| C1["Current Speed (c)"]
        C1 --> DS["Effective Speed = v + c"]
    end
    subgraph Upstream["Upstream Motion"]
        B2["Boat Speed (v)"] -->|"-"| C2["Current Speed (c)"]
        C2 --> US["Effective Speed = v - c"]
    end