WORD PROBLEMS

Overview

Word problems are the AMC10's way of testing whether you can read mathematics hidden inside everyday language. The core skill is translation: identifying what quantity each phrase represents, writing an equation (or system) that captures every constraint, and then solving that equation efficiently. Almost every algebra topic — linear equations, ratios, rates, percentages — shows up wrapped in a word-problem costume.

The AMC10 favors several recurring word-problem types: rate and work problems (two workers or pipes finishing a job together), mixture problems (blending two solutions or combining items at different unit costs), and age problems (relating ages now and at some past or future time). For each type a single organizing equation template handles most variants, so recognizing the type is half the battle.

Weighted averages complete the picture. Whenever a problem asks for an overall average of groups with different sizes or weights, you must weight each sub-average by its count or fraction — never simply average the averages. This single mistake is responsible for a large share of wrong answers on AMC word problems.

Key facts

• Rate formula: $d = rt$ (distance = rate × time), rearranged as $r = d/t$ or $t = d/r$. For opposite-direction problems add rates; for same-direction problems subtract them.
• Combined work rate: if worker A takes $a$ hours alone and worker B takes $b$ hours alone, together their rate is $\tfrac{1}{a} + \tfrac{1}{b}$ jobs per hour, so the combined time is $\dfrac{ab}{a+b}$.
• Mixture equation: (amount of substance in solution 1) + (amount in solution 2) = (amount in final mixture). In symbols: $c_1 V_1 + c_2 V_2 = c_f (V_1 + V_2)$.
• Age problems: define a variable for someone's current age, express all other ages in terms of it, then translate the future or past relationship into a single equation.
• Weighted average: $\bar{x} = \dfrac{w_1 x_1 + w_2 x_2 + \cdots + w_n x_n}{w_1 + w_2 + \cdots + w_n}$. Weights can be counts, volumes, or any meaningful measure of size.
• Round-trip average speed: for equal distances at speeds $u$ and $v$, the average speed is the harmonic mean $\dfrac{2uv}{u+v}$, not the arithmetic mean.

Worked example 1

A faucet fills a bathtub in 24 minutes. The drain empties it in 40 minutes. If both are open, how long does it take to fill the tub?

Identify rates: the faucet adds $\tfrac{1}{24}$ of the tub per minute; the drain removes $\tfrac{1}{40}$ per minute. With both open, the net fill rate is

$\frac{1}{24} - \frac{1}{40} = \frac{5}{120} - \frac{3}{120} = \frac{2}{120} = \frac{1}{60}.$

So the tub fills at $\tfrac{1}{60}$ per minute, meaning it takes 60 minutes.

Why this works: we treat "fraction of job completed" as the quantity, so rates simply add (or subtract). The time is just the reciprocal of the net rate.

Worked example 2

A chemist has 12 liters of a 20% acid solution. She wants to create a 30% acid solution by adding pure acid. How many liters of pure acid must she add?

Let $x$ = liters of pure acid added. Pure acid is a 100% acid solution.

Before mixing: $0.20 \times 12 = 2.4$ liters of acid.
After mixing: total volume is $12 + x$, and the acid amount is $2.4 + x$.

Setting up the concentration equation:

$\frac{2.4 + x}{12 + x} = 0.30.$

Cross-multiply: $2.4 + x = 0.30(12 + x) = 3.6 + 0.30x$.

$0.70x = 1.2 \implies x = \frac{1.2}{0.70} = \frac{12}{7}.$

So she must add $\dfrac{12}{7} \approx 1.71$ liters of pure acid.

Check: acid in final mix = $2.4 + \tfrac{12}{7} = \tfrac{16.8 + 12}{7} = \tfrac{28.8}{7}$. Total volume = $12 + \tfrac{12}{7} = \tfrac{96}{7}$. Concentration = $\tfrac{28.8/7}{96/7} = \tfrac{28.8}{96} = 0.30$. ✓

Common traps

• Averaging the averages. If Group 1 averages 70 and Group 2 averages 90, the combined average is NOT 80 unless both groups are the same size. Always weight by group size.
• Mixing up "together" and "alone" rates. The combined time $ab/(a+b)$ is always less than either individual time. If your answer is larger than the smaller of the two individual times, you made an error.
• Forgetting the drain (or opposing force). In work problems, re-read carefully for phrases like "the drain is open" or "one machine undoes part of the work." Subtract opposing rates.
• Setting up the age equation in the wrong time frame. If a problem says "in 5 years, the father will be twice the son's age," write $(f + 5) = 2(s + 5)$, not $f = 2s + 5$ or $f + 5 = 2s$.
• Using arithmetic mean for round-trip speed. If you drive from A to B at 60 mph and return at 40 mph, the average speed is $2(60)(40)/(60+40) = 48$ mph, not $50$ mph.