<h3>The Main Idea</h3> <p><strong>Rates add, times don't.</strong></p> <p>When two workers or machines work together, you add their rates (jobs per hour), not their times. This is the most common mistake — you can't just add hours together.</p> <h3>What is a Combined Work Problem?</h3> <p>These problems involve two or more workers/machines completing a task together. You're usually asked: "How long does it take working together?" or "How much does each contribute?"</p> <h3>The Rate Approach</h3> <ol> <li><strong>Convert time to rate</strong> — If someone finishes in 6 hours, their rate is 1/6 job per hour.</li> <li><strong>Add the rates</strong> — Combined rate = Rate A + Rate B</li> <li><strong>Convert back to time</strong> — Time = 1 ÷ Combined rate</li> </ol> <h3>Key Formula</h3> <p>If A takes <em>a</em> hours alone and B takes <em>b</em> hours alone:</p> $$\frac{1}{t} = \frac{1}{a} + \frac{1}{b}$$ <p>Where <em>t</em> is the time working together.</p> <h3>Shortcut Formula</h3> <p>For two workers:</p> $$t = \frac{a \times b}{a + b}$$ <p>Product over sum — quick when you have two simple times.</p> <h3>Common Variations</h3> <ul> <li><strong>Finding individual time</strong> — Given combined time and one person's time, find the other's.</li> <li><strong>Partial work</strong> — One person works alone for a while, then they work together.</li> <li><strong>Three or more workers</strong> — Just add all the rates: 1/a + 1/b + 1/c</li> </ul> <h3>Common Pitfalls</h3> <ul> <li><strong>Adding times directly</strong> — WRONG: "3 hours + 6 hours = 9 hours together." Times don't add!</li> <li><strong>Forgetting to flip</strong> — After adding rates, flip to get time.</li> <li><strong>Mixing units</strong> — Keep everything in the same unit (hours or minutes, not both).</li> </ul>
Practice This Goal<h3>The Main Idea</h3> <p><strong>The amount of pure substance stays trackable.</strong></p> <p>In mixture problems, track the actual amount of the "stuff" (salt, acid, alcohol) — not just the percentages. Concentration × Volume = Amount of pure substance.</p> <h3>What is a Mixture Problem?</h3> <p>These problems involve combining solutions of different concentrations, or adding/removing liquid to change a concentration. You're usually finding the final concentration or how much to add.</p> <h3>The Core Equation</h3> $$\text{Concentration} \times \text{Volume} = \text{Amount of Pure Substance}$$ <p>This works for any mixture — salt water, acid solutions, alcohol, etc.</p> <h3>Setting Up the Problem</h3> <ol> <li><strong>Identify what's being mixed</strong> — Two solutions? Adding water? Adding pure substance?</li> <li><strong>Calculate pure amounts</strong> — For each component: concentration × volume</li> <li><strong>Add the amounts</strong> — Pure substance from A + Pure substance from B = Total pure</li> <li><strong>Find final concentration</strong> — Total pure ÷ Total volume = Final concentration</li> </ol> <h3>Common Scenarios</h3> <ul> <li><strong>Mixing two solutions</strong> — Amount₁ + Amount₂ = Total Amount</li> <li><strong>Adding water (diluting)</strong> — Water has 0% concentration, but adds to total volume</li> <li><strong>Adding pure substance</strong> — Pure substance is 100% concentration</li> <li><strong>Evaporation</strong> — Volume decreases, but pure amount stays the same</li> </ul> <h3>Common Pitfalls</h3> <ul> <li><strong>Averaging percentages</strong> — WRONG: "20% + 40% = 30%." Only works if volumes are equal!</li> <li><strong>Forgetting that water is 0%</strong> — Adding water adds volume but zero pure substance.</li> <li><strong>Mixing up what you're solving for</strong> — Read carefully: final concentration? Amount to add?</li> </ul>
Practice This Goal<h3>The Main Idea</h3> <p><strong>If it's a right triangle, use a² + b² = c².</strong></p> <p>The Pythagorean Theorem relates the three sides of a right triangle. The key is recognizing when a problem involves a right angle.</p> <h3>The Theorem</h3> $$a^2 + b^2 = c^2$$ <p>Where <em>c</em> is the hypotenuse (longest side, opposite the right angle) and <em>a</em> and <em>b</em> are the legs.</p> <h3>When to Use It</h3> <p>Look for these right-angle scenarios:</p> <ul> <li><strong>Ladders against walls</strong> — Ground and wall form a right angle</li> <li><strong>Diagonals of rectangles</strong> — Rectangle corners are 90°</li> <li><strong>Distance problems</strong> — North/South and East/West are perpendicular</li> <li><strong>Shadows and heights</strong> — Object and ground form a right angle</li> <li><strong>"Straight line distance"</strong> — Often implies a right triangle</li> </ul> <h3>Solving Steps</h3> <ol> <li><strong>Draw the triangle</strong> — Label the right angle and all sides.</li> <li><strong>Identify hypotenuse vs legs</strong> — Hypotenuse is always opposite the right angle.</li> <li><strong>Set up the equation</strong> — Plug known values into a² + b² = c².</li> <li><strong>Solve for the unknown</strong> — Square root at the end if finding a side.</li> </ol> <h3>Common Pythagorean Triples</h3> <p>Memorize these — they appear often and save calculation time:</p> <ul> <li><strong>3-4-5</strong> (and multiples: 6-8-10, 9-12-15, 12-16-20)</li> <li><strong>5-12-13</strong> (and 10-24-26)</li> <li><strong>8-15-17</strong></li> <li><strong>7-24-25</strong></li> </ul> <h3>Common Pitfalls</h3> <ul> <li><strong>Mixing up legs and hypotenuse</strong> — c² is always the LONGEST side, alone on one side of the equation.</li> <li><strong>Forgetting the square root</strong> — After finding c², you still need to take √c².</li> <li><strong>Not recognizing the right angle</strong> — Walls, floors, perpendicular paths = right angles.</li> </ul>
Practice This Goal