Write rational expressions from context

1. Use distance = rate × time: 240 = r \cdot t.
2. Solve for time by dividing both sides by r: t = \dfrac{240}{r}.

Answer: \dfrac{240}{r}

Check: If r = 60 mph, the time is \dfrac{240}{60} = 4 hours, and 60 \times 4 = 240 miles ✓. A faster speed (larger r) gives a smaller time, which makes sense.

Tip: Time is always distance on top, speed on the bottom. Test any choice with a friendly speed; 240r would give 14,400 hours.

1. The total cost stays $90.
2. The number of people changes from n to n + 2.
3. Each person's share = total ÷ number of people = \dfrac{90}{n + 2}.

Answer: \dfrac{90}{n + 2}

Check: If n = 4, there are now 6 people, and each pays \dfrac{90}{6} = \$15. The expression gives \dfrac{90}{4 + 2} = 15 ✓.

Tip: A change in the number of people belongs in the denominator, because it changes what you divide by.

1. Time = (amount of work) ÷ (rate of work).
2. The amount of work is 450 pages. The rate is p pages per minute.
3. So the time is \dfrac{450}{p} minutes.
4. Units check: \dfrac{\text{pages}}{\text{pages per minute}} = \text{minutes} ✓.

Answer: \dfrac{450}{p}

Check: Try p = 30: \dfrac{450}{30} = 15 minutes. And 30 pages each minute for 15 minutes is 30 \times 15 = 450 pages ✓. The expression \dfrac{p}{450} would give \tfrac{1}{15} of a minute — far too fast.

Tip: When you are unsure how to write an expression, replace the variable with a simple number, solve that easy problem, and then put the variable back where the number was.

1. Flying into a headwind slows the plane down, so its ground speed is 180 - w miles per hour.
2. Time = distance ÷ speed.
3. So the flight time is \dfrac{450}{180 - w} hours.

Answer: \dfrac{450}{180 - w}

Check: Try w = 30: ground speed 150 mph, time \dfrac{450}{150} = 3 hours. With no wind it would take \dfrac{450}{180} = 2.5 hours, so the headwind makes the trip longer ✓.

Tip: A headwind or an upstream current is subtracted from the speed; a tailwind or a downstream current is added. The change goes inside the denominator, with the speed.

1. The total cost stays the same: $1,500.
2. One family leaves, so the number of families sharing is f - 1.
3. Each family's share = total ÷ number of families = \dfrac{1500}{f - 1}.

Answer: \dfrac{1500}{f - 1}

Check: Suppose f = 6. Then 5 families pay \dfrac{1500}{5} = 300 dollars each, and 5 \times 300 = 1500 ✓. With 6 families each would pay $250, so the share goes up when someone leaves — which matches the expression.

Tip: A change in the number of people changes the denominator. Write the new count first, in parentheses, then divide the total by it.

1. Fixed cost for the day: $350 (it does not depend on m).
2. Ingredient cost for m meals: 2.50m dollars.
3. Total daily cost: 350 + 2.50m.
4. Average cost per meal = total ÷ number of meals = \dfrac{350 + 2.50m}{m}.

Answer: \dfrac{350 + 2.50m}{m}

Check: For m = 100 meals: \dfrac{350 + 250}{100} = \dfrac{600}{100} = 6 dollars per meal. In words: $3.50 of fixed cost per meal plus $2.50 of ingredients is $6 ✓.

Tip: Average cost = (fixed cost + cost per item × number of items) ÷ number of items. The per-item cost is multiplied by the variable; the fixed cost is not.

1. Upstream means against the current, so the current slows the boat: speed = 15 - c mph.
2. Time = distance ÷ speed = \dfrac{36}{15 - c}.

Answer: \dfrac{36}{15 - c}

Check: If c = 3, the boat moves at 12 mph upstream, and 36 \div 12 = 3 hours. The expression gives \dfrac{36}{15 - 3} = 3 ✓.

Tip: Upstream: subtract the current. Downstream: add the current. Then divide the distance by that speed.

1. New points: 95 on each of n quizzes = 95n.
2. New total: 410 + 95n.
3. New number of quizzes: 5 + n.
4. Average = \dfrac{410 + 95n}{5 + n}.

Answer: \dfrac{410 + 95n}{5 + n}

Check: If n = 1: total = 505 over 6 quizzes, average \approx 84.2. The expression gives \dfrac{505}{6} \approx 84.2 ✓.

Tip: When new items are added to an average, update the numerator and the denominator.

1. The numerator, 480 + 15n, is the total cost of the order.
2. The term 15n grows with the number of jerseys, so $15 is the cost of each jersey.
3. The term 480 does not change with n. It is a one-time (fixed) cost, such as a design or setup fee.
4. The denominator, n, is the number of jerseys, so dividing gives the average cost per jersey.

Answer: a fixed cost of $480 that is paid no matter how many jerseys are ordered

Check: Order 10 jerseys: total 480 + 150 = 630. Order 20 jerseys: total 480 + 300 = 780. The difference is $150 for 10 more jerseys ($15 each), while the $480 is in both totals ✓.

Tip: In an expression like \dfrac{F + vn}{n}, the term with the variable is the per-item part, and the plain number is the fixed part.

1. Older mixer: \dfrac{1}{5} of a batch per hour, so in 2 hours: \dfrac{2}{5}.
2. Newer mixer: \dfrac{1}{h} of a batch per hour, so in 2 hours: \dfrac{2}{h}.
3. Together: \dfrac{2}{5} + \dfrac{2}{h}.
4. Common denominator 5h: \dfrac{2h}{5h} + \dfrac{10}{5h} = \dfrac{2h + 10}{5h}.

Answer: \dfrac{2h + 10}{5h}

Check: If h = 5 (both mixers equal), each makes \tfrac{2}{5} in 2 hours, together \tfrac{4}{5}. The expression: \dfrac{10 + 10}{25} = \dfrac{20}{25} = \dfrac{4}{5} ✓.

Tip: Work-rate problems add rates (fractions of a job per hour), never times. Work done = rate × time.

1. Highway time: \dfrac{15}{r} hours.
2. Town speed: 25 mph slower, so r - 25 mph.
3. Town time: \dfrac{6}{r - 25} hours.
4. Total time: \dfrac{15}{r} + \dfrac{6}{r - 25}.

Answer: \dfrac{15}{r} + \dfrac{6}{r - 25}

Check: Try r = 60: highway \dfrac{15}{60} = 0.25 h; town speed 35 mph, time \dfrac{6}{35} \approx 0.17 h; total about 0.42 h (25 minutes). Choice B would give \dfrac{21}{95} \approx 0.22 h, which is less than the highway part alone — impossible.

Tip: Each leg of a trip has its own distance and its own speed, so it needs its own fraction. Combine legs by adding times.

1. Acid in the original solution: 30\% of 20 liters = 0.30 \times 20 = 6 liters.
2. Adding pure water does not change the amount of acid: still 6 liters.
3. Total volume after adding water: 20 + x liters.
4. Fraction that is acid = \dfrac{\text{acid}}{\text{total}} = \dfrac{6}{20 + x}.

Answer: \dfrac{6}{20 + x}

Check: If she adds x = 10 liters of water: \dfrac{6}{30} = 0.2 = 20\%. The solution got weaker (from 30% to 20%), which makes sense for added water ✓.

Tip: Concentration = (amount of the ingredient) ÷ (total amount). Adding water changes only the denominator; adding the ingredient itself changes both the numerator and the denominator.

1. Old time: \dfrac{30}{r}. New time: \dfrac{30}{r + 10}.
2. Time saved = \dfrac{30}{r} - \dfrac{30}{r + 10}.
3. LCD = r(r + 10): \dfrac{30(r + 10) - 30r}{r(r + 10)}.
4. Numerator: 30r + 300 - 30r = 300.
5. Time saved = \dfrac{300}{r(r + 10)}.

Answer: \dfrac{300}{r(r + 10)}

Check: Let r = 50: old time = 0.6 h, new time = 0.5 h, saved = 0.1 h. Expression: \dfrac{300}{50 \cdot 60} = \dfrac{300}{3000} = 0.1 ✓.

Tip: "Time saved," "how much more," and "how much less" are differences of two rational expressions. Write each one first, then subtract.

1. Price per bottle, 24-pack: \dfrac{d}{24}.
2. Price per bottle, 36-pack: \dfrac{d + 3}{36}.
3. Difference: \dfrac{d}{24} - \dfrac{d + 3}{36}.
4. Common denominator 72: \dfrac{3d}{72} - \dfrac{2(d + 3)}{72} = \dfrac{3d - 2d - 6}{72}.
5. Simplify: \dfrac{d - 6}{72}.

Answer: \dfrac{d - 6}{72}

Check: Let d = 12: 24-pack is \tfrac{12}{24} = 0.50 per bottle; 36-pack is \tfrac{15}{36} \approx 0.4167. Difference \approx 0.0833. Expression: \dfrac{12 - 6}{72} = \dfrac{6}{72} \approx 0.0833 ✓.

Tip: When you subtract a fraction whose numerator is a sum, put it in parentheses: -2(d + 3) = -2d - 6. Forgetting to distribute the minus sign is the most common error.

1. x + 2 is the number of hours Tom needs to sort the whole shipment alone (2 hours longer than Grace).
2. If a whole job takes x + 2 hours, then in 1 hour you finish \dfrac{1}{x + 2} of it.
3. So \dfrac{1}{x + 2} is the fraction of the shipment Tom sorts in one hour (Tom's work rate).
4. Likewise, \dfrac{1}{x} is Grace's fraction per hour, and the sum is their combined fraction per hour.

Answer: the fraction of the shipment Tom sorts in one hour

Check: Try x = 4: Tom needs 6 hours, so in 1 hour he sorts \tfrac{1}{6} of the shipment, and \dfrac{1}{4 + 2} = \dfrac{1}{6} ✓.

Tip: In work problems, a time looks like x or x + 2; a rate looks like 1 over a time. Read which one an expression shows before you answer.

1. Downstream speed: 5 + c, time \dfrac{4}{5 + c}.
2. Upstream speed: 5 - c, time \dfrac{4}{5 - c}.
3. Total: \dfrac{4}{5 + c} + \dfrac{4}{5 - c}.
4. Common denominator (5 + c)(5 - c) = 25 - c^2: \frac{4(5 - c) + 4(5 + c)}{25 - c^2} = \frac{20 - 4c + 20 + 4c}{25 - c^2} = \frac{40}{25 - c^2}

Answer: \dfrac{40}{25 - c^2}

Check: Let c = 1: downstream \tfrac{4}{6} \approx 0.667 h, upstream \tfrac{4}{4} = 1 h, total \approx 1.667 h. Expression: \dfrac{40}{24} \approx 1.667 ✓.

Tip: With a current, the trip with the current and the trip against it get different denominators. The current does not cancel out: \dfrac{40}{25 - c^2} is longer than the still-water time \dfrac{8}{5}.

1. Total miles: 120 + x.
2. Gallons in the city: \dfrac{120}{20} = 6 gallons.
3. Gallons on the highway: \dfrac{x}{30} gallons.
4. Total gallons: 6 + \dfrac{x}{30}.
5. Overall mileage = \dfrac{\text{total miles}}{\text{total gallons}} = \dfrac{120 + x}{6 + \dfrac{x}{30}}.

Answer: \dfrac{120 + x}{6 + \dfrac{x}{30}}

Check: Let x = 180: total miles 300; gallons 6 + 6 = 12; mileage \dfrac{300}{12} = 25 mpg ✓. Now let x = 30: \dfrac{150}{7} \approx 21.4 mpg — closer to the city figure because most driving was in the city. A fixed answer like 25 cannot be right for every x.

Tip: Miles per gallon is a rate. To combine two rates, add the amounts (miles) and add the bases (gallons) separately; never average the two rates.

1. Share with n members: \dfrac{1200}{n}.
2. Share with n + 5 members: \dfrac{1200}{n + 5}.
3. The share goes down, so subtract: \dfrac{1200}{n} - \dfrac{1200}{n + 5}.
4. Common denominator n(n + 5): \dfrac{1200(n + 5) - 1200n}{n(n + 5)} = \dfrac{1200n + 6000 - 1200n}{n(n + 5)}.
5. Simplify: \dfrac{6000}{n(n + 5)}.

Answer: \dfrac{6000}{n(n + 5)}

Check: Let n = 20: 20 members pay $60 each; 25 members pay $48 each; savings $12. Expression: \dfrac{6000}{20 \cdot 25} = \dfrac{6000}{500} = 12 ✓.

Tip: "How much less" means (old amount) − (new amount). Keep the order so the answer comes out positive.

1. Water still needed: 18{,}000 - 3{,}000 = 15{,}000 gallons.
2. Time in minutes: \dfrac{15000}{g}.
3. There are 60 minutes in an hour, so divide by 60: \dfrac{15000}{60g} hours.
4. Simplify: \dfrac{15000}{60} = 250, so the time is \dfrac{250}{g} hours.

Answer: \dfrac{250}{g}

Check: Let g = 10 gallons per minute: \dfrac{15000}{10} = 1500 minutes, and 1500 \div 60 = 25 hours. Expression: \dfrac{250}{10} = 25 ✓.

Tip: Watch the units in the question. If the rate is per minute but the answer must be in hours, divide the minutes by 60 (an hour is a bigger unit, so the number gets smaller).

1. A: Speed x + 3 mph over 60 miles. Time = distance ÷ speed = \dfrac{60}{x + 3}. Matches.
2. B: Driving 60 miles at x mph takes \dfrac{60}{x} hours; adding a 3-hour stop gives \dfrac{60}{x} + 3. Does not match.
3. C: A $60 bill plus a $3 tip shared by x people: \dfrac{63}{x}. Does not match.
4. D: x + 3 miles at 60 mph takes \dfrac{x + 3}{60} hours. Does not match (upside down).

Answer: A — the number of hours to travel 60 miles at a speed 3 mph faster than x mph

Check: Let x = 9: at 12 mph, 60 miles take 5 hours, and \dfrac{60}{9 + 3} = 5 ✓. Choice B would give \dfrac{60}{9} + 3 \approx 9.67 hours.

Tip: Where the "+3" sits tells the story. Inside the denominator, it changes the rate or group size; outside the fraction, it adds to the answer; in the numerator, it adds to the total.

1. Lena's time: x hours, so her rate is \dfrac{1}{x} set per hour.
2. Omar's time: x + 3 hours, so his rate is \dfrac{1}{x + 3}.
3. Combined rate: \dfrac{1}{x} + \dfrac{1}{x + 3} = \dfrac{(x + 3) + x}{x(x + 3)} = \dfrac{2x + 3}{x(x + 3)}.
4. Time together = reciprocal: \dfrac{x(x + 3)}{2x + 3}.

Answer: \dfrac{x(x + 3)}{2x + 3}

Check: Let x = 3: Lena takes 3 h, Omar 6 h. Rates \tfrac{1}{3} + \tfrac{1}{6} = \tfrac{1}{2}, so together they take 2 h. Expression: \dfrac{3 \cdot 6}{9} = 2 ✓.

Tip: Two-step work problems: add rates, then flip. If the choice you pick is a rate (a fraction of a job per hour), you have stopped one step early.

1. Cost per cake: 7.50 + 0.50 = 8 dollars.
2. Total cost for c cakes: 180 + 8c.
3. Average cost per cake: \dfrac{180 + 8c}{c}.
4. Set equal to 12: \dfrac{180 + 8c}{c} = 12.
5. Multiply by c: 180 + 8c = 12c.
6. Subtract 8c: 180 = 4c, so c = 45.

Answer: \dfrac{180 + 8c}{c}; 45 cakes

Check: \dfrac{180 + 8(45)}{45} = \dfrac{180 + 360}{45} = \dfrac{540}{45} = 12 ✓.

Tip: In a cost table, sort each line into fixed (paid once) or per item (multiplied by the variable). Add all the per-item lines before you write the expression.

1. Route A time: \dfrac{50}{r}.
2. Route B speed: r - 15, so time: \dfrac{40}{r - 15}.
3. How much longer B takes: \dfrac{40}{r - 15} - \dfrac{50}{r}.
4. Common denominator r(r - 15): \frac{40r - 50(r - 15)}{r(r - 15)} = \frac{40r - 50r + 750}{r(r - 15)} = \frac{750 - 10r}{r(r - 15)}

Answer: \dfrac{750 - 10r}{r(r - 15)}

Check: Let r = 55: Route A \dfrac{50}{55} \approx 0.909 h; Route B \dfrac{40}{40} = 1 h; difference \approx 0.091 h. Expression: \dfrac{750 - 550}{55 \cdot 40} = \dfrac{200}{2200} \approx 0.091 ✓.

Tip: When you subtract 50(r - 15), the minus sign reaches both terms: -50r + 750. Test your final expression with one easy number to catch sign slips.