Ratios, Rates & Proportions — AMC 8 & Math Kangaroo Lesson with Worked Examples

About this topic

Two cups of flour make six cookies. Six cups make… eighteen. You did not set up an equation. You felt the pattern: triple the flour, triple the cookies. That feeling is the whole lesson.

A ratio compares two amounts by division. Writing '2 : 3' (say 'two to three') tells you the shape of a comparison — for every 2 boys, 3 girls — not the actual count. A rate is a ratio with units: miles per hour, dollars per pound, dimples per second. A proportion says two ratios are equal, and chasing the missing piece of one is the most common move in contest rate problems.

Get this and you get a lot for free: a percent is a ratio out of 100, a map scale is a ratio, and the entire $D = S \times T$ family is one rate wearing different costumes. Nine chapters: ratio parts, proportions and percents, the $D = S \times T$ formula, average speed, unit conversion, relative speed, reading graphs, exponential growth, and work-rate.

CHAPTER 1

Ratios — parts of a whole

THEORY

Here is a recipe for trail mix: 2 scoops peanuts to 3 scoops raisins. Now picture three different bowls a friend made:

4 peanuts, 6 raisins
20 peanuts, 30 raisins
200 peanuts, 300 raisins

Different sizes — but bite for bite they taste identical. Each is the same 2 : 3 recipe, scaled up. A ratio doesn't pin down the counts. It pins down the flavor.

So how do you find the real counts? Think in parts. Each number in the ratio is a count of equal-sized parts. 2 : 3 means 5 parts in all.

RATIO PARTS

A ratio a : b has a + b total parts.

Total amount ÷ total parts = the size of one part.

Then each side's count = (its parts) × (one part).

Walkthrough. 30 students split 2 : 3, boys to girls.

Total parts: 2 + 3 = 5.
One part: 30 ÷ 5 = 6 students.
Boys = 2 × 6 = 12. Girls = 3 × 6 = 18.
Check: 12 + 18 = 30 ✓.

Three-way ratios work the same. A 1 : 2 : 3 split of 18 cookies has 6 parts, so one part = 3, and the three people get 3, 6, and 9.

The one-step shortcut: each side is a fraction OF THE WHOLE

Once you see the parts, you can skip computing 'one part' entirely. In 2 : 3, boys are 2 of the 5 parts — so boys are simply $\tfrac{2}{5}$ of everyone, and girls are $\tfrac{3}{5}$. Multiply that fraction by the total in one shot.

PART-TO-WHOLE

In a ratio a : b, the first side is a / (a + b) of the whole and the second is b / (a + b).

Each side's count = (its fraction) × (the total). One multiply, no 'find one part' detour.

So those 30 students: girls = $\tfrac{3}{5} \times 30 = 18$ — done in a single step. A 1 : 4 rope of 10 ft has a long piece that is $\tfrac{4}{5}$ of it = 8 ft. The 'one part' method and this fraction method are the same arithmetic; reach for whichever reads faster. The fraction form is also the bridge to percents in the next chapter — a percent is just that part-to-whole fraction forced onto 100.

Framing inspired by AoPS Prealgebra.

Bar models make the parts you can see

Draw each part as one equal bar — the same bars from the Algebra lesson. This is where they pay off.

Boys to girls is 3 : 5, and there are 12 more girls than boys. How many students in all?

Three equal bars for boys, five for girls. The girls have 2 extra bars — and those 2 bars are the 12 extra girls. So one bar = 12 ÷ 2 = 6. Total = 8 bars = 8 × 6 = 48.

THE MOVE: Add the parts. Find one part. Everything else is multiplication.

Joining two ratios that share a quantity

Now a move that unlocks a whole class of #16-level problems. You're handed two separate ratios that overlap on one quantity, and asked about the two that don't touch. Say a paint mix gives blue : white = 2 : 3 and, separately, white : red = 4 : 5. What is blue : red?

You can't just line them up — the white is written as 3 in one ratio and 4 in the other, and those are supposed to be the same stuff. The fix: force the shared term to agree. Scale each ratio so the white matches. The smallest value both 3 and 4 divide into is their LCM, 12:

blue : white = 2 : 3, multiply by 4 → 8 : 12.
white : red = 4 : 5, multiply by 3 → 12 : 15.

Now both call the white 12, so they snap into one chain:

Reading off the two ends: blue : red = 8 : 15. The shared white was the hinge — once it agreed, the outer two locked into place.

THE MOVE: Two ratios sharing one quantity → scale each so the shared number becomes their LCM, then read the chain.

Framing inspired by AoPS Prealgebra.

🎯 Try it

The ratio of 8th-graders to 6th-graders is 5 : 3, and the ratio of 8th-graders to 7th-graders is 8 : 5. What is the smallest total number of students who could be in the project?

Walkthrough: The 8th-graders are the shared hinge — their count must work in both ratios, so it's a multiple of 5 AND of 8. Smallest is lcm(5, 8) = 40. Then 6th = 40 × 3/5 = 24 and 7th = 40 × 5/8 = 25. Total = 40 + 24 + 25 = 89. (Smallest because we used the smallest shared count. This is amc8-2013-16, answer E.)

🎯 Try it

In sea water the ratio of salt to fresh water is 7 : 193. How many kilograms of salt are in 1000 kg of sea water?

Walkthrough: Total parts = 7 + 193 = 200. One part = 1000 ÷ 200 = 5 kg. Salt = 7 parts = 7 × 5 = 35 kg. (This is kangaroo-2013-kadett-03, answer A.)

THE TRICK

If a problem only asks for the difference between the two sides, don't compute both. The difference equals (difference in part counts) × (one part). For 3 : 5 with one part = 6, the gap is (5 − 3) × 6 = 12.

WORKED EXAMPLE

PROBLEM · 2020 #1

Luka is making lemonade to sell at a school fundraiser. His recipe requires 4 times as much water as sugar and twice as much sugar as lemon juice. He uses 3 cups of lemon juice. How many cups of water does he need?

A) 6 B) 8 C) 12 D) 18 E) 24

Luka's lemonade uses 4 times as much water as sugar, and twice as much sugar as lemon juice. He uses 3 cups of lemon juice. How much water?

Resist setting up three variables. Chain the scalings instead. Water is 4× the sugar, and sugar is 2× the lemon — so water is 4 × 2 = 8× the lemon juice.

Water = 8 × 3 = 24 cups (choice E).

One jump, not two. Any 'A is k times B, B is m times C' chain collapses to 'A is k·m times C'.

The temptation is to write $s$, $w$, $\ell$ and grind. But 'twice' then 'four times' is multiply-then-multiply. Collapse the chain first; then it's one multiplication on a real number you were handed.

Answer: E — 24 cups.

RULE OF THUMB

Add all parts to get the total parts. Total amount ÷ total parts = one part. Multiply each part count. For chained 'k times' descriptions, multiply the factors into one.

MORE LIKE THIS

2009 · #2 On average, for every 4 sports cars sold at the local dealership, 7 sedans are sold. The dealership predicts that it will sell 28 sports...

On average, for every 4 sports cars sold at the local dealership, 7 sedans are sold. The dealership predicts that it will sell 28 sports cars next month. How many sedans does it expect to sell?

Show answer

Answer: D — 49 sedans.

Show hints

Hint 1 of 2

Think of the cars in repeating BATCHES: every batch is 4 sports cars + 7 sedans. The whole question is just 'how many batches?'

Still stuck? Show hint 2 →

Hint 2 of 2

Find the scale factor (what turns 4 into 28), then apply that SAME factor to 7. This is what 'keeping a ratio' means.

Show solution

Approach: scale the ratio by a single factor

How many batches of 4 sports cars make 28? 28 ÷ 4 = 7 batches.
Each batch also has 7 sedans, so 7 batches give 7 × 7 = 49 sedans.
You'll see it again as: any 'A is to B' ratio scaled to a new amount — find the multiplier on one quantity, reuse it on the other. No cross-multiplying needed when the numbers divide nicely.

Another way — proportion (cross-multiply):

Set 4/7 = 28/x. Cross-multiply: 4x = 7 × 28 = 196.
x = 49. (Slower here, but the go-to when the scale factor isn't a whole number.)

1989 · #9 There are 2 boys for every 3 girls in Ms. Johnson's math class. If there are 30 students in her class, what percent of them are boys?

There are 2 boys for every 3 girls in Ms. Johnson's math class. If there are 30 students in her class, what percent of them are boys?

Show answer

Answer: C — 40%.

Show hints

Hint 1 of 3

A ratio of 2 boys to 3 girls means the class splits into 2 + 3 = 5 equal-size groups. Boys aren't 2 out of 3 — they're 2 out of how many?

Still stuck? Show hint 2 →

Hint 2 of 3

Turn a ratio into a fraction-of-the-whole by adding the parts: boys are 2 of the 5 total parts.

Still stuck? Show hint 3 →

Hint 3 of 3

You don't even need the 30: boys are 2/5 of the class, and 2/5 is already a percent in disguise.

Show solution

Approach: add the ratio parts, then take the fraction of the whole

The ratio 2 : 3 means 2 + 3 = 5 equal parts make up the class, and boys fill 2 of them — so boys are 2/5 of everyone.
2/5 = 40/100 = 40%. (Checking with the count: 30 ÷ 5 = 6 per part, boys = 2×6 = 12, and 12/30 = 40%.)
Trap to avoid: 60% is the boys-to-girls comparison (2 is 2/3 of 3) — but the question asks boys as a slice of the whole class, which is 2/5. Always ask 'fraction of WHAT?' before converting to a percent.

2019 · #27 Ria and Flora compare their savings and find that they are in the ratio 5 : 3. Then Ria buys a tablet for 160 €. The ratio of their...

Ria and Flora compare their savings and find that they are in the ratio 5 : 3. Then Ria buys a tablet for 160 €. The ratio of their savings now changes to 3 : 5. How much money did Ria have before she bought the tablet?

Show answer

Answer: C — 250 €

Show hints

Hint 1 of 2

Write both savings using the 5 : 3 ratio with one unknown.

Still stuck? Show hint 2 →

Hint 2 of 2

Subtract 160 from Ria's and set the new ratio to 3 : 5.

Show solution

Approach: set the changed ratio equal to 3 : 5

Let savings be 5x (Ria) and 3x (Flora).
After Ria spends 160: (5x − 160) : 3x = 3 : 5, so 5(5x − 160) = 9x.
That gives 25x − 800 = 9x, 16x = 800, x = 50, so Ria had 5x = 250 Euros.

CHAPTER 2

Proportions & percents — same ratio over here as over there

THEORY

Watch the pattern, then we'll name it. Each row keeps the same recipe; the cookie count just rides along:

cups of flour	cookies	cookies ÷ cups
2	6	3
4	12	3
6	?	3

The right column never changes. That's what a proportion is: two fractions that must stay equal. The whole game is find the missing piece. Since cookies ÷ cups = 3, the missing count is 6 × 3 = 18.

The classic move is cross-multiplication. Picture an X drawn across two equal fractions: each diagonal gives a product, and the two products are equal.

Multiply along the diagonals:

4 × ? = 6 × 6 ⇒ ? = 36 ÷ 4 = 9 cookies

Why is that allowed? Cross-multiplication isn't a magic trick — it's just clearing the denominators. Start from $\tfrac{a}{b} = \tfrac{c}{d}$ and multiply both sides by $b$ and by $d$ (the same legal move on each side keeps it balanced). The $b$ on the left cancels and the $d$ on the right cancels, leaving $a \cdot d = b \cdot c$. That's the whole proof: two fractions are equal exactly when their cross-products are. So you may always trade a proportion for one tidy equation with no fractions.

WHEN TO REACH FOR A PROPORTION

Any time the problem says 'at this rate' or 'in the same ratio':

Recipes — double the flour, double the cookies.
Map scales — 1 inch = 50 miles, so 4.2 inches = 210 miles.
Scale models — a 1 : 20 replica is the real thing ÷ 20.
Per-unit prices — 3 candies for $1, 9 candies for $3.

THE MOVE: Line up the same units top-with-top, then cross-multiply.

The trap: when MORE means LESS (inverse proportion)

Not every scaling story is a proportion. If 6 painters take 4 hours to paint a house, do 12 painters take 8 hours? No — more painters means faster, not slower.

Quick check: 'If I DOUBLE one, does the other DOUBLE (direct) or HALVE (inverse)?'

Type	What stays constant	Set up
Direct	RATIO (a/b)	`a/b = c/d` → cross-multiply
Inverse	PRODUCT (a·b)	`a · b = c · d`

🎯 Try it

For every 4 sports cars sold, 7 sedans are sold. The dealer predicts 28 sports cars next month. How many sedans?

Walkthrough: 28 sports cars is 28 ÷ 4 = 7 times the recipe. So sedans = 7 × 7 = 49. (Direct: more sports cars, more sedans. This is amc8-2009-02, answer D.)

The differences are proportional too

Here is a move most kids never learn. If a/b = c/d, then the gaps follow the same ratio: (a−c)/(b−d) equals it as well. Subtract top-from-top and bottom-from-bottom, and the leftover keeps the recipe.

Why it helps: when a problem hands you a difference instead of a total, you can ride the difference straight to the answer. A 5ft pole casts an 8ft shadow at noon. A flagpole stands 15ft taller — how much longer is its shadow? Height-to-shadow is 5 : 8, and the extra 15ft of height carries an extra shadow in that same 5 : 8 ratio: 15 is 3× the 5, so the extra shadow is 3 × 8 = 24ft longer. You never needed either full height.

Same ratio, three places: the tops, the bottoms, AND their difference. Pick whichever the problem gives you.

Framing inspired by Competition Math for Middle School (AoPS).

Percent is just a ratio out of 100

You already met this idea in the intro; here is the engine. A percent is a part-to-whole ratio where the whole is forced to 100. So 45% is the fraction $\tfrac{45}{100}$, full stop. That single fact lets one comparison wear three outfits:

ratio	fraction of whole	percent
girls : boys = 3 : 2	girls are 3/5	60% girls

All one statement. The ratio names the parts; the fraction picks 'out of the whole'; the percent rescales that whole to 100. To turn any fraction into a percent, push the bottom to 100 (a proportion): $\tfrac{3}{16} = \tfrac{x}{100}$ cross-multiplies to $16x = 300$, so $x = 18.75\%$. Or divide and slide the decimal: $3 \div 16 = 0.1875 = 18.75\%$.

Going the other way — a percent OF a number — turn the percent into a decimal and multiply: 15% of 80 is $0.15 \times 80 = 12$. Or use number sense: 15% is 3 out of every 20, and 80 is four 20s, so it's 4 × 3 = 12. Friendly anchors to keep ready: 10% (move the decimal one spot), 25% ($\div 4$), 50% ($\div 2$).

THE MOVE: Read every percent as 'out of 100'. To find a percent of something, multiply by its decimal.

🎯 Try it

There are 2 boys for every 3 girls in a class of 30. What percent are boys?

Walkthrough: 2 + 3 = 5 parts, and boys fill 2 of them — so boys are 2/5 of the WHOLE class, not 2/3. $\tfrac{2}{5} = \tfrac{40}{100} =$ 40%. (The 30 is a decoy: the fraction alone gives the percent. Trap: 60% compares boys to girls, not boys to everyone. This is ajhsme-1989-09, answer C.)

Framing inspired by Competition Math for Middle School (AoPS).

Percent change — and the trap that −20% then +20% does NOT cancel

Percent change is always measured against the starting amount: $\text{percent change} = \dfrac{\text{change}}{\text{original}}$. A price drops from $50 to $40? The change is $10, and $\tfrac{10}{50} = 20\%$ off. Shortcut: you now pay $\tfrac{40}{50} = 80\%$ of the old price, so you saved the other 20%.

Now the trap, and it is the most-tested percent idea on the whole contest. A $100 jacket is marked down 20%, then later marked up 20%. Back to $100? Your instinct screams yes. Resist it.

Down 20%: $100 \times 0.80 = 80$.
Up 20%: $80 \times 1.20 = 96$.

You land at $96, a 4% loss. The +20% was taken on the smaller $80, so it adds back fewer dollars than the −20% removed. Percents don't measure from the same place twice.

The clean way to chain percents is multipliers: 'down 20%' is $\times 0.80$, 'up 20%' is $\times 1.20$. String them: $0.80 \times 1.20 = 0.96$, a net 4% drop — no matter the starting price. A raise of 10% four years running isn't +40%; it's $1.1^4 \approx 1.464$, a 46.4% rise.

THE MOVE: Turn each percent change into a multiplier (down 30% = ×0.70, up 30% = ×1.30) and MULTIPLY them. Never add the percents.

Each percent change is a multiplier. Chain them by multiplying; a drop and an equal-size rise never get you home.

Percent-change framing inspired by Competition Math for Middle School (AoPS).

THE TRICK

For 'more workers, less time' problems, set up a · b = c · d, not a / b = c / d. The direction of the relationship is the entire decision — ask yourself the 'double or halve' question before you write anything down.

WATCH OUT

Bogus solution

It takes 6 painters 4 hours to paint a house. The crew grows to 9 painters. More painters, so set up a proportion: $\tfrac{6}{4}=\tfrac{9}{x}$. Cross-multiply: $6x=36$, so $x=6$ hours.

Why it breaks: a proportion assumes that when one quantity goes up the other goes up too. But more painters means the job finishes sooner, not later — 9 painters can't take longer than 6 did. This is inverse, not direct.

The fix: For inverse pairs, the product stays fixed, not the ratio. The job is $6\times 4 = 24$ painter-hours of work. With 9 painters: $24\div 9 = 2\tfrac23$ hours. Always ask first: 'double one — does the other double (direct) or halve (inverse)?'

WORKED EXAMPLE

PROBLEM · 1985 #6

A stack of paper containing 500 sheets is 5 cm thick. Approximately how many sheets of this type of paper would there be in a stack 7.5 cm high?

A) 250 B) 550 C) 667 D) 750 E) 1250

500 sheets of paper stack 5 cm high. How many sheets in a 7.5 cm stack?

Per-unit path (lightest). 500 sheets ÷ 5 cm = 100 sheets per cm. Then 100 × 7.5 = 750 sheets (choice D).

Same answer as the proportion 500/5 = x/7.5, but you compute the rate once and finish with a single multiply.

Taller stack means more sheets — direct, so it's a true proportion. Reach for 'sheets per cm' first; the per-unit rate turns the whole thing into one multiplication.

Answer: D — 750.

RULE OF THUMB

Direct: a/b = c/d, cross-multiply (or find the per-unit rate). Inverse: a·b = c·d. Always ask 'double one — does the other double or halve?' first.

MORE LIKE THIS

2018 · #1 An amusement park has a collection of scale models, with a ratio of 1 : 20, of buildings and other sights from around the country. The...

An amusement park has a collection of scale models, with a ratio of 1 : 20, of buildings and other sights from around the country. The height of the United States Capitol is 289 feet. What is the height in feet of its replica at this park, rounded to the nearest whole number?

Show answer

Answer: A — 14 feet.

Show hints

Hint 1 of 2

A scale of 1 : 20 means "the model is the smaller one." The real building is the big side (20), the replica is the small side (1) — so which way does 289 go, up or down?

Still stuck? Show hint 2 →

Hint 2 of 2

The technique: a ratio 1 : k shrinks the real size by dividing by k. Map the bigger number to the bigger part of the ratio so you never flip it by accident.

Show solution

Approach: divide by the scale factor

The replica is the "1" side and the real Capitol is the "20" side, so the replica is 1/20 of 289 — smaller, which is the sanity check that we're dividing (a model should be tiny).
289 ÷ 20 = 14.45, which rounds to 14 feet.
You'll see it again: any scale-model or map problem is just multiply or divide by the scale factor — the only decision is which way, and matching big-to-big settles it.

1989 · #9 There are 2 boys for every 3 girls in Ms. Johnson's math class. If there are 30 students in her class, what percent of them are boys?

There are 2 boys for every 3 girls in Ms. Johnson's math class. If there are 30 students in her class, what percent of them are boys?

Show answer

Answer: C — 40%.

Show hints

Hint 1 of 3

A ratio of 2 boys to 3 girls means the class splits into 2 + 3 = 5 equal-size groups. Boys aren't 2 out of 3 — they're 2 out of how many?

Still stuck? Show hint 2 →

Hint 2 of 3

Turn a ratio into a fraction-of-the-whole by adding the parts: boys are 2 of the 5 total parts.

Still stuck? Show hint 3 →

Hint 3 of 3

You don't even need the 30: boys are 2/5 of the class, and 2/5 is already a percent in disguise.

Show solution

Approach: add the ratio parts, then take the fraction of the whole

The ratio 2 : 3 means 2 + 3 = 5 equal parts make up the class, and boys fill 2 of them — so boys are 2/5 of everyone.
2/5 = 40/100 = 40%. (Checking with the count: 30 ÷ 5 = 6 per part, boys = 2×6 = 12, and 12/30 = 40%.)
Trap to avoid: 60% is the boys-to-girls comparison (2 is 2/3 of 3) — but the question asks boys as a slice of the whole class, which is 2/5. Always ask 'fraction of WHAT?' before converting to a percent.

2009 · #9 A lift can carry either 12 adults or 20 children. What is the maximum number of children that could travel in the lift with 9 adults?

A lift can carry either 12 adults or 20 children. What is the maximum number of children that could travel in the lift with 9 adults?

Show answer

Answer: C — 5

Show hints

Hint 1 of 2

Think in fractions of capacity: 12 adults fill the lift, so 9 adults fill 9/12 of it.

Still stuck? Show hint 2 →

Hint 2 of 2

Apply the leftover fraction of capacity to the 20-children full load.

Show solution

Approach: convert adults to a fraction of capacity

The full lift holds 12 adults, so 9 adults take up 9/12 = 3/4 of the capacity.
The remaining 1/4 holds 1/4 x 20 = 5 children.

CHAPTER 3

D = S × T — the master rate formula

THEORY

Walk for 2 hours at 3 miles per hour. How far did you go? Six miles — you multiplied without thinking. Now you know one rate formula; the other two are the same fact rearranged.

DISTANCE–SPEED–TIME

Distance = Speed × Time

Speed = Distance ÷ Time Time = Distance ÷ Speed

One picture stores all three. Write D on top, S and T on the bottom of a triangle. Cover the letter you want — what's left is the formula.

Cover D → S × T. Cover S → D / T. Cover T → D / S.

Every time, name which two of (D, S, T) you were handed, then solve for the third.

THE MOVE: Match the time unit to the speed unit FIRST, then plug in.

Units bite here. If speed is in mph but time is in minutes, convert one. 30 minutes at 60 mph is $(\tfrac12 \text{ hr}) \times 60 = 30$ miles — not 1800.

For two-leg trips (walk then run, drive then bike), find each leg's distance separately and add. Do not average the speeds — chapter 4 shows why that fails.

Clock arithmetic — the easiest points kids throw away

The T in $D = S \times T$ is often a clock time, and clocks are sneaky: they roll over at 60, not 100, and flip am/pm at 12. More test-takers miss a plain 'how long was the trip?' than miss the speed formula itself. Slow down here.

How long? A bus leaves at 11:36 am and arrives at 2:23 pm. Don't subtract like plain numbers — the minutes will betray you. Hop it in pieces:

11:36 → 12:00 is 24 min (60 − 36).
12:00 → 2:00 is 2 hr.
2:00 → 2:23 is 23 min.
Total: 2 hr + 24 + 23 = 2 hr 47 min.

What time? A 2½-hour test starts at 9:47 am. Add whole hours first, then minutes: 9:47 + 2 hr = 11:47, then + 30 min → 12:17 pm. (47 + 30 = 77 rolled past 60, so an extra hour kicked in — that roll-over is exactly where points leak.)

Split the jump at every hour line and every 12 o'clock. Add hours, then minutes, and watch the roll-over at 60.

(Clock-arithmetic examples adapted from Problem Solving via the AMC, Australian Maths Trust.)

🎯 Try it

A motorcyclist covers 28 km in 30 minutes. What is the average speed in km/h?

Walkthrough: The answer is per hour, so match units: 30 min = ½ hr. Speed = D ÷ T = 28 ÷ ½ = 28 × 2 = 56 km/h. (Half an hour means double the distance per full hour. This is kangaroo-2011-benjamin-02, answer C.)

THE TRICK

When time is in minutes and speed is in mph, convert time to hours by dividing by 60 before you do anything. Match the time unit to the speed unit and $D = S \times T$ works straight off.

WORKED EXAMPLE

PROBLEM · 2018 #6

On a trip to the beach, Anh traveled 50 miles on the highway and 10 miles on a coastal access road. He drove three times as fast on the highway as on the coastal road. If Anh spent 30 minutes driving on the coastal road, how many minutes did his entire trip take?

A) 50 B) 70 C) 80 D) 90 E) 100

Anh drove 50 miles on the highway and 10 miles on a coastal road, going three times as fast on the highway. He spent 30 minutes on the coastal road. How many minutes was the whole trip?

Two legs — handle them one at a time.

Coastal: 10 miles in 30 minutes. That fixes the coastal speed and the coastal time (30 min, given).
Highway speed is 3× the coastal speed. The highway is 50 miles = 5× the coastal 10 miles. Going 3× as fast over 5× the distance takes $\tfrac{5}{3}$ of the coastal time: $\tfrac{5}{3} \times 30 = 50$ minutes.
Total: 30 + 50 = 80 minutes (choice C).

No speeds in mph needed — just the ratio of distances against the ratio of speeds.

You never have to name the coastal speed. Time = distance ÷ speed, so the time ratio is (distance ratio) ÷ (speed ratio) = 5 ÷ 3. Multiply the known 30 minutes by that. Light path beats grinding out mph.

Answer: C — 80 minutes.

RULE OF THUMB

Know which two of (D, S, T) you're given. Convert times to one unit (usually hours). For two-leg trips, do each leg separately, then add — never average the speeds.

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2016 · #4 When Cheenu was a boy he could run 15 miles in 3 hours and 30 minutes. As an old man he can now walk 10 miles in 4 hours. How many...

When Cheenu was a boy he could run 15 miles in 3 hours and 30 minutes. As an old man he can now walk 10 miles in 4 hours. How many minutes longer does it take for him to travel a mile now compared to when he was a boy?

Show answer

Answer: B — 10 minutes longer.

Show hints

Hint 1 of 2

The question asks about ONE mile, but the two trips are different lengths — comparing the raw trip times would be unfair. Shrink each trip down to a per-mile rate so they're on equal footing.

Still stuck? Show hint 2 →

Hint 2 of 2

The unit that matches the question is minutes-per-mile: divide each trip's total minutes by its miles, then the answer is just the gap between the two paces.

Show solution

Approach: reduce each trip to minutes-per-mile, then subtract

Boy: 3 h 30 min = 210 minutes for 15 miles ⇒ 210 ÷ 15 = 14 minutes per mile.
Old man: 4 h = 240 minutes for 10 miles ⇒ 240 ÷ 10 = 24 minutes per mile.
He's slower now, so it takes 24 − 14 = 10 extra minutes per mile.
Sanity check: he covers fewer miles in more time as an old man, so the pace MUST be slower — a positive difference is expected, and 10 is small enough to be one mile's worth (not the whole trip).
You'll see this again as: any rate comparison — convert both to the SAME unit the question asks about (here, minutes per mile) before comparing; never compare totals of different sizes.

2014 · #17 George walks 1 mile to school. He leaves home at the same time each day, walks at a steady speed of 3 miles per hour, and arrives just...

George walks 1 mile to school. He leaves home at the same time each day, walks at a steady speed of 3 miles per hour, and arrives just as school begins. Today he was distracted by the pleasant weather and walked the first 12 mile at a speed of only 2 miles per hour. At how many miles per hour must George run the last 12 mile in order to arrive just as school begins today?

Show answer

Answer: B — 6 mph.

Show hints

Hint 1 of 2

Time is the fixed budget here, not speed. He must arrive at the usual moment, so his total travel time is locked at the normal value — figure that out first.

Still stuck? Show hint 2 →

Hint 2 of 2

Subtract the time the slow first half ate up; whatever's left is all he has to cover the last half-mile. Speed = distance ÷ that leftover time.

Show solution

Approach: time is the fixed budget — subtract what's spent

Normal trip: 1 mile at 3 mph takes 1/3 hr. That's the total time budget he must hit today.
Slow first half: (1/2 mile) ÷ (2 mph) = 1/4 hr used up.
Time left for the second half-mile: 1/3 − 1/4 = 4/12 − 3/12 = 1/12 hr.
Required speed = (1/2 mile) ÷ (1/12 hr) = 6 mph.
Why this transfers: in "arrive on time" problems, hold time constant and treat it as a budget. Don't average the speeds — dawdling on the first half costs disproportionately more time, which is why he must nearly triple his pace, not just speed up a little.

2026 · #5 Casey went on a road trip that covered 100 miles, stopping only for a lunch break along the way. The trip took 3 hours in total and her...

Casey went on a road trip that covered 100 miles, stopping only for a lunch break along the way. The trip took 3 hours in total and her average speed while driving was 40 miles per hour. In minutes, how long was the lunch break?

Show answer

Answer: B — 30 minutes.

Show hints

Hint 1 of 2

The 40 mph only describes the part where she's actually moving — the 3 hours also hides the lunch stop. Which piece can you compute directly?

Still stuck? Show hint 2 →

Hint 2 of 2

Time = distance ÷ speed gives the driving time only. Subtract that from the 3 total hours and what's left is the break.

Show solution

Approach: the 3 hours is driving + break; only driving obeys distance ÷ speed

The 40 mph is her speed while driving, so distance ÷ speed gives only the driving time, not the whole trip. Find that first: 100 ÷ 40 = 2.5 hours.
Whatever's left of the 3 total hours is the lunch break: 3 − 2.5 = 0.5 hour = 30 minutes.
Sanity check: 30 minutes of lunch is reasonable, and 2.5 h of driving at 40 mph really does cover 100 miles. The reusable idea: always separate ‘moving time’ from total time before using rate = distance ÷ time — the rate only describes the moving part.

★ MINI-QUIZ

Ratios, proportions, D = ST

Three problems on parts-of-a-whole and the master rate formula.

2020 · #1 Luka is making lemonade to sell at a school fundraiser. His recipe requires 4 times as much water as sugar and twice as much sugar as...

Show answer

Answer: E — 24 cups.

Show hints

Hint 1 of 2

Lemon juice is the “smallest” ingredient and everything is measured against it. So instead of two separate steps, ask: how many times bigger is water than lemon juice?

Still stuck? Show hint 2 →

Hint 2 of 2

When one thing scales another which scales a third, the scale factors multiply. Water is 4× sugar and sugar is 2× lemon, so water is 4 × 2 = 8 times the lemon juice.

Show solution

Approach: multiply the scale factors into one jump

Chained scalings multiply: water is 4× sugar and sugar is 2× lemon, so water is 4 × 2 = 8 times the lemon juice — one jump instead of two.
With 3 cups of lemon juice, water = 8 × 3 = 24 cups.
You'll see this again as: any “A is k times B, B is m times C” chain collapses to “A is k·m times C.” Gear ratios and unit conversions work the same way.

Another way — one step at a time: lemon → sugar → water (MAA):

Sugar is twice the lemon juice: 2 × 3 = 6 cups.
Water is four times the sugar: 4 × 6 = 24 cups.

2023 · #5 A lake contains 250 trout, along with a variety of other fish. When a marine biologist catches and releases a sample of 180 fish from...

A lake contains 250 trout, along with a variety of other fish. When a marine biologist catches and releases a sample of 180 fish from the lake, 30 are identified as trout. Assume the ratio of trout to the total number of fish is the same in both the sample and the lake. How many fish are there in the lake?

Show answer

Answer: B — 1500 fish.

Show hints

Hint 1 of 2

Think of the net of 180 fish as a shrunk-down copy of the whole lake — same recipe, smaller pot. So the trout fraction matches.

Still stuck? Show hint 2 →

Hint 2 of 2

Find the trout fraction in the sample (30 out of 180), then apply that same fraction to the 250 known trout to recover the whole.

Show solution

Approach: the trout fraction is the same in sample and lake

The sample is a tiny scale model of the lake: the trout fraction in your net should match the trout fraction in the whole lake. So find that one fraction and apply it.
In the sample, 30 of 180 are trout: 30 ÷ 180 = 16. That clean fraction is the heart of the problem — trout are 1 in every 6 fish.
So the 250 real trout are 16 of the whole lake, meaning the lake holds 250 × 6 = 1500 fish. This transfers to every ‘capture sample’ (or poll, or survey): part-of-sample = part-of-whole.

Another way — scale the whole sample up:

The lake has 250 trout but the sample only caught 30 — so the lake is 250 ÷ 30 = 253 times as ‘trout-rich’ as the sample.
Everything scales by that same factor, so total fish = 180 × 253 = 60 × 25 = 1500.

2002 · #24 Miki has a dozen oranges of the same size and a dozen pears of the same size. Miki uses her juicer to extract 8 ounces of pear juice...

Miki has a dozen oranges of the same size and a dozen pears of the same size. Miki uses her juicer to extract 8 ounces of pear juice from 3 pears and 8 ounces of orange juice from 2 oranges. She makes a pear-orange juice blend from an equal number of pears and oranges. What percent of the blend is pear juice?

Show answer

Answer: B — 40%.

Show hints

Hint 1 of 2

The "dozen" is pure bait — she uses an *equal* number of each fruit, so however many that is divides out. All that matters is the juice from *one* pear versus *one* orange.

Still stuck? Show hint 2 →

Hint 2 of 2

Pears: 8 oz from 3, so 8/3 oz each. Oranges: 8 oz from 2, so 4 oz each. The blend's pear share is just one pear's juice over (one pear + one orange).

Show solution

Approach: compare juice per fruit

Since she blends *equal counts* of each fruit, the common count cancels — work per fruit. One pear yields 8/3 oz; one orange yields 8/2 = 4 oz.
Pear-to-orange juice is 8/3 : 4, and clearing the 3 gives 8 : 12 = 2 : 3.
Pear's share = 2 ÷ (2 + 3) = 2/5 = 40%.
*Worth keeping:* when two things are mixed in equal *counts*, the actual count is irrelevant — reduce to one of each and compare. Chasing the dozen (32 oz vs 48 oz) gives the same answer with bigger numbers.

Another way — scale up to the full dozen:

12 pears give 12 × 8/3 = 32 oz; 12 oranges give 12 × 4 = 48 oz.
Pear fraction = 32 ÷ (32 + 48) = 32/80 = 2/5 = 40% — same ratio, just unscaled.

CHAPTER 4

Average speed — never average the speeds

THEORY

Drive 60 mph for an hour, then 30 mph for another hour. Average speed? 45 — equal times, so the plain average works. Now change one word: drive 60 miles at 60 mph, then 60 miles back at 30 mph. Same two speeds. The average is not 45. It's 40. Watch why.

AVERAGE SPEED — ONE FORMULA, ALWAYS

average speed = TOTAL distance ÷ TOTAL time

Never average two speeds directly unless you spent equal TIME at each.

THE MOVE: Ignore the speeds. Add all the distance, add all the time, divide once.

Round-trip shortcut

For a round trip where each leg is the same distance at speeds a and b:

average = 2ab / (a + b)

(This is the harmonic mean.) For 60 and 30: 2·60·30 / 90 = 40 ✓.

Two speeds	Simple avg (TRAP)	Harmonic mean (CORRECT)
60 & 30 mph	45	40
40 & 60 mph	50	48
3 & 6 mph	4.5	4
10 & 90 mph	50	18 (slow really dominates!)

The harmonic mean is ALWAYS below the plain average. The bigger the speed gap, the harder it gets pulled down.

Pick your own numbers — the 'convenient units' trick

When a problem gives only fractions and percents with no actual distance or time, your instinct is to write $x$ and $d$ and grind. There's a faster way: you pick a friendly number for the missing piece. An average-speed answer can't depend on how long the trip really was — so make it whatever is easiest.

The trip splits into thirds. A plane flies 800 km/h for the first third of its TIME and averages 700 km/h over the whole trip. How fast was the rest?

The trip is in thirds, so the friendliest total time is 3 hours (one tidy hour per third):

Whole trip: 3 hr × 700 = 2100 km.
First third: 1 hr × 800 = 800 km.
Remaining 2 hr cover 2100 − 800 = 1300 km.
Speed for the rest: 1300 ÷ 2 = 650 km/h.

No variable, no equation — the 3 turned every fraction into whole hours. Pick the total to match the denominators: thirds → use 3; quarters → use 4; 'half the way' → use 2.

When only the shape of a trip matters, not its real size, pick the size yourself — and pick it to kill the fractions.

(Convenient-units idea adapted from Problem Solving via the AMC, Australian Maths Trust.)

🎯 Try it

Bonnie drives 50 miles to Temple at 60 mph, then buses the same 50 miles back at 40 mph. Average speed for the whole round trip, in mph?

Walkthrough: Equal distances, so use the harmonic mean: $\tfrac{2 \cdot 60 \cdot 40}{60 + 40} = \tfrac{4800}{100} = $ 48 mph — below the plain 50, as it must be. (This is amc8-2009-14, answer B.)

THE TRICK

'Equal distance each leg' → the average is always less than the plain average (use total÷total, or the harmonic mean). 'Equal time each leg' → the plain average is correct.

WATCH OUT

Bogus solution

Maya drives 30 miles to the lake at 60 mph, then drives the same 30 miles home at 20 mph. She went half the trip at 60 and half at 20, so her average speed is just the average of the two: $(60 + 20)\div 2 = 40$ mph.

Why it breaks: 'half the trip' means half the distance, not half the time — and the slow 20-mph leg eats far more clock time, so it should count for more, not the same.

The fix: Add the real time. Out: 30÷60 = ½ hr. Back: 30÷20 = 1½ hr. Total 60 miles in 2 hours = 30 mph, not 40. (Check with the harmonic mean: $\tfrac{2\cdot 60\cdot 20}{80} = 30$.) The plain average is only right when the two legs take equal time.

Framing inspired by AoPS Prealgebra.

WORKED EXAMPLE

PROBLEM · 1992 #18

On a trip, a car traveled 80 miles in an hour and a half, then was stopped in traffic for 30 minutes, then traveled 100 miles during the next 2 hours. What was the car's average speed in miles per hour for the 4-hour trip?

A) 45 B) 50 C) 60 D) 75 E) 90

A car goes 80 miles in 1.5 hours, sits in traffic for 0.5 hours, then drives 100 more miles in 2 hours. Average speed for the whole 4-hour trip?

Different times and a stop — go straight to the definition.

Total distance = 80 + 100 = 180 miles.
Total time = 1.5 + 0.5 + 2 = 4 hours (the 30-minute stop counts!).
Average = 180 ÷ 4 = 45 mph (choice A).

Don't drop the stop. Forget it and you'd get 180 ÷ 3.5 ≈ 51. The stop is the whole point — it's time on the clock at speed 0, and it pulls the average down.

The setters parked that stop in the middle on purpose. Total distance over total elapsed time beats every trap at once: it refuses to average speeds and it refuses to ignore the standstill.

Answer: A — 45.

RULE OF THUMB

Average speed = total distance ÷ total time. Never average two speeds unless the times are equal. Equal-distance round trip: 2ab/(a+b).

MORE LIKE THIS

2008 · #5 Barney Schwinn notices that the odometer on his bicycle reads 1441, a palindrome, because it reads the same forward and backward. After...

Barney Schwinn notices that the odometer on his bicycle reads 1441, a palindrome, because it reads the same forward and backward. After riding 4 more hours that day and 6 the next, he notices that the odometer shows another palindrome, 1661. What was his average speed in miles per hour?

Show answer

Answer: E — 22 mph.

Show hints

Hint 1 of 2

"Palindrome" is just flavor — all you need is the two odometer readings and the total hours.

Still stuck? Show hint 2 →

Hint 2 of 2

Average speed always means total distance ÷ total time, no matter how the trip was split up.

Show solution

Approach: total distance ÷ total time

Distance is just how far the odometer moved: 1661 − 1441 = 220 miles. Don't be distracted by the palindrome story.
Total time is 4 + 6 = 10 hours, so average speed = 220 ÷ 10 = 22 mph.
Why this transfers: average speed is never the average of two speeds — it's always all the miles over all the hours.

2011 · #9 Carmen takes a long bike ride on a hilly highway. The graph indicates the miles traveled during the time of her ride. What is Carmen's...

Carmen takes a long bike ride on a hilly highway. The graph indicates the miles traveled during the time of her ride. What is Carmen's average speed for her entire ride in miles per hour?

Show answer

Answer: E — 5 mph.

Show hints

Hint 1 of 2

Average speed doesn't care about the hills, the slow stretches, or the bends in the graph — only where the ride started and where it ended. Read just the final point.

Still stuck? Show hint 2 →

Hint 2 of 2

Average speed = total distance ÷ total time. The far-right end of the curve gives you both at once.

Show solution

Approach: average speed depends only on the endpoints

The curve ends at 35 miles after 7 hours — that's the whole ride: 35 miles in 7 hours.
Average speed = total distance ÷ total time = 35 ÷ 7 = 5 mph.
Why this transfers: "average speed" is always end-distance over end-time. The wiggly middle of a distance-time graph is a distraction — never average the steeper and flatter parts.

2026 · #5 Casey went on a road trip that covered 100 miles, stopping only for a lunch break along the way. The trip took 3 hours in total and her...

Show answer

Answer: B — 30 minutes.

Show hints

Hint 1 of 2

The 40 mph only describes the part where she's actually moving — the 3 hours also hides the lunch stop. Which piece can you compute directly?

Still stuck? Show hint 2 →

Hint 2 of 2

Time = distance ÷ speed gives the driving time only. Subtract that from the 3 total hours and what's left is the break.

Show solution

Approach: the 3 hours is driving + break; only driving obeys distance ÷ speed

The 40 mph is her speed while driving, so distance ÷ speed gives only the driving time, not the whole trip. Find that first: 100 ÷ 40 = 2.5 hours.
Whatever's left of the 3 total hours is the lunch break: 3 − 2.5 = 0.5 hour = 30 minutes.
Sanity check: 30 minutes of lunch is reasonable, and 2.5 h of driving at 40 mph really does cover 100 miles. The reusable idea: always separate ‘moving time’ from total time before using rate = distance ÷ time — the rate only describes the moving part.

CHAPTER 5

Unit conversion — multiply by 1

THEORY

You already convert units in your head: 2 hours is 120 minutes — you multiplied by 60 because '1 hour = 60 minutes'. The grown-up version writes that '60' as a fraction equal to 1 and lets the units cancel like algebra. Once you trust the units, you never again guess whether to multiply or divide.

Multiply by a fraction whose top and bottom are the same quantity in different units (like 5280 ft / 1 mi). That fraction equals 1, so you change the look, not the value. This is the factor-label method: write the units, let them cancel, and the numbers fall out.

FACTOR-LABEL RECIPE

Write the starting value with units (e.g., 60 mi/hr).
Multiply by conversion fractions, each equal to 1 (e.g., 5280 ft / 1 mi).
Put the unit you DON'T want in the OPPOSITE part of the fraction so it cancels.
The unit you DO want should survive.
Multiply the numerators, divide by the denominators — the surviving units are your answer's units.

THE MOVE: Write the units on every number. Let them cancel; the answer's units prove you set it up right.

Conversion facts to know cold

From	To	Multiply by
1 mile	feet	5280
1 mile	yards	1760
1 yard	feet	3
1 foot	inches	12
1 hour	minutes	60
1 minute	seconds	60
1 hour	seconds	3600
1 mph	ft/s	≈ 1.467 (so 60 mph = 88 ft/s)
1 km	meters	1000
1 lb	ounces	16

If the final units come out wrong, you flipped a fraction. Turn that one fraction upside down and re-cancel.

🎯 Try it

In Kangaroo Town, distance is measured in 'hops'. Klaus needs 12 minutes per hop. How many hops can he cover in 12 hours?

Walkthrough: Convert hours to minutes: 12 hr × 60 = 720 minutes. Each hop eats 12 minutes, so 720 ÷ 12 = 60 hops. (Watch the units — mixing hours and minutes is the trap. This is kangaroo-2025-kadett-05, answer A.)

THE TRICK

Don't memorize 'when do I multiply by 60 vs. divide.' Write the conversion factors with units and let cancellation decide. The units are guardrails — trust them.

WORKED EXAMPLE

PROBLEM · 1993 #8

To control her blood pressure, Jill's grandmother takes one half of a pill every other day. If one supply of medicine contains 60 pills, then the supply would last approximately

A) 1 month B) 4 months C) 6 months D) 8 months E) 1 year

Jill's grandmother takes ½ pill every other day. A bottle holds 60 pills. About how many months will it last?

Stack the conversions and let them cancel:

60 pills × (1 dose / ½ pill) × (2 days / 1 dose) × (1 month / 30 days)

60 × 2 × 2 = 240 days.
240 days × (1 month / 30 days) ≈ 8 months (choice D).

The units are the proof: started in 'pills', ended in 'months', everything between cancelled.

'Every other day' hides a conversion (1 dose ↔ 2 days). 'Half a pill per dose' hides another (1 dose ↔ ½ pill). Writing each as a fraction with units drags the hidden conversions into the open so cancellation can finish.

Answer: D — 8 months.

RULE OF THUMB

Multiply by 1 = (target unit) / (source unit). Units cancel like algebra; the number falls out. Wrong units = flip a fraction.

MORE LIKE THIS

1994 · #5 Given that 1 mile = 8 furlongs and 1 furlong = 40 rods, the number of rods in one mile is

Given that 1 mile = 8 furlongs and 1 furlong = 40 rods, the number of rods in one mile is

Show answer

Answer: B — 320.

Show hints

Hint 1 of 2

The miles–rods jump is too big to do directly, but you have a stepping-stone (furlongs) that connects them — hop mile → furlong → rod.

Still stuck? Show hint 2 →

Hint 2 of 2

Each hop multiplies: stack the two conversion factors together. (Ignore the decoy numbers 660, 1760, 5280 — those are real mile facts but for feet/yards, not this puzzle's furlongs and rods.)

Show solution

Approach: chain the conversions

Bridge through furlongs: 1 mile is 8 furlongs, and each of those 8 furlongs is 40 rods.
So 1 mile = 8 × 40 = 320 rods.
Why multiply (not add): each furlong unpacks into 40 rods, and you have 8 furlongs, so it's 8 groups of 40 — that's multiplication. This 'chain the units' move handles any multi-step conversion.

2022 · #7 When the World Wide Web first became popular in the 1990s, download speeds reached a maximum of about 56 kilobits per second....

When the World Wide Web first became popular in the 1990s, download speeds reached a maximum of about 56 kilobits per second. Approximately how many minutes would the download of a 4.2-megabyte song have taken at that speed? (Note that there are 8000 kilobits in a megabyte.)

Show answer

Answer: B — 10 minutes.

Show hints

Hint 1 of 2

The speed is in kilobits per second but the song is in megabytes — the units don't match. Fix that first; everything else is one division.

Still stuck? Show hint 2 →

Hint 2 of 2

Convert the song to kilobits (4.2 × 8000), then time = size ÷ speed, and finish by turning seconds into minutes.

Show solution

Approach: make the units match the speed, then divide

Insight: don't divide yet — the answer choices span 0.6 to 36000, so a units slip is the whole danger. The speed is in kilobits/sec, so put the song in kilobits too: 4.2 × 8000 = 33,600 kilobits.
Time = 33,600 ÷ 56 = 600 seconds = 10 minutes.
Lighter path: avoid the big division by reshaping the size as 56 × (something): 33,600 = 56 × 600, so it's 600 seconds straight off — then ÷60 for minutes.

2019 · #2 Ten quarters of an hour correspond to how many hours?

Ten quarters of an hour correspond to how many hours?

Show answer

Answer: E — 2½

Show hints

Hint 1 of 2

A quarter of an hour is 15 minutes.

Still stuck? Show hint 2 →

Hint 2 of 2

Turn the total minutes back into hours.

Show solution

Approach: convert quarters to minutes then to hours

Ten quarter-hours are 10 × 15 = 150 minutes.
150 minutes ÷ 60 = 2.5 hours.
So the answer is 2½ hours.

CHAPTER 6

Relative speed — meet, chase, lap

THEORY

Two cars head straight at each other on a 100-mile road, one at 40 mph and one at 60 mph. How long until they meet? You could track each car's position and solve. Or you could notice the only thing that matters: how fast the gap between them shrinks. That number is the closing speed, and it makes these problems collapse.

Case 1: heading TOWARD each other (speeds ADD)

Closing speed = 40 + 60 = 100 mph. Time to meet = 100 mi ÷ 100 mph = 1 hour.

Why add? Every mile the gap loses is shared: A eats 40 of it per hour, B eats 60, so the gap drops 100 miles each hour.

Case 2: same direction, faster behind (speeds SUBTRACT)

Closing speed = 8 − 4 = 4 mph. Time = 1 mi ÷ 4 mph = ¼ hr = 15 minutes.

Why subtract? Bob moves at 8, but Alice keeps fleeing at 4, carrying 4 miles of progress away each hour. Bob only nets 4 mph of catching up.

RELATIVE-SPEED SUMMARY

Situation	Closing speed	Time
Toward each other	v₁ + v₂	gap ÷ (v₁ + v₂)
Same direction (faster behind)	v_fast − v_slow	gap ÷ (v_fast − v_slow)
Round track, same direction	v_fast − v_slow	track length ÷ that (next lap)
Round track, opposite direction	v₁ + v₂	track length ÷ that (next meet)

THE MOVE: Don't track two movers. Track the gap, and how fast it closes.

How many do you pass? (the train trap)

This one fools almost everyone. Trains leave City B for City A every hour on the hour; each trip takes 3 hours. You leave A at noon and reach B at 3:00. How many B-to-A trains do you pass on the way?

Your gut says 3 — the ones that leave while you travel (12, 1, 2). Wrong. You also pass trains already on the track when you started.

The 10:00 train from B is still rolling toward you at noon. So is the 11:00 train. You pass both, plus the 12, 1, and 2 o'clock trains. That's five: 10, 11, 12, 1, 2. (The 9:00 train reaches A as you leave and the 3:00 train reaches B as you arrive — you meet those at the stations, not on the road.)

Count every train on the route during your trip, not only the ones that leave during it.

(Train-passing problem adapted from Problem Solving via the AMC, Australian Maths Trust.)

🎯 Try it

Two runners share a 720 m round track, going opposite ways. One laps in 4 min, the other in 5 min. How many metres does the slower runner cover between two consecutive meetings?

Walkthrough: Speeds: 720/4 = 180 and 720/5 = 144 m/min. Opposite ways, so closing speed = 180 + 144 = 324 m/min. They meet every 720 ÷ 324 min. The slower runner covers 144 × (720/324) = 320 m. (This is kangaroo-2017-kadett-28, answer E.)

THE TRICK

Same direction → subtract speeds. Opposite directions → add. The real trap is not seeing that 'catching up' or 'meeting' is a relative-speed problem at all — the instant you spot a gap, find how fast it closes.

WATCH OUT

Bogus solution

Two cars start 120 miles apart and drive toward each other, one at 50 mph and one at 70 mph. Their average speed is $(50+70)\div 2 = 60$ mph, so to cross the 120-mile gap takes $120\div 60 = 2$ hours.

Why it breaks: the gap isn't being closed by one car going 60 mph — it's being closed by both cars at once, each eating into it. Averaging the speeds throws away one of the two cars and makes the gap close half as fast as it really does.

The fix: Track the gap, not the cars. Heading toward each other, the gap shrinks by 50 miles AND another 70 miles every hour — so it closes at 50 + 70 = 120 mph. Time to close 120 miles = 120÷120 = 1 hour. Add the speeds when they approach; subtract when one chases the other.

WORKED EXAMPLE

PROBLEM · 2010 #8

As Emily is riding her bicycle on a long straight road, she spots Emerson skating in the same direction 1/2 mile in front of her. After she passes him, she can see him in her rear mirror until he is 1/2 mile behind her. Emily rides at a constant rate of 12 miles per hour, and Emerson skates at a constant rate of 8 miles per hour. For how many minutes can Emily see Emerson?

A) 6 B) 8 C) 12 D) 15 E) 16

Emily (12 mph) and Emerson (8 mph) ride the same direction. When Emily spots him he is ½ mile ahead; she loses him from her mirror when he is ½ mile behind. How long was he in view?

The gap swings from ½ mile ahead to ½ mile behind — a relative shift of 1 mile, not ½. (Picture Emily still and Emerson drifting backward past her: he must cover a full mile relative to her.)

Closing speed = 12 − 8 = 4 mph.
Time = 1 mile ÷ 4 mph = ¼ hour = 15 minutes (choice D).

The trap is reading '½ mile' once and treating it as a single ½-mile chase. It's two halves stitched together: ½ mile to catch up, ½ mile to pull away — both at the same 4 mph relative speed. So 1 mile ÷ 4 mph = 15 minutes, not 7.5.

Answer: D — 15 minutes.

RULE OF THUMB

Toward each other: speeds add. Same direction: speeds subtract. Time to meet/catch = gap ÷ relative speed.

MORE LIKE THIS

2006 · #13 Cassie leaves Escanaba at 8:30 AM heading for Marquette on her bike. She bikes at a uniform rate of 12 miles per hour. Brian leaves...

Cassie leaves Escanaba at 8:30 AM heading for Marquette on her bike. She bikes at a uniform rate of 12 miles per hour. Brian leaves Marquette at 9:00 AM heading for Escanaba on his bike. He bikes at a uniform rate of 16 miles per hour. They both bike on the same 62-mile route between Escanaba and Marquette. At what time in the morning do they meet?

Show answer

Answer: D — 11:00 AM.

Show hints

Hint 1 of 2

Two riders heading toward each other close the gap between them at the SUM of their speeds — think of the empty road shrinking, not the riders moving. But first deal with the half-hour head start so you start both clocks together.

Still stuck? Show hint 2 →

Hint 2 of 2

Closing speed: when two things approach each other, add their speeds; the gap divided by that combined speed is the time to meet.

Show solution

Approach: settle the head start, then close the gap at combined speed

Cassie rides alone from 8:30 to 9:00, half an hour, covering ½ × 12 = 6 miles. So at 9:00 the gap between them is 62 − 6 = 56 miles.
From 9:00 they ride toward each other, closing the gap at 12 + 16 = 28 mph.
Time to close 56 miles: 56 ÷ 28 = 2 hours. They meet at 9:00 + 2:00 = 11:00 AM.
Why add the speeds: in one hour Cassie eats 12 miles of road and Brian eats 16, so 28 miles of gap vanish per hour regardless of where they are. This "combined speed" trick turns every meeting problem into one simple division — just remember to handle any head start first.

2024 · #29 Anne drives from point A to point B and then immediately back to A. Benni drives from point B to point A and then immediately back to B....

Anne drives from point A to point B and then immediately back to A. Benni drives from point B to point A and then immediately back to B. They drive on the same road, start at the same time, and both drive at constant speed. Anne’s speed is three times as high as Benni’s speed. They meet for the first time 15 minutes after they start. How long after the start will they meet for the second time?

Show answer

Answer: C — 30 min

Show hints

Hint 1 of 2

At the first meeting the two together have covered the whole road once, fixing the road length in their speeds.

Still stuck? Show hint 2 →

Hint 2 of 2

Track who turns around when; the second meeting comes after Anne (the faster one) catches Benni from behind.

Show solution

Approach: track positions through the turnarounds to the second meeting

With speeds 3v and v meeting head-on after 15 min, the road length is 4v·15 = 60v.
Anne reaches the far end at 20 min and turns back; Benni is still heading the same way.
Anne then closes the 20v gap at relative speed 2v, taking 10 more min, so they meet at 30 min.

1995 · #25 Buses from Dallas to Houston leave every hour on the hour. Buses from Houston to Dallas leave every hour on the half hour. The trip from...

Buses from Dallas to Houston leave every hour on the hour. Buses from Houston to Dallas leave every hour on the half hour. The trip from one city to the other takes 5 hours. Assuming the buses travel on the same highway, how many Dallas-bound buses does a Houston-bound bus pass on the highway (not in the station)?

Show answer

Answer: D — 10.

Show hints

Hint 1 of 3

You don't have to track positions or speeds. The only thing that lets two buses meet on the road is that they are on the road AT THE SAME TIME — so this is really a question about overlapping time windows, not motion.

Still stuck? Show hint 2 →

Hint 2 of 3

Pin down ONE Houston-bound bus and its 5-hour window. Then an oncoming bus is passed exactly when its own 5-hour window overlaps that one — so count the oncoming departure times whose trips overlap.

Still stuck? Show hint 3 →

Hint 3 of 3

An oncoming bus already on the road when yours starts still counts (you pass it later on), and one that starts before you finish counts too. So look both BACKWARD and forward from your window — don't just count buses that leave during your trip.

Show solution

Approach: forget speed — count oncoming buses whose road-time overlaps yours

The freeing insight: two buses on the same highway pass each other if and only if they share the road at the same moment. Speeds and exact meeting points never matter — only whether their time-on-road windows overlap. So pick one bus and compare windows.
Pin your Houston-bound bus: say it leaves Dallas at 12:00 and (5-hour trip) arrives Houston at 17:00, so it owns the road-window 12:00–17:00.
A Dallas-bound bus that left Houston at time t owns the window (t, t+5). It overlaps yours when it hasn't already arrived (t + 5 > 12:00, i.e. t > 7:00) AND it has already left (t < 17:00). So the meeting condition is simply 7:00 < t < 17:00.
Houston buses leave on the half hour, so the departures in that range are 7:30, 8:30, 9:30, … , 16:30 — that's 10 buses (one every hour across a 10-hour span).
The trap this catches: only counting buses that leave during your own 12:00–17:00 trip gives 5 or 6 and misses the ones already underway when you start — that's why the overlap test must reach back to 7:00. Sanity check: the early 7:30 bus reaches Dallas at 12:30 (just after you set out, so you do pass it on the road), and the late 16:30 bus is still rolling when you arrive — both genuine highway meetings, not station ones.
Why this transfers: whenever you're asked 'how many of these cross paths,' translate each traveler into a time interval and count overlapping intervals — the geometry of who-is-where dissolves into simple interval arithmetic.

Another way — draw it as a distance–time picture:

Sketch time across the bottom and distance (Dallas at the bottom, Houston at the top) up the side. Your bus is one slanted line going up; every Dallas-bound bus is a slanted line going down, starting on the half hours.
Two lines crossing = a pass. Your line spans the 5-hour width, and a down-line crosses it exactly when it starts within 5 hours before you finish and ends within 5 hours after you begin — the same 7:00-to-17:00 window. Counting the crossing lines gives 10.

★ MINI-QUIZ

Speeds and conversions

Three problems on average speed, unit conversion, and relative speed.

2026 · #5 Casey went on a road trip that covered 100 miles, stopping only for a lunch break along the way. The trip took 3 hours in total and her...

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Answer: B — 30 minutes.

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Hint 1 of 2

The 40 mph only describes the part where she's actually moving — the 3 hours also hides the lunch stop. Which piece can you compute directly?

Still stuck? Show hint 2 →

Hint 2 of 2

Time = distance ÷ speed gives the driving time only. Subtract that from the 3 total hours and what's left is the break.

Show solution

Approach: the 3 hours is driving + break; only driving obeys distance ÷ speed

The 40 mph is her speed while driving, so distance ÷ speed gives only the driving time, not the whole trip. Find that first: 100 ÷ 40 = 2.5 hours.
Whatever's left of the 3 total hours is the lunch break: 3 − 2.5 = 0.5 hour = 30 minutes.
Sanity check: 30 minutes of lunch is reasonable, and 2.5 h of driving at 40 mph really does cover 100 miles. The reusable idea: always separate ‘moving time’ from total time before using rate = distance ÷ time — the rate only describes the moving part.

2014 · #17 George walks 1 mile to school. He leaves home at the same time each day, walks at a steady speed of 3 miles per hour, and arrives just...

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Answer: B — 6 mph.

Show hints

Hint 1 of 2

Time is the fixed budget here, not speed. He must arrive at the usual moment, so his total travel time is locked at the normal value — figure that out first.

Still stuck? Show hint 2 →

Hint 2 of 2

Subtract the time the slow first half ate up; whatever's left is all he has to cover the last half-mile. Speed = distance ÷ that leftover time.

Show solution

Approach: time is the fixed budget — subtract what's spent

Normal trip: 1 mile at 3 mph takes 1/3 hr. That's the total time budget he must hit today.
Slow first half: (1/2 mile) ÷ (2 mph) = 1/4 hr used up.
Time left for the second half-mile: 1/3 − 1/4 = 4/12 − 3/12 = 1/12 hr.
Required speed = (1/2 mile) ÷ (1/12 hr) = 6 mph.
Why this transfers: in "arrive on time" problems, hold time constant and treat it as a budget. Don't average the speeds — dawdling on the first half costs disproportionately more time, which is why he must nearly triple his pace, not just speed up a little.

2006 · #13 Cassie leaves Escanaba at 8:30 AM heading for Marquette on her bike. She bikes at a uniform rate of 12 miles per hour. Brian leaves...

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Answer: D — 11:00 AM.

Show hints

Hint 1 of 2

Still stuck? Show hint 2 →

Hint 2 of 2

Closing speed: when two things approach each other, add their speeds; the gap divided by that combined speed is the time to meet.

Show solution

Approach: settle the head start, then close the gap at combined speed

Cassie rides alone from 8:30 to 9:00, half an hour, covering ½ × 12 = 6 miles. So at 9:00 the gap between them is 62 − 6 = 56 miles.
From 9:00 they ride toward each other, closing the gap at 12 + 16 = 28 mph.
Time to close 56 miles: 56 ÷ 28 = 2 hours. They meet at 9:00 + 2:00 = 11:00 AM.
Why add the speeds: in one hour Cassie eats 12 miles of road and Brian eats 16, so 28 miles of gap vanish per hour regardless of where they are. This "combined speed" trick turns every meeting problem into one simple division — just remember to handle any head start first.

CHAPTER 7

Reading rates from graphs

THEORY

A graph is a rate problem with the numbers hidden in the lines. The skill is knowing what to read where. Before touching a graph, ask one question: does the height show a running total, or a single moment? The answer changes everything.

Cumulative total (dollars spent vs. month, miles ridden vs. time): the slope is the rate; the difference of two heights is the amount over that stretch.
Distance vs. time: slope = speed. Steeper means faster.
Speed vs. time: the area underneath = distance covered.
Pie chart: each slice's angle ÷ 360° = its fraction of the whole.

The cumulative-graph trap — read TWO heights, then subtract

Here is the question setters reach for most: a graph shows total dollars saved by the end of each month. 'How much was saved during summer (June, July, August)?' The wrong move is to read August's bar and call it the answer — but that bar already includes every month before June. On a running-total graph, a single height is everything-so-far, not the slice you want.

So you read the height at end-of-August, read the height at end-of-May, and subtract. The gap between those two heights is the summer total — the graph already added the three months for you.

THE MOVE: On a cumulative graph, the answer is a DIFFERENCE of two heights — never a single reading.

🎯 Try it

On a distance-vs-time graph, a rider's line goes from (0 hr, 0 mi) up to (2 hr, 10 mi) in a straight line. What is the average speed in mph?

Walkthrough: Slope = rise ÷ run = 10 miles ÷ 2 hours = 5 mph. On a distance-vs-time graph, slope is speed.

THE TRICK

Cumulative graph → your answer is a difference of two heights. Per-period bar chart → read or sum directly. Pie chart → (slice fraction) × total. Decide which kind you're looking at before you read a single number.

WORKED EXAMPLE

PROBLEM · 2011 #9

Carmen takes a long bike ride on a hilly highway. The graph indicates the miles traveled during the time of her ride. What is Carmen's average speed for her entire ride in miles per hour?

A) 2 B) 2.5 C) 4 D) 4.5 E) 5

Carmen's bike ride is drawn as miles traveled vs. time. The question asks for her average speed over the whole ride.

Don't fuss over the hills and dips in the middle — average speed only needs the two endpoints.

Read total distance at the end of the ride and the total time across.
The line runs from the start to (the end time, the end distance); average speed is the overall slope = total miles ÷ total hours.
That slope works out to 5 mph (choice E).

Every wiggle in between is a distraction. Average speed = (final height) ÷ (final time), full stop.

The graph tempts you to measure each segment's speed and average them — the chapter-4 trap in disguise. Average speed is total distance over total time, which on a distance-vs-time graph is the straight slope from first point to last.

Answer: E — 5 mph.

RULE OF THUMB

Decide cumulative vs. per-period first. Cumulative answers are differences of heights; distance-vs-time slope is speed; speed-vs-time area is distance.

MORE LIKE THIS

1999 · #4 (figure problem)

Show answer

Answer: A — About 15 miles.

Show hints

Hint 1 of 2

The word "more" means a gap, and a gap on a graph is a vertical distance. Go to Hours = 4 and look at how far apart the two lines sit there.

Still stuck? Show hint 2 →

Hint 2 of 2

Read each line's height at the 4-hour line, then subtract — you don't need the actual mileages anywhere else.

Show solution

Approach: the 'how much more' is the vertical gap at 4 hours

Slide up the vertical line at Hours = 4. Alberto's line is at about 60 miles, Bjorn's at about 45.
"How many more" is the gap between them: 60 − 45 = 15 miles.
Reading tip you'll reuse: a difference question on a graph is always a vertical gap at one chosen x-value — find that x, then measure straight up between the curves. Steeper line = faster rider, which is why Alberto pulls ahead.

2005 · #17 The results of a cross-country team's training run are graphed below. Which student has the greatest average speed?

The results of a cross-country team's training run are graphed below. Which student has the greatest average speed?

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Answer: E — Evelyn.

Show hints

Hint 1 of 2

Speed = distance ÷ time. On a distance-vs-time graph, that ratio for each runner is the steepness of the line from the origin O out to their dot.

Still stuck? Show hint 2 →

Hint 2 of 2

You don't need numbers off the axes. Just eyeball which dot's line from O tilts up most steeply — high distance for little time.

Show solution

Approach: fastest = steepest line from the origin

Average speed is distance ÷ time, which is exactly the slope of the segment joining O to a runner's dot. Steeper line = more distance per unit time = faster.
Evelyn's dot sits high (large distance) and far left (small time), so her line from O is the steepest ⇒ Evelyn.
Watch the trap: Carla is the highest dot, but she took the most time, so she isn't fastest — 'farthest' is not 'fastest.' Only the slope from the origin tells you speed.

CHAPTER 8

Exponential growth — when a quantity multiplies each step

THEORY

Fold a piece of paper in half, then again, then again. Three folds: 8 layers. Ten folds: over a thousand. The layer count doesn't add — it doubles every fold, and doubling sprints away from adding faster than your eyes expect. That's exponential growth, and contests love it.

Two ways a quantity can grow:

Linear: the SAME amount is ADDED each step. (Save $5 every week.)
Exponential: the value is MULTIPLIED by the same factor each step. (Bacteria double each hour; savings earn 10% a year.)

They start close, then exponential rockets up. Watch $1 over 10 steps:

Both start at $1. By step 10, linear sits at $11 (still hugging the axis); exponential has hit $1024. The gap is 93×.

EXPONENTIAL FORMULA

Start V₀, multiplier r per step, after n steps:

V(n) = V₀ × rⁿ

r > 1 grows. 0 < r < 1 shrinks (0.9 = lose 10% each step).

THE MOVE: Spot the word 'each' (each year, each step). 'Adds the same' = linear; 'multiplies by the same' = exponential.

Growth factors to recognize on sight

Phrase	Multiplier per step	5 steps gives …
doubles each year	× 2	× 32
triples each year	× 3	× 243
grows 50% each year	× 1.5	× 7.59
grows 10% each year	× 1.1	× 1.61 (not 1.5!)
shrinks 10% each year	× 0.9	× 0.59
half-life: halves each step	× 0.5	× 0.03 (1/32)

The doubling-jar trap — work BACKWARD from full

The most famous doubling puzzle, and almost everyone trips. A bug culture doubles every minute. Dropped in a jar at 3:00, the jar is exactly full at 3:12. When was it half full?

Your instinct screams '3:06 — halfway through the time!' Resist it. Half the time is not half the amount when something doubles.

Run the tape backward. Forward it doubles each minute, so backward it halves each minute. Full at 3:12 means half that one minute earlier — half full at 3:11.

Full at 3:12 ⇒ half full at 3:11 (one step back) ⇒ quarter full at 3:10 (two steps back).

That second line is the real contest move: 'quarter full' is just two doublings short of full. You never needed a formula — you counted steps backward from the end.

When something doubles each step, the second-to-last step is already HALF done. Count backward from the finish, not forward from the start.

(Doubling-jar idea adapted from Problem Solving via the AMC, Australian Maths Trust.)

🎯 Try it

This one is LINEAR, not exponential — spot the difference. In 1980, CO₂ was 338 ppm, rising about 1.515 ppm each year. What is the expected level in 2030 (nearest whole number)?

Walkthrough: 'Increases by the same amount each year' is ADDING, so it's linear, not ×. Years: 2030 − 1980 = 50. Rise = 50 × 1.515 = 75.75 ≈ 76. New level = 338 + 76 = 414 ppm. (This is amc8-2024-10, answer B.)

THE TRICK

For 'doubles every N hours': start at 1 at hour 0, and after kN hours you have $2^k$. Mind the off-by-one — at hour 0 it's 1, at hour N it's 2, at hour 2N it's 4. And before you multiply, double-check the word: 'adds the same each time' is linear, not exponential.

WORKED EXAMPLE

PROBLEM · 1998 #16

Nisos Isles. In 1998 the islands have 200 people, and the population triples every 25 years. Estimate the year in which the population will be about 6000.

A) 2050 B) 2075 C) 2100 D) 2125 E) 2150

In 1998 the Nisos Isles have 200 people, and the population triples every 25 years. About when will it reach 6000?

This 'when does it reach a target?' question is a count-the-multiplications problem — no logarithms.

Target as a multiple of the start: 6000 ÷ 200 = 30×.
Count triplings: ×3, ×9, ×27 — three triplings give 27, close to 30. (A fourth jumps to 81, way past.) So three 25-year steps.
3 × 25 = 75 years after 1998 → about 2075 (choice B).

Memorize the powers of 3 (3, 9, 27, 81) and you read this like a clock.

Don't solve $200 \cdot 3^n \ge 6000$ with logs. Divide target by start to get '30 times', then count how many ×3 jumps clear it. That count, times the step length, is the time. Same skill as a logarithm — done by hand.

Answer: B — About 2075.

RULE OF THUMB

Exponential = ×r each step; after n steps it's V₀ × r^n. Don't confuse it with linear (adds a fixed amount). To find WHEN it reaches a target, divide target by start and count the multiplications.

MORE LIKE THIS

1998 · #15 Nisos Isles. In 1998 the islands have 200 people, and the population triples every 25 years. Estimate the population in the year 2050.

Nisos Isles. In 1998 the islands have 200 people, and the population triples every 25 years. Estimate the population in the year 2050.

Show answer

Answer: D — About 2000.

Show hints

Hint 1 of 2

'Triples every 25 years' means the clock matters, not a formula. How many full 25-year steps fit between 1998 and 2050? Each step multiplies the people by 3.

Still stuck? Show hint 2 →

Hint 2 of 2

Two steps means multiply by 3, then by 3 again — don't add the steps, stack the triplings.

Show solution

Approach: count the 25-year steps, multiply by 3 for each

From 1998, the tripling moments land at 2023 (one step) and 2048 (two steps) — and 2048 is essentially 2050. So two full triplings have happened.
Each tripling is a ×3: 200 → 600 → 1800. (Growth multiplies, it doesn't add — that's the whole flavor of these problems.)
1800 is closest to 2000.
Why this transfers: for repeated-multiplying growth, count how many doubling/tripling periods have passed, then multiply that many times. Adding the periods instead of multiplying is the classic mistake.

2024 · #10 In January 1980 the Mauna Loa Observatory recorded carbon dioxide CO2 levels of 338 ppm (parts per million). Over the years the average...

In January 1980 the Mauna Loa Observatory recorded carbon dioxide CO2 levels of 338 ppm (parts per million). Over the years the average CO2 reading has increased by about 1.515 ppm each year. What is the expected CO2 level in ppm in January 2030? Round your answer to the nearest integer.

Show answer

Answer: B — 414 ppm.

Show hints

Hint 1 of 2

"Same amount every year" is the signal for one move: total rise = rate × number of years. Don't add year by year.

Still stuck? Show hint 2 →

Hint 2 of 2

Technique — linear growth: increase = (yearly rate) × (years elapsed), then add to the starting level. Count the years from 1980 to 2030 first.

Show solution

Approach: rate × time, then add to start

A constant yearly increase means the rise is just rate × time — no need to step through 50 separate years. Years elapsed: 2030 − 1980 = 50.
Total increase: 50 × 1.515 = 75.75 ≈ 76 ppm.
New level: 338 + 76 = 414. Sanity check: ~1.5 ppm/yr over 50 years is roughly 75, landing just above 338+75 = 413 — only 414 is in range, so estimation alone nails the choice.

CHAPTER 9

Work-rate problems — when rates add

THEORY

One hose fills a tank in 4 hours. A second fills the same tank in 6 hours. Both run at once — how long? Almost every kid's first guess is to average 4 and 6 to get 5. It's wrong, and it's wrong in a way worth feeling: two hoses pouring together should beat the faster hose alone (4 hours), so any answer above 4 can't be right. The fix is one idea: rates add; times don't.

RATES ADD; TIMES DON'T

If A finishes alone in t_a, A's rate is 1/t_a jobs per unit time. Same for B at 1/t_b. Together the rates add:

combined rate = 1/t_a + 1/t_b

Combined time is the reciprocal:

t = 1 / (1/t_a + 1/t_b) = (t_a · t_b) / (t_a + t_b)

Why rates add. 'Jobs per hour' behaves like a speed. Each worker pours their own fraction of the tank every hour, and fractions of the same tank add. The hours each worker would need on their own are not the thing that combines — the per-hour fractions are.

Together they fill 5/12 of the tank each hour, so the whole job takes 12/5 = 2.4 hours — under the 4 hours the faster hose needs alone. ✓

Worked walkthrough. Hose A: 4 hours. Hose B: 6 hours. Both at once.

A's rate: 1/4 tank/hr. B's rate: 1/6 tank/hr.
Combined: 1/4 + 1/6 = 3/12 + 2/12 = 5/12 tank/hr.
Time: 1 ÷ (5/12) = 12/5 = 2.4 hours.

THE MOVE: Turn each time into a 'per hour' rate, ADD the rates, then flip to get the time.

Identical workers shortcut. If k identical workers each take t alone, together they take t/k. (Their rates just multiply by k.)

Work = Rate × Time — the same formula as $D = S \times T$, only the story changes. 'Job' replaces 'distance', 'jobs per hour' replaces 'mph'. The triangle still works: cover what you want.

🎯 Try it

Faucet A fills a tub in 6 minutes; faucet B fills it in 12 minutes. Running together, how many minutes to fill the tub?

Walkthrough: Rates add: 1/6 + 1/12 = 2/12 + 1/12 = 3/12 = 1/4 tub per minute. Flip it: 1 ÷ (1/4) = 4 minutes — faster than A alone (6 min), as it must be.

THE TRICK

For 'A alone takes a, B alone takes b, how long together?', memorize ab/(a+b) — the same harmonic-mean shape from chapter 4. For 3+ workers, add all the rates 1/a + 1/b + 1/c, then flip.

Trap reframe. 'A and B together take 3 hours, A alone takes 5; how long for B alone?' Set B's rate as the unknown: 1/5 + 1/x = 1/3, solve for x.

WATCH OUT

Bogus solution

Tom paints a fence in 6 hours; Huck paints the same fence in 5 hours. Working together, they'll take the average of their times: $(6 + 5)\div 2 = 5.5$ hours.

Why it breaks: two people working together must finish faster than the quicker one alone — so any answer above 5 hours is impossible on its face. You can't average the times.

The fix: Rates add, times don't. Tom does $\tfrac16$ of the fence per hour, Huck $\tfrac15$; together $\tfrac16+\tfrac15=\tfrac{11}{30}$ of the fence each hour. Flip it: $1\div\tfrac{11}{30}=\tfrac{30}{11}\approx 2.7$ hours — under 5, as it must be.

Framing inspired by AoPS Prealgebra.

WORKED EXAMPLE

PROBLEM · 2009 #6

Steve's empty swimming pool will hold 24,000 gallons of water when full. It will be filled by 4 hoses, each of which supplies 2.5 gallons of water per minute. How many hours will it take to fill Steve's pool?

A) 40 B) 42 C) 44 D) 46 E) 48

Steve's pool holds 24,000 gallons. Four hoses each pour 2.5 gallons/min. How many hours to fill it?

The hoses are identical, so don't bother with 1/a + 1/b — just add the rates by multiplying.

Combined rate: 4 × 2.5 = 10 gallons/min.
Time in minutes: 24,000 ÷ 10 = 2400 min.
In hours: 2400 ÷ 60 = 40 hours (choice A).

The deeper formula $1/(1/t_a + 1/t_b + \ldots)$ collapses to t/k when all k rates match — which is exactly this.

Pin the unit down first. Hoses are gallons per minute; the question wants hours. Convert once, at the start or the end — doing it in one spot keeps the minutes and hours from tangling.

Answer: A — 40 hours.

RULE OF THUMB

Rates add; times don't. Two workers (times a, b): combined time ab/(a+b). k identical workers (each time t): t/k. Sanity check: the combined time is always below the fastest worker alone.

MORE LIKE THIS

2009 · #6 Steve's empty swimming pool will hold 24,000 gallons of water when full. It will be filled by 4 hoses, each of which supplies 2.5...

Show answer

Answer: A — 40 hours.

Show hints

Hint 1 of 2

Four hoses running at once act like one big hose — their rates simply ADD. The answer wants hours, so save yourself a conversion by building the combined rate in gallons-per-HOUR from the start.

Still stuck? Show hint 2 →

Hint 2 of 2

Once you have a single combined rate, time = total amount ÷ rate.

Show solution

Approach: combine rates, then divide total by rate

One hose: 2.5 gal/min. Working together, rates add: 4 × 2.5 = 10 gal/min. Since the answer is in hours, scale up now: 10 gal/min × 60 = 600 gal/hour.
Time = total ÷ rate = 24,000 ÷ 600 = 40 hours.
Why this transfers: whenever several workers/pipes/machines run simultaneously, add their individual rates into one combined rate — then it's a single division. Converting units BEFORE dividing (here min→hr) avoids a clumsy 2,400-minute intermediate.

2010 · #14 In order to sew together three short strips of cloth to get one long strip, Cathy needs 18 minutes. How much time does she need to sew...

In order to sew together three short strips of cloth to get one long strip, Cathy needs 18 minutes. How much time does she need to sew together a really long piece consisting of six short strips?

Show answer

Answer: D — 45 minutes

Show hints

Hint 1 of 2

Joining strips needs one fewer seam than the number of strips.

Still stuck? Show hint 2 →

Hint 2 of 2

Find the time per seam from the three-strip case, then count seams for six strips.

Show solution

Approach: count seams, scale by time per seam

Three strips need 2 seams and take 18 min, so each seam takes 9 min.
Six strips need 5 seams: 5×9 = 45 minutes.

1994 · #14 Two children at a time can play pairball. For 90 minutes, with only two children playing at a time, five children take turns so that...

Two children at a time can play pairball. For 90 minutes, with only two children playing at a time, five children take turns so that each one plays the same amount of time. The number of minutes each child plays is

Show answer

Answer: E — 36.

Show hints

Hint 1 of 2

Two kids are always playing, so the game produces TWO 'player-slots' every minute — count the total slot-time first, then hand it out fairly.

Still stuck? Show hint 2 →

Hint 2 of 2

Total playing-time available = 2 slots × 90 minutes. Share that pot equally among the 5 children.

Show solution

Approach: total child-minutes, shared equally

Each minute fills 2 playing spots, and the game runs 90 minutes, so there are 2 × 90 = 180 'child-minutes' of play to give out.
Five children split it evenly: 180 ÷ 5 = 36 minutes each.
The key move is counting total work in 'person-units' (here child-minutes) before dividing — the same trick behind 'if 3 painters take 4 hours, that's 12 painter-hours of work.' Reality check: 36 < 90, which makes sense since nobody plays the whole time.

⬢ FINAL TEST

Stretch test

Five harder rate problems combining D=ST with average-speed, relative motion, and trade-chains.

1999 · #22 In a far-off land three fish can be traded for two loaves of bread, and a loaf of bread can be traded for four bags of rice. How many...

In a far-off land three fish can be traded for two loaves of bread, and a loaf of bread can be traded for four bags of rice. How many bags of rice is one fish worth?

Show answer

Answer: D — 2⅔ bags.

Show hints

Hint 1 of 2

Bread is the middle-man currency. Pick one good to measure everything in — rice — and convert the bread in the fish trade into rice. Then the fish↔rice rate falls right out.

Still stuck? Show hint 2 →

Hint 2 of 2

Chain the rates so the middle unit (bread) cancels: fish→bread→rice. This is exactly how unit conversions work in science too.

Show solution

Approach: chain the trades through a common unit so bread cancels

Express the fish deal in rice. 3 fish = 2 loaves, and each loaf = 4 bags, so 2 loaves = 8 bags. Thus 3 fish = 8 bags of rice.
Divide to get one fish: 8 ÷ 3 = 2⅔ bags of rice.
The transferable move: when trades link A→B→C, convert through the shared item so it cancels — just like converting hours→minutes→seconds. (Sanity check: a fish is worth a bit more than half a loaf, and a loaf is 4 bags, so 2-and-a-bit bags per fish feels right.)

Another way — scale to whole pieces (avoid fractions of a fish):

To dodge thirds, work with 3 fish at once. 3 fish trade for 2 loaves, and 2 loaves = 8 bags of rice, so 3 fish ↔ 8 bags.
Reading the ratio 3 fish : 8 bags, one fish is 8/3 = 2⅔ bags.
Bundling to the smallest whole quantities (here 3 fish) keeps the arithmetic clean and only divides once at the very end.

2002 · #24 Miki has a dozen oranges of the same size and a dozen pears of the same size. Miki uses her juicer to extract 8 ounces of pear juice...

Show answer

Answer: B — 40%.

Show hints

Hint 1 of 2

The "dozen" is pure bait — she uses an *equal* number of each fruit, so however many that is divides out. All that matters is the juice from *one* pear versus *one* orange.

Still stuck? Show hint 2 →

Hint 2 of 2

Pears: 8 oz from 3, so 8/3 oz each. Oranges: 8 oz from 2, so 4 oz each. The blend's pear share is just one pear's juice over (one pear + one orange).

Show solution

Approach: compare juice per fruit

Since she blends *equal counts* of each fruit, the common count cancels — work per fruit. One pear yields 8/3 oz; one orange yields 8/2 = 4 oz.
Pear-to-orange juice is 8/3 : 4, and clearing the 3 gives 8 : 12 = 2 : 3.
Pear's share = 2 ÷ (2 + 3) = 2/5 = 40%.
*Worth keeping:* when two things are mixed in equal *counts*, the actual count is irrelevant — reduce to one of each and compare. Chasing the dozen (32 oz vs 48 oz) gives the same answer with bigger numbers.

Another way — scale up to the full dozen:

12 pears give 12 × 8/3 = 32 oz; 12 oranges give 12 × 4 = 48 oz.
Pear fraction = 32 ÷ (32 + 48) = 32/80 = 2/5 = 40% — same ratio, just unscaled.

2022 · #22 A bus takes 2 minutes to drive from one stop to the next, and waits 1 minute at each stop to let passengers board. Zia takes 5 minutes...

A bus takes 2 minutes to drive from one stop to the next, and waits 1 minute at each stop to let passengers board. Zia takes 5 minutes to walk from one bus stop to the next. As Zia reaches a bus stop, if the bus is at the previous stop or has already left the previous stop, then she will wait for the bus. Otherwise she will start walking toward the next stop. Suppose the bus and Zia start at the same time toward the library, with the bus 3 stops behind. After how many minutes will Zia board the bus?

Show answer

Answer: A — 17 minutes.

Show hints

Hint 1 of 2

Zia only makes a decision at the instants she reaches a stop — every 5 minutes. So you don't need a continuous chase; just check the situation at t = 5, 10, 15, … and see where the bus is each time.

Still stuck? Show hint 2 →

Hint 2 of 2

The bus's rhythm is 2 min driving + 1 min waiting = 3 min per stop. Track which stop each of them is at on Zia's 5-minute marks.

Show solution

Approach: only check the few moments Zia can act — her 5-minute arrivals

Insight: a step-by-step chase looks messy, but Zia only chooses to wait-or-walk when she arrives at a stop — at t = 5, 10, 15. Sampling just those moments (the bus runs on a tidy 3-min-per-stop cycle: 2 driving + 1 waiting) makes it a 3-line simulation. Number the stops from the bus's start (stop 0); at t = 0, Zia is at stop 3, bus at stop 0.
t = 5: Zia at stop 4. Bus took 5 min → finished stop 1 (arrived at 2 min, left at 3 min, arrived at stop 2 at 5 min). Bus is at stop 2 — not yet at the previous stop (3), so Zia walks on.
t = 10: Zia at stop 5. Bus: from t = 5 (at stop 2) waits 1 min (leaves at 6), drives 2 min to stop 3 (arrives at 8), waits till 9, drives to stop 4 (arrives at 11). So at t = 10, bus is mid-drive between stops 3 and 4 — not at the previous stop (4), so Zia walks on.
t = 15: Zia at stop 6. Bus: arrives at stop 4 at 11, waits till 12, drives to stop 5 (arrives 14, waits till 15). At t = 15, bus is at stop 5 — the previous stop. Zia waits.
Bus leaves stop 5 at t = 15 and drives 2 min to stop 6: arrives at t = 17.
You'll see this again: when one mover only acts at fixed intervals, you don't have to track time continuously — jump straight to those decision instants and read off the other mover's state. Discretizing turns a chase into a short table.

2024 · #29 Anne drives from point A to point B and then immediately back to A. Benni drives from point B to point A and then immediately back to B....

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Answer: C — 30 min

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Hint 1 of 2

At the first meeting the two together have covered the whole road once, fixing the road length in their speeds.

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Hint 2 of 2

Track who turns around when; the second meeting comes after Anne (the faster one) catches Benni from behind.

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Approach: track positions through the turnarounds to the second meeting

With speeds 3v and v meeting head-on after 15 min, the road length is 4v·15 = 60v.
Anne reaches the far end at 20 min and turns back; Benni is still heading the same way.
Anne then closes the 20v gap at relative speed 2v, taking 10 more min, so they meet at 30 min.

2017 · #28 Two runners are training at the same time on a 720 m long, round running track. They run with constant speed in opposite directions. The...

Two runners are training at the same time on a 720 m long, round running track. They run with constant speed in opposite directions. The first runner needs four minutes for one lap, the second five minutes. How many metres does the second runner run between two consecutive meetings of the two runners?

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Answer: E — 320

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Hint 1 of 2

Running opposite ways, the two runners' speeds add when finding how fast the gap to the next meeting closes.

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Hint 2 of 2

Find the time between meetings, then multiply by the slower runner's speed.

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Approach: closing speed gives the meeting interval

Speeds: 720/4 = 180 m/min and 720/5 = 144 m/min; closing speed = 180 + 144 = 324 m/min.
They meet every 720 ÷ 324 minutes; in that time the second runner covers 144 × 720/324.
That equals 320 m.

🚀 STRETCH

Stretch practice — beyond AMC 8

4 bonus problems on Ratios, Rates & Proportions. These are typed-answer (no multiple choice) and tilt harder — closer to early AMC 10. Try the ones that look fun.

Stretch · #2 A fly and a jogger start 12 km apart. The jogger runs straight toward the fly's starting spot at 4 km per hour. Meanwhile the fly zooms...

A fly and a jogger start 12 km apart. The jogger runs straight toward the fly's starting spot at 4 km per hour. Meanwhile the fly zooms back and forth at 6 km per hour: it flies to the jogger, turns around, flies back, turns around again, and keeps doing this. The fly keeps flying until the jogger reaches the fly's starting spot. How far does the fly travel in total?

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Answer: 18 km

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Hint 1 of 4

The fly's back-and-forth path looks scary, but you only need the TOTAL distance it flies. And distance = speed multiplied by time. So the real question is: how long is the fly in the air?

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Hint 2 of 4

Stop trying to follow the fly. Watch the jogger instead. The flying stops the exact moment the jogger finishes the 12 km. So all you need is the jogger's travel time.

Still stuck? Show hint 3 →

Hint 3 of 4

How long does the jogger take to go 12 km at 4 km per hour? Divide: 12 divided by 4.

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Approach: Find the time, not the path

The trap is adding up all the tiny back-and-forth flights. The clever move is to think about TIME, not the fly's messy path.
The fly stops flying when the jogger covers the 12 km. At 4 km/hr that takes $\dfrac{12}{4}=3$ hours.
The fly was flying that entire 3 hours at 6 km/hr, so it covers $6\times3=18$ km.
The fly travels 18 km. (Notice the puzzle has to tell us when the flying stops, or the answer wouldn't be clear — spotting that hidden assumption is part of careful problem solving.)

Stretch · #6 Two hikers need to reach a village 12 km away as fast as possible. They walk at 4 km/h and ride a bike at 12 km/h, but they have only...

Two hikers need to reach a village 12 km away as fast as possible. They walk at 4 km/h and ride a bike at 12 km/h, but they have only one bike, carrying one rider. Using the smartest plan, what is the shortest time (in hours) for BOTH of them to arrive?

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Answer: 2 hours (average speed 6 km/h)

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Hint 1 of 4

The two hikers are identical, so the fair, symmetric plan is to share the bike equally and have both arrive at the same moment.

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Hint 2 of 4

Hiker 1 rides, then leaves the bike on the road and walks the rest. Hiker 2 walks until he reaches the parked bike, then rides. Where should the bike be dropped so they tie?

Still stuck? Show hint 3 →

Hint 3 of 4

Drop the bike exactly at the halfway point, 6 km. Then each rides 6 km and walks 6 km. Compute the time for one hiker: $\tfrac{\text{riding}}{12} + \tfrac{\text{walking}}{4}$.

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Approach: Symmetric bike-sharing: each rides half, walks half

Since both hikers have the same speeds, the smartest plan is symmetric: split the biking equally so they arrive together.
Hiker 1 rides the first 6 km, leaves the bike, and walks the last 6 km. Hiker 2 walks the first 6 km, finds the parked bike, and rides the last 6 km. Each rides 6 km and walks 6 km, so by symmetry they arrive together.
For either hiker: $T = \tfrac{6}{12} + \tfrac{6}{4} = 0.5 + 1.5 = 2$ hours.
So both arrive in 2 hours. (Average speed $= \tfrac{12}{2} = 6$ km/h, the harmonic mean of 4 and 12.)

Stretch · #17 Two trains are $200$ miles apart on the same track, heading toward each other. One goes $60$ mph, the other $40$ mph. A fly starts...

Two trains are $200$ miles apart on the same track, heading toward each other. One goes $60$ mph, the other $40$ mph. A fly starts on the front of the slower train and flies back and forth between the two trains at $240$ mph, turning around instantly each time it reaches a train, until the trains crash and squash it. How far does the fly travel in total?

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Answer: 480 miles

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Hint 1 of 3

Adding up the fly's endless shrinking back-and-forth trips is a nightmare. Ask the key question: what single thing, times the fly's speed, gives its total distance?

Still stuck? Show hint 2 →

Hint 2 of 3

Distance = speed $\times$ time, and the fly is flying the WHOLE time until the trains crash. So the real question is: how long until they crash?

Still stuck? Show hint 3 →

Hint 3 of 3

The two trains close the $200$-mile gap together at $60 + 40 = 100$ mph. Find how long that takes, then multiply by the fly's $240$ mph.

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Approach: Asking the key question — find the time, not the zig-zags

Don't add up the fly's zig-zags. Ask the key question: how long does the fly fly? Its distance is just speed times time, and it flies until the trains crash.
The trains approach each other at a combined speed of $60 + 40 = 100$ mph. To close a $200$-mile gap at $100$ mph takes time $= \frac{200}{100} = 2$ hours.
The fly flies for those whole $2$ hours at $240$ mph: $240 \times 2 = 480$ miles.

Stretch · #5 Anton and Ben start running toward each other from the two ends of a long straight path (Anton from end A, Ben from end B). Each runs at...

Anton and Ben start running toward each other from the two ends of a long straight path (Anton from end A, Ben from end B). Each runs at his own steady speed. They first meet 800 m from Ben's end. They keep going, reach the far ends, instantly turn around, and meet a second time 400 m from Anton's end. How long is the path (in meters)?

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Answer: 2000 m

Show hints

Hint 1 of 4

Don't track each runner separately at first — track the TOTAL distance the two of them run together.

Still stuck? Show hint 2 →

Hint 2 of 4

At the FIRST meeting, the two of them together have covered the path exactly once (their two pieces fill it).

Still stuck? Show hint 3 →

Hint 3 of 4

At the SECOND meeting, together they have covered the path exactly three times. (Draw it: each one went to the far end and partway back.)

Show solution

Approach: Combined distance triples between the two meetings

Because both run steadily, their distances grow in the same proportion. At the first meeting they have together run one full path length; at the second meeting they have together run three path lengths (each finishes the path and comes partway back).
So the combined distance tripled, and since both run steadily, each runner's own distance also triples.
Ben ran 800 m to the first meeting (it was 800 m from his end B), so by the second meeting Ben has run $3 \times 800 = 2400$ m. The second meeting is 400 m from Anton's end A, meaning Ben ran the whole path and came back 400 m, so path $= 2400 - 400 = 2000$ m.
The path is 2000 m long. (Check Anton: he ran $2000 - 800 = 1200$ m to the first meeting, then $3 \times 1200 = 3600$ m by the second, which is one path plus 1600 m back, leaving him $2000 - 1600 = 400$ m from end A. It fits.)

APPENDIX

Rates quick-reference

Memorize these

FORMULAS / FACTS TO KNOW COLD

D = S × T (and its two cousins). Same shape as Work = Rate × Time.
Ratio parts: a : b has a + b parts; one part = total ÷ total parts. Each side is also a fraction of the whole: a/(a+b) and b/(a+b) — multiply by the total in one step.
Join two ratios on a shared term: scale each so the shared quantity becomes their LCM, then read the chain (blue:white 2:3 and white:red 4:5 → white = 12 → blue : red = 8 : 15).
Average speed = total distance ÷ total time. NEVER the average of the speeds (unless the times are equal). This is the marquee trap.
Factor-label conversion: multiply by fractions equal to 1 (5280 ft / 1 mi); unwanted units cancel, the wanted unit survives.
1 hour = 3600 seconds = 60 minutes.
1 mile = 5280 feet. 1 km = 1000 m. 1 yard = 3 feet. 1 foot = 12 inches.
1 mph ≈ 1.467 ft/s (so 60 mph = 88 ft/s).
Equal-distance round trip avg speed: 2ab / (a+b) (harmonic mean).
Equal-time legs avg speed: (a+b)/2 (simple average).
Work-rates ADD; times don't. A alone a, B alone b → together T = ab / (a+b); for 3+ workers add all rates 1/a + 1/b + … and flip. k identical workers each taking t: t/k.
Same-direction closing speed: faster − slower.
Opposite-direction closing speed: sum of speeds.
Exponential growth: V(n) = V₀ · rⁿ. Compound interest: V₀ · (1 + p/100)ⁿ.
Percent = ratio out of 100. Percent of change = change ÷ original. Chain percent changes by MULTIPLYING multipliers (down 20% = ×0.80), never by adding.

Common traps

Averaging speeds for equal-distance trips — THE marquee trap. The slow leg eats more time, so the average is pulled toward the slow speed. Always use total distance ÷ total time (60 & 30 mph round trip averages 40, not 45).
Averaging speeds to close a gap. Two cars approaching each other close the gap at the SUM of their speeds, not the average — both eat into it at once.
Lining up two ratios without matching the shared term. blue:white 2:3 and white:red 4:5 can't chain until you scale both so white agrees (=12).
Forgetting to convert minutes to hours before applying D = S × T with mph. (Wrong final units = a flipped conversion fraction.)
Inverse vs. direct proportion. 'More workers, less time' is inverse (product fixed); 'more time, more distance' is direct (ratio fixed).
Averaging times for work-rate problems. Rates add, times don't — two helpers must finish FASTER than the quicker one alone.
Cumulative vs. per-period graphs. Cumulative answers are differences of two heights; per-period answers are direct reads.
Linear vs. exponential. 'Adds the same each year' is linear; 'multiplies each year' is exponential — a 10% raise four times compounds to 46.4%, not 40%.
Doubling-jar 'halfway in time' error. If something doubles each step and is full at step 12, it was half full at step 11, not at the time-midpoint — count backward from the finish.
Counting only the vehicles that depart during your trip. You also pass those already en route when you started (the train-passing trap).
Dropping a stop or a head start from total time in an average-speed problem.
Assuming −p% then +p% cancels. It never does: −20% then +20% lands at ×0.96, a 4% loss, because the rise is taken on a smaller amount.
Reading 'percent of the whole' as 'percent of the other part'. A 2 : 3 split makes boys 2/5 (40%) of the class, not 2/3.

Warm-ups

Drill these:

60 mph for 90 minutes = how many miles? (90)
Faucet A fills a tub in 6 min, B in 12 min; together: 1/6 + 1/12 = 1/4 → 4 min.
Mix 30% of 10 L with 70% of 20 L. Combined %? ((0.3·10 + 0.7·20) / 30 ≈ 56.7%)
4 painters take 9 hours; how long for 6? (Inverse: 4·9 = 36 painter-hours; ÷6 = 6 hours.)
Round trip 60 mph and 30 mph (equal distance) average? (40 mph, via 2·60·30/90.)

Want to climb higher? — advanced rate ideas (#22–#25 territory)

Pick convenient numbers. When only the shape of a trip is given (fractions, percents, no real distance/time), choose the total yourself to kill the denominators — thirds → use 3, quarters → use 4. The answer can't depend on the real size.
Round-track repeated meetings. Opposite ways: they meet every track ÷ (v₁ + v₂). Same way: the faster laps the slower every track ÷ (v_fast − v_slow).
Three-worker together-time (alone a, b, c): T = 1 / (1/a + 1/b + 1/c). Same pattern, three terms.
Filling AND draining at once. Fill rate 1/a, drain rate 1/d give net rate 1/a − 1/d — the drain subtracts.
Mixture problems. Mixing x liters of A% with y liters of B% gives concentration (xA + yB) / (x + y) — a weighted average.
Train passes a person vs. a station. Past a point: train travels its own length. Past a station: train length + station length. Time = length ÷ train speed.
Continuous compound interest (rare here): e^rt grows faster than discrete (1+r)^t for the same r, t.

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