Ratios, Rates & Proportions — Same ratio in two places.

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.

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.

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.

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.

Framing inspired by AoPS Prealgebra.

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

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:

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.

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.

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.

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:

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% (\(\div 2\)).

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.

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).

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.

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.

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.)

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.

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 ✓.

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.)

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.

Conversion facts to know cold

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)

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)

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.

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.)

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.

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×.

Growth factors to recognize on sight

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.

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.)

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.

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.

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