Ratios, Rates & Two-Stage Word Problems

By the end of this chapter you will:

1. Set up and scale ratios fluently
2. Handle two-stage problems (before / after a change) by anchoring on the unchanged quantity
3. Use $d = rt$ for speed-time-distance, including the average-speed trap
4. Solve coin / mixture problems with two simultaneous constraints

3.0 The BC gap

BC baseline: G6 introduces ratios (part-part, part-whole, equivalent) — but only the introductory mechanics. There is no two-stage manipulation, no $d = rt$ formula, no average-speed problems, no mixture/coin word problems. G5 mentions duration / elapsed time but doesn't teach the base-60 carry/borrow that Gauss requires. This chapter fills all four gaps.

3.1 Ratio basics

A ratio $a : b$ is the same information as the fraction $\dfrac{a}{b}$. Scaling preserves the ratio: $3 : 5 = 6 : 10 = 30 : 50$.

Setup move: if quantities are in ratio $a : b$, write them as $ak$ and $bk$ for some scaling factor $k$.

Example

Two numbers are in ratio $3 : 7$ and sum to $50$. Find the smaller.

Write them as $3k$ and $7k$. Then $3k + 7k = 50 \Rightarrow k = 5$. Smaller $= 3 \cdot 5 = 15$.

3.2 Two-stage ratio problems

The Gauss flavor: a ratio is given, an event changes one quantity, a new ratio is reported, and you must reconstruct.

Example — paradigm problem

Red and blue marbles are in ratio $3 : 5$. After 6 red marbles are added, the ratio becomes $3 : 4$. How many blue marbles?

Anchor on the unchanged quantity — blue.

Original: $3k$ red, $5k$ blue.

After: $3k + 6$ red, still $5k$ blue.

Ratio: $\dfrac{3k+6}{5k} = \dfrac{3}{4}$.

Cross-multiply: $4(3k + 6) = 3 \cdot 5k \Rightarrow 12k + 24 = 15k \Rightarrow k = 8$.

Blue marbles $= 5k = \mathbf{40}$.

🔑 Anchor strategy: identify the quantity that doesn't change. Write the new ratio using that fixed anchor and one unknown. The equation falls out.

3.3 Speed, time, distance — $d = rt$

The three forms:

• $d = rt$ (distance = rate × time)
• $r = d / t$
• $t = d / r$

2024 Q9 — division word problem

Olivia cuts a $42$ cm string into $2$ cm pieces. Jeff cuts a $42$ cm string into $3$ cm pieces. How many more pieces does Olivia have than Jeff?

Olivia: $42 / 2 = 21$ pieces. Jeff: $42 / 3 = 14$. Difference: $21 - 14 = \mathbf{7}$. Answer (A). ✅

The average-speed trap

A car travels from A to B at $60$ km/h and returns at $30$ km/h. What is the average speed?

Wrong: $\dfrac{60 + 30}{2} = 45$.

Right: average speed = $\dfrac{\text{total distance}}{\text{total time}}$. If $A \to B$ is $d$ km:

• Time outbound: $d / 60$ h
• Time return: $d / 30$ h
• Total time: $\dfrac{d}{60} + \dfrac{d}{30} = \dfrac{d + 2d}{60} = \dfrac{3d}{60} = \dfrac{d}{20}$ h
• Total distance: $2d$ km

Average speed $= \dfrac{2d}{d/20} = 40$ km/h.

🔑 Average speed is NOT the average of the speeds. It's harmonic-mean-like. Always use total distance over total time.

3.4 Mixture & money problems

The pattern: two types (cheap/expensive coins, low-/high-% solutions), two facts (count total, value total), set up two equations.

Example

A piggy bank has dimes ($\$0.10$) and quarters ($\$0.25$), $24$ coins totaling $\$3.60$. How many quarters?

Let $q$ = quarters, $d$ = dimes.

$$q + d = 24 \qquad 0.25q + 0.10d = 3.60$$

From the first: $d = 24 - q$. Substitute:

$$0.25q + 0.10(24 - q) = 3.60 \Rightarrow 0.15q + 2.40 = 3.60 \Rightarrow q = 8$$

3.5 Time arithmetic — base-60

Time problems aren't decimal arithmetic — hours and minutes carry at $60$, not $10$.

2024 Q7 — worked in full

Katie completed two laps. The first lap took $3$ min $45$ s. The second took $4$ min $35$ s. What was her total time?

Add minutes and seconds separately, then carry if seconds $\ge 60$.

• Seconds: $45 + 35 = 80 = 60 + 20$ → write $20$ s, carry $1$ min.
• Minutes: $3 + 4 + 1 = 8$ min.

Total: 8 min 20 s. Answer (D). ✅

🔑 The carry on time problems is $60$, not $10$. Same idea for hours: $h:m:s$ carries every $60$ on $s$ and $m$, every $24$ on $h$ for calendar problems.

Elapsed time — the subtraction case

A movie starts at $4{:}35$ pm and ends at $7{:}20$ pm. How long is it?

Borrow $1$ hour from the $7$ to lend $60$ minutes to the $20$:

• $7{:}20 \to 6{:}80$
• $6{:}80 - 4{:}35 = 2{:}45$

Movie length: 2 h 45 min.

⚠️ BC G5 covers "duration / elapsed time" informally but does NOT teach base-60 carry/borrow. Gauss expects fluency with this.

3.6 Trap Alerts ⚠️

1. Ratio total ≠ sum. "Boys : girls = 3 : 5, class of 24" means $3k + 5k = 24$, so $k = 3$ — not 24 boys.
2. Average speed harmonic-mean trap. Always go via $\dfrac{\text{total distance}}{\text{total time}}$.
3. Two-stage: identify the anchor. If both quantities change, you'll need two variables and two equations.

3.7 Mnemonic

"Scale by $k$, anchor what doesn't move, total over total for averages."

Practice Set

1. (Part A) The ratio of dogs to cats is $2 : 3$, and there are $12$ dogs. How many cats?
2. (Part B) A car drives $180$ km at $60$ km/h, then $120$ km at $80$ km/h. Average speed?
3. (Part B) Apples and oranges are in ratio $4 : 7$. After 9 more apples arrive, the ratio is $5 : 7$. How many oranges?
4. (Part C) Jamal walks at $5$ km/h; his friend cycles at $20$ km/h. They start $30$ km apart heading toward each other. How long until they meet?

Answers: 1) 18; 2) $\dfrac{300}{180/60 + 120/80} = \dfrac{300}{3 + 1.5} = 66.\overline{6}$ km/h; 3) Anchor oranges: $7k$. Apples $4k \to 4k+9$. So $\dfrac{4k+9}{7k} = \dfrac{5}{7} \Rightarrow 7(4k+9) = 35k \Rightarrow 63 = 7k \Rightarrow k = 9$, oranges = 63; 4) Combined speed 25 km/h, time = $\frac{30}{25}$ = 1.2 h = 1 h 12 min.

End of chapter. Next: Algebra Readiness — Variables, Substitution & Cryptarithms.