Fractions, Decimals & Percent — Competition Techniques

By the end of this chapter you will:

1. Convert between fractions, decimals, and percents at sight
2. Compose percent-of-percent chains correctly
3. Read a circle graph and extract the right slice
4. Use the complement ("one minus") to dodge messy direct counts

2.0 The BC gap

BC baseline: G5 covers equivalent fractions and decimals to thousandths. G6 adds whole-number percents and discount problems (one-step). What BC does not cover: percent-of-percent chains, weighted-region probability, and complement-counting tricks. Gauss compounds the basics: $20\%$ of $50\%$ of $80$, or "Brett pours half out, then Juanita pours $20\%$ of her glass in". These chains are where students lose points.

2.1 Fluency table — memorize these

2.2 Circle graphs — 2024 Q4

Students were asked to choose their favourite school day. Results: Monday 15%, Tuesday 10%, Wednesday 25%, Thursday 20%, Friday 30%. Which day was chosen by exactly one-quarter of the students?

One-quarter $= 25\% \rightarrow$ Wednesday. Answer (C). ✅

🔑 Match the fraction to a familiar percent. $\tfrac{1}{4} = 25\%$; $\tfrac{1}{5} = 20\%$; $\tfrac{3}{10} = 30\%$.

2.3 Percent-of-percent chains — 2024 Q16

Brett and Juanita each have $300$ mL in a glass. Brett pours half of his out. Then Juanita pours $20\%$ of her water into Brett's glass. What volume is in Brett's glass now?

Step 1 — Brett pours half out: $300 \cdot \tfrac{1}{2} = 150$ mL left.

Step 2 — Juanita pours $20\%$ of her $300$ mL: $300 \cdot 0.20 = 60$ mL transferred.

Step 3 — Brett's glass now: $150 + 60 = \mathbf{210}$ mL. Answer (A). ✅

⚠️ Trap: Juanita's 20% is of her current amount ($300$ mL, untouched), not of Brett's $150$. Read carefully — "her water".

2.4 Complement: "1 minus the easy case"

A spinner is split into 12 unshaded and 3 shaded sections, with each unshaded $3 \times$ the size of each shaded. P(shaded) = ? (2024 Q17)

Let one shaded sector be $1$ unit; each unshaded is $3$ units.

Total area = $3 \cdot 1 + 12 \cdot 3 = 3 + 36 = 39$ units.

$$P(\text{shaded}) = \frac{3}{39} = \frac{1}{13}$$

Answer (D). ✅

🔑 Probability with weighted regions = $\dfrac{\text{shaded area}}{\text{total area}}$, NOT $\dfrac{\#\text{shaded slices}}{\#\text{slices}}$. The slice count is irrelevant when slices differ in size.

2.5 Trap Alerts ⚠️

1. 20% off then 20% more ≠ original. $100 \to 80 \to 96$, not $100$. Successive percent changes don't add, they multiply.
2. "% of A" vs "% of B" — read which number the percent applies to.
3. Equal-size assumption — only valid when explicitly stated. Spinners look equal in the diagram even when they aren't.

2.6 Mnemonic

"Convert, chain, complement, check the base."

• Convert to whichever form is easiest (fraction beats decimal in Gauss)
• Chain percentages by multiplying the multipliers ($\times 0.80 \times 0.80 = \times 0.64$)
• Complement for "at least" / hard regions
• Check the base — what does the percent apply to?

Practice Set

1. (Part A) $\tfrac{2}{5}$ of $150$ is ____.
2. (Part B) After two consecutive $10\%$ price increases, a $\$200$ item costs ____.
3. (Part B) A circle graph has slices $30\%, 25\%, x\%, 15\%, 10\%$. Find $x$.
4. (Part C) Of $400$ apples, $25\%$ are red, $40\%$ are green, the rest are yellow. If $1/3$ of the green ones are sold, how many apples remain in total?

Answers: 1) 60; 2) $\$242$; 3) 20; 4) $400 - \frac{1}{3} \cdot 160 = 400 - 53.\overline{3}$, rounded depending on integer constraint — assume integer apples → 347.

End of chapter. Next: Ratios, Rates & Two-Stage Word Problems.