13.1 Time, Percentages & Ratios
Key Takeaways
- Convert mixed times to the same unit (usually decimal hours) before adding: ½ hour = 0.5, 1¾ hours = 1.75, then add whole hours.
- Engine 4 hazmat scene time: 0.5 + 1.75 + 3.0 = 5.25 hours (not 5.75 or 6.00).
- A percent increase multiplies the base by (1 + rate): $1,600,000 × 1.20 = $1,920,000; 20% of the base alone is only the increase, not the new total.
- A ratio of candidates to positions is total candidates ÷ positions: 1,600 ÷ 80 = 20, written as 20 to 1.
- Common traps: adding fractions incorrectly, treating the percent change as the new total, and reversing ratio order (positions to candidates).
13.1 Time, Percentages & Ratios
Quick Answer: Convert every time piece to the same unit (decimal hours), then add. A 20% increase means new total = old total × 1.20, not “add 20” or keep only the 20% piece. A ratio of candidates to jobs is candidates ÷ jobs (for example, 1,600 ÷ 80 = 20 to 1).
Section Twelve of the official 2026 Firefighter Recruit Exam Study Packet opens with three problem types you will also see on general aptitude tests: mixed-time addition, percent increase, and simple ratios. The numbers are fire-service themed (hazmat scene time, apparatus budget, recruit competition), but the methods are pure arithmetic. Master the method once; the story is only packaging.
Method 1 — Adding mixed times (fractions → decimals)
Goal
Find total on-scene time when the packet gives pieces in fractions and whole hours.
Step-by-step method
- List every time segment exactly as given (do not skip “assist in cleanup” or similar trailing pieces).
- Convert every fraction to a decimal hour (or convert everything to minutes—pick one system and stick to it).
- Add the decimals (or minutes).
- Match the answer choices (usually decimal hours).
Fraction → decimal hour conversions you must know cold
| Fraction of an hour | Minutes | Decimal hours |
|---|---|---|
| ¼ hour | 15 min | 0.25 |
| ½ hour | 30 min | 0.50 |
| ¾ hour | 45 min | 0.75 |
| 1¼ hours | 75 min | 1.25 |
| 1½ hours | 90 min | 1.50 |
| 1¾ hours | 105 min | 1.75 |
| 2¼ hours | 135 min | 2.25 |
Worked example — Engine 4 hazmat (packet Q1)
Given: Engine 4 spent ½ hour identifying the spilled product, one and three-quarter hours containing the product, and three hours assisting cleanup.
Find: Total time on the scene.
Work:
| Segment | Given | Decimal hours |
|---|---|---|
| Identify product | ½ hour | 0.50 |
| Contain product | 1¾ hours | 1.75 |
| Clean up | 3 hours | 3.00 |
| Total | 0.50 + 1.75 + 3.00 = 5.25 |
Answer: 5.25 hours.
Why the other choices look tempting
| Wrong choice | How people get it | Trap name |
|---|---|---|
| 5.75 hours | 0.5 + 2.25 + 3 (misread 1¾ as 2¼) or 0.75 + 2 + 3 | Bad fraction conversion |
| 6.00 hours | Rough “about half + almost 2 + 3 ≈ 6” | Rounding instead of calculating |
| 6.75 hours | 0.5 + 1.75 + 3 + 1.5 (double-counted a segment) or treated ¾ as extra hours | Adding a phantom piece |
Minute check (optional): 30 + 105 + 180 = 315 minutes → 315 ÷ 60 = 5.25 hours. Same answer; use this if decimal addition feels shaky under time pressure.
Practice pattern (same method, new numbers)
Engine spends ¾ hour size-up, 2½ hours overhaul, and 1 hour rehab. Total = 0.75 + 2.50 + 1.00 = 4.25 hours.
Method 2 — Percent increase (and decrease)
Goal
Find the new amount after a stated percent change to a budget, inventory, or count.
Formulas (memorize both)
-
New total after an increase of p%:
new = original × (1 + p/100)
Example: 20% increase → multiply by 1.20. -
Amount of the increase alone:
increase = original × (p/100)
Example: 20% of $1,600,000 = $320,000 (this is not the new budget). -
New total after a decrease of p%:
new = original × (1 − p/100)
Example: 15% cut → multiply by 0.85.
Worked example — apparatus budget (packet Q2)
Given: Budget for new apparatus was $1,600,000 in 2001. Increased by 20% for 2002.
Find: 2002 budget.
Work (preferred one-step method):
$1,600,000 × 1.20 = $1,920,000
Work (two-step method—same answer):
20% of $1,600,000 = 0.20 × $1,600,000 = $320,000 (the increase only)
New budget = $1,600,000 + $320,000 = $1,920,000
Answer: $1,920,000.
Distractor map (packet-style)
| Choice | What it really is | Why it is wrong for “new budget” |
|---|---|---|
| $320,000 | 20% of the base only | That is the increase, not the new total |
| $660,000 | Unrelated or partial arithmetic | No valid percent path from $1,600,000 at 20% |
| $1,300,000 | Looks like a decrease or a mis-subtraction | 20% increase cannot lower the base |
| $1,920,000 | Base × 1.20 | Correct |
Partial-percent traps
- “Increased by 20%” ≠ “is 20% of.”
“Is 20% of” would be $320,000. “Increased by 20%” is $1,920,000. - Do not add 20 to the number ($1,600,020)—percent means per hundred of the base.
- Compound wording: “increased by 20% and then by another 10%” is not a single 30% step; apply 1.20 then 1.10 (or multiply by 1.20 × 1.10 = 1.32). The packet’s Q2 is a single 20% step only.
Quick percent check table
| Original | Change | Multiplier | New total |
|---|---|---|---|
| $1,600,000 | +20% | ×1.20 | $1,920,000 |
| $1,600,000 | −20% | ×0.80 | $1,280,000 |
| $500,000 | +10% | ×1.10 | $550,000 |
| $80,000 | +25% | ×1.25 | $100,000 |
Method 3 — Simple ratios (part-to-part)
Goal
Express “A to B” as a reduced ratio, often ending in “… to 1.”
Step-by-step method
- Identify numerator group and denominator group from the question wording.
“Ratio of candidates to positions” → candidates on top, positions on bottom. - Divide:
ratio = first group ÷ second group. - If asked for “___ to 1,” report the quotient as that number (round only if the problem requires it; packet Q3 divides evenly).
Worked example — recruit competition (packet Q3)
Given: 1,600 candidates competed for 80 Firefighter Recruit positions.
Find: Ratio of candidates to positions ___ to 1.
Work:
1,600 ÷ 80 = 20
So the ratio is 20 to 1 (twenty candidates per one position).
Answer: 20.
Order matters
| Question wording | Calculation | Result |
|---|---|---|
| Candidates to positions | 1600 ÷ 80 | 20 to 1 |
| Positions to candidates | 80 ÷ 1600 | 0.05 to 1 (or 1 to 20) |
If you reverse the division you may land on a distractor like 10, 15, or 25 after a secondary arithmetic error—always re-read “X to Y” before dividing.
Reducing a ratio (when numbers are not already “to 1”)
If the question said “candidates to positions as a simplified ratio,” you could also write 1600:80, divide both by 80 → 20:1. Same result.
Extra practice (ratio)
240 applicants for 12 openings → 240 ÷ 12 = 20 to 1 (same structure).
900 applicants for 45 openings → 900 ÷ 45 = 20 to 1.
1,500 applicants for 50 openings → 1,500 ÷ 50 = 30 to 1.
Exam workflow for Section 13.1 problems
- Label the type: time add, percent change, or ratio.
- Convert first (fractions → decimals; percent → multiplier).
- Compute once on scratch paper; do not chain mental shortcuts that skip the conversion.
- Sanity-check:
- Total time must exceed every individual segment.
- New budget after an increase must be larger than the old budget.
- Candidates-to-positions ratio for a competitive hire is almost always greater than 1.
- Eliminate distractors that equal the change only (percent) or a reversed ratio.
Key formulas card
| Skill | Formula |
|---|---|
| Mixed time total | Σ (each segment in decimal hours) |
| Percent increase total | original × (1 + p/100) |
| Percent increase amount | original × (p/100) |
| Ratio A to B (as n to 1) | A ÷ B |
Lock these three methods and you own the first block of the packet math section. The next section builds unit rates (cost per gallon and miles per gallon) on the same “divide carefully, label units” habit.
Engine 4 spent 1/2 hour identifying a spilled product, 1¾ hours containing it, and 3 hours assisting cleanup. How much total time did Engine 4 spend on the scene?
A fire department’s apparatus purchase budget was $1,600,000 and then increased by 20%. What is the new budget?
Sixteen hundred candidates competed for 80 Firefighter Recruit positions. The ratio of candidates to positions is ___ to 1.
Which calculation correctly finds a new total after a 20% increase on a $50,000 training budget?