Introduction
Problems about clocks that run slow or fast require converting between the faulty clock's reading and the true time. This pattern is important because many aptitude questions ask when a faulty clock will be correct, what the true time is when the clock shows a certain reading, or how long it takes to gain or lose a given amount.
Pattern: Clock Running Slow / Fast
Pattern
Key concept: Treat the faulty clock as running at a steady rate and use direct proportion (or linear interpolation) to convert elapsed times or errors.
Useful forms (keep these intuitive):
- Rate form (per-hour): Compute minutes gained or lost per hour, then scale to required hours:
Gain/Loss in required time = (Given gain/loss × Desired time) ÷ Given time. - Elapsed-time ratio (simple): If the clock runs uniformly, use the ratio of clock-interval to true-interval to scale times:
True elapsed = Clock elapsed × (True time interval ÷ Clock time interval). - Interpolation (when two errors are known): If a clock has error E₁ at true time T₁ and E₂ at true time T₂, then the time when error equals any intermediate value E is:
T = T₁ + (E - E₁) × (T₂ - T₁) ÷ (E₂ - E₁).
Step-by-Step Example
Question
A clock is 5 minutes slow at the true time 8:00 AM, and 5 minutes fast at the true time 8:00 PM the same day. When was the clock exactly correct?
Solution
-
Step 1: State the readings and errors clearly
At the true time 8:00 AM, the faulty clock shows 7:55 AM → error = -5 minutes (clock behind). At the true time 8:00 PM, the faulty clock shows 8:05 PM → error = +5 minutes (clock ahead). -
Step 2: Compute total change in error and interval
Total change in error = +5 - (-5) = 10 minutes. Total true-time interval = 8:00 AM → 8:00 PM = 12 hours. -
Step 3: Use linear interpolation to find time to zero error
Time from 8:00 AM to correctness = (initial error magnitude ÷ total error change) × total interval = (5 ÷ 10) × 12 hours = 6 hours. -
Final Answer:
2:00 PM -
Quick Check:
After 6 hours the error will have moved halfway from -5 to +5 (i.e., to 0). 8:00 AM + 6 hours = 2:00 PM ✅
Quick Variations
1. Find true time when clock shows a reading: Compute clock's elapsed from a known reset, convert to true elapsed using the rate, then add to known true time.
2. Find when the clock will be correct: Use interpolation between two known error points (as above).
3. Find time to gain/lose X minutes: Convert the given gain/loss into a per-hour rate, then divide X by that rate.
Trick to Always Use
- Step 1 → Always express errors as (Clock - True). Use signs: fast = +, slow = -.
- Step 2 → Prefer per-hour rates for scaling and interpolation for “when correct” problems.
Summary
Summary
- Clarify reference frames: always state whether a time is the true time or the clock's shown time.
- Use proportional scaling (gain/loss per period) to compute totals over longer intervals.
- Use linear interpolation between two known error points to find intermediate times (e.g., when error = 0).
- Track signs: add gain for fast clocks, subtract for slow clocks when converting between true and clock readings.
Example to remember:
If a clock is 5 minutes slow at 8:00 AM and 5 minutes fast at 8:00 PM, it will be correct at 2:00 PM (6 hours after 8:00 AM).
Hint: Take half the total time when slow and fast magnitudes are equal.
Common Mistakes: Mixing up total change and initial error.