It is between 3 P.M. and 4 P.M. and the distance between the hour and the minute hand of clock is 18 minute spaces. What time does the clock show?

### Concept & Relative Speed In a clock, the minute hand moves faster than the hour hand. The minute hand travels 60 minute spaces in 60 minutes. The hour hand travels 5 minute spaces in 60 minutes. Therefore, the minute hand gains $60 - 5 = 55$ minute spaces on the hour hand in 60 actual minutes. $$\text{Time taken} = \frac{60}{55} \times (\text{Minute spaces to be gained})$$ ### Step-by-Step Solution 1. **Analyze the initial position (at 3:00 P.M.):** At exactly 3:00, the hour hand is at 3 and the minute hand is at 12. The minute hand is $5 \times 3 = 15$ minute spaces *behind* the hour hand. 2. **Determine required relative movement:** The problem states the hands are 18 minute spaces apart. Case 1: Minute hand is 18 spaces *behind*. (Not possible, as it starts 15 behind and moves faster, so it only gets closer until they meet). Case 2: Minute hand is 18 spaces *ahead*. To be 18 spaces ahead, it must first cover the 15-space gap to catch up to the hour hand, and then travel an additional 18 spaces ahead of it. Total minute spaces to be gained = $15 + 18 = 33 \text{ spaces}$. 3. **Calculate the time taken:** Using the relative speed formula: $$\text{Time} = \left(\frac{60}{55}\right) \times 33$$ $$\text{Time} = \left(\frac{12}{11}\right) \times 33$$ $$\text{Time} = 12 \times 3 = 36 \text{ minutes}$$ 4. **Final Time:** 36 minutes past 3:00 P.M. is 3:36 P.M. ### Exam Strategy & Shortcut Use the fixed multiplier $\frac{12}{11}$ for all "space gaining" problems. Spaces to gain = Initial gap (15) + Desired gap (18) = 33. Time = $\frac{12}{11} \times 33 = 36$. ### Common Pitfall Forgetting to account for the starting gap at 3:00. Students sometimes just calculate time for 18 spaces, yielding an incorrect answer. ### Final Answer Therefore, the correct answer is **3.36 P.M.**