How much does a watch losse per day if its hands coincide every 64 minutes?

### Concept & Coincidence Interval In a correct clock, the hands coincide every $\frac{720}{11}$ minutes (or $65 \frac{5}{11}$ minutes). If a clock's hands coincide every $x$ minutes, the total time gained or lost in a day (24 hours) is calculated by comparing this interval to the standard interval over the course of a day (1440 minutes). $$ \text{Gain/Loss in a day} = \left( \frac{720}{11} - x \right) \times \left( \frac{24 \times 60}{x} \right) \text{ minutes} $$ ### Step-by-Step Solution * Given: The interval $x = 64$ minutes. Since $64 < 65 \frac{5}{11}$, the clock is actually running fast (gaining time), despite the question's wording ("losse"). We calculate the magnitude of this shift. * Substitute $x = 64$ into the formula: $\text{Gain/Loss} = \left( \frac{720}{11} - 64 \right) \times \left( \frac{1440}{64} \right)$ * Simplify the first bracket: $\frac{720 - (64 \times 11)}{11} = \frac{720 - 704}{11} = \frac{16}{11}$ * Simplify the second bracket: $\frac{1440}{64} = \frac{45}{2}$ * Multiply the two parts: $\frac{16}{11} \times \frac{45}{2} = \frac{8 \times 45}{11} = \frac{360}{11}$ * Convert to a mixed fraction: $\frac{360}{11} = 32 \frac{8}{11}$ minutes. ### Exam Strategy & Shortcut Memorize the true coincidence time: $65 \frac{5}{11}$ minutes. The difference for one coincidence is $1 \frac{5}{11} = \frac{16}{11}$ minutes. Since it coincides every 64 minutes, in a full day (1440 minutes), it coincides $1440 / 64 = 45/2$ times. Multiply the difference per coincidence by the number of coincidences: $\frac{16}{11} \times \frac{45}{2} = \frac{360}{11} = 32 \frac{8}{11}$. ### Common Pitfall Confusing whether the clock gains or loses. If the interval is less than $65 \frac{5}{11}$ mins, the clock is fast (gaining). If it's more, it's slow (losing). Also, arithmetic errors when subtracting 64 from $\frac{720}{11}$. ### Final Answer Therefore, the correct answer is **$32 \frac{8}{11}$ minutes**.