Question

A clock gains 10 minutes a day. The clock was corrected at 6:00 am. What will be the correct time when the clock shows 11:00 am the following day?

• A. 10{:}50 AM
• B. 10{:}45 AM
• C. 10{:}48 AM
• D. 10{:}25 AM

Hint:

For fast/slow clocks, relate shown time to real time via a constant ratio, then multiply the shown interval by the inverse ratio to get the real interval.

Answer 1

The Correct Option is C

Step 1: Rate of gain.
Clock gains \(10\) minutes in \(24\) real hours.
So \(\dfrac{\text{shown time}}{\text{real time}}=\dfrac{24\text{ h }10\text{ m}}{24\text{ h}}=\dfrac{1450}{1440}=\dfrac{145}{144}\).
Step 2: Shown elapsed time to 11:00 AM next day.
From 6{:}00 AM (corrected) to 11{:}00 AM next day on the \emph{fast} clock is \(29\) hours shown.
Step 3: Convert to real elapsed time.
\[ \text{Real elapsed} = 29 \times \frac{144}{145}\ \text{hours} = \frac{4176}{145}\ \text{hours} = 28.8\ \text{hours} = 28\ \text{hours }48\ \text{minutes}. \] Step 4: Add to the start time.
\(6{:}00\ \text{AM} + 28\ \text{h }48\ \text{m} = 10{:}48\ \text{AM (next day)}.\)
\[ \boxed{10{:}48\ \text{AM}} \]