Question

A watch which gains 5 seconds in 3 minutes was set right at 7 a.m. In the afternoon of the same day, when the watch indicated quarter past 4 o'clock, the true time is:

• A. \(59 \tfrac{7}{12}\) min. past 3
• B. 4 p.m.
• C. \(58 \tfrac{7}{11}\) min. past 3
• D. \(2 \tfrac{3}{11}\) min. past 4

Hint:

Set up the ratio \(\text{watch}:\text{real}=37:36\) for a watch gaining \(5\) s per \(3\) min; multiply the shown interval by \(\tfrac{36}{37}\).

Answer 1

The Correct Option is B

Step 1: Find the rate of the faulty watch.
Gains \(5\) s in \(3\) min \(=180\) s \(\Rightarrow\) in real \(36\) s, watch shows \(37\) s.
Thus \(\dfrac{\text{watch time}}{\text{real time}}=\dfrac{37}{36}\).
Step 2: Convert indicated time gap to real time.
From \(7{:}00\) a.m. to \(4{:}15\) p.m. (watch) \(=9\) h \(15\) min \(=555\) min \(=33{,}300\) s.
Real elapsed \(=33{,}300\times \dfrac{36}{37}=32{,}400\) s \(=540\) min \(=9\) h. 

Step 3: Add to the start time.
\(7{:}00\) a.m. + \(9\) h \(= \boxed{4{:}00\ \text{p.m.}}\).

Answer 2

Instead of scaling by \(37/36\) directly, work out the watch's gain per real hour and check each option against the resulting real time.

1. Find the hourly gain: \(5\) s per \(3\) min \(=100\) s per hour, so in one real hour the watch shows \(1\) hour \(100\) s, i.e. a ratio of \(3700\text{ s}/3600\text{ s}=37/36\).
2. Elapsed watch time: from \(7\!:\!00\) a.m. to \(4\!:\!15\) p.m. is \(9\) h \(15\) min \(=9.25\) h on the watch.
3. Solve for real hours \(R\): \(R\times\dfrac{37}{36}=9.25\ \Rightarrow\ R=9.25\times\dfrac{36}{37}=\dfrac{333}{37}=9\) real hours exactly.
4. Check the options: \(7{:}00\) a.m. \(+\,9\) h \(=4{:}00\) p.m., matching option (b) exactly; the other three options (times expressed in minutes past 3 or past 4) do not correspond to a whole \(9\)-hour real gap.

The real time is confirmed as \(4\) p.m.

the correct answer is 4 p.m.