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:

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Set up the ratio \(\text{watch}:\text{real}=37:36\) for a watch gaining \(5\) s per \(3\) min; multiply the shown interval by \(\tfrac{36}{37}\).
Updated On: Aug 20, 2025
  • \(59 \tfrac{7}{12}\) min. past 3
  • 4 p.m.
  • \(58 \tfrac{7}{11}\) min. past 3
  • \(2 \tfrac{3}{11}\) min. past 4
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The Correct Option is B

Solution and Explanation

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 \(= \boxed{4{:}00\ \text{p.m.}}\).
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