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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:

  1. $59 \frac{7}{12}\;\text{min. past 3}$
  2. $4 \;\text{p.m.}$
  3. $58 \frac{7}{11}\;\text{min. past 3}$
  4. $2 \frac{3}{11}\;\text{min. past 4}$

 

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 7 min. past 3
12
B. 4 p.m.
C.
58 7 min. past 3
11
D.
2 3 min. past 4
11 

2 Answers

3 3 votes
3 min of right clock = $(3 + \frac{1}{12}) = \frac{37}{12}$ min of wrong clock

3/60 hr of right = 37/(12*60) hr of wrong

and wrong clock run 9hr 15 min from 7am to 4:15pm

so (9 + 15/60) = 37/4 hr of wrong = (3/60) * 3 * 60 = 9hr of right clock.

So right clock show 7am + 9hr = 4pm
0 0 votes
A watch gains $5$ seconds in $3$ minutes.

Since $3$ minutes $=180$ seconds,

\[
\text{Gain rate}=\frac{5}{180}
\]

For $1$ hour:

\[
1 \text{ hour } = 3600 \text{ seconds}
\]

Gain in $1$ hour:

\[
3600 \times \frac{5}{180} = 100 \text{ seconds}
\]

Thus, for every $3600$ seconds, the watch gains $100$ seconds.

From $7:00$ AM to $4:00$ PM, the elapsed time is:

\[
9 \text{ hours}
\]

Total gain in $9$ hours:

\[
9 \times 100 = 900 \text{ seconds}
\]

Now consider the additional $15$ minutes.

\[
15 \text{ minutes} = 900 \text{ seconds}
\]

Gain in $15$ minutes:

\[
900 \times \frac{5}{180} = 25 \text{ seconds}
\]

Therefore, total gain:

\[
900 + 25 = 925 \text{ seconds}
\]

Convert $925$ seconds into minutes:

\[
925 = 15 \text{ minutes } 25 \text{ seconds}
\]

Since the watch is ahead, subtract this from the indicated time:

\[
4:15:00 - 0:15:25 = 3:59:35
\]

Thus, the true time is approximately

\[
\boxed{4:00 \text{ PM}}
\]
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