Question

A watch, which gains uniformly, is 3 min slow at 12 noon on Sunday and is 5 min 36 seconds fast at 4 pm on the
next Sunday. At what time it was correct?
(a) 12 am on same day
(b) 12:00 pm on Monday
(c) 12 am on Tuesday
(d) 12:00 am on Wednesday​

Answer 1

Answer:

3 min + 5 +36/60 (changing in min and overall lose/gain of time)

= 43/5 min

Total Hours in 7 days of 4 p.m next Sunday to 12 noonSunday

=24×7 +4=172 hours

43/5 ---= 172

3 --= 60 hours =2 days+ 12 hours

12 noon of Sunday+ 2 days + 12 hours =12 a.m. Tuesday (c)

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Answer 2

Answer:

The Clock was correct at 12 a.m. on Tuesday.

So, Option (c) is the correct answer.

Step-by-step explanation:

Total (extra) gain = $= 3 \: min + 5 \: min \: 36 \: sec = 8 \: min \: 36 \: sec$

Clock covers the extra time (516 sec) in 7 days and 2 hrs.

So Clock moves 516 sec in 170 hrs

⇒ So 180 sec covers in ⇒ $\frac{170 \times 60 \times 60}{516} \times 180 = 213,488.372 \: sec$

$= (213,488.372 \div 3600) = 59.30232 \: hrs$

⇒ 2 days and 11.30 hrs So clock was correct at approximately 12 a.m. on Tuesday.

Ans (C)