Question

A watch which gains uniformly ,is 5 min,slow at 8 o'clock in the morning on sunday and it is 5 min 48 sec.fast at 8 p.m on following sunday. when was it correct?

Answer 1

Hi there !

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Solution :

On the first Sunday morning at 8 : 00 a.m. , the watch is 5 min slow, and next Sunday at 8 : 00 p.m. , it becomes 5 min 48 secs fast.

So, the watch gains

$\mathsf{5 + 5 \frac{48}{60} \: min \: in \: a \: time \: of \: (7 \times 24) + 12} \\ \\ \mathsf{ \implies{180 \: hours.}}$

Hence,

To show the correct time, it must have to gain 5 min .

$\mathsf{ \frac{54}{5} \: min \rightarrow 180 \: hours.}$

5 min --) are gained in ( 180 × 5 / 54 × 5)

=> 83 hours 20 min = 3 days 11 hours 20 min.

Watch is correct 3 days 11 hours 20 min. after 8 am of Sunday.

Hence,

The watch will be correct at 20 min past at 7 p.m. on Wednesday.


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Thanks for the question !

Answer 2

$\mathbb{ \huge{ \fcolorbox{green}{pink}{SOLUTION:-}}}$

$\mathbb{ \bold{ \fcolorbox{green}{pink}{METHOD: OF: SOLUTION :-:-}}}$

Question:-

A watch which gains uniformly ,is 5 min,slow at 8 o'clock in the morning on sunday and it is 5 min 48 sec.fast at 8 p.m on following sunday. when was it correct?

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Time from 8 AM on Sunday to 8 PM. on following Sunday =) 7 days 12 hr

=) 180 hours.

The watch gains (5 + 29/5) minute  / 54/5 minutes..

To show the Exactly tine it has to gain 5 minutes.

$\implies \frac{5}{ \frac{54}{5} }   \times 180 \\  \\  \\  \implies \frac{25}{54}  \times 180 \\  \\  \\  \\  \implies \frac{4500}{54}  = 83 \frac{1}{3}hrs$

Expand into The simplest form:-

$\implies83  \bold{\frac{1}{3}hrs = 72hrs + 11 \frac{1}{3}}$

$\implies \bold{ \fbox{3days + 11hr + 20minutes}}$

Exapand into Days after given as per as QUESTION;

After Sunday ,The  Correct time will be Wednesday at 7:20 pm.

$\mathbb{CONCLUSION}$

$\mathsf{ \huge{7 \ratio20pm}}$