Slow-running watch

In every minute passing in real life, the time passing in the watch is only (60-10) seconds or 50 seconds. Therefore, when n minutes had passed in real life, 50n seconds had passed in the watch. If ''a'' is the number of seconds that had passed in the watch,

a = 50n

The watch has recorded 630 seconds as the total time travel of the bus, so a = 630.

630 = 50n

630/50 = n

12.6 = n

Therefore, in real life 12.6 minutes had passed. Since 12.6 minutes equals 12.6(60) or 756 seconds, therefore, 756 seconds is the answer.

The bus trip took 10 minutes 30 seconds of time according to her watch. This is equal to 630 seconds (10*60+ 30). Her watch was running at 50 seconds for every 60 real time seconds (10 seconds less per minute (60-10 = 50)). This can be expressed as running at 50/60, simplified as 5/6, of real time. So, if the trip took 639 seconds at 5/6 speed, we can re-arrange, giving:

639/5 *6 = 756 seconds.

Since her watch goes 10 seconds further behind after every minute in real time, this means that at a one minute interval in real time her watch records 50 seconds.

i.e ${1min}={50s}$

According to her watch she spent 10 minutes 30 seconds from where she boarded the bus to the resort

converting that time into seconds we get a total of 630 seconds

But we know that ${1 min}={50 s}$

thus 630 seconds equals $\frac{1\times630}{50}$

which gives ${12.6mins}$ in real time

converting this to seconds i.e. ${12.6\times60}$

we get the required answer of $\boxed{756s}$

For every 60 seconds gone by, Mei's watch will go 50 seconds. After 630 of her seconds, we simply set up a proportion to solve. 50 / 60 = 630 / x. Solving for x, we get 756.

From $10:00:00 AM$ to $10:10:30 AM$ there are $630$ seconds. Her minutes are of $50$ seconds in her clock, so there's a total (in minutes) of $630/50 = 12.6$ minutes. Now, each real minute has $60$ seconds, so $12.6 * 60 = \boxed{756}$ .

If the REAL time is 60 seconds, then the SLOWED time is 50 seconds.

So we conclude that:

$\frac{REAL}{SLOWED}=\frac{60}{50}=\frac{6}{5}$

We knew that according to her watch she used 630 seconds, which is SLOWED,

$\frac{REAL}{SLOWED}=\frac{6}{5}$

$\frac{REAL}{630}=\frac{6}{5}$

$REAL=756$

As for 60 seconds in real time the watch completed only 50 seconds, we know that only 5 6th of the real time is completed by the watch per each minute. hence the watch reading falls back by 10 seconds with each passing minute. (Eg: for 60sec Real time -> 50 sec watch & then for 120 sec real time -> 100 sec watch time) thus for finding how much real time has passed for 630 seconds of watch time, (50/60)*x=630. And we get x as 756

Mei's watch 50 sec=6o sec actual = = 1 = 60|5o=1.2 sec actual.. eq.(1) now 10;10;30-10;10;00=630 sec 1.2*630= 756 ans

10*60+30=630 so 630x60/50=756

<p>10:10:30 - 10:00:00 = 00:10:30</p>

<p>10 min * 60 sec + 30 sec = 630 sec "watch duration"</p>

<p>630 sec / 50 sec = 12.6 min</p>

<p>12.6 min * 60 sec = 756 sec "actual duration"</p>

To solve this problem, we must make use of the given fact that for every minute in actual time , the watch gets behind by ten seconds . Which means, for every minute, there will be a difference in time between the watch and real time; first equal to 10, then 20, then 30, and so on.

Thus, we can conclude the watch counts 50 seconds for each minute instead of 60 seconds. Therefore, the minute count of the watch is (5/6)th of a minute, instead of (6/6)th of a minute.

Let 'm' represent one minute in actual time. Total time elapsed in watch is 10min & 30sec = 10.5 min.

(5/6)m=10.5

m=63/5 (in minutes)

Thus, no. of seconds in 63/5 minutes = (63/5)*60= \boxed{756}

Let the time taken in reality be "x" min ,then for each min the delay is by 10 sec therefore the time fed in the watch is 50 sec , its like a man's mind read 60 sec as 50 sec only always (each min is 50 sec for him ) . Now as she has travelled for "x" min , so 50 x = 630( 630 sec is the time in her watch ) So x= 63/5 min = (63/5) 60 sec ( why I have multiplied 50 and x ?...., becoz as the man can only remembers 50 sec for 60 sec , so in a long run i.e after x min he will remember only 50x sec ...).

For every every min(60 seconds) Mei's watch records 50 seconds passed, therefore actual time passed= (6/5) * time passed according to her watch. Here trip took 10*60+30 seconds= 630. therefore, 6/5 * 630 =756sceonds.

1 minute in her watch = 50 seconds every 6 seconds her watch got behind 1 second = 5 seconds that means the actual time = 6/5 the watch = 1.2 the watch the bus take 10 minutes 30 seconds = 630 seconds the actual time = 1.2 x 630 = 756 seconds

According to the problem, for every minute, the clock runs 10 secs slower. Therefore at end of every 50 secs in the slower clock, the actual clock would be ahead by 10 secs. So after every 5 secs in the slower clock, the actual clock is ahead by 1 sec. Which implies the actual time is equal to 630+(630/5)=756.

10.10.30 - 10.00.00 = 10.30 which is 630 seconds. Assume that the real time (in seconds) is x.

\­( 630 = x - \frac{x}{60} \times 10)

$630 = x - \frac{10x}{60}$

$630 + \frac{10x}{60} = x$

$630 + \frac{x}{6} = x$

$5y = 3780$

$y = \boxed{756}$

ratio time in watch : real time = (60 - 10) second : 60 second = 5 : 6 according time in watch has passed 630 seconds, so that real time is (630 : 5) x 6 = 756