In the days before watches were invented, clocks were valuable items. There was a man who had one clock in his house. It kept good time, but

I am sorry Alan but I am still really struggling with this 

I wind up my clock, and as I walk out my door the clock reads 12:00

I know it is slow so the real time is 12:00   +   X minutes

I walk to my friends place  I don't know how long I take but I will call it T minutes

When I arrive his clock reads  12:40         

So

X+T=40 minutes

x=40 minutes - T

I chat for 10 minutes and then return home

I have been away from home for   T+T+10  and my clock tells me that this adds up to 70 minutes

So

2T+10=70

2T=60

T=30 minutes.

I know that  X=40-T  so X=10.   The clock is 10 minutes slow!

WOW I worked it out - Now i just have to work out what you did in your answer  

His clock says t1 when he leaves.  The time on his friends clock when he arrives is T, so his apparent journey time is  δt = T - t1.  He stays for a known time δT.

Assuming his apparent return journey time is also δt then his apparent round trip takes 2δt + δT or 2(T - t1) + δT. His clock now reads t2.

Assuming his clock runs at the right speed the interval Δt = t2-t1 is the true interval (though neither t1 nor t2 show the true time).

half the difference between the true interval and the apparent interval gives the required correction:

correction = (2(T-t1) + δT - Δt)/2 

All the values on the RHS are known, so he knows the size of the correction.  If it is positive the apparent interval is longer than the true interval, which means his clock is behind true time.  If it is negative the apparent interval is shorter than the true interval and his clock is ahead of true time.

So he knows by how much and in what direction to adjust his clock.

.

I am having a problem getting my head around this Alan  

Maybe you could do it with some pretend times     

Example:  

                                     Time by his clock                         His friend's clock (true time)

Sets off                                   12.00                                         12.10     (say)

Arrives at friend's place             12.30                                         12.40

( a 30min walk)                               Apparent journey time δt = 40 mins 

Leaves friend's place                 12.40                                          12.50

                                                                      δT = 10 mins

Arrives home                            13.10                                          13.20

                                        True journey time Δt = 13.10-12.00 (= 13.20 - 12.10) = 70 mins

Correction = (2*δt + δT - Δt)/2 = (2*40 + 10 - 70)/2 = 20/2 = 10 minutes

Correction is positive so he must add 10 minutes to the time his clock is showing.

.

But his clock was stopped!  

 Did he wind it up before he left to start it again!       (no one told me that   )

Well done Melody!  You got it so you don't need to worry about my generalisation!

Thanks Alan.

That was a great question.   It would have been better if I knew the clock was rewound before the man left but I guess the question did not say that it was not rewound :)   I am not very good at thinking between the lines  :))

I have added this one to the "great answers to learn from" sticky thread     

(under puzzles)

Well, Melody, you have been working very hard for the past year and a half. Maybe you need to unwind a little. :)

Yes, perhaps I do               LOL  

Yeah, Melody....I like the way you presented that answer, too........it makes a lot of sense........and this problem doesn't seem as daunting as presented......

Remember.....a watched pot never boils.....but an unwatched clock leads to math problems........!!!!!