Rajesh went out between 5 and 6 for a work. He came to home between 6 and 7. When he came he saw that the hands have exactly changed places. When did Rajesh went out?

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Rajesh went out between 5 and 6 for a work. He came to home between 6 and 7. When he came he saw that the hands have exactly changed places. When did Rajesh went out?

Mathematics
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well by changing hands, what do you mean, does it mean, interchanging position of hour and minute hand?
I also don't know what does the book mean by it but I do have the answer. Ok! Let us take the meaning as specified by you @ghazi ... what answer you arrived ?

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well i consider as interchanging of minute and hour hands , which means both will acquire new position if turned by 180 degrees , lets say its 6 o clock then minute hand is at 12 and hour is at 6 and if hands are interchanged then it will be 12.30 , thats what i concluded
not 12.30 but a bit less, or lets say its 5 past 7 then if hands are interchanged , it will be 1.35 , thats what i can think of , but i guess here book says about mirror image like 5.25 and its mirror image is 6.40 , which satisfies your question
No but that is not the answer...
I think the question is wrong,,,,,
changing hands is confusing me , i think its 5.30 and then 6.25
like her left at 5.30 and if hands are changed it will be 6.25
*he
yes, clocks used to have hands on them to tell the time with :)
|dw:1360761423073:dw|
unfortunately its obsolete and now digital watch is preferred lol for higher precision and accuracy
so i guess i was correct lol
5.30 to 6.30 is a suitable result
but then we are talking about angles .... and not time itself lol
Well I don't think tht question means for mirror images..
exactly, i am trying to think of mathematical explanation and @mathslover we didnt use mirror image , we just interchanged the position of hands in your given time interval, 5 at 6 and 6 at 5 ;)
|dw:1360761555731:dw| we have different angular speeds to acowingspan for
oh sorry
its okay but try to think that mathematically :) that would be fun
lol; so wither account is a bad word? or was my typo a bad wrod
wingspan stands for male genital lol
i was horrified by your word , what you wanted to write lol hehe
the hour hand has a constant speed of 360 per hour; the minute hand has a speed of 360/60 per hour. if you can narrow down how to account for a general position you can use the speeds to find a related rates problem maybe .... or it could be one of the harry and sally paint a room things
we know the time had to be less than an hour, since the minute hand didnt make it all the way around ....
lets assume 50 minutes just to cover the span from the 5 to the 7 and my speeds are off
minute hand is 360 per hour; and hour hand is 360+360/12
minute hand moves faster than hour, minute hand completes 360 degrees in one hour and hour hand covers only 30 degree in one hour
...lol, its been so long since i looked at a clock i spose
lets assume for a minute that im an idiot ;)
^ http://www.edunetworking.com/PlayVideo.aspx?u=fFWIDQNe0nQ&Class=CAT%20&Subject=Logical%20Reasoning
so a tortise and a hare are chasing ech other around a track at defined speeds. how long does it take the tortise to reach where the hare was, at the same time that the hare reaches the position that the tortise was ....
but the hare slept on its way :O
lazy hares :)
hehe
alright fellas ;) i guess we are done ;) lets go back to get lazy
also need to determine the angle of separation at a given time .... im sure someones figured this out in like 1587bc
|dw:1360762766659:dw| for a given degree alpha we need to determine when "equivalent" distances get traveled at a given speed; when alpha = 360-alpha does that make any sense?
the time interval is the same for each distance traveled \[hour: \frac{30^o}{60sec}t=\alpha^o\\minute:\frac{360^o}{60sec}t=360^o-\alpha\] 2 equations in 2 unknowns right?
\[\frac{390^o}{60}t=360^o\] \[t=60sec~\frac{360^o}{390^o}=55\frac{5}{13}sec\]
\[\alpha=\frac{30^o}{60sec}*\frac{720}{13}sec=(\frac{360}{13})^o\] at what time between 5 and 6 are the hands that many degrees apart? id start at noon and do some additional calculations
or start at 5 :)
|dw:1360764111867:dw| \[\frac{360^o}{60}t-(150^o+\frac{30^o}{60}t)=(\frac{360}{13})^o\] solve for t
and im writing sec for minutes for some reason ... ; so i get 5:32.31 as a starting time
at any rate, i got a good idea, might have the speeds off tho so dbl chk if your interested in it:)

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