mathslover Group Title 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? one year ago one year ago

1. mathslover

2. ghazi

well by changing hands, what do you mean, does it mean, interchanging position of hour and minute hand?

3. mathslover

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 ?

4. ghazi

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

5. ghazi

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

6. mathslover

No but that is not the answer...

7. mathslover

I think the question is wrong,,,,,

8. ghazi

changing hands is confusing me , i think its 5.30 and then 6.25

9. ghazi

like her left at 5.30 and if hands are changed it will be 6.25

10. ghazi

*he

11. amistre64

yes, clocks used to have hands on them to tell the time with :)

12. amistre64

|dw:1360761423073:dw|

13. ghazi

unfortunately its obsolete and now digital watch is preferred lol for higher precision and accuracy

14. ghazi

so i guess i was correct lol

15. amistre64

5.30 to 6.30 is a suitable result

16. amistre64

but then we are talking about angles .... and not time itself lol

17. mathslover

Well I don't think tht question means for mirror images..

18. ghazi

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 ;)

19. amistre64

|dw:1360761555731:dw| we have different angular speeds to acowingspan for

20. mathslover

oh sorry

21. ghazi

its okay but try to think that mathematically :) that would be fun

22. amistre64

lol; so wither account is a bad word? or was my typo a bad wrod

23. ghazi

wingspan stands for male genital lol

24. ghazi

i was horrified by your word , what you wanted to write lol hehe

25. amistre64

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

26. amistre64

we know the time had to be less than an hour, since the minute hand didnt make it all the way around ....

27. amistre64

lets assume 50 minutes just to cover the span from the 5 to the 7 and my speeds are off

28. amistre64

minute hand is 360 per hour; and hour hand is 360+360/12

29. ghazi

minute hand moves faster than hour, minute hand completes 360 degrees in one hour and hour hand covers only 30 degree in one hour

30. amistre64

...lol, its been so long since i looked at a clock i spose

31. amistre64

lets assume for a minute that im an idiot ;)

32. mathslover
33. amistre64

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 ....

34. AravindG

but the hare slept on its way :O

35. amistre64

lazy hares :)

36. mathslover

hehe

37. ghazi

alright fellas ;) i guess we are done ;) lets go back to get lazy

38. amistre64

also need to determine the angle of separation at a given time .... im sure someones figured this out in like 1587bc

39. amistre64

|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?

40. amistre64

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?

41. amistre64

$\frac{390^o}{60}t=360^o$ $t=60sec~\frac{360^o}{390^o}=55\frac{5}{13}sec$

42. amistre64

$\alpha=\frac{30^o}{60sec}*\frac{720}{13}sec=(\frac{360}{13})^o$ at what time between 5 and 6 are the hands that many degrees apart? id start at noon and do some additional calculations

43. amistre64

or start at 5 :)

44. amistre64

|dw:1360764111867:dw| $\frac{360^o}{60}t-(150^o+\frac{30^o}{60}t)=(\frac{360}{13})^o$ solve for t

45. amistre64

and im writing sec for minutes for some reason ... ; so i get 5:32.31 as a starting time

46. amistre64

at any rate, i got a good idea, might have the speeds off tho so dbl chk if your interested in it:)