Help me pleas to solve this...

When does The scorpions of clock meet ?

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- October 9th 2006, 11:59 AMMahaTRICKy
**Help me pleas to solve this...**

When does The scorpions of clock meet ? - October 9th 2006, 12:24 PMQuick
um, what do you mean "scorpions of clock"

Maybe you mean "hands" such as, at 12:00 both hands point towards 12

at 1:05 both hands point towards 1

at 2:10 both hands point towards 2

at 3:15 both hands point towards 3

at 4:20 both hands point towards 4

at 5:25 both hands point towards 5

at 6:30 both hands point towards 6

at 7:35 both hands point towards 7

at 8:40 both hands point towards 8

at 9:45 both hands point towards 9

at 10:50 both hands point towards 10

at 11:55 both hands point towards 11

Or maybe you mean something totally different... - October 9th 2006, 01:14 PMMaha
hi

that what i mean

but your Answer is Wrong

i sad that for my Teacher , he sad we can find it by angle

because when it at 1 :05

the hand of hours Far slightly ( not exactly on 1)

i hope u Understand what i whant to say - October 9th 2006, 05:07 PMQuick
I do, and you're right, I was wrong :o (which I've been lately :()

Yes, well they meet at 12:00 (obvious)

Next, you know that they'll meet again when the hour hands in the 1-2 hours section.

So I'm going to use a system of equations to find that time...

The hour hand=(angle per minute)minutes + the starting angle.

The starting angle is the angle between the one and the twelve. Now notice the clock is 360 degrees. Notice that the 1 is 1/12 of that 360 degrees, thus the angle is**360*1/12=360/12=30**

The hour hand moves 1/60 of the way between the 1 and the 2 per minute, and the distance between the 1 and the 2 is 30 degrees. Thus it moves**1/60*30=30/60=1/2**degrees per minute.

thus the equation is:**x = (1/2)m + 30**

Now the equation for the minute hand is**y = 6m + 0**

Now we want to know when x=y, so:**(1/2)m + 30 = 6m**

Then:**30=(6-1/2)m**

Then:**30 = (11/2)m**

divide:**60/11=m**

So**60/11**past 1 is when they will be the same (now try for all others) - October 9th 2006, 05:17 PMThePerfectHacker
I do not know what the question is but this is my favorite thing to use when working with these clocks problems.

If the hands art both upon 12 then,

The minute hand moves 6 degrees per minute.

The hour hand moves .5 degrees per minute.

Thus the difference is 5.5 degree per minute.

Thus, after t=>0 minutes the distance between them is 5.5t (of course if they do not travel all the way around).

If you do consider the difficulty when they travel all the way around then the difference is (5.5t)_{360} that means the remainder that is left after subtracting 360 sufficiently enough times.

For example (750)_{360}=30 because we subtract 360 twice and have a number less then 360 which we need. - October 10th 2006, 08:04 AMearboth
- October 15th 2006, 07:50 PMMaha
**Quick**

ThePerfectHacker

earboth

thanx for help me :)

i solve it With your help :o

and i found These answers

they meet at 12 then ~=

m~= 5 ==> 1-2

m~= 11 ==> 2-3

m~= 16.3 ==> 3-4

m~= 21.8 ==>4-5

m~= 27.2 ==>5-6

m~= 32.8 ==>6-7

m~= 38 ==>7-8

m~= 43.6 ==>8-9

m~= 49 ==>9-10

m~= 54.5 ==>10-11

m~= 60 ==>11-12