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# Clock Rates

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The rate of the minute hand is 1 space per minute, and the rate of the hour hand is 5 spaces per 60 minutes or 1/12 space per minute.

Diogenes held his lantern high so he could see clearly. It was between 4 and 5 o'clock, and the big hand was exactly 5 minutes ahead of the little hand what time was it?

Feb 9, 2016

#5
+10

Sorry, Got confused.

For every 12 spaces the minute hand moves, the hour hand moves 1

We are trying to find the minutes, because we know that it is 4 o'clock

Now, because it is between 4 and 5 o'clock the hour hand is between 4 and 5 on the clock

That means that the hour hand is already starting 20 spaces ahead of the minute hand, If we start at 4 o'clock

Hence the equation h    (let h=the hour hand) =1/12x     (moves 1/12 of a space per minute) +20    (starting at the 20th space)

Now that the minute had moves 1 space per minute, we can say that m   (let m = minute hand) = x (Starts at 0)

h=1/12x+20

m=x

We want the time when m is 5 ahead of h

m=h+5

Now substitute

x=1/12x+25

11x/12=25

11x=25*12

25*12 = 300

300/11 = 27.2727272727272727

Now lets check

(27.2727272727272727)/12+20 = 22.272727272727272727 = h

27.2727272727272727 = m

So, m-h should equal 5

Lets find out

27.2727272727272727-22.2727272727272727 = 5

So the time is 4 o'clock, 27 minutes, and 27 seconds

Feb 9, 2016

#1
+5

Let B = the time of the big hand

Let L = the time of the little hand

B = 1/12x+20

L = x

B=L-5

1/12x+20=x-5

25=11x/12

25*12 = 300

300/11 = 27.2727272727272727

x=27.27

4:22.27

Feb 9, 2016
#2
0

The big hand is the minute hand and the little hand is the hour hand. The hour hand moves at 1/12 space per minute. And besides that, whenever someone other than a moderator helps me, I don't understand what they are trying to do. Thanks for trying though mate.

Feb 9, 2016
#3
+5

I meant the big hand as in meaning the hour hand

and little hand as the minute hand

Feb 9, 2016
#4
0

It cannot be worked like that. The big hand is ahead by 5 spaces, the big hand is the minute hand, so the minute hand is ahead, not the hour hand. Thank you for trying, but I think I need to wait for a moderator.

Feb 9, 2016
#5
+10

Sorry, Got confused.

For every 12 spaces the minute hand moves, the hour hand moves 1

We are trying to find the minutes, because we know that it is 4 o'clock

Now, because it is between 4 and 5 o'clock the hour hand is between 4 and 5 on the clock

That means that the hour hand is already starting 20 spaces ahead of the minute hand, If we start at 4 o'clock

Hence the equation h    (let h=the hour hand) =1/12x     (moves 1/12 of a space per minute) +20    (starting at the 20th space)

Now that the minute had moves 1 space per minute, we can say that m   (let m = minute hand) = x (Starts at 0)

h=1/12x+20

m=x

We want the time when m is 5 ahead of h

m=h+5

Now substitute

x=1/12x+25

11x/12=25

11x=25*12

25*12 = 300

300/11 = 27.2727272727272727

Now lets check

(27.2727272727272727)/12+20 = 22.272727272727272727 = h

27.2727272727272727 = m

So, m-h should equal 5

Lets find out

27.2727272727272727-22.2727272727272727 = 5

So the time is 4 o'clock, 27 minutes, and 27 seconds

SpawnofAngel Feb 9, 2016
#6
+5

Thanks mate Feb 9, 2016
#7
+10

Here's another way to look at it.

The number of degrees the minute hand moves in one minute  = 6°

The number of degrees the hour hand moves in one minute  = .5°

But.....at 4PM,  the hour hand is already 120°  ahead of the minute hand.

So....the minute hand must travel 120° plus another 30° relative to the hour hand's movement in some time t - in minutes -  in order to be 5 minutes ahead of the hour hand. So, we're trying to solve this equation:

6t - .5t  = 150

5.5t = 150  divide both sides by 5.5

But .27  of a minute ≈ 16.2 sec

So.....the minute hand will be 5 minutes ahead at approx. 27 minutes, 16.2 seconds after 4 PM   Feb 9, 2016
edited by CPhill  Feb 12, 2016