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?
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
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
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.
I meant the big hand as in meaning the hour hand
and little hand as the minute hand
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.
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
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
t = about 27.27 minutes
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