clocks hands
- Thread starter shahar
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What do you think, and, more importantly, why do you think that?
shahar
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Because...
the left = minutes, the right = hours
1) 0-12
2) 5-1
3) 10-2
4) 15-3
5) 20-4
6) 25-5
7) 30-6
8) 35-7
9) 40-8
10) 45-9
11) 50-10
12) 55-11
13) 0-12
Now, I continue...
14) 5-1
15) 10-2
16) 15-3
17) 20-4
18) 25-5
19) 30-6
20) 35-7
21) 40-8
22) 45-9
23) 50-10
24) 55-11
25) 0-12
There is 25 possibilities.
Am I right?!
Is there a way to solve the problem by using algebraic way or by using algebraic equation?!
the left = minutes, the right = hours
1) 0-12
2) 5-1
3) 10-2
4) 15-3
5) 20-4
6) 25-5
7) 30-6
8) 35-7
9) 40-8
10) 45-9
11) 50-10
12) 55-11
13) 0-12
Now, I continue...
14) 5-1
15) 10-2
16) 15-3
17) 20-4
18) 25-5
19) 30-6
20) 35-7
21) 40-8
22) 45-9
23) 50-10
24) 55-11
25) 0-12
There is 25 possibilities.
Am I right?!
Is there a way to solve the problem by using algebraic way or by using algebraic equation?!
Last edited:
Dr.Peterson
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First, there's an interpretation issue: Are you going to count both the midnight at which a day starts, and the midnight at which it ends? Someone could want to; but the question doesn't say to start at midnight, but asks about ANY period of 24 hours, so you can think of it as, say, from 12:01 AM to 12:01 AM the next day. That will force you to include only one of them.
Now, figure out the time from one such meeting to the next. To do this, you might calculate the rate (degrees per hour) of motion of each hand, and write an equation for when the positions will next be equal, if they start together. (This will be a little tricky, but persevere!)
Then you can just divide 24 hours by this time.
Please show your work, so we can check it.
There are other ways, some more intuitive, but this is the most algebraic.
(By the way, the word should be "analog"; look up "analogical" to see why.)
Dr.Peterson
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Here what I use:
the hours = a + 360k
the minutes = 5a + 360k
How it combine to one formula?!
This means nothing without explanation. What do your variables a and k mean? What does each expression represent? (They are not hours and minutes; perhaps the number of degrees each hand has moved?) And are you doing what I suggested, or something else entirely?
Here is a little more about my method: If we start with the hands together, then since the minute hand moves faster, it will go around once to get back to where it started, plus however many degrees the hour hand has moved. If the time until they meet again is t hours, how far has each hand moved in that time?
shahar
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O.K.
I have the answer.
There is 2 times that hands are not exactly one on another.
So the answer is 24 - 2 = 22.
So, this is a tricky question.
The answer is correct because there are 2 times that the hands is not exactly one on another and the minutes hand "pass" a little the hours hand.
I have the answer.
There is 2 times that hands are not exactly one on another.
So the answer is 24 - 2 = 22.
So, this is a tricky question.
The answer is correct because there are 2 times that the hands is not exactly one on another and the minutes hand "pass" a little the hours hand.
Dr.Peterson
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The answer of 22 is correct, but I don't follow your reasoning. There are infinitely many times when the hands are not together!
You asked for an algebraic method; have you tried my suggestion?
For a non-algebraic solution, you might observe that the times when hands are together are at 12:00, 1:something, 2:something, and so on, where the "somethings" (minutes) increase each time. But when you reach "11:something", the "something" is actually 60, so the time is 12:00 and we are back to the start of the next 12-hour cycle. Thus there are only 11 times in each 12 hour period. It is not that the hands are not quite together at 12:00, but that that time must not be counted twice.
shahar
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I don't understand your method
Can you write it by more simple words.
I read your reply and I don't understand.
O.K?
Why dividing by 24? I know there are 24 hours in equation?
Or:
What to write on the nominator?
I don't understand your method to the solution.
Can you write it by more simple words.
I read your reply and I don't understand.
O.K?
Why dividing by 24? I know there are 24 hours in equation?
Or:
What to write on the nominator?
Dr.Peterson
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I didn't show you an equation, or divide anything by 24, or have a denominator. What are you talking about?
Previously, I mentioned dividing 24 hours by the time between instances of the hands being together. Is that what you mean?
If you are asking about the algebraic approach I have suggested, please show me whatever steps you have taken. How fast does each hand move? How far has each hand moved after t hours (relative to the starting point, which you can take to be the top)? Then you can write an equation.
If you're asking about my non-algebraic approach, ask me specific questions line by line, so I can see where you are misunderstanding. It is notoriously hard to explain in simple words!
Perhaps you'll get your answer at that other site...under "policer" ?
http://mathhelpforum.com/algebra/281875-clock-hands.html
http://mathhelpforum.com/algebra/281875-clock-hands.html
Last edited:
Steven G
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I knew that Denis would get his hands in this one.
Steven G
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Hmm, why would the answer be 25 since it is odd. Do you really think that the 1st 12 hours will have more/less situations you're looking for than the last 12 hours? I don't think so.
Ya, used 'em to wind the clock....
Consider the hour starting at 11 on the dot. The hour hand is on 11 and the minute hand is on 12. When will the minute hand overtakes the hour hand? Not until 12 on the dot, which is in the NEXT hour. In other words, the hour starting exactly at 11 has no moment when the hands of the clock are perfectly superimposed. This is not true of the other hours. They each have one position of perfect superposition. In short, a clock has only 11 positions of perfect superposition.
Now let's take Dr. P's algebraic solution. The minute hand moves at a rate of 360 degrees per hour. The hour hand moves at a rate of 30 degrees per hour. Starting from one perfect superposition to the next, p degrees further along, the hour hand must take t hours, where t = p / 30. During those some t hours, the minute hand must go 360 + p degrees. In other words, t is the time in hours required to reach perfect superposition from the previuos perfect superposition.
\(\displaystyle \therefore \dfrac{p}{30} = t = \dfrac{360 + p}{360} \implies 12p = 360 + p \implies\)
\(\displaystyle p = \dfrac{360}{11} \implies t = \dfrac{\dfrac{360}{11}}{30} = \dfrac{12}{11} \text { hours.}\)
Now \(\displaystyle \dfrac{24}{\dfrac{12}{11}} = 24 * \dfrac{11}{12} = 2 * 11 = 22.\)
Now let's take Dr. P's algebraic solution. The minute hand moves at a rate of 360 degrees per hour. The hour hand moves at a rate of 30 degrees per hour. Starting from one perfect superposition to the next, p degrees further along, the hour hand must take t hours, where t = p / 30. During those some t hours, the minute hand must go 360 + p degrees. In other words, t is the time in hours required to reach perfect superposition from the previuos perfect superposition.
\(\displaystyle \therefore \dfrac{p}{30} = t = \dfrac{360 + p}{360} \implies 12p = 360 + p \implies\)
\(\displaystyle p = \dfrac{360}{11} \implies t = \dfrac{\dfrac{360}{11}}{30} = \dfrac{12}{11} \text { hours.}\)
Now \(\displaystyle \dfrac{24}{\dfrac{12}{11}} = 24 * \dfrac{11}{12} = 2 * 11 = 22.\)