Clock question.

[helping]

At how many minutes after noon do the hour hand and minute hand of an analog clock first meet again? Explain your answer to the nearest whole number.

Using basic knowledge of a clock I got 65. Is there also an algebraic equation that can be set up to answer this question?

The explanation is as follows: At noon both the hour hand and the minute hand are at 12. As the minute hand starts moving, so does the hour hand, although at a much slower pace. After 60 minutes, the minute hand would once again be on 12 and the hour hand would now be on 1. After 5 more minutes , the minute hand will meet the hour hand once again - at this time right after 1.

Writing the above I realized that the hour hand moves at 1/12 of the speed of the minute hand. Still not sure how to use it to set up the equation.

 

[soroban]

Hello, helping!

At how many minutes after noon do the hour hand and minute hand of an analog clock first meet again?
Explain your answer to the nearest whole number.
. . A strange request ... anyone could guess the answer.

Using basic knowledge of a clock I got 65.
Is there also an algebraic equation that can be set up to answer this question?

There are several ways to set up this problem.
Here is one of them.

The minute hand moves 360^o in 60 minutes: \(\displaystyle 6^o\) per minute.
The hour hand moves \(\displaystyle 30^o\) in 60 minutes: \(\displaystyle \frac{1}{2}^o\) per minute.

At noon, the hands are together at "12": both at \(\displaystyle 0^o.\)

After an hour, the minute hand is back at "12";
. . the hour hand has moved \(\displaystyle 30^o\) to "1".

In the next \(\displaystyle t\) minutes, the minute hand moves from \(\displaystyle 0^o\) to \(\displaystyle 6t\) degrees.
At the same time, the hand moves from \(\displaystyle 30^o\) to \(\displaystyle 30\!+\!\frac{1}{2}\!t\) degrees.

At that time, the two hands are together again.
.. The angles are equal: .\(\displaystyle 0 + 6t \:=\:30 + \frac{1}{2}\!t\)
. . \(\displaystyle \frac{11}{2}t \:=\:30 \quad\Rightarrow\quad t \:=\:\frac{60}{11}\) minutes.

The hands are together at \(\displaystyle 5\!\frac{5}{11}\) minutes after 1:00 o'clock.

Total time: .\(\displaystyle 60 + 5\!\frac{5}{11} \;\approx\;65\) minutes.

 

[HallsofIvy]

In 65 minutes, the minute hand will have gone around the clock dial once and then to the "1 hour" position. The hour hand will have moved to the "1 hour" position in 60 minute and then moved 5/60= 1/12 of the way toward the "2 hour" position. So, no, the hour and minute hands will not quite be in the same position 65 minutes after noon (1:05) but that is correct to the "nearest integer" minute.

Let \(\displaystyle \theta\) be the angle (in degrees) the minute and hours hands make with the vertical when they match again. The hour hand goes round the entire dial, 360 degrees, in 12 hours so it will go through \(\displaystyle \theta\) degrees in h hours when \(\displaystyle \frac{\theta}{360}= \frac{h}{12}\) or \(\displaystyle h= \frac{12}{360}\theta= \frac{1}{30}\theta\) hours which is the same as \(\displaystyle \frac{60}{30}\theta= 2\theta\) minutes.

The minute hand goes around the 360 degrees in one hour so it will go through \(\displaystyle \theta\) degrees in m minutes when \(\displaystyle \frac{m}{60}= \frac{\theta}{360}\) or \(\displaystyle m= \frac{\theta}{6}\). But the minute hand has to go all the way around and then catch up to the hour hand- it has to go \(\displaystyle 60+ \frac{\theta}{6}\). You can solve the equation \(\displaystyle m= 2\theta= 60+ \frac{\theta}{6}\) for \(\displaystyle \theta\), then find the number of minutes after noon. I get 65 minutes and 27.3 seconds to the nearest tenth of a second (1:05:27.3).

 

[helping]

Ever heard of google, Helping?
http://www.google.ca/#hl=en&sclient...470c902fdd37b2&bpcl=37643589&biw=1024&bih=571

Yes, sure, but this particular question is not there. Also, I already got the answer using basic knowledge of the clock. I just wanted to know how to do it using an equation.

Both solutions offered appear a bit difficult to understand for an 8th grader who just started learning Algebra. Is there a simpler way to set up an equation?

Really wanted to avoid using degrees as that is usually associated with measuring angles.

 

[helping]

How hard did you look? Here:
http://puzzles.nigelcoldwell.co.uk/thirtyfive.htm

Looked pretty hard, but didn't see that on the first or second page of google results.

I think it's just a play on words...

What did you type in the search box on google to get the above page as one of the results.

Thanks

 

[HallsofIvy]

What is "just a play on words"?