Clock Problem

[88bobcat]

Problem 08G-68

A clock face reads exactly 8:47. How long will it take the minute hand to align with the hour hand? Answer is to be expressed in minutes.


The answer key states 62.1 min. I've tried working the problem several different ways, and I continually calculate 59.4 min. as my answer.

Initial hour-hand position: 263.5 degrees from 12:00
Initial minute-hand position: 282 degrees from 12:00

[unparseable or potentially dangerous latex formula]

[unparseable or potentially dangerous latex formula]


Am I making a gross error in my evaluation, or is there something about the problem I'm not understanding? Can anyone shed any light on this for me, please?

Thanks in advance.

[AuSmith]

There are [unparseable or potentially dangerous latex formula] hours on a clock.

[calcguy]

Here is the way I recommend you work this kind of clock problem.

Move the minute hand to the straight up (12) position either clockwise or counterclockwise, whichever way is necessary so the hands do not cross over each other. So from 8:47, you must move it 13 minutes clockwise to 9:00.

Now let x = the number of minutes after 9:00 till the hands align. Therefore x = 45 min + (x/60)∙5 min because (x/60) is the fraction of the "distance" in minutes the hour hand must move from 9 to 10.

Solve for x and you get 49.1 min. Finally, add the 13 min you initially moved the hand to get 49.1 + 13 = 62.1 min.

[88bobcat]

[AuSmith]

I find it easier to find the relative rates between the hands. Then, you don't have to worry about what position's where, but only the gap. Some people on our team memorized the rates between hands on the clock. But, you can quickly derive them... just know that there are [unparseable or potentially dangerous latex formula] between the numbers and [unparseable or potentially dangerous latex formula] between notches. Therefore,

hour hand goes @ [unparseable or potentially dangerous latex formula]

minute hand goes @ [unparseable or potentially dangerous latex formula]

second hand goes @ [unparseable or potentially dangerous latex formula]

The min and hour hand go at [unparseable or potentially dangerous latex formula] relative to each other, and the minute hand is already ahead. It needs to bridge the gap of [unparseable or potentially dangerous latex formula]... just divide.

[88bobcat]

[hansond10]

What I always do on these sorts of problems is use:

abs(60h-11m)/2 = angle between hands.

Now, since in this problem, we know that it will be some time after 8:47, you can plug in 8 for the h (hour) and figure out the minute time that will correspond to the angle =0.

abs(60*8-11m)/2=0
abs(480-11m)=0
480-11m=0
480=11m
m=480/11=43.636363...

Now, since that time as already passed, we have to move on to the 9:00 hour.

abs(60*9-11m)/2=0
abs(540-11m)=0
540-11m=0
540=11m
m=49.090909...

Finally, since the time started at 8:47, it required 13 minutes to get to 9:00, in addition to the 49.09 minutes after 9:00 that it took for the minute and hour hand to align, to get a final answer of 62.090909... or 62.1

Does all of that make sense? Since the problems are changing quickly to make them harder (i.e. making the minute and hour hand at a 40 degree angle), I especially like the abs(60h-11m)/2 formula to get the angle because it is very versatile.

[austorior]

im not sure if this is the same thing as the others, but I just calculate the ang between two hands (18.5) and then do 18.5+5.5x-360=0 and solve for x. 5.5 is the rate stated by above.