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Shortly after 10 'o clock, the hands of a clock made a 90 degree angle. What's the time?

 Jun 16, 2021
 #1
avatar+121006 
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Let  M  be  the  number of  minutes  after  10

The hour  hand  moves   (1/2)°  every minute

The  minute  hand  moves  6°  every  minute

 

Call   10 o'clock   =  0°

So....the  minute  hand is  at  60°   at 10

 

So.....we  just need to solve  this

 

6M  - (1/2)M  +  60 =   90

 

5.5M     =   30

 

M =  30  / 5.5  ≈  5.45  minutes  after  10   =  5  + .45 *60   ≈  5 min 27 sec  after 10

 

 

cool cool cool

 Jun 16, 2021
 #2
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That doesn't make sense! If the minute moves 30 degrees per 5 minute interval, then 30*10 = 300 degrees should be the degrees the minute hand travels until 10!

 Jun 16, 2021
 #3
avatar+1452 
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Shortly after 10 'o clock, the hands of a clock made a 90-degree angle. What's the time?

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Here's another way to solve this problem.

 

Arc from 10 to 12 = 60º

 

Let x be the arc-path of minute hand in degrees.

 

60 + x - (x/12) = 90

 

x = 360/11 = 32.72727272º     (minute hand arc)

 

x/12 = 30/11 = 2.72727272º     (hour hand arc)

 Jun 16, 2021
 #4
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This is amazing. I just posted an answer to this clock problem on a more recent posting of the problem. I had not seen Phil's solution, but I could have believably copied his verbatim. The same thing happened on a regular decagon problem, even though our solutions were not quite so similar. This is fun;  post problems as many times as your heart desires, even though I am getting dizzy looking through the solutions. Just kidding.  Please have mercy. I really think only one posting should be allowed at one time, so no time is wasted and the better solutions can be compared.By the way, I tried to sign up today, but the activation email never arrived. Still waiting patiently.

 Jun 17, 2021

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