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Showing posts with label problem. Show all posts
Showing posts with label problem. Show all posts

Friday, April 4, 2014

Bell


A first bell ring at every 6 minutes, a second bell rings every 7 minutes and a third bell rings every 1 hour, if all the bells start ringing at 12 noon today when the next day will all ring at the same time?

Solution:


-The 1st two bells will ring at the same time every 6*7 = 42 mn

-The 3rd bell ring once every hour, or once every 60mn

-What is the smallest multiple of 42 that is evenly divided by 60?
(10*42)/60 = 420/60 =7

-So, every 420 min, all 3 bells ring together
420/60 = 7h

-12 noon : 12pm + 7h = 7pm is when they all ring together
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Kim Chan
11:23 AM

Tuesday, April 1, 2014

Find the value of M

M is a positive number that follow these conditions:
  • M is a number that 80<M<100 
  • if M divided by 4 the remainder is 3 
  • and if M divided by 5 the remainder is 1.

Find the value of M?



Solution:


  • M divided by 4 gives remainder 3. So M can be written as 4r + 3 for some integer r
  • M divided by 5 gives remainder 1. So M can be written as 5*t + 1 for some integer t
So M=4*r + 3 or M=5*t + 1

  • But  80<M<100 
Thus 80<4*r + 3<100 , 77<4r<97 , 19.25<r<24.25 so r={20,21,22,23,24}
  • r=20 => M=4*20+3=83 
  • r=21 => M=4*21+3=87
  • r=22 => M=4*22+3=91
  • r=23 => M=4*23+3=95
  • r=24 => M=4*24+3=99
Only r=22 that M=91 can be written as 91=5*18+1  so the only one value of M=91 that apply to these two forms.




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Kim Chan
3:36 PM
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