problem

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

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.