3 strings of a musical instrument vibrate 6,8 and 12 times a second respectively.if all the 3 begin to vibrate simultaneously,find the shortest time interval before all 3 vibrate together again?
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Source: Beat The GMAT — Problem Solving |
Find LCM of ( 6,8,12)
gmatmachoman wrote:Find LCM of ( 6,8,12)
Not really...
We are in fact given the frequencies and not the time-periods of the 3 strings. Hence, we first need to take reciprocals of 6, 8 and 12 to get the time-periods of the 3 strings as 1/6, 1/8, and 1/12 in seconds. Now comes the time to take the LCM of 1/6, 1/8, and 1/12 and answer it in seconds.
LCM of 1/6, 1/8, and 1/12 = LCM of 1, 1, 1/HCF of 6, 8, 12 = [spoiler]½[/spoiler] a second is the shortest time interval before all 3 vibrate together again.
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Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
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Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
let's do it in another way :
1/6 x a = t
1/8 x b = t
1/12 x c = t
when t is the time we will get the first string together .
we get :
b = 4/3 x a
c = 2 x a
the first integer a we give us an integer b is : a = 3
so we get :
t = 1/6 x 3 => t= 0.5 seconde
1/6 x a = t
1/8 x b = t
1/12 x c = t
when t is the time we will get the first string together .
we get :
b = 4/3 x a
c = 2 x a
the first integer a we give us an integer b is : a = 3
so we get :
t = 1/6 x 3 => t= 0.5 seconde
