Dana has 3 sets of lights
1st set stays lit for 11 sec and then shuts off for 1 sec.
2nd set stays lit for 7 sec and then shuts off for 1 sec
3rd set stays lit for 4 sec and then shuts off for 1 sec.
If the 3 sets of lights are plugged in at the same time and begin the light cycle together, after how many seconds will all 3 sets of lights first be off??
2006-12-21 15:49:06 · 11 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Every 12
Every 8
Every 5
24, 36, 48, 60, 72, 84, 96,
120? after 119...
Again, I will point out that the answer is 'after 119 seconds'...
LOL! @ Juan (below) & his addition of 'try halving the answer 480 that I gave (confidently) & which was wrong...then halve it again... you get 120... which is one quarter of the answer that I first gave as the final answer... act like this procedure of halving, then quartering your first answer is standard practice...'
THE LIGHTS WILL GO ON AGAIN AFTER 120 SECONDS, the lights GO OFF AFTER 119 SECONDS!!!
I'd appreciate any explanation of why it's 308...
If anyone is working on the assumption that 11 by 7 by 4 is the path to anywhere but complete f***ing stupidity, then I've got some news...
2006-12-21 15:52:51 · answer #1 · answered by K V 3 · 2⤊ 1⤋
One complete cycle for 1st set is 12 sec, 8 sec for 2nd, and 5 sec for 3rd. So to find when all 3 will turn off simultaneously, you need to find the LCM:
12 = 2*2*3
8 = 2*2*2
5 = 5
_____________
LCM = 2*2*2*3*5 = 120
So the first time all 3 sets of lights will turn off simultaneously is after 120 seconds.
2006-12-22 00:00:22 · answer #2 · answered by euclidjr 2 · 2⤊ 1⤋
120 secs
11+1=12
7+1=8
4+1=5
multiples of all 3 numbers
2006-12-21 23:58:03 · answer #3 · answered by out of it 2 · 1⤊ 1⤋
1st set is off 1/12 of the time.
2nd set is off 1/8 of the time.
3rd set is off 1/5 of the time.
1/12 x 1/8 x 1/5 = 1/480
480 seconds.... try halving it: 240 seconds is a multiple of all 3 numbers. Half it again, 120 also works. Every 120 seconds the light will be dark. (2 minutes).
2006-12-21 23:53:42 · answer #4 · answered by MisterRE 3 · 0⤊ 1⤋
This is a total guess, but an educated one.
All you have to do is take the lowest common multiple (LCM) of 11, 7, and 4. The LCM is 308.
Therefore, after 308 seconds, all 3 sets of lights will first be off.
Why I'm not 100% sure the answer is 308 seconds is because I'm not sure if the fact that they shut off for 1 second comes into play at all.
2006-12-21 23:57:03 · answer #5 · answered by Puggy 7 · 0⤊ 3⤋
120
2006-12-22 02:20:08 · answer #6 · answered by Léon 1 · 0⤊ 1⤋
Just find the LCM of 12,8, and 5, and subtract 1.
119 s, or 1 min, 59 s
2006-12-25 23:50:59 · answer #7 · answered by _anonymous_ 4 · 0⤊ 0⤋
308 seconds. or 5 min, 8 seconds.
2006-12-21 23:53:02 · answer #8 · answered by Kelly A 4 · 0⤊ 3⤋
308 seconds into the cycle.
2006-12-21 23:52:32 · answer #9 · answered by master_furches 2 · 0⤊ 3⤋
3 seconds? I don't know, this question is too hard!
2006-12-21 23:51:28 · answer #10 · answered by ? 3 · 0⤊ 3⤋