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17 Mar 2021, 08:00
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D
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Difficulty:
85%
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Question Stats:
60% (03:15) correct
40%
(03:01)
wrong
based on 98
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Marie has two electronic beepers: one beeps 5 times in every 90 seconds and the other one beeps 6 times in every 96 seconds. If both the beepers are started simultaneously, how many times will they beep together in next 3 hours 35 mins and 16 seconds?
A. 86
B. 87
C. 88
D. 89
E. 90
A. 86
B. 87
C. 88
D. 89
E. 90
Marie has two electronic beepers: one beeps 5 times in every 90 seconds and the other one beeps 6 times in every 96 seconds. If both the beepers are started simultaneously, how many times will they beep together in next 3 hours 35 mins and 16 seconds?
Beeper 1 = 90/5 = 18 , every +18 seconds is going to beep
Beeper 2 = 96/6 = 16, every +16 seconds is going to beep
When are they going to beep together?
we can check the LCM (least common multiple) of those two: 18 = 2*3*3 , 16 = 2*2*2*2, the LCM is 2*2*2*2*3*3 =144
So, every 146 seconds they're going to beep together.
Now everything in seconds: 3*60*60 +35*60 + 16 = 12916
If in every 146 seconds they beep together, we can divide 12916/144 = 89 something.
So 89 times.
Answer is D.
Beeper 1 = 90/5 = 18 , every +18 seconds is going to beep
Beeper 2 = 96/6 = 16, every +16 seconds is going to beep
When are they going to beep together?
we can check the LCM (least common multiple) of those two: 18 = 2*3*3 , 16 = 2*2*2*2, the LCM is 2*2*2*2*3*3 =144
So, every 146 seconds they're going to beep together.
Now everything in seconds: 3*60*60 +35*60 + 16 = 12916
If in every 146 seconds they beep together, we can divide 12916/144 = 89 something.
So 89 times.
Answer is D.
17 Mar 2021, 20:48
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Bunuel
So, one beeps every 18 seconds and other every 16 seconds.
LCM(18,16) = LCM(\(3^2*2\),\(2^4\)) = \(3^2*2^4\)
Total time = 3*60*60 + 35*60 = 60*215 = \(2^2*3*5^2*43\)
Times they beep together = \(\frac{2^2*3*5^2*43}{3^2*2^4} = 89.5\)
i.e. 89 times
Answer D.
