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?

Could someone solve the above problem and explain why my approach is wrong? The correct answer is 89.

**My Approach:**

I treated this as a work rate problem because I saw that the rates for the 2 beepers were given.

work = rate * time

Since the beepers are working simultaneously, I counted the final rate as = rate of beeper 1 + rate of beeper 2.

so rate = 5/90 + 6/96 = 110/864 beeps per second

now time would be = (3 * 60 * 60)+(35 * 60)=16 = 12916 seconds

so work = (110 * 12916)/864 = 1644 beeps

My answer is obviously wrong but is it because it is incorrect to look at this problem as a work rate question because the beepers arenâ€™t â€śworking togetherâ€ť? I saw another answer somewhere online regarding LCM but I donâ€™t understand how do we know that we need to find LCM.

Hey,

I guess the keyword in the question is â€śtogether,â€ť which should trigger you to use LCM.

Hereâ€™s the solution:

5\frac{beep}{90\ sec} = \frac{5}{90}\frac{beep}{sec}

So the first beeper takes \frac{90}{5}=18 seconds to beep.

Similarly, the second beeper takes \frac{96}{6} = 16 seconds to beep.

LCM(18, 16) = 144

So these two beepers beep simultaneously every 144 seconds.

3 hours 35 minutes and 16 seconds is equal to 12916 seconds.

\frac{12916}{144} = 89.69444\dots

Now the question is, do they beep at the moment we start them? If so, we will have 1+89 simultaneous beeps. Otherwise, 89.

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Thank you for the reply! This might be a stupid question but why is the work rate approach wrong? Iâ€™m trying to understand what I misunderstood/did wrong so I can work on it.

I suppose that the work rate approach would be useful when weâ€™re looking at some kind of work being done collectively by multiple agents. But in this case, thereâ€™s no â€śworkâ€ť per se, just two beepers working independently, and the fact that beeper B starts simultaneously with beeper A, doesnâ€™t affect beeper Aâ€™s speed or progress, or vice versa.

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Thank you for your explanation!