Question 1197195
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If Machine A makes a yo-yo every five minutes and Machine B takes ten minutes to make a yo-yo, 
how many hours would it take them working together to make 20 yo−yos?
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<pre>
We assume that both machines started simultaneously.


Every 10 minutes, machine A produces 2 yo-yo; every 10 minutes machine B produces 1 yo-yo.

So, two machines produce 3 yo-yo every 10 minutes.


Divide 20 by 3 with a remainder: 20/3 = 6 + 2/3.

It tells you that in 6*10 = 60 minutes two machines will produce 6*3 = 18 yo-yos,
which is still less than 20 yo-yous.


In the next 5 minutes, machine A will produce the 19-th yo-yo.
In the next 5 minutes, machine A will produce the 20-th yo-yo, and machine B will complete the 21-th yo-yo.


Thus 20 yo-yo (and even 21 (!) yo-yo) will be completed 70 minutes after start moment.


70 minutes is the <U>minimal time</U> to wait to have 20 yo-yos completed.
</pre>

Solved, with full explanations.



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The solution by @josgarithmetic is totally wrong, since it uses wrong idea that the production units 


go continuously in time, while the process is in discrete units.



May god saves you from making such mistakes and from solving the problem by the method as @josgarithmetic does it.



Would you solve it at the exam or at the Math competition in a way as @josgarithmetic does it, the lowest possible score is guaranteed to you.