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Im2bz2p345
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Hi,

If you have to go by equation route, then your equation is correct:

(1/x+1/y)=1/T

Solving this, T= xy/(x+y)

Alternatively, by my explanation:

Jointly, they do (1/x+1/y) in 1 hour
Therefore, they will do the whole(1) in 1/(1/x+1/y)=xy/(x+y) which can be called T hours.

Suggest, try plugging in numbers in place of x,y and then it will be clear.



Regards

Argha
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Im2bz2p345
Can someone please explain how the formula \(\frac{xy}{x+y}\) is derived?

It is shown here: the-discreet-charm-of-the-ds-126962-20.html#p1039633

The solution given states "sum of the rates equal to the combined rate or reciprocal of total time": \(\frac{1}{x}+\frac{1}{y}=\frac{1}{T}\) --> \(T=\frac{xy}{x+y}\)).

I know the basic combined worked formula, but when you solve for T, shouldn't it be \(T=\frac{x+y}{xy}\)?

Thanks for all help in advance in understanding this. I tried to search around, but couldn't find any place that derived this formula.

~ Im2bz2p345 :)


Rate of work is additive, time is not. This means that if a machine makes 2 nuts in a minute and another makes 5 nuts in a minute, together, they make 2+5 = 7 nuts in a minute. So you added the two rates to get the combined rate. On the other hand, time is not. If a machine takes 3 hrs to complete a work and another takes 4 hrs, total time taken by both of them together to complete the work is not 3+4 = 7 hrs. They will take less than 3 hrs to complete the work together.

Usually you are given the time taken by the two machines independently and you need to find the time taken by them together.

How do you find rate when time is given e.g. a machine takes 3 hrs to complete a particular work. Rate of work is work done per hour i.e. (1/3)rd of the work per hour.
Rate = Work/Time
So if there is 1 work, Rate = 1/TIme

That is why, given one machine takes x hrs and another takes y hrs, rate of one machine is 1/x and another is 1/y
Total rate = 1/x + 1/y = 1/Total time

(x+y)/xy = 1/T
T = xy/(x+y) (Cross multiply)
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Thank you so much for your followups Argha & Karishma!

I think that I was just confusing myself when I looked at the equation for the first time. Your explanations helped me better understand the concept.

~ Im2bz2p345 :)

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