yuri wrote:guys, I don't understand how you get this:
(1/L) + (1/J) = 1
So, (1/J) = (x-3)/x
I end up with something else when subbing.
Guess ur doing something wrong then ..
Let me give it a try .. here is the question ..
Lindsay can paint 1/x of a certain room in 20 minutes. What fraction of the same room can Joseph paint in 20 minutes if the two of them can paint the room in an hour, working together at their respective rates?
let x = 10 .. that means Lindsay can paint 1/10th of the room in 20 minutes ... so in 1 hr she can paint 3/10th of the room (bcoz 20min *3=60min=1hr)
Now, let us assume that painting the room requires 100 units of work to be done , if Lindsay does 3/10th of that work in one hour ( that is 3/10*100 = 30 units of work ) then Joseph does the rest = 100-30 = 70 units of work .. i.e Joseph does 7/10 th of the total work in one hour .. So in 20 minutes Joseph can do 1/3rd of the work he does in one hour that is he can do 7/30 of the work .. so that is our answer ..
Now look at the given answer choices and substitute x = 10 in them and check for which one we get the answer as 7/30, .. it should be (x-3) / 3x ..
PS: - u dont have to choose 100 as the unit of work that was done to keep the solution simple .. Regarding your question as to how we get the
fraction of work done by Lindsay+ fraction of work done by Joseph = 1
Let the amount of work to be done be "x" units .. Lindsay does 3/10th of it and Joseph does the rest i.e. 7/10th ..
The addition of the total work done by both of them should be equal to "x" .. So 3/10*x + 7/10*x = x .. Since "x" is common it can be canceled out and then we have 3/10+7/10 = 1 ... that is fraction of work done by Lindsay+Fraction of work done by Joseph should be equal to 1
