Men, Work and Time 10 men can do a piece of work in X days and 20 women can do the same work in X-5 days. 30 men can do the same work in Y days and 80 women can do that work in Y-5 days. Find X and Y
I wish to understand the solution of this question. I know the answer, however I wish to understand, how are those answers obtained. I do it like this: 10 men X days 1 work, so 1 man 1 day = 1/(10*X) work. Please explain me, how to proceed thereafter by using this technique. 
 A: In man-days we have $10X=30Y$ and therefore $X=3Y$.
In woman-days we have $20(X-5)=80(Y-5)$ and so $X=4Y-15$.
Combining the two equations we obtain $X=45,Y=15$.
A: Let $M$ and $W$ represent the amount of the job done by one man and one woman in one day, respectively.  Since all of these problems talk about one job being done by teams of people in a given number of days, we have
$$10MX=1\\20W(X-5)=1\\30MY=1\\80W(Y-5)=1$$
From the first and third equations, we have $10MX=30MY$, which simplifies to $X=3Y$.  Substituting that into the second equation and setting that equal to the fourth equation, we get $$20W(3Y-5)=80W(Y-5)\\3Y-5=4(Y-5)=4Y-20\\Y=15\\X=3(15)=45$$
A: Just find how long a man can do the work for, assuming all people work at the same rate of course.
Then if $10$ men take $x$ days, them one man will take $10x$ days, and if $20$ women take $x-5$ days, then one woman will take $20(x-5)$ days.
Also, one man takes $30y$ days and one woman takes $80(y-5)$ days. Thus it must be the case that $$10x=30y$$ and $$20(x-5)=80(y-5).$$ This is the system you want to solve for $x$ and $y.$
