Let Barney's regular hourly wage be \(x\), then his Saturday wage will be \(2x\) and Alan's hourly wage will be \(1.5x\);

Let the # of hours Barney worked Monday through Friday be \(m\) and on Saturday be \(n\) and the # of hours Alan worked Monday through Friday be \(p\) and on Saturday be \(q\);

Given: \(xm+2xn=1.5x(p+q)\) and \(m+n=p+q\).

\(xm+2xn=1.5x(p+q)\) --> \(m+2n=1.5(m+n)\) --> \(m=n\) --> Barney worked the equal # of hours Monday-Friday and on Saturday.

The above directly tells us that II must be true (as Barney worked total non-zero # of hours and he worked an integer # of hours on any given day then he must have been worked at least one hour on Saturday.)

As for I: Alan may have worked ALL his hours Monday through Friday so in this case this statement is not true (p=total>m). Alan also may have worked all his hours on Saturday. Or algebraically: there are any distribution possible between p and q, p=0 and q=total or p=total and q=0 or any other;

The above means that III is also not always true: if Alan worked all his hours on Saturday then he made all his money on Saturday thus he made more money on Saturday than Barney did.

Answer: B (II only).

But the above can also be done with much less algebra:

As Alan and Barney worked the same # of hours and earned the same amount of money, then their hourly average wages must have been the same: (average wage)=(total amount earned)/(# of hours worked). Now, Alan has constant hourly wage which is \(1.5*x\) and Barney's average (\(\frac{xm+2xn}{m+n}\)) to be equal to this he must have been worked the equal # of hours Monday-Friday and on Saturday, so \(m=n\).

Hope it's clear.