Let's note:
x=the number of bags of cement the can be safely lifted on the elevator in the same trip.

We have: \(\displaystyle{245}+{95}{x}\le{3000}\)

We solve the inequality: \(\displaystyle{245}+{95}{x}-{245}\le{3000}-{245}\)

\(\displaystyle{95}{x}{<}{2755}\)</span>

\(\displaystyle{x}\le\frac{{2755}}{{95}}\)

\(\displaystyle{x}\le{29}\)

We have: \(\displaystyle{245}+{95}{x}\le{3000}\)

We solve the inequality: \(\displaystyle{245}+{95}{x}-{245}\le{3000}-{245}\)

\(\displaystyle{95}{x}{<}{2755}\)</span>

\(\displaystyle{x}\le\frac{{2755}}{{95}}\)

\(\displaystyle{x}\le{29}\)