Question

what is the sum of all integer values of $x$ such that $\frac{31}{90} < \frac{x}{100} < \frac{41}{110}$ is true?

Answer 1

First, we can simplify the inequalities by cross-multiplying:
\begin{align*}
\frac{31}{90} < \frac{x}{100} &\iff 3100 < 90x < 31\cdot 100 \
\frac{x}{100} < \frac{41}{110} &\iff 100x < 41\cdot 110
\end{align*}

Solving the first inequality, we have $3100/90 < x < 31\cdot 100/90$, or $31/9 < x < 310/9$. Multiplying both sides by $9$ gives $31 < 9x < 310$, so $x$ can be any integer between $32$ and $34$ inclusive.

Solving the second inequality, we have $x < 41\cdot 110/100$, or $x < 45.1$. Since $x$ is an integer, the largest possible value of $x$ that satisfies this inequality is $45-1=44$.

Therefore, the possible values of $x$ are $32, 33, 34,$ and $44$, and their sum is $32+33+34+44=\boxed{143}$.