Q.The marks obtained by a student in two tests are 65 and $x$. Find the range of values of $x$ so that the average of the two marks is at least 60 and at most 80.

[!TLDR] Translate "average is at least 60 and at most 80" into a compound inequality on $x$ and solve. [!ANSWER] $55 \le x \le 95$.

The average of the two marks $65$ and $x$ is $\dfrac{65+x}{2}$. The condition "at least 60 and at most 80" translates to the compound inequality $60 \le \dfrac{65+x}{2} \le 80$. Multiply all three parts by $2$ (positive, no flip): $120 \le 65+x \le 160$. Subtract $65$ from all three parts: $55 \le x \le 95$. So the second test mark $x$ must lie between $55$ and $95$ (inclusive) for the average to fall in the required range. Check: if $x=55$, average $=\dfrac{65+55}{2}=60$ ✓ (boundary met); if $x=95$, average $=\dfrac{65+95}{2}=80$ ✓. [!ANSWER] $55 \le x \le 95$.

Write the average as an algebraic expression in $x$, set it between the two given bounds as a compound inequality, then multiply by 2 and subtract 65 across all three parts.

A frequent mistake is forgetting to multiply the average expression's denominator through all three parts of the compound inequality, mixing up the bounds on $x$ with the bounds on the average itself.