SAT Question 54cb53cf: Percentages answer and explanation with Desmos solution

Let $x$ represent the number of zebras in the population in $2014$ and let $y$ represent the number of zebras in the population in $2018$. It's given that the number of zebras in this population in $2018$ was $1.27$ times the number of zebras in this population in $2014$. It follows that the equation $y=1.27x$ represents this situation. Dividing both sides of this equation by $1.27$ yields $\frac{y}{1.27}=x$, or $\left(\frac{1}{1.27}\right)y=x$. Therefore, the number of zebras in this population in $2014$ is $\frac{1}{1.27}$ times the number of zebras in this population in $2018$. If the number of zebras in this population in $2014$ is $p\%$ of the number of zebras in this population in $2018$, then $x=\frac{p}{100}y$. It follows that $\frac{1}{1.27}=\frac{p}{100}$, or $\frac{100}{1.27}=p$, which means $p$ is approximately equal to $78.74$. Therefore, the value of $p$, to the nearest whole number, is $79$.