Street Fighter

From the question, we can deduce that

$\frac{dx}{dt}=-k_{1}y^{2}$

$\frac{dy}{dt}=-k_{2}x$

Where $t$ denotes time and $k_{1}$ and $k_{2}$ are constants.

Dividing the two equations, we get

$\frac{dy}{dx}=\frac{x}{y^{2}}(\frac{k_{2}}{k_{1}})$

For the sake of simplification, let $r$ be $\frac{k_{2}}{k_{1}}$ . Arranging this equation, we get

$y^{2}dy=rxdx$

Then we take the integral of both sides.

$\frac{y^{3}}{3}+C=r\frac{x^{2}}{2}$

Multiply each side by $6$ :

$2y^{3}+C=3rx^{2}$

Given the initial conditions $x=y=1$ , we can find $C$ in terms of $r$ .

$2+C=3r$

$C=3r-2$

We are also given that at a point in time, $x=0.25$ and $y=0.5$ . Substituting these values in, we can find $r$ .

$2(0.5)^{3}+3r-2=3r(0.25)^{2}$

$\frac{1}{4}+3r-2=\frac{3}{16}r$

$\frac{45}{16}r=\frac{7}{4}$

$r=\frac{28}{45}$

Now substitute $x=0$ .

$2y^{3}+3(\frac{28}{45})-2=0$

$y^{3}=\frac{1}{15}$

$y=0.4054801...=40.54801\%$

Giving the answer of $\boxed{4055}$ .