imc

We can set up a system of equations where $a$ is the sets of twins, $b$ is the sets of triplets, and $c$ is the sets of quadruplets. $$\begin{array}{rl}2a+3b+4c& =1000\\ b& =4c\\ a& =3b\end{array}$$ Solving for $c$ and $a$ in the second and third equations and substituting into the first equation yields $$\begin{array}{rl}2(3b)+3b+4(0.25b)& =1000\\ 6b+3b+b& =1000\\ b& =100\end{array}$$ Since we are trying to find the number of babies and NOT the number of sets of quadruplets, the solution is not $c$, but rather $4c$. Therefore, we strategically use the second initial equation to realize that $b$ $=$ $4c$, leaving us with the number of babies born as quadruplets equal to $\boxed{100}$.