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By the Cauchy-Schwarz inequality, $$(a^2+b^2+c^2)(x^2+y^2+z^2)\ge (ax+by+cz{)}^2.$$This becomes $25\cdot 36\ge 30^2.$ Since $25\cdot 36=900=30^2,$ we get equality.

From the equality condition in Cauchy-Schwarz, $$\frac{a}{x}=\frac{b}{y}=\frac{c}{z}.$$Let $$k=\frac{a}{x}=\frac{b}{y}=\frac{c}{z}.$$Then $a=kx,$ $b=ky$ and $c=kz,$ so $$a^2+b^2+c^2=k^2{x}^2+k^2{y}^2+k^2{z}^2=k^2(x^2+y^2+z^2).$$Then $36k^2=25,$ so $k^2=\frac{25}{36},$ which means $k=\frac{5}{6}.$ Therefore, $$\frac{a+b+c}{x+y+z}=\boxed{\frac{5}{6}}.$$