Two isosceles triangles each have at least one angle that measures $70^{\circ}$. In the first triangle, the measure in degrees of each of the remaining two angles is even. In the second triangle, the measure in degrees of each of the remaining two angles is odd. In the first triangle, the sum of the equal angles is $S$. In the second triangle, the sum of the equal angles is $T$. The value of $S+T$ is
(A) $280^{\circ}$
(B) $250^{\circ}$
(C) $220^{\circ}$
(D) $200^{\circ}$
(E) $300^{\circ}$
Isosceles triangles have two equal angles, and so the possibilities for these two triangles are:
1) The two equal angles are each equal to $70^{\circ}$, or
2) The two equal angles are each not equal to $70^{\circ}$.

(We note that a triangle can not have three angles measuring $70^{\circ}$ since the sum of the three angles would be $3 \times 70^{\circ}=210^{\circ}$, which is greater than $180^{\circ}$.)

If the two equal angles are each equal to $70^{\circ}$, then the measure of the third angle is $180^{\circ}-2 \times 70^{\circ}=180^{\circ}-140^{\circ}=40^{\circ}$.

If the two equal angles are each not equal to $70^{\circ}$, then the sum of the measures of the two equal angles is $180^{\circ}-70^{\circ}=110^{\circ}$, and so the measure of each of the equal angles is half of $110^{\circ}$ or $55^{\circ}$.

We note that in the first triangle, the measure of each of the two remaining angles $\left(70^{\circ}\right.$ and $40^{\circ}$ ) is even, and in the second triangle, the measure of each of the two remaining angles ( $55^{\circ}$ and $55^{\circ}$ ) is odd.

In the first triangle, the sum of the two equal angles is $S=70^{\circ}+70^{\circ}=140^{\circ}$.

In the second triangle, the sum of the two equal angles is $T=55^{\circ}+55^{\circ}=110^{\circ}$.

The value of $S+T$ is $140^{\circ}+110^{\circ}=250^{\circ}$.

ANSWER: (B)