Solution
Given : $$31x+31y=403$$
=> $$31(x+y)=403$$
=> $$(x+y)=\frac{403}{31}=13$$
Dividing both sides by 2, we get :
$$\therefore$$ Required average = $$\frac{(x+y)}{2}$$
= $$\frac{13}{2}=6.5$$
=> Ans - (C)
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