1. If 31x + 31y = 403, then what is the average of x and y?
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By: anil on 05 May 2019 02.06 am
Given : $$31x+31y=403$$ => $$31(x+y)=403$$ => $$(x+y)=frac{403}{31}=13$$ Dividing both sides by 2, we get : $$ herefore$$ Required average = $$frac{(x+y)}{2}$$ = $$frac{13}{2}=6.5$$ => Ans - (C)
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