Question 10

If 31x + 31y = 403, then what is the average of x and y?

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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