ACE test series

Question

Please  explain the solution  Ans iis given - C

Comments

some part is missing ?

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answer will depend on database instance....say in 2nd case if i put NULL at such a places where RATING are high ...OPTION C can never be correct one ....such question need an instance of DB ...i think so ......

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I think ans is B because they said COUNT not COUNT(*) , as except COUNT(*) all other aggregate function ignore NULL values therefore when sum is divided by count it is equal to avg.

Please correct me if my logic is wrong.