Question

help only question number 21.​

Answer 1

since numbers adjacent to 5 have sum as 9

confirmed that 5 isn't in the middle cause then the sum of rest would exceed 9

5 has to be somewhere at the sides

CASE 1: if its between 1,2 then middle number would be 6

CASE 2:if its between 2,4 middle number would be 3  (this isn't possible as 3 is already placed)

CASE 3: if its between 4,3 middle number would be 2 (this isn't possible as 2 is already placed)

CASE 4:if its between 3,1 middle number would be 5 (this isn't possible as 5 is already placed)

therefore only 1 satisfies

so 5 is between 1,2 and 6 is in middle

so the sum would be

5+7+8+9=29

option E