Xavier Aptitude Test 2022 Paper

Given that $X, Z$ are positive $Y$ is negative and $W$ can be either positive or zero or negative. The given conditions are : $W^4 + X^3 + Y^2 + Z \le 4$ $X^3 + Z \ge 2$ $W^4 + Y^2 \le 2$ $Y^2 + Z \ge 3$ For $W^4 + Y^2 \le 2$ : Since $Y$ is negative but $Y^2$ is always positive and must be less than $2$ because is a non negative value. Hence$Y = -1$ is the only possibility. For $W$ this can take any value among $-1, 0, 1$ For $Y^2 + Z \ge 3$ : Since $Y = -1, Z$ must be at least equal to $2$ so the value of $Y^2 + Z \ge 3$ is greater than $2$.$X$ is a positive value and must at least be equal to $1$. The condition $W^2 + X^2 + Y^2$ here has all the independent values: $X^2, Y^2, Z^2, W^2$ are non negative. For $W^4 + X^3 + Y^2 + Z \le 4$ : Since the value of $Z$ is at least equal to $2$ the value of is equal to $1$.Since $X$ is a positive number in order to have the condition of $W^4 + X^3 + Y^2 + Z \le 4$ satisfied. The value of $Z$ must be the minimum possible so that $X^3 + Y^2 + Z$ to have a value equal to $4$ when $X$ takes the minimum possible positive value equal to $1$. Hence $X$ must be $1$. $W$ must be equal to $0$ so that : $W^4 + X^3 + Y^2 + Z \le 4 =$ The sum $= (0+1+1+2) = 4$. The only possible case- $W^2 + X^2 + Y^2 + Z^2$ $= (0+1+1+4)$ $= 6.$