xat 2022 Question 22

Question

Wilma, Xavier, Yaska and Zakir are four young friends, who have a passion for integers. One day, each of them selects one integer and writes it on a wall. The writing on the wall shows that Xavier and Zakir picked positive integers, Yaska picked a negative one, while Wilma’s integer is either negative, zero or positive. If their integers are denoted by the first letters of their respective names, the following is true:
$W^{4}+X^{3}+Y^{2}+Z\leq4$
$X^{3}+Z\geq2$
$W^{4}+Y^{2}\leq2$
$Y^{2}+Z\geq3$
Given the above, which of these can $W^{2}+X^{2}+Y^{2}+Z^{2}$ possibly evaluate to?

Accepted Answer

Given that X, Z are positive Y is negative and W can be either positive or zero or negative.

The given conditions are :

$W^{4}+X^{3}+Y^{2}+Z\leq4$

$X^{3}+Z\geq2$

$W^{4}+Y^{2}\leq2$

$Y^{2}+Z\geq3$

For $W^4+\ Y^2\ \le\ 2$ . Since Y is negative $but\ Y^2$ is always positive and must be less than 2 because $W^4$ is a nonnegative value. Hence Y = -1 is the only possibility. For W this can take any value among -1, 0, 1.

$Y^2+Z\ \ge\ 3$ . 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: $W^{2}+X^{2}+Y^{2}+Z^{2}$ here has all the independent values: $X^2,\ Y^2,\ Z^2,\ W^2$ are nonnegative.

$W^{4}+X^{3}+Y^{2}+Z\leq4$ :

Since the value of Z is at least equal to 2 the value of $Y^2$ is equal to 1.

Since X is a positive number in order to have the condition of $W^{4}+X^{3}+Y^{2}+Z\leq4$ 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 :

$W^{4}+X^{3}+Y^{2}+Z\leq4$ . = The sum = (0+1+1+2) = 4. The only possible case.

The value of $W^{2}+X^{2}+Y^{2}+Z^{2}$ = (0+1+1+4) = 6.

Question 22.

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