Question:

Wilma, Xavier, Yaska and Zakir are four young friends, who have a passion for integers. One day, each of them selects one inWilma, 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:
W4 + X3 + Y2 + Z\(≤\) 4, 
X3 + Z \(≥\) 2, 
W4 + Y2 \(≤\) 2, 
Y2 + Z \(≥\) 3 
Given the above, which of these can W2 + X2 + Y2 + Z2 possibly evaluate to?

Updated On: Aug 22, 2025
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The Correct Option is B

Solution and Explanation

To solve the given problem, let's analyze the constraints step by step:

1. \(W^4 + X^3 + Y^2 + Z \leq 4\)

2. \(X^3 + Z \geq 2\)

3. \(W^4 + Y^2 \leq 2\)

4. \(Y^2 + Z \geq 3\) 

Given that Xavier and Zakir picked positive integers and Yaska picked a negative integer, Wilma's integer can be negative, zero or positive. We must identify reasonable values for \(W\), \(X\), \(Y\), and \(Z\) that fulfill all constraints and then evaluate \(W^2 + X^2 + Y^2 + Z^2\).

Step 1: Assume \(Y = -1\). Then \(Y^2 = 1\).

Step 2: From constraint 4, \(1 + Z \geq 3 \Rightarrow Z \geq 2\), thus the lowest integer for \(Z\) satisfying this is \(Z = 2\).

Step 3: With \(Z = 2\), constraint 2 becomes \(X^3 + 2 \geq 2 \Rightarrow X^3 \geq 0\). Since \(X\) is a positive integer, \(X = 1\) is the smallest possibility.

Step 4: Substitute \(X = 1\), \(Y = -1\), and \(Z = 2\) into constraint 1: \(W^4 + 1 + 1 + 2 \leq 4 \Rightarrow W^4 \leq 0\). Therefore, \(W = 0\).

Given \(W = 0\), \(X = 1\), \(Y = -1\), \(Z = 2\), calculate \(W^2 + X^2 + Y^2 + Z^2 = 0^2 + 1^2 + (-1)^2 + 2^2 = 0 + 1 + 1 + 4 = 6\).

The possible value is thus: 6.

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