Question 69

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?

Solution

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.

Video Solution

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