CAT Quantitative Aptitude Questions | CAT Algebra - Inequalities questions

The question is about maximum and minimun value. Our task is to maximize and minimize the given inequality. Inequalities are crucial to understand many topics that are tested in the CAT exam. Having a good foundation in this subject can help us tackle questions in Coordinate Geometry, Functions, and most importantly in Algebra. A range of CAT questions can be asked based on this simple concept.

Question 24: What are the maximum and minimum possible values for \\frac{|x+y|}{|x|+|y|}\\) + \\frac{|z+y|}{|z|+|y|}\\) + \\frac{|z+x|}{|x|+|z|}\\)?

1. 3 and 1
2. 3 and 0
3. 4 and 0
4. 4 and 1

Video Explanation

Explanatory Answer

Method of solving this CAT Question from Algebra - Inequalities : The Question might seem very simple, but there is a catch!!

We know that |x + y| < |x| + |y|.
So, each of these fractions lies between 0 and 1. So, all three added together should go from 0 to 3?

Is that the case? Is 2IIM running questions that are this simple?

When would this be 3? If x, y and z all have the same sign. Each fraction would be 1 and we would get 3 overall. Spot on! So, the maximum value is 3.
When can this go to zero.
When x=-y this fraction goes to 0. When x and y have opposite signs, the first term would go to zero. Likewise for the second and third terms as well.

So, what is the catch.

Among x, y, and z at least two will have the same sign.
So, of the three terms maximum of two can go to zero. One will be +1.
So, the minimum total overall = 1 (not 0)

The question is "What are the maximum and minimum possible values for \\frac{|x+y|}{|x|+|y|}\\) + \\frac{|z+y|}{|z|+|y|}\\) + \\frac{|z+x|}{|x|+|z|}\\)?"

Hence the answer is "3 and 1"

Choice A is the correct answer.