While most equations indicate a clear relationship between two statements, inequality problems ask test takers to answer questions based on limited information. And by “limited information,” inequality problems merely tell you that one side of the equation is larger (or smaller) than the other.

With that in mind, here are a few tips to help you tackle even the trickiest inequality questions on the CAT.

## 1. Use your algebra skills to solve inequality

It’s easy to forget that inequality questions can be solved using the same rules you would use to solve any problem using the algebra skills you already have in your toolbox.

Let’s take a look at the example below:

5 – 5x > 20

If we were to simplify this equation under the assumption that the > sign was actually an equal sign, we would use the common denominator of 5 to simplify the equation. While the above example is fairly simple, the message is clear: don’t be afraid to use the algebra skills you used in previous sections to help you solve inequalities.

## 2. When multiplying or dividing by a negative, flip the sign

While you can solve an inequality using the same pieces as a traditional formula, there is one major exception. When you find yourself multiplying or dividing by a negative on each side, you’ll need to flip the inequality sign.

Although this is typically taught as a rule, it’s important to understand why it’s a rule. If you were looking at a statement that told you x > 5, it’s clear that anything greater than 5 would be true. However, if you were to multiply both sides by -1, you would end up with a number that is less than -5, which illustrates why the sign needs to be flipped.

## 3. Be careful when multiplying or dividing if you don’t know the sign

Although the example above is fairly simple, it’s important to note that you should never reduce an inequality by a variable if you don’t know whether it’s positive or negative. This is important for one reason: without this knowledge, you won’t know if you have to flip the inequality’s sign.

If a statement on the CAT indicates that x/y > 4, you would typically multiply both sides by y to determine if x > y. However, it’s unclear if y is a positive or negative variable, making it difficult to determine if you need to flip the sign. When you come across statements like these on the CAT, don’t be tempted to multiply by a variable you can’t determine is either positive or negative.

*Here are a set of 3 interesting CAT level Inequalities questions with detailed solutions for the same*.

## Question 1

*How many positive integer values can x take that satisfy the inequality: (x – 8) (x – 10) (x – 12)…….(x – 100) < 0*?

## Solution

Let us try out a few values to see if that gives us anything.

*When x = 8, 10, 12, ….100* this goes to zero. So, these cannot be counted.*When x = 101, 102 or beyond*, all the terms are positive, so the product will be positive.

Straight-away we are down to numbers 1, 2, 3, …7 and then odd numbers from there to 99.

Let us substitute* x =1*

All the individual terms are negative. There are totally 47 terms in this list (How? Figure that out). Product of 47 negative terms will be negative. So, *x = 1 *works. So will *x =2, 3, 4, 5, 6, and 7*.

## What else we can latch on to?

Remember, product of an odd number of negative terms is negative; product of even number of negative terms is positive. Now, this idea sets up the rest of the question and is useful when it comes to inequalities questions.

When *x = 9*, there is one positive terms and 46 negative terms. So, the product will be positive.

When *x = 11*, there are two positive terms and 45 negative terms. So, the product will be negative.

When *x = 13*, there are three positive terms and 44 negative terms. So, the product will be positive and so on.

Essentially, alternate odd numbers need to be counted, starting from 11.

So, the numbers that will work for this inequality are 1, 2, 3, 4, 5, 6, 7…and then 11, 15, 19, 23, 27, 31,….. and so.

What will be the last term on this list? 99, because when *x = 99*, there are 46 positive terms and 1 negative term.

So, we need to figure out how many terms are there in the list 11, 15, 19,….99. These can be written as:

4 * 2 + 3,

4 * 3 + 3,

4 * 4 + 3

…

4 * 24 + 3

A set of 23 terms. So, total number of values = 23 + 7 = 30.

*30 positive integer values of x exist satisfying the condition. *

## Question 2

*Find the range of x for which (x + 2) (x + 5) > 40.*

## Solution

*First Method*

We can expand and simplify this algebraically as *x^2 + 7x + 10 > 40* or *x^2 + 7x – 30 > 0(x + 10) (x -3) > 0*

The roots are -10 and +3.

=> x should lie outside the roots.

Now, what is this based on? There is a simple thumb rule for solving quadratic inequality, which is that for any quadratic inequality* ax ^{2}+ bx + c < 0,* factorize it as

*a(x – p) ( x – q) < 0*

Whenever a is greater than 0, the above inequality will hold good if x lies between p and q. *a(x – p) (x – q)* will be greater than 0, whenever x does not lie between p and q. In other words x should lie in the range ( -infinity, p) or (q, infinity)

Now, coming back to the question’s solution:

*(x+10) (x -3) > 0* or,* x < -10 or x > 3*

*Second Method*

5 * 8 = 40, -8 * -5 = 40; this essentially means if *x + 2 > 5* this will hold good => *x > 3*

If *x+ 2 < -8* also, this will hold good => *x < -10*.

The first method is far more robust.

## Question 3

*Find the range of x where ||x – 3| – 4| > 3*.

## Solution

If we have an inequality *|y| > 3*, this will be satisfied if => *y > 3* or* y < -3*.

Hence, the above inequality simplifies to two inequalities:

1. *| x – 3| – 4 > 3*

*| x – 3| – 4 > 3 => | x – 3 | > 7*

*x – 3 > 7* or *x – 3 < -7*

Or, *x > 10* or *x < -4*

x lies outside of -4 and 10.

*Or, x can lie in the range ( – infinity, -4) or ( 10, infinity) *

#### 2. *|x -3| – 4 < -3*

*|x – 3 | – 4 < – 3*

=> *| x – 3 | < 1*

=> *-1 < x – 3 < 1* or x lies between 2 and 4

The *final solution is given by the range( – infinity, -4) or (2,4) or ( 10, infinity)*

Best wishes for CAT!!

*Rajesh Balasubramanian** takes the CAT every year and is a 4-time CAT 100 percentiler. He likes few things more than teaching Math and insists to this day that he is a better teacher than exam-taker.*

Abhishek Mukherjee works for 2IIM. Apart from solving interesting math questions he likes to eat biriyani and watch movies.

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