Inequalities in CAT Quant: The Sign-Line Method for Ranges

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Inequality questions in CAT Quant tend to go wrong in a predictable way. The aspirant expands the brackets, moves everything to one side, and then tries to remember which way the sign flips and where the solution set starts and stops. Two minutes later they have an answer that is off by one interval, or one boundary point.

A number line fixes most of this. Instead of manipulating the expression, you mark the points where it changes sign and read the answer off the picture. This guide shows a sign-line method that works for products, ratios and modulus questions, with worked examples you can check yourself.

Why Algebra Alone Fails on CAT Inequalities

The trouble with inequalities is that the usual equation habits do not carry over. You can cancel a common factor in an equation. In an inequality, cancelling a factor whose sign you do not know can silently delete part of the answer, or flip the direction. You can square both sides of an equation. In an inequality, squaring can create solutions that were never there.

A sign line removes the need for these moves. You are not transforming the inequality. You are asking a simpler question: for which values of x is this expression positive, negative or zero? Once you know that, any inequality about its sign is answered by looking at the line.

This is the same idea behind the wavy curve approach, but applied in a way that handles modulus and ratios without a separate rule for each. Our longer reference on inequality properties and the wavy curve method covers the properties. This post is about using the line quickly under time pressure.

The Sign-Line Method: Five Steps for Any Inequality

The method has five steps and they work in the same order every time. Memorise the steps, not the cases, because the cases are only variations on the same five moves.

The fifth step is the one that catches people. A factor like (x-2) squared never changes sign at 2, because a square is never negative. So you cross that point without flipping, although the expression is zero there. This is also why the point 2 can be a solution of a non-strict inequality even when the neighbouring regions are not.

Products and Powers: Reading the Line

Take (x-1)(x-3)(x+2)≥0. The critical points are -2, 1 and 3. The leading coefficient is positive, so the region to the right of 3 is positive. Crossing 3, the sign flips, so the region between 1 and 3 is negative. Crossing 1, it flips again, so the region between -2 and 1 is positive. To the left of -2 it is negative.

The inequality asks for values where the expression is positive or zero. That gives the interval from -2 to 1, including both ends, and everything from 3 onwards, including 3. Written out, the solution is -2≤x≤1 or x≥3. You can check one value from each region: at x=0 the expression is (-1)(-3)(2)=6, which is positive, and at x=2 it is (1)(-1)(4)=-4, which is negative.

Now a repeated factor. Take (x-2)²(x+1)<0. The critical points are -1 and 2. The rightmost region is positive. Crossing 2 does not change the sign because the power is even. Crossing -1 does flip it, so the region to the left of -1 is negative. The solution is x<-1. The point x=2 gives zero, which fails the strict inequality, and the sign line shows the region around it is positive anyway.

Expression	Critical points	Solution
(x-1)(x-3)(x+2)≥0	-2, 1, 3	-2≤x≤1 or x≥3
(x-2)²(x+1)<0	-1, 2 (even power)	x<-1
(x+1)/(x-4)≤0	-1 closed, 4 open	-1≤x<4
|x-3|+|x+1|<6	-1, 3 (modulus breaks)	-2<x<4

Ratios: Keep the Denominator Open

A ratio behaves like a product for the purpose of signs, with one extra rule. The denominator can never be zero, so its critical points are always open, even when the inequality is non-strict. Take (x+1)/(x-4)≤0. The critical points are -1 and 4. The point -1 makes the numerator zero, so it can be included. The point 4 makes the denominator zero, so it must be excluded.

The rightmost region, beyond 4, is positive. Crossing 4, the sign flips, so the region between -1 and 4 is negative. Crossing -1, it flips again, so the region to the left is positive. We want negative or zero, so the answer is -1≤x<4, a closed end at -1 and an open end at 4.

The trap here is multiplying both sides by (x-4) to clear the fraction. That is legitimate only if you know the sign of (x-4), and you do not. The sign line never asks you to know it in advance. It builds the sign of the whole expression from the sign of each factor.

Modulus: Split at Every Break Point

A modulus is a piecewise function. The expression |x-a| equals x-a when x is at least a, and a-x otherwise. So a modulus question is really several simple inequalities, one for each region between the break points, and you stitch the results together.

The single modulus case is quick. |2x-5|>3 means 2x-5 is either above 3 or below -3. The first gives x>4 and the second gives x<1. The solution is x<1 or x>4. For the opposite case, |x-a|<b, the solution is a-b<x<a+b, one connected interval.

Two moduli need regions. Take |x-3|+|x+1|<6. The break points are -1 and 3. For x below -1, the expression becomes (3-x)+(-x-1)=2-2x, and 2-2x<6 gives x>-2, so the region contributes -2<x<-1. For x between -1 and 3, the expression is (3-x)+(x+1)=4, which is always less than 6, so the whole region works. For x above 3, it becomes 2x-2, and 2x-2<6 gives x<4, so this region contributes 3<x<4. Joining the three pieces gives -2<x<4.

The check that matters is that each region's answer is intersected with the region itself. In the first region, x>-2 alone is not the answer, because the region is x<-1. The answer is the overlap. Forgetting this is the commonest reason modulus answers come out too large.

Counting Integers and Ranges from the Line

CAT often wraps an inequality inside a counting question. How many integers satisfy it, what is the sum of the integer solutions, or what is the largest value of x that works. The sign line gives you the answer without extra algebra, because the intervals are already laid out.

For -2≤x≤1 or x≥3, the integers in the first interval are -2, -1, 0 and 1, which is four integers. The second interval is unbounded, so the count of integer solutions is infinite unless the question adds a bound. Read the question for the bound before counting, because a missing bound changes the answer from a count to an unbounded set.

Watch the endpoints when you count. If an interval is open at an end, the endpoint itself does not count, even when it is an integer. For -1≤x<4, the integers are -1, 0, 1, 2 and 3, which is five. The open end at 4 removes 4 from the list. Those small endpoint decisions are where marks leak in TITA questions, since there are no options to nudge you.

The same structure helps with functions and graphs, where modulus and domain questions use identical reasoning. Our guide on functions, modulus and domain in CAT is a useful companion once the sign line feels natural.

Build the Habit with Short, Repeated Sessions

You will not remember five steps from reading about them. Do ten inequality questions in one sitting, draw a sign line for every one, and write the endpoint reason beside each. Then repeat on another day with new questions. By the third session, the steps stop feeling like steps.

Keep a note of the questions where you got the interval right but the endpoint wrong, and the ones where you flipped the wrong way across an even power. Those two errors account for most mistakes, and seeing them written down makes you watch for them. If you want the formula-level reminders in one place, keep them in your CAT formula sheets and cheatsheets.

If you are short of time, a CAT study planner can set aside a few short slots across the week for this chapter, so it does not get crowded out by mocks. The chapter is small, and a little steady practice goes a long way.