Modulus Part 01 for CAT Exam 2026 | CAT Quants Preparation | MBA PROPELLER 2026

Modulus (Absolute Value) for CAT Quantitative Aptitude

The lesson introduces modulus as absolute value and develops a distance-based approach to solving linear modulus equations and inequalities. It also covers sums and comparisons of moduli.

1. Meaning and basic properties

The modulus of a number is its distance from zero, so it is always nonnegative:

• (|-9|=9)
• (|5|=5)

For an expression (u):

[ |u|= \begin{cases} u, & u\ge 0,\ -u, & u<0. \end{cases} ]

For expressions containing variables, the sign inside the modulus may depend on the value of the variable. Expressions that are always nonnegative, such as (x^2), do not require a sign check.

Modulus is also symmetric:

[ |a-b|=|b-a|, ]

because distance does not depend on direction.

2. Modulus as distance on a number line

The central method is to interpret a modulus as a distance:

• (|a-b|) is the distance between (a) and (b) on a number line.
• For example, (|4-9|=5), the distance between 4 and 9.
• A sum inside the modulus can often be rewritten as a difference: (|4+9|=|4-(-9)|), the distance between 4 and (-9).

For example, to solve (|x-9|=4), find the points 4 units from 9. The solutions are (x=5) and (x=13).

For (|2x-9|=7), treat the expression as the distance between (2x) and 9:

[ 2x=9+7 \quad \text{or} \quad 2x=9-7. ]

Thus, (2x=16) or (2x=2), giving (x=8) or (x=1).

The same reasoning applies to expressions containing a plus sign. For example, (|2x+9|=9) can be interpreted as the distance between (2x) and (-9).

3. Solving linear modulus equations

For an equation of the form (|ax-b|=c):

1. Identify the quantities whose distance is being measured: (ax) and (b).
2. Move (c) units in both directions from the known point to find the possible values of (ax).
3. Solve each resulting linear equation for (x).
4. Check whether the question asks for the solutions, their sum, or another quantity.

The two possibilities arise because the point can be either to the left or to the right of the reference point.

4. Linear modulus inequalities

Inequalities can be understood using the same distance picture:

• (|x-a|>c): (x) is more than (c) units from (a), so the solution lies outside the boundary points.
• (|x-a|<c): (x) is less than (c) units from (a), so the solution lies between the boundary points.

For example, (|x-9|>11) means (x) is more than 11 units from 9. The boundary points are (-2) and 20, so the solution is

[ x<-2 \quad \text{or} \quad x>20. ]

When the expression inside the modulus has a coefficient, first find the range for that expression and then convert it to a range for (x). For example, in (|3x-7|>11), the boundary values for (3x) are (-4) and 18. Dividing by 3 gives the boundary values (-\frac{4}{3}) and 6, so

[ x<-\frac{4}{3} \quad \text{or} \quad x>6. ]

5. A sum of two moduli: (|x-a|+|x-b|)

This is a commonly tested type of expression:

• Mark the critical points (a) and (b) on a number line.
• Between them, the sum of the distances is constant and equals the distance between the critical points, (|b-a|).
• This constant is the minimum value of the expression.
• Outside the interval, the sum increases as (x) moves farther away.

For example, (|x-4|+|x-9|) has a minimum value of (9-4=5) for every (x) between 4 and 9, including the endpoints. Therefore, the equation

[ |x-4|+|x-9|=5 ]

has infinitely many solutions in that interval.

More generally:

• If the right-hand side equals the minimum, there are infinitely many solutions between the critical points.
• If it is greater than the minimum, there are two solutions, one on either side of the interval.
• If it is less than the minimum, there are no solutions.

If (x) moves (t) units outside the interval, the sum increases by (2t). This can help solve equations quickly without expanding every case.

If the expression involves a substitution such as (t=2^x), solve first in terms of (t), then check which resulting values are possible. Since (2^x) is positive, a negative value of (t) cannot be a solution.

6. Finding minima for sums of several moduli

A sum of absolute-value terms represents the sum of distances to several points:

• Place all critical points on a number line.
• With an odd number of points, the sum is minimized at the middle point.
• With an even number of points, the minimum is attained throughout the interval between the two middle points.
• Substitute a convenient point in the minimizing location to calculate the minimum value.

The lesson illustrates this with sets of points such as (10,30,50,80,90) and (10,30,40,50,80,90). The key is to choose a location that minimizes the total distance to all the points.

7. Comparing two moduli

An inequality comparing two distances can be interpreted as asking which reference point is closer:

• (|x-a|<|x-b|) means (x) is closer to (a) than to (b).
• The boundary between the two regions is the midpoint (\frac{a+b}{2}).
• The solution lies on the side of the midpoint that is closer to the point in the smaller distance.

For example, (|x-40|<|x-10|) means (x) is closer to 40 than to 10. The midpoint is 25, so the solution is (x>25).

When combining two such conditions, find the ranges for each condition and intersect them. If the question asks for integer solutions, count the integers in the resulting interval.

8. Exam-solving guidance

• Draw a number line and identify critical points rather than opening every modulus mechanically.
• Use the distance method as the main tool for single linear moduli. Graph-based and traditional sign-case methods may be useful for more complicated expressions.
• Practise assigned questions, review notes and practice sheets, and analyze mock-test errors to determine whether the issue was interpretation, equation setup, or calculation.
• Focus on steady improvement rather than worrying immediately about a target score.
• Be alert to fake messages or payment requests impersonating course faculty, and use official channels.

Speakers and sources featured

• Primary speaker: An unnamed male instructor teaching CAT quantitative aptitude. The subtitles do not provide his name.
• Students and viewers: The instructor refers to questions and responses from students in the live class, but no separate student speaker is clearly identifiable.
• Other sources mentioned: CAT questions from 2021 and 2022, and Vineet Sir in connection with classes and video solutions.