Modulus questions in CAT Quant can seem challenging because the value of an expression inside the modulus depends on whether it is positive or negative. Removing the modulus directly without checking this condition can result in incorrect solutions. Questions involving equations, inequalities, parameters, and multiple modulus terms can therefore require careful case analysis.
The key to solving these questions is to understand modulus as absolute value and interpret it as distance on a number line. Once this idea is clear, expressions such as (|x-9|), equations such as (|x-9|=4), and inequalities such as (|x-a|>b) become easier to solve. The same approach can also be applied to two-modulus expressions and maximum or minimum problems.
The modulus, or absolute value, of a number represents its distance from zero on the number line. Since distance cannot be negative, the value of a modulus is always non-negative.
For any real number x,
\[ |x| = \begin{cases} x, & x\geq0\\ -x, & x<0 \end{cases} \]
| Value of x | Modulus |x| | What Happens? |
| x>0 | x | The positive value remains unchanged. |
| x=0 | 0 | Zero remains zero. |
| x<0 | -x | The negative sign is removed. |
For example:
|7|=7
because 7 is positive.
Similarly,
|-7|=-(-7)=7
because -7 is negative, so its sign changes.
Therefore, for every real number x,
\[ \boxed{|x|\geq0} \]
|
Inside the Modulus |
How to Remove the Modulus |
Example |
|
Positive |
Keep the expression unchanged |
( |
|
Zero |
Keep the value as zero |
( |
|
Negative |
Change its sign |
( |
This makes the basic rule easy to remember: a modulus always gives a non-negative value.
The modulus cannot always be removed directly; first check whether the expression inside it is positive, negative, zero, or always non-negative.
|
Situation |
Condition / Example |
Result |
|
Expression can be positive or negative |
( |
x-1 |
|
Expression is positive |
x-1, when x1 |
( |
|
Expression is negative |
x-1, when x<1 |
( |
|
Expression is zero |
x-1=0 |
( |
|
Expression is always non-negative |
x20 for every real x |
(\boxed{ |
|
Even power |
x2n0 |
(\boxed{ |
|
Two quantities are reversed |
( |
a-b |
The most useful interpretation of modulus in CAT Quant is distance. The expression |a-b| represents the distance between a and b on the number line, regardless of their order.
|
Concept |
Interpretation / Result |
|
Distance between two numbers |
7-3 |
|
Reversing the order |
3-7 |
|
Distance property |
a-b |
|
Distance of x from a fixed number |
x-9 |
|
x=12 |
12-9 |
|
x=6 |
6-9 |
|
x=9 |
9-9 |
Key idea: Whenever you see \(\boxed{|x-a|}\), think “distance of x from a”.
For broader CAT Quant practice, the CAT Quantitative Aptitude section covers other important concepts that require similar calculation and logical interpretation.
A linear modulus equation can be understood as a distance equation: if the distance of x from a number is fixed, x can lie on either side of that number.
For example,
|x-9|=4
means that x is 4 units away from 9. Therefore,
x=9-4=5
or
x=9+4=13
Hence,
\[ \boxed{x=5,13} \]
In general, for
|x-a|=b,b0
the two possible solutions are
\[ \boxed{x=a-b\quad\text{or}\quad x=a+b} \]
|
Equation |
Distance Interpretation |
Solutions |
|
( |
x-9 |
=4) |
|
( |
x-3 |
=7) |
|
( |
x+2 |
=5) |
Key idea: For \(\boxed{|x-a|=b}\), simply find the two numbers that are b units away from a.
The same distance approach works when the expression inside the modulus contains a coefficient.
Consider:
|2x-9|=7
This means:
2x-9=7
or
2x-9=-7
Solving both cases:
2x=16x=8
and
2x=2x=1
Therefore,
\[ \boxed{x=1,8} \]
A useful CAT approach is to first find the two possible values of the entire expression inside the modulus, and then solve for x.
An expression involving addition can often be rewritten into subtraction form before applying the distance interpretation.
For example,
|x+5|
can be written as:
|x-(-5)|
Therefore, it represents the distance between x and -5.
Similarly,
|2x+7|=|2x-(-7)|
This makes the critical point immediately visible:
2x=-7
or
x=-72
The same idea works whenever the expression can be rewritten as:
|x-a|
or as a transformed version of it.
Modulus inequalities can also be understood through distance.
For:
|x-a|>b
the distance of x from a must be greater than b. Therefore, x lies outside the two critical points:
a-b,a+b
Hence,
\[ \boxed{|x-a|>b\Rightarrow x<a-b\ \text{or}\ x>a+b} \]
For:
|x-a|<b
the distance from a must be less than b. Therefore, x lies between the two critical points:
\[ \boxed{a-b<x<a+b} \]
|
Modulus inequality |
Meaning |
Solution |
|
( |
x-a |
<b) |
|
( |
x-a |
\leq b) |
|
( |
x-a |
>b) |
|
( |
x-a |
\geq b) |
Practising different Quant topics alongside modulus can help improve speed and accuracy. The MBA Quantitative Aptitude section includes additional topics for quantitative practice.
Consider:
|3x-7|>11
First consider the expression inside the modulus:
3x-7>11
or
3x-7<-11
Solving the two inequalities:
3x>18x>6
and
3x<-4x<-43
Therefore,
\[ \boxed{x<-\frac43\quad\text{or}\quad x>6} \]
The same result can be understood by first identifying the range of 3x-7:
3x-7<-11or3x-7>11
and then converting that range into the corresponding values of x.
CAT questions may ask for the sum or product of the possible values instead of asking for the values individually.
Consider:
|x-7|=5
The two solutions are:
x=7-5=2
and
x=7+5=12
Therefore,
\[ x_1+x_2=2+12=\boxed{14} \]
and
\[ x_1x_2=2(12)=\boxed{24} \]
For the general equation
|x-a|=b,
the two solutions are:
x1=a-b,x2=a+b
Therefore,
\[ \boxed{x_1+x_2=2a} \]
and
\[ \boxed{x_1x_2=a^2-b^2} \]
These relationships can save time when the question asks only for the sum or product.
For additional practice with calculation-based Quant concepts, the Top Videos on Average Quantitative Aptitude for CAT Exam 2026 provide another useful learning resource.
When an expression contains two modulus terms, such as |x-a|+|x-b|, interpret each term as a distance on the number line and analyse the intervals formed by a and b.
A particularly important CAT type involves two modulus terms:
|x-a|+|x-b|
The critical points are:
x=a,x=b
These two points divide the number line into three regions:
x<a,axb,x>b
assuming a<b.
The expression behaves differently in each region, but the distance interpretation makes the central region especially important.
The expression |x-a|+|x-b| represents the total distance from x to a and from x to b, so its value is smallest when x lies between a and b.
Suppose:
a<b
The expression
|x-a|+|x-b|
represents the total distance from x to a and from x to b.
When x lies between a and b:
|x-a|=x-a
and
|x-b|=b-x
Therefore,
|x-a|+|x-b|
=(x-a)+(b-x)
=b-a
Thus, throughout the interval
\[ \boxed{a\leq x\leq b}, \]
the expression remains constant.
Its value is:
\[ \boxed{b-a=|a-b|} \]
This is the minimum value of the expression.
|
Region |
Expression |
Value/Behaviour |
|
x<a |
(a-x)+(b-x) |
Increases as x moves left |
|
axb |
(x-a)+(b-x) |
Constant at b-a |
|
x>b |
(x-a)+(x-b) |
Increases as x moves right |
The important CAT observation is: \[ \boxed{\min\left(|x-a|+|x-b|\right)=|a-b|} \] and this minimum is achieved for every x between a and b. Therefore, there are infinitely many real values of x for which the minimum is attained.
Modulus questions often become lengthy because of unnecessary casework. Avoid these common errors:
Removing the modulus without checking the sign: |x-1|x-1 for every x.
Forgetting the negative case: If the inside quantity is negative, its sign must be reversed.
Ignoring the distance interpretation: Expressions such as |x-a| are often faster to solve geometrically.
Missing the critical points: In |x-a|+|x-b|, the important points are a and b.
Confusing minimum value with minimum point: The minimum of |x-a|+|x-b| occurs for every x between a and b, not at only one point.
Ignoring the required type of solution: Check whether the question asks for real, integer or positive values.
Solving the modulus before simplifying the expression: If the inside contains a repeated or transformed expression, temporarily substitute it with a variable.
A wider CAT preparation plan can help bring together Quant practice with the other sections of the examination.
Modulus questions in CAT Quant become easier when absolute value is understood as distance on the number line. Start by checking the sign of the expression inside the modulus, then use the appropriate rule for equations, inequalities, or multiple modulus terms. For expressions such as |x-a|, think of the distance of x from a, while for two-modulus expressions, identify the critical points and work through the relevant intervals.