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Modulus — PYPs

12 solved Modulus previous year questions (PYQs) from past year papers — attempt each and check the answer.

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Q1:ipmat indore 2022QA › ModulusMediumSA · TITA

The area enclosed by

2|x|+3|y| \leq 6

is ____________ sq. units.

Enter your answer to attempt

The Setup: The inequality

2|x| + 3|y| \le 6

forms a geometric shape (a rhombus) bounded by absolute value constraints. We must calculate its total enclosed area. Step 1: Find the intercepts of the boundary lines. Consider the boundary equation in the first quadrant where

x \ge 0

and

y \ge 0

2x + 3y = 6

Find the

x

-intercept (set

y = 0

2x = 6 \implies x = 3

. Find the

y

-intercept (set

x = 0

3y = 6 \implies y = 2

. The vertices of the full rhombus across all four quadrants are

(\pm 3, 0)

and

(0, \pm 2)

. Step 2: Calculate the area of the bounded shape. The area of a rhombus is given by

\frac{1}{2} \times d_1 \times d_2

, where

d_1

and

d_2

are the lengths of the diagonals. The horizontal diagonal connects

(-3, 0)

(3, 0)

, so

d_1 = 6

. The vertical diagonal connects

(0, -2)

(0, 2)

, so

d_2 = 4

\text{Area} = \frac{1}{2} (6)(4) = 12

Final Answer: 12

Q2:ipmat indore 2025QA › ModulusMediumSA · TITA

If a, b, c are three distinct natural numbers, all less than 100, such that

|a-b|+|b-c|=|c-a|

, then the maximum possible value of b is

Enter your answer to attempt

The Setup: This is a classic absolute value distance problem masquerading as pure algebra. We need to decode the 1D geometry hidden in the equation and min-max the constraints like we are optimizing a character build. No cap, this looks intimidating, but it's actually free points once you see the matrix. Step 1: Decoding the Modulus (The Geometry Meta) Let's look at the core equation:

|a-b| + |b-c| = |c-a|

In 1D geometry,

|x-y|

represents the absolute distance between point

x

and point

y

on a number line. The equation is literally saying: "The distance from

a

b

, plus the distance from

b

c

, equals the total distance from

a

c

." For this to hold true, point

b

must lie exactly between

a

and

c

on the number line.

Condition	Mathematical Meaning	Valid Range for $b$
Case 1	$c$ is the maximum	$a < b < c$
Case 2	$a$ is the maximum	$c < b < a$

In both scenarios,

b

is sandwiched in the middle. It can *never* be the largest of the three numbers. Step 2: Pushing to the Limit (Max Stats) We are given strict constraints for our variables: * They are natural numbers (positive integers: 1, 2, 3...). * They are distinct (no duplicates allowed). * They are all less than 100. To maximize

b

, we need to drag the entire sequence as high up the number line as possible. The absolute ceiling is 100, but the numbers must be strictly *less* than 100. So, our absolute maximum possible integer in this domain is 99. Let's assign 99 to our top-end variable (let's say

a = 99

). Since

b

must be strictly less than the top-end variable (

b < a

), the largest possible integer we can assign to

b

without hitting 99 is 98. Then

c

just has to be any natural number less than 98 (e.g.,

c = 97

). Step 3: The Audit (Double Check Protocol) Running the numbers back to ensure we aren't throwing. Let

a = 99

b = 98

, and

c = 97

. (All distinct natural numbers

< 100

. Checked.) Plug them into the original equation:

|99 - 98| + |98 - 97| = |97 - 99|

|1| + |1| = |-2|

1 + 1 = 2

2 = 2

The logic is flawlessly validated. The absolute ceiling for

b

is locked at 98. Final Answer: 98

Q3:ipmat indore 2020QA › ModulusMediumSA · TITA

The minimum value of

f(x)=|3-x|+|2+x|+|5-x|

is equal to __________.

Enter your answer to attempt

The Setup: We are tasked with finding the global minimum of the function

f(x)=|3-x|+|2+x|+|5-x|

. In absolute value geometry, this represents the sum of distances from a coordinate

x

to three specific points on a number line, and the minimum lands at the median of those points. Rather than just quoting that rule, we can lock the answer down with a two-line proof. Step 1: Standardize the terms. First, use the property

|a-b|=|b-a|

to rewrite the function so

x

is always first. This prevents sign errors.

f(x)=|x-3|+|x-(-2)|+|x-5|

The roots (the points on the number line) are exactly

3

-2

and

5

. In ascending order:

-2,\ 3,\ 5

- so the median is

3

. Step 2: Bound the outer pair. For the two extreme points, the triangle inequality gives a floor that no

x

can beat:

|x+2|+|x-5|\geq |(x+2)-(x-5)|=7

with equality for every

x

in the interval

[-2,5]

. Step 3: Bound the middle term. Obviously

|x-3|\geq 0

, with equality only at

x=3

. Adding the two bounds gives

f(x)\geq 7+0=7

for all

x

. Step 4: Show the floor is actually reached. The two equality conditions must hold at once, and they do:

x=3

lies inside

[-2,5]

. Substituting:

f(3)=|3-3|+|3-(-2)|+|3-5|=0+5+2=7

The bound is achieved, so

7

is the true minimum and not merely a lower estimate. Final Answer: 7

Q4:ipmat indore 2022QA › ModulusMediumSA · TITA

Given that

f(x)=|x|+2|x-1|+|x-2|+|x-4|+|x-6|+2|x-10|, x \in(-\infty, \infty)

the minimum value of

f(x)

is _________.

Enter your answer to attempt

The Setup: We must find the minimum value of a sum of absolute value functions

f(x)=|x|+2|x-1|+|x-2|+|x-4|+|x-6|+2|x-10|

. Step 1: Translate the function into a median problem. The minimum value of a sum of absolute deviations

\sum c_i |x - a_i|

occurs at the median of the data points

a_i

, weighted by their frequencies

c_i

. List out the points based on their respective coefficients: *

0

(Frequency 1) *

1

(Frequency 2) *

2

(Frequency 1) *

4

(Frequency 1) *

6

(Frequency 1) *

10

(Frequency 2) The complete ordered data set contains

8

points:

\{0, 1, 1, 2, 4, 6, 10, 10\}

. Step 2: Identify the median of the set. Since there is an even number of data points (

n=8

), the minimum is achieved anywhere along the interval between the 4th and 5th values. The 4th point is

2

and the 5th point is

4

. Any

x \in [2, 4]

minimizes

f(x)

. Step 3: Calculate the minimum value

f(x)

. Substitute any value from the valid interval (e.g.,

x = 2

) into

f(x)

f(2) = |2-0| + 2|2-1| + |2-2| + |2-4| + |2-6| + 2|2-10|

f(2) = 2 + 2(1) + 0 + |-2| + |-4| + 2|-8|

f(2) = 2 + 2 + 0 + 2 + 4 + 16 = 26

Final Answer: 26

Q5:ipmat indore 2024QA › ModulusMediumSA · TITA

The number of pairs

(x, y)

of integers satisfying the inequality

|x - 5| + |y - 5| \leq 6

is:

Enter your answer to attempt

The Setup: We must find the number of integer coordinate pairs

(x,y)

that satisfy the absolute value inequality

|x-5|+|y-5| \le 6

. Step 1: Translate the bounded region to center it at the origin. Let

u = x - 5

and

v = y - 5

. Since

x

and

y

are elements of

\mathbb{Z}

u

and

v

must also be elements of

\mathbb{Z}

. Substitute into the inequality:

|u| + |v| \le 6

This bounded region forms a solid square rotated

45^{\circ}

on the Cartesian plane. Step 2: Apply the lattice point summation formula for Manhattan boundaries. The exact number of integer coordinate pairs

(u, v)

satisfying

|u| + |v| \le k

for an integer

k \ge 0

is governed by the discrete sequence formula

2k(k+1) + 1

. Step 3: Evaluate the formula for the target boundary distance

k = 6

\text{Total Pairs} = 2(6)(6+1) + 1

\text{Total Pairs} = 2(6)(7) + 1 = 84 + 1 = 85

Because the translation mapping

(x,y) \rightarrow (u,v)

is a bijective

1:1

map, the count remains identical for the original uncentered inequality. Final Answer: 85

Q6:ipmat indore 2023QA › ModulusMediumSA · TITA

The length of the line segment joining the two intersection points of the curves

y = 4970 - |x|

and

y = x ^ 2

Enter your answer to attempt

The Setup: We must find the geometric length of the line segment connecting the two intersection points of the absolute value function

y = 4970 - |x|

and the parabola

y = x^2

. Step 1: Set the equations equal to find intersection points.

x^2 = 4970 - |x|

Rearrange to form a quadratic in terms of

|x|

, recognizing that

x^2 = |x|^2

|x|^2 + |x| - 4970 = 0

Step 2: Factor the quadratic equation. We need two integers that multiply to

-4970

and add to

1

. Note that

4970 = 10 \times 497 = 10 \times 7 \times 71 = 70 \times 71

(|x| + 71)(|x| - 70) = 0

Step 3: Solve for

x

and locate the coordinates. Since absolute value

|x|

must be non-negative,

|x| = -71

is rejected.

|x| = 70 \implies x = 70 \text{ or } x = -70

Because the equations are perfectly symmetric across the y-axis, the segment connecting the two points is perfectly horizontal. Step 4: Calculate the distance. The length of a horizontal line segment is simply the absolute difference between the

x

-coordinates.

\text{Length} = |70 - (-70)| = 140

Final Answer: 140

Q7:ipmat indore 2024QA › ModulusEasyMCQ · MCQ

|x+1| + (y+2)^2 = 0

and

ax - 3ay = 1

, then the value of

a

A $\frac{1}{5}$
B $\frac{1}{2}$
C $\frac{1}{7}$
D2

Pick an option to attempt

The Setup: We are given the equations

|x+1|+(y+2)^{2}=0

and

ax-3ay=1

. We must determine the value of

a

. Step 1: Analyze the non-negative components of the first equation. The absolute value function

|x+1|

and the squared term

(y+2)^2

must both be strictly greater than or equal to zero for all real numbers. For their sum to exactly equal zero, each individual term must independently be zero:

|x+1| = 0 \implies x = -1

(y+2)^2 = 0 \implies y = -2

Step 2: Substitute the coordinates into the second linear equation. Substitute

x = -1

and

y = -2

into

ax - 3ay = 1

a(-1) - 3a(-2) = 1

-a + 6a = 1

5a = 1 \implies a = \frac{1}{5}

Final Answer:

\frac{1}{5}

Q8:ipmat indore 2024QA › ModulusEasyMCQ · MCQ

The number of real solutions of the equation

x^2 - 10|x| - 56 = 0

Pick an option to attempt

The Setup: We need to find the number of real solutions for the absolute value quadratic equation

x^{2}-10|x|-56=0

. Step 1: Use substitution to simplify the quadratic. Recognize that for any real number,

x^2 = |x|^2

. Rewrite the equation in terms of

|x|

|x|^2 - 10|x| - 56 = 0

Step 2: Factor the quadratic equation. We look for two numbers that multiply to

-56

and add to

-10

. These are

-14

and

+4

(|x| - 14)(|x| + 4) = 0

Step 3: Solve for

|x|

and determine real values for

x

. This yields two potential cases for

|x|

: Case 1:

|x| - 14 = 0 \implies |x| = 14 \implies x = 14 \text{ or } x = -14

Case 2:

|x| + 4 = 0 \implies |x| = -4

Because the absolute value of a real number cannot be negative, Case 2 produces no real solutions. This leaves exactly

2

valid real solutions from Case 1. Final Answer: 2

Q9:ipmat indore 2022QA › ModulusMediumMCQ · MCQ

The sum of the squares of all the roots of the equation

x^{2}+|x+4|+|x-4|-35=0

A74
B175
C50
D148

Pick an option to attempt

The Setup: The equation

x^{2}+|x+4|+|x-4|-35=0

relies on absolute value boundaries. We must partition the domain to find all valid real roots, then calculate the sum of their squares. Step 1: Notice the inherent symmetry. Since

|x+4| + |x-4|

is an even function, if

x

is a valid root,

-x

is also a root. Step 2: Evaluate Case 1 (

x \ge 4

). Both absolute values evaluate to their positive counterparts:

x^2 + (x + 4) + (x - 4) - 35 = 0

x^2 + 2x - 35 = 0

(x + 7)(x - 5) = 0

The potential roots are

-7

and

5

. Because our domain is strictly

x \ge 4

x = 5

is a valid root. Step 3: Evaluate Case 2 (

-4 < x < 4

). The first term remains positive, but the second evaluates to its negative counterpart:

x^2 + (x + 4) + (4 - x) - 35 = 0

x^2 + 8 - 35 = 0

x^2 - 27 = 0 \implies x = \pm\sqrt{27} \approx \pm 5.196

Because

\pm 5.196

falls strictly outside the bounded domain of

(-4, 4)

, no valid roots exist in this region. Step 4: Evaluate Case 3 (

x \le -4

). Both absolute values flip sign:

x^2 - (x + 4) - (x - 4) - 35 = 0

x^2 - 2x - 35 = 0

(x - 7)(x + 5) = 0

The potential roots are

7

and

-5

. Because our domain is strictly

x \le -4

x = -5

is a valid root. This agrees with the symmetry noted in Step 1. The two valid real roots are

5

and

-5

. Step 5: Calculate the sum of squares.

\text{Sum} = (5)^2 + (-5)^2 = 25 + 25 = 50

Final Answer: 50

Q10:ipmat indore 2019QA › ModulusEasyMCQ · MCQ

The area enclosed by the curve

2|x| + 3|y| = 6

A12 square units
B3 square units
C4 square units
D24 square units

Pick an option to attempt

The Setup: Absolute values mirror the graph across both axes, so this curve is a quadrilateral centred at the origin. Find the intercepts, confirm the shape, and the area follows from one formula. Step 1: Find the intercepts. Setting

y=0

2|x|=6 \implies |x|=3 \implies x=\pm 3

Setting

x=0

3|y|=6 \implies |y|=2 \implies y=\pm 2

Step 2: Identify the shape. In each quadrant the equation is linear - for instance

2x+3y=6

when both are positive - so the curve is four straight segments joining

(3,0)

(0,2)

(-3,0)

and

(0,-2)

. Each side runs between an

x

-intercept and a

y

-intercept, so every side has length

\sqrt{3^2+2^2}=\sqrt{13}

: all four equal, making it a genuine rhombus with perpendicular diagonals along the axes. Step 3: Measure the diagonals.

d_1=3-(-3)=6, \qquad d_2=2-(-2)=4

Step 4: Compute the area. For a rhombus (indeed any quadrilateral with perpendicular diagonals), the area is half the product of the diagonals:

\text{Area}=\frac{1}{2}d_1d_2=\frac{1}{2}(6)(4)=12

Equivalently, the first-quadrant piece is a right triangle of legs 3 and 2, area

3

, and symmetry gives four of them:

4\times 3=12

. Two routes, same number. Final Answer: 12 square units

Q11:ipmat indore 2019QA › ModulusEasyMCQ · MCQ

For

a > b > c > 0

, the minimum value of the function

f(x) = |x - a| + |x - b| + |x - c|

A $2a - b - c$
B $a + b - 2c$
C $a + b + c$
D $a - c$

Pick an option to attempt

The Setup: The sum of distances from a moving point

x

to three fixed points

c<b<a

on a number line. The median minimises it - but rather than quote that rule, we can prove it in two lines and get the exact minimum for free. Step 1: Bound the outer pair. By the triangle inequality, for any real

x

|x-a|+|x-c|\geq |(x-c)-(x-a)|=a-c

with equality precisely when

x

lies between

c

and

a

, i.e.

x\in[c,a]

. Step 2: Bound the middle term. Trivially

|x-b|\geq 0

, with equality only at

x=b

. Step 3: Add the bounds. For every real

x

f(x)=|x-a|+|x-b|+|x-c|\geq (a-c)+0=a-c

Step 4: Show the floor is reached. The two equality conditions must hold simultaneously, and they can:

x=b

satisfies the second exactly, and since

c<b<a

it also lies inside

[c,a]

, satisfying the first. Substituting

x=b

, and using

a>b

|b-a|=a-b

, and

b>c

|b-c|=b-c

f(b)=(a-b)+0+(b-c)=a-c

The

b

terms cancel, which is why the answer depends only on the outermost two points - the middle value

b

affects where the minimum occurs but not what it is. The bound is attained, so it is the true minimum rather than merely a lower estimate. Final Answer:

a - c

Q12:ipmat indore 2026QA › ModulusMediumSA · TITA

n

is an integer such that

\frac{|n+6|-|n-3|}{\sqrt{100-n^3}} \geq 0

, then the number of possible values of

n

is ___

Enter your answer to attempt

The Setup: We've got a rational inequality with a strict vibe check on the domain. The denominator has a square root, meaning it must be strictly positive (no imaginary numbers, and definitely no dividing by zero). Once we establish the domain boundaries, we let the numerator 1v1 itself to find the overlapping integer set. Math and logic double-verified. Step 1: Check the denominator's HP (Domain Constraints). For the expression to even exist in the real number universe, the stuff inside the square root must be strictly greater than 0.

100 - n^3 > 0

n^3 < 100

Since

n

is an integer, let's find the max level cap for

n

4^3 = 64

(Passes)

5^3 = 125

(Fails) So, our upper bound is locked in:

n \leq 4

. Step 2: Isolate the numerator. Because a square root (when real and non-zero) always outputs a positive value, the denominator is essentially just a positive buff. It doesn't affect the

\geq 0

sign of the whole fraction. The numerator alone dictates if the whole expression is positive or zero.

|n+6| - |n-3| \geq 0

|n+6| \geq |n-3|

Step 3: Square up. Since both sides of the inequality are absolute values (meaning they are inherently non-negative), squaring both sides is perfectly safe and won't illegally flip the inequality sign. It's the cleanest way to strip the absolute value armor.

(n+6)^2 \geq (n-3)^2

n^2 + 12n + 36 \geq n^2 - 6n + 9

Cancel out the

n^2

terms and group the rest:

18n \geq -27

n \geq -\frac{3}{2}

n \geq -1.5

Step 4: Find the overlapping integer zone. Since

n

must be an integer, the lowest value it can take based on Step 3 is

-1

. Combining this with our domain cap from Step 1 (

n \leq 4

), we get our final bounded region:

-1 \leq n \leq 4

The valid integer roster is

\{-1, 0, 1, 2, 3, 4\}

. Step 5: Count the squad. Counting the integers in that set gives us 6 unique values. Final Answer: 6

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