IPMAT Indore 2025 QA PYQs

Q1:ipmat indore 2025QA › Time & WorkMediumSA · TITA

Monica, who is 18 years old, is one-third the age of her father. The age at which she will be half the age of her father is

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The Setup: We are looking at a classic linear age progression problem. Since this is a TITA (Type In The Answer) question, there are no multiple-choice crutches to save us—we have to raw-dog the math and lock in the exact value. We need to establish the current timeline and project it into the future where the age ratio shifts from one-third to one-half.

Step 1: Establish the Current Era
Monica is currently 18 years old. Since she is explicitly stated to be one-third her dad's age, we multiply her age by 3 to lock in his current stats.
$18 \times 3 = 54$
Her dad is currently 54 years old.

Step 2: Set Up the Future Timeline
Let $x$ be the number of years it takes for this timeline shift to happen. Fast forward $x$ years: Monica will be $18+x$ years old, and her dad will level up to $54+x$ years old. The problem states that at this point, her age will be exactly half of his.
$18+x = \frac{1}{2}(54+x)$

Step 3: Solve the Equation (No Cap)
Multiply both sides by 2 to clear the fraction and avoid messy calculations.
$2(18+x) = 54+x$
$36+2x = 54+x$
Now, isolate $x$ by subtracting $x$ from both sides, and moving the 36 over.
$x = 18$
It will take exactly 18 years for this ratio to hit.

Step 4: Calculate the Final Age
Don't get baited by the $x$ value. The question asks for her age when this happens, not how many years it takes. Add the 18 years to her current age of 18.
$18+18 = 36$
Monica will be 36 years old when she is half her father's age.

Final Answer: 36

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

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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 $a$ to $b$ , plus the distance from $b$ to $c$ , equals the total distance from $a$ to $c$ ."
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 2025QA › Progression & SeriesMediumSA · TITA

If the sum of the first 21 terms of the sequence

\ln(\frac{a}{b}), \ln(\frac{a}{b\sqrt{b}}), \ln(\frac{a}{b^2}) \dots

is

\ln(\frac{a^m}{b^n})

, then the value of

m+n

is

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The Setup:
At first glance, this looks like a cursed sequence of logarithms, but if we peel back the formatting, it is literally just an Arithmetic Progression (AP) in disguise. We just need to unpack the logs, find the common difference to establish our AP meta, calculate the sum of the first 21 terms, and then repackage it to find our $m$ and $n$ values.

Step 1: Unpacking the Logs (The Vibe Check)
Let's analyze the first three terms to figure out exactly what kind of progression we are dealing with. Using the logarithm quotient rule ( $\ln(x/y) = \ln(x) - \ln(y)$ ) and exponent rules, we can break them down:
* Term 1 ( $T_1$ ): $\ln(\frac{a}{b}) = \ln(a) - \ln(b)$
* Term 2 ( $T_2$ ): $\ln(\frac{a}{b\sqrt{b}}) = \ln(\frac{a}{b^{1.5}}) = \ln(a) - 1.5\ln(b)$
* Term 3 ( $T_3$ ): $\ln(\frac{a}{b^2}) = \ln(a) - 2\ln(b)$

Step 2: **Finding the Common Difference ( $D$ )**
To confirm this is a standard AP and to find our step size, we subtract $T_1$ from $T_2$ :
$D = T_2 - T_1$
$D = (\ln(a) - 1.5\ln(b)) - (\ln(a) - \ln(b))$
$D = -0.5\ln(b)$
The sequence is linearly dropping by $0.5\ln(b)$ every single term. This confirms we are working with an AP where the first term $A = \ln(a) - \ln(b)$ and the common difference $D = -0.5\ln(b)$ .

Step 3: The Carry (Summing 21 Terms)
Now we deploy the standard AP sum formula for 21 terms:
$S_n = \frac{n}{2} [2A + (n-1)D]$
Plug in our $n = 21$ , $A$ , and $D$ values:
$S_{21} = \frac{21}{2} [2(\ln(a) - \ln(b)) + (21-1)(-0.5\ln(b))]$
$S_{21} = \frac{21}{2} [2\ln(a) - 2\ln(b) + 20(-0.5\ln(b))]$
$S_{21} = \frac{21}{2} [2\ln(a) - 2\ln(b) - 10\ln(b)]$
Combine the $\ln(b)$ terms to clean up the bracket:
$S_{21} = \frac{21}{2} [2\ln(a) - 12\ln(b)]$
Factor out the 2 inside the bracket to cancel the denominator and nerf the equation:
$S_{21} = 21 [\ln(a) - 6\ln(b)]$

Step 4: Repackaging the File
The question demands the final answer in the format $\ln(\frac{a^m}{b^n})$ . We need to compress our result back into a single log expression using power rules ( $k\ln(x) = \ln(x^k)$ ).
First, distribute the 21:
$S_{21} = 21\ln(a) - 126\ln(b)$
Now apply the power rule to push the coefficients back into the logs:
$S_{21} = \ln(a^{21}) - \ln(b^{126})$
Finally, use the quotient rule to merge them into one fraction:
$S_{21} = \ln\left(\frac{a^{21}}{b^{126}}\right)$

Step 5: The Audit (Double Check Protocol)
Let's match our final expression against the target form $\ln(\frac{a^m}{b^n})$ .
By direct comparison:
$m = 21$
$n = 126$
The prompt asks for the value of $m + n$ .
$m + n = 21 + 126 = 147$
Math is verified, rules applied correctly, logic is absolutely flawless.

Final Answer: 147

Q4:ipmat indore 2025QA › Set TheoryMediumSA · TITA

English exam and Math exam were conducted separately for a class of 120 students. The number of students who did not appear for the English exam is twice the number of students who did not appear for the Math exam. The number of students who passed the Math exam is twice the number of students who appeared but failed the English exam. If the number of students who passed the English exam is twice the number of students who appeared but failed the Math exam, then the number of students who appeared but failed the English exam is

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The Setup:
We are tackling a system of linear equations disguised as a set theory and word problem. It’s all about mapping out the states (Passed, Failed, Did Not Appear) for two separate exams. We just need to define our variables, translate the English into algebra, and solve the system without getting lost in the sauce.

Step 1: Defining the Variables (The Roster)
Let's define the base variables using the Math exam stats to keep it clean.
Let $x$ = Students who did not appear for Math.
Let $z$ = Students who appeared but failed Math.
Let $y$ = Students who appeared but failed English (our target variable).

Now we translate the constraints for the English exam and the rest of the Math exam based on the prompt:
* "Did not appear for English is twice did not appear for Math" $\rightarrow 2x$
* "Passed Math is twice appeared but failed English" $\rightarrow 2y$
* "Passed English is twice appeared but failed Math" $\rightarrow 2z$

Step 2: Building the State Equations
For both exams, the total class size is strictly locked at 120.
Total = (Passed) + (Failed) + (Did Not Appear)

Math Exam Equation:
$2y (\text{Passed}) + z (\text{Failed}) + x (\text{Did Not Appear}) = 120$
$x + 2y + z = 120$

English Exam Equation:
$2z (\text{Passed}) + y (\text{Failed}) + 2x (\text{Did Not Appear}) = 120$
$2x + y + 2z = 120$

Step 3: Solving the System (Speedrun Strat)
We have a 2-equation system and need to isolate $y$ . Let's use the elimination method.
Take the Math equation and multiply the entire thing by 2 to align the $x$ and $z$ coefficients:
$2(x + 2y + z) = 2(120)$
$2x + 4y + 2z = 240$

Now, subtract the English equation from this scaled-up Math equation:
$(2x + 4y + 2z) - (2x + y + 2z) = 240 - 120$
The $x$ and $z$ terms completely cancel each other out, leaving us with a clean isolation:
$3y = 120$
$y = 40$

Step 4: The Audit (Double Check Protocol)
Let's run the numbers back to guarantee the logic holds.
If $y = 40$ , we plug it into the Math equation:
$x + 2(40) + z = 120 \rightarrow x + z = 40$
If we plug it into the English equation:
$2x + 40 + 2z = 120 \rightarrow 2x + 2z = 80 \rightarrow x + z = 40$
The system is completely consistent. Because $x + z = 40$ in both cases, the exact distribution of $x$ and $z$ doesn't even matter (e.g., it could be $x=20, z=20$ , or $x=10, z=30$ ). The value for $y$ is strictly locked at 40 no matter what. No cap, the solution is mathematically bulletproof.

Final Answer: 40

Q5:ipmat indore 2025QA › Matrices & DeterminantsMediumSA · TITA

If

A=\begin{bmatrix} 2 & n \\ 4 & 1 \end{bmatrix}

such that

A^3=27\begin{bmatrix} 4 & q \\ p & r \end{bmatrix}

, then

p+q+r

equals

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The Setup: We are literally entering the Matrix for this one. This is a linear algebra scaling problem where we need to find the cube of a $2 \times 2$ matrix, equate it to a scaled target matrix, and extract the hidden variables. I have fully audited the matrix multiplication below to ensure zero dropped frames.

Step 1: Squaring the Matrix (Level 1)
First, we need to find $A^2$ by multiplying matrix $A$ by itself. We use the standard row-by-column dot product method.
$A^2 = \begin{bmatrix} 2 & n \\ 4 & 1 \end{bmatrix} \begin{bmatrix} 2 & n \\ 4 & 1 \end{bmatrix}$
$A^2 = \begin{bmatrix} (2)(2)+(n)(4) & (2)(n)+(n)(1) \\ (4)(2)+(1)(4) & (4)(n)+(1)(1) \end{bmatrix}$
$A^2 = \begin{bmatrix} 4+4n & 3n \\ 12 & 4n+1 \end{bmatrix}$

Step 2: Cubing the Matrix (Boss Phase)
Now we multiply our $A^2$ result by $A$ to reach the final form, $A^3$ .
$A^3 = \begin{bmatrix} 4+4n & 3n \\ 12 & 4n+1 \end{bmatrix} \begin{bmatrix} 2 & n \\ 4 & 1 \end{bmatrix}$
Let's calculate each element one by one to avoid throwing:
* Top-Left: $(4+4n)(2) + (3n)(4) = 8 + 8n + 12n = 8+20n$
* Top-Right: $(4+4n)(n) + (3n)(1) = 4n^2 + 4n + 3n = 4n^2+7n$
* Bottom-Left: $(12)(2) + (4n+1)(4) = 24 + 16n + 4 = 16n+28$
* Bottom-Right: $(12)(n) + (4n+1)(1) = 12n + 4n + 1 = 16n+1$

So, our fully loaded $A^3$ matrix is:
$A^3 = \begin{bmatrix} 8+20n & 4n^2+7n \\ 16n+28 & 16n+1 \end{bmatrix}$

Step 3: Decoding the Target Matrix
The problem gives us the target state for $A^3$ . Let's distribute that scalar multiplier (27) into the matrix so we can match it element for element.
$A^3 = 27\begin{bmatrix} 4 & q \\ p & r \end{bmatrix} = \begin{bmatrix} 108 & 27q \\ 27p & 27r \end{bmatrix}$

Step 4: Extracting the Lore (Solving for Variables)
We now set the elements of our calculated $A^3$ equal to the target matrix. The top-left element is the key to unlocking the entire board because it only relies on $n$ .
$8+20n = 108$
$20n = 100$
$n = 5$

Now substitute $n=5$ into the other elements to find $p$ , $q$ , and $r$ :
* **Top-Right ( $q$ ):** $4(5)^2 + 7(5) = 100 + 35 = 135$
$27q = 135 \rightarrow q = 5$
* **Bottom-Left ( $p$ ):** $16(5) + 28 = 80 + 28 = 108$
$27p = 108 \rightarrow p = 4$
* **Bottom-Right ( $r$ ):** $16(5) + 1 = 80 + 1 = 81$
$27r = 81 \rightarrow r = 3$

Step 5: The Final Calculation
The question asks for the sum of $p+q+r$ . Just add up our freshly extracted stats.
$4 + 5 + 3 = 12$

Step 6: The Audit (Double Check Protocol)
Let's run a determinant check to confirm the matrix integrity.
If $n=5$ , matrix $A = \begin{bmatrix} 2 & 5 \\ 4 & 1 \end{bmatrix}$ .
The determinant of $A$ is $(2)(1) - (5)(4) = -18$ .
Using determinant properties, $\det(A^3) = (\det(A))^3 = (-18)^3 = -5832$ .
Now let's check the determinant of our target matrix $27\begin{bmatrix} 4 & 5 \\ 4 & 3 \end{bmatrix}$ :
$\det = 27^2 \times ((4)(3) - (5)(4)) = 729 \times (12 - 20) = 729 \times (-8) = -5832$ .
The determinants match perfectly. The math is completely flawless.

Final Answer: 12

Q6:ipmat indore 2025QA › LogarithmsMediumSA · TITA

If

\log_{3}(x^{2}-1)

,

\log_{3}(2x^{2}+1)

and

\log_{3}(6x^{2}+3)

are the first three terms of an arithmetic progression, then the sum of the next three terms of the progression is

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The Setup:
This question is trying to gatekeep arithmetic progressions behind a wall of logarithms. But if we use basic log properties, we can crack the code and reveal a super simple sequence underneath. I've audited the math to ensure it is fully locked in.

Step 1: The Arithmetic Progression Meta
If three terms $A$ , $B$ , and $C$ are in an AP, then the middle term is the average of the other two, meaning $2B = A + C$ . Let's plug our log terms into this core rule.
$2\log_3(2x^2+1) = \log_3(x^2-1) + \log_3(6x^2+3)$

Step 2: Deploying Log Rules
Use the power rule on the left side ( $k\log(x) = \log(x^k)$ ) and the product rule on the right side ( $\log(x) + \log(y) = \log(xy)$ ).
$\log_3((2x^2+1)^2) = \log_3((x^2-1)(6x^2+3))$
Since the bases are the exact same, we can just drop the logs entirely and work with the pure algebra.
$(2x^2+1)^2 = (x^2-1)(6x^2+3)$

Step 3: Nerfing the Polynomial
Before we expand this and accidentally create a massive, toxic degree-4 polynomial, notice that we can factor a 3 out of the right side to find a matching term.
$(2x^2+1)^2 = 3(x^2-1)(2x^2+1)$
Since $\log_3(x^2-1)$ exists in the prompt, $x^2-1 > 0$ , which means $x^2 > 1$ . Therefore, $(2x^2+1)$ is definitely a positive non-zero number. We can safely divide both sides by $(2x^2+1)$ without losing any valid roots.
$2x^2+1 = 3(x^2-1)$
$2x^2+1 = 3x^2-3$
$x^2 = 4$

Step 4: Revealing the Sequence
Now plug $x^2 = 4$ back into the original three log terms to see what this AP actually looks like in practice.
* Term 1: $\log_3(4-1) = \log_3(3) = 1$
* Term 2: $\log_3(2(4)+1) = \log_3(9) = 2$
* Term 3: $\log_3(6(4)+3) = \log_3(27) = 3$
The progression is literally just 1, 2, 3... Absolute baby numbers.

Step 5: The Final Carry
The question asks for the sum of the *next* three terms of this progression.
The sequence continues: 4, 5, 6.
$4 + 5 + 6 = 15$

Step 6: The Audit (Double Check Protocol)
Let's run it back to verify.
If $x^2 = 4$ , our sequence is $\log_3(3)$ , $\log_3(9)$ , $\log_3(27)$ .
These evaluate to $1, 2, 3$ . This is a valid AP with a common difference of $1$ .
The next three terms are $4, 5, 6$ .
Sum = $15$ .
The math is completely flawless. Zero errors.

Final Answer: 15

Q7:ipmat indore 2025QA › CirclesMediumSA · TITA

A circle of radius 13 cm touches the adjacent sides AB and BC of a square ABCD at M and N, respectively. If

AB=18

cm and the circle intersects the other two sides CD and DA at P and Q, respectively, then the area, in sq. cm, of triangle PMD is

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The Setup: We are dropping into a coordinate geometry map for this one. We have a square bounding box and a circle spawning inside it, touching the walls. By setting a strategic origin point, we can turn this messy geometry problem into a pure algebra speedrun. I have fully audited the coordinates to ensure the hitbox interactions are completely accurate.

Step 1: Setting the Origin (The Spawn Point)
Let's anchor the square's corner $B$ exactly at the origin $(0,0)$ .
Since the square's side length is $AB = 18$ cm, we can lock in the corner coordinates:
* $B = (0,0)$
* $C = (18,0)$
* $A = (0,18)$
* $D = (18,18)$

Step 2: Defining the Circle's Hitbox
The circle has a radius $r = 13$ cm. It touches side $AB$ (which sits on the y-axis) and side $BC$ (which sits on the x-axis).
Because it is perfectly tangent to both axes in the positive quadrant, its center $O$ must be at $(r, r)$ , which is $(13, 13)$ .
The equation of the circle is locked in:
$(x-13)^2 + (y-13)^2 = 13^2$
$(x-13)^2 + (y-13)^2 = 169$

Step 3: Locating the Intersections
The circle touches $AB$ at $M$ , so $M$ is simply the y-intercept at $(0, 13)$ .
The circle intersects $CD$ at point $P$ . Side $CD$ is the vertical line $x = 18$ .
Substitute $x = 18$ into the circle's equation to find $P$ 's exact y-coordinate:
$(18-13)^2 + (y-13)^2 = 169$
$5^2 + (y-13)^2 = 169$
$25 + (y-13)^2 = 169$
$(y-13)^2 = 144$
$y-13 = \pm 12$
Since $P$ must lie on the physical segment $CD$ , its y-coordinate must be between 0 and 18. So we take the subtraction route: $y = 13 - 12 = 1$ .
Our point $P$ is officially at $(18, 1)$ .

Step 4: Calculating the Triangle Area
We need the area of $\triangle PMD$ . Let's pull the coordinates of our vertices:
* $M = (0, 13)$
* $P = (18, 1)$
* $D = (18, 18)$

Notice that $P$ and $D$ both lie on the exact same vertical line ( $x = 18$ ). We can treat the segment $PD$ as the base of our triangle to completely cheese the calculation.
Base length $PD = 18 - 1 = 17$ units.
The height of the triangle is the perpendicular horizontal distance from the base line ( $x=18$ ) back to the third point $M$ (at $x=0$ ).
Height $= 18 - 0 = 18$ units.
$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$
$\text{Area} = \frac{1}{2} \times 17 \times 18$
$\text{Area} = 17 \times 9 = 153$

Step 5: The Audit (Double Check Protocol)
Running the shoelace formula (determinant method) to double-check the area calculation and guarantee zero errors.
$\text{Area} = 0.5 \times |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$
$\text{Area} = 0.5 \times |0(1-18) + 18(18-13) + 18(13-1)|$
$\text{Area} = 0.5 \times |0 + 18(5) + 18(12)|$
$\text{Area} = 0.5 \times |90 + 216|$
$\text{Area} = 0.5 \times 306 = 153$
Both methods yield exactly 153. The geometry holds up perfectly. The math is flawless.

Final Answer: 153

Q8:ipmat indore 2025QA › Linear EquationsEasySA · TITA

Monica, who is 18 years old, is one-third the age of her father. The age at which she will be half the age of her father is

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The Setup: We are running it back with the Monica age progression problem. This was literally the tutorial boss, but now it is officially slotted as Q11 in this specific paper's roster. We need to project the current timeline into the future where the age ratio shifts from one-third to one-half. I have double-checked the logic to guarantee zero errors.

Step 1: Establish the Current Era
Monica is currently 18 years old. Since she is explicitly stated to be one-third her dad's age, we multiply her age by 3 to lock in his current stats.
$18 \times 3 = 54$
Her dad is currently 54 years old.

Step 2: Set Up the Future Timeline
Let $x$ be the number of years it takes for this timeline shift to happen. Fast forward $x$ years: Monica will be $18+x$ years old, and her dad will level up to $54+x$ years old. The problem states that at this point, her age will be exactly half of his.
$18+x = \frac{1}{2}(54+x)$

Step 3: Solve the Equation (No Cap)
Multiply both sides by 2 to clear the fraction and avoid messy calculations.
$2(18+x) = 54+x$
$36+2x = 54+x$
Now, isolate $x$ by subtracting $x$ from both sides, and moving the 36 over.
$x = 18$
It will take exactly 18 years for this ratio to hit.

Step 4: Calculate the Final Age
Don't get baited by the $x$ value. The question asks for her age when this happens, not how many years it takes. Add the 18 years to her current age of 18.
$18+18 = 36$

Step 5: The Audit (Double Check Protocol)
Let's run the numbers back.
Current ages: Monica is 18, Dad is 54. ( $18 \times 3 = 54$ . Checked).
In 18 years: Monica will be 36, Dad will be 72.
Is 36 exactly half of 72? Yes.
The math is completely flawless.

Final Answer: 36

Q9:ipmat indore 2025QA › Integral SolutionsMediumSA · TITA

If

m

and

n

are two positive integers such that

7m+11n=200

, then the minimum possible value of

m+n

is

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The Setup: We are dropping into a linear Diophantine equation problem. We need to find positive integer pairs that satisfy a fixed budget equation ( $7m + 11n = 200$ ) and then optimize for the minimum sum of $m + n$ . I have fully audited the math logic below to ensure zero errors.

Step 1: Analyzing the Modulo Constraints
We are given the equation:
$7m + 11n = 200$
Since $m$ and $n$ must be positive integers, we can isolate one of the variables to analyze its boundaries. Let's isolate $7m$ :
$7m = 200 - 11n$
$m = \frac{200 - 11n}{7}$

For $m$ to be a positive integer, $200 - 11n$ must be a positive multiple of 7. Let's look at $200 \pmod 7$ :
$200 \div 7 = 28 \text{ with a remainder of } 4$
So, $200 \equiv 4 \pmod 7$ .
We also know that $11 \equiv 4 \pmod 7$ . Therefore:
$200 - 11n \equiv 4 - 4n \pmod 7 \equiv 0 \pmod 7$
$4n \equiv 4 \pmod 7$
$n \equiv 1 \pmod 7$

Step 2: Finding the Valid Integer Pairs
Since $n \equiv 1 \pmod 7$ , the possible positive values for $n$ are in the sequence $1, 8, 15, 22, \dots$ (keeping $11n < 200$ , so $n$ maxes out around $18$ ). Let's test these values to find our $(m, n)$ pairs:
* **Case 1 ( $n = 1$ ):**
$7m + 11(1) = 200 \rightarrow 7m = 189 \rightarrow m = 27$
Pair: $(27, 1) \rightarrow m + n = 27 + 1 = 28$
* **Case 2 ( $n = 8$ ):**
$7m + 11(8) = 200 \rightarrow 7m = 200 - 88 = 112 \rightarrow m = 16$
Pair: $(16, 8) \rightarrow m + n = 16 + 8 = 24$
* **Case 3 ( $n = 15$ ):**
$7m + 11(15) = 200 \rightarrow 7m = 200 - 165 = 35 \rightarrow m = 5$
Pair: $(5, 15) \rightarrow m + n = 5 + 15 = 20$

If we try the next step ( $n = 22$ ), $11(22) = 242$ , which exceeds 200. Thus, our search space is completely exhausted.

Step 3: Finding the Minimum Sum
Let's evaluate the sums of all valid pairs we found:
* $27 + 1 = 28$
* $16 + 8 = 24$
* $5 + 15 = 20$

The absolute minimum possible value for $m + n$ is 20.

Step 4: The Audit (Double Check Protocol)
Let's run it back to verify.
If $m = 5$ and $n = 15$ :
$7(5) + 11(15) = 35 + 165 = 200$
Both $m$ and $n$ are positive integers. The equation is fully satisfied, and the sum $5 + 15 = 20$ is the lowest possible outcome among all valid configurations. The math is completely flawless.

Final Answer: 20

Q10:ipmat indore 2025QA › FactorisationEasySA · TITA

The number of factors of

3^{5}\times5^{8}\times7^{2}

that are perfect squares is

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The Setup: We are looking at a classic number theory factorisation problem. To find how many factors of a large prime-factorised number are perfect squares, we need to analyze the exponents and use combinatorics to lock in the exact count. I have fully audited the math logic below to ensure zero errors.

Step 1: The Perfect Square Meta
A number is a perfect square if and only if all the exponents in its prime factorization are even integers.
Our target number is given in its prime-factorised form:
$N = 3^5 \times 5^8 \times 7^2$
Any factor of $N$ will take the general form $3^a \times 5^b \times 7^c$ , where:
* $0 \le a \le 5$
* $0 \le b \le 8$
* $0 \le c \le 2$

For this factor to be a perfect square, the exponents $a$ , $b$ , and $c$ must all be even numbers.

Step 2: Selecting Even Exponents
Let's find the valid choices for each exponent based on our constraints:
* **For prime base 3 ( $0 \le a \le 5$ ):** The even integers in this range are $0, 2, \text{ and } 4$ . That gives us 3 choices.
* **For prime base 5 ( $0 \le b \le 8$ ):** The even integers in this range are $0, 2, 4, 6, \text{ and } 8$ . That gives us 5 choices.
* **For prime base 7 ( $0 \le c \le 2$ ):** The even integers in this range are $0 \text{ and } 2$ . That gives us 2 choices.

Step 3: Applying the Fundamental Counting Principle
To find the total number of unique perfect square factors, we multiply the number of valid choices for each prime base together:
$\text{Total Perfect Square Factors} = 3 \times 5 \times 2$
$\text{Total} = 30$

Step 4: The Audit (Double Check Protocol)
Let's run it back to verify.
Could $a$ be 6? No, because $a \le 5$ .
Could $b$ be 10? No, because $b \le 8$ .
Are all selected exponents ( $0, 2, 4$ for base 3; $0, 2, 4, 6, 8$ for base 5; $0, 2$ for base 7) strictly even? Yes.
Multiplying independent choices ( $3 \times 5 \times 2 = 30$ ) is the textbook method for finding restricted factor counts. The math is completely flawless.

Final Answer: 30

Q11:ipmat indore 2025QA › RemainderEasySA · TITA

If the polynomial

ax^{2}+bx+5

leaves a remainder 3 when divided by

x-1

, and a remainder 2 when divided by

x+1

, then

2b-4a

equals

Enter your answer to attempt

The Setup: We are using the Remainder Theorem, which is the ultimate shortcut for polynomials. Instead of doing long division, we just plug the roots of the divisors directly into the function to lock in our remainders.

Step 1: Setting up the Polynomial Function
Let our polynomial be $P(x) = ax^2 + bx + 5$ .
According to the Remainder Theorem:
* When divided by $x - 1$ , the root is $x = 1$ , and the remainder is $P(1) = 3$ .
* When divided by $x + 1$ , the root is $x = -1$ , and the remainder is $P(-1) = 2$ .

Step 2: Building the Equations
Let's evaluate $P(1)$ and $P(-1)$ explicitly:
For $x = 1$ :
$a(1)^2 + b(1) + 5 = 3$
$a + b + 5 = 3$
$a + b = -2$

For $x = -1$ :
$a(-1)^2 + b(-1) + 5 = 2$
$a - b + 5 = 2$
$a - b = -3$

Step 3: Solving the System
We now have our system of linear equations:
1. $a + b = -2$
2. $a - b = -3$

To find $2b$ , subtract the second equation from the first:
$(a + b) - (a - b) = -2 - (-3)$
$2b = 1$

To find $4a$ , first add the two equations together to get $2a$ :
$(a + b) + (a - b) = -2 + (-3)$
$2a = -5$
Multiply by 2 to lock in our $4a$ value:
$4a = -10$

Step 4: The Final Calculation
The question asks for the exact value of $2b - 4a$ . Plug in our extracted values:
$2b - 4a = 1 - (-10)$
$1 + 10 = 11$

Step 5: The Audit (Double Check Protocol)
Let's run it back to verify.
If $2b = 1$ , then $b = 0.5$ .
If $2a = -5$ , then $a = -2.5$ .
Does $a + b = -2$ ? $(-2.5) + 0.5 = -2$ (Checked).
Does $a - b = -3$ ? $(-2.5) - 0.5 = -3$ (Checked).
Calculate $2(0.5) - 4(-2.5) = 1 + 10 = 11$ .
The math is completely flawless.

Final Answer: 11

Q12:ipmat indore 2025QA › Progression & SeriesMediumMCQ · MCQ

Given that

1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\frac{1}{4^{2}}+...=\frac{\pi^{2}}{6}

, the value of

1+\frac{1}{3^{2}}+\frac{1}{5^{2}}+\frac{1}{7^{2}}+...

is

A $\frac{\pi^{2}}{6}-1$
B $\frac{\pi}{6}$
C $\frac{\pi^{2}}{12}$
D $\frac{\pi^{2}}{8}$

Pick an option to attempt

The Setup: This is a classic infinite series manipulation. We are taking the legendary Basel problem (solved by Euler, the absolute GOAT of 18th-century mathematics) and splitting it into its odd and even components. Think of it like taking a full, mixed studio track and isolating just the vocals. We know the total sum, so we can isolate the odd-denominator series by mathematically filtering out the even-denominator series.

Step 1: Define the master series.
Let the total sum be represented by $S$ . The problem gives us the total value of this series right out of the gate:
$S = 1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\frac{1}{4^{2}}+... = \frac{\pi^{2}}{6}$
This series contains both the odd terms (which we want) and the even terms. Let's split them into two distinct squads:
$S = S_{odd} + S_{even}$

Step 2: Isolate and analyze the even terms.
The even terms form their own infinite sub-series:
$S_{even} = \frac{1}{2^{2}}+\frac{1}{4^{2}}+\frac{1}{6^{2}}+...$
Notice that every denominator here has a common factor of $2^2$ . If we factor out $\frac{1}{2^2}$ (which is just $\frac{1}{4}$ ), watch what happens:
$S_{even} = \frac{1}{4}\left(1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+...\right)$
Bro, look closely at what is inside the parentheses—it is literally just our original series $S$ hitting a respawn. So, we can rewrite this as:
$S_{even} = \frac{1}{4}S$

Step 3: Solve for the odd terms.
Now we take this finding and plug it back into our main equation from Step 1.
$S = S_{odd} + \frac{1}{4}S$
To get $S_{odd}$ by itself, just subtract $\frac{1}{4}S$ from both sides.
$S_{odd} = S - \frac{1}{4}S$
$S_{odd} = \frac{3}{4}S$

Step 4: Calculate the final value.
Now for the final boss phase. Plug in Euler's iconic value for $S$ to see exactly what the odd terms equal.
$S_{odd} = \frac{3}{4}\left(\frac{\pi^{2}}{6}\right)$
Simplify the fraction by cross-canceling the $3$ and the $6$ , and the math is mathing perfectly:
$S_{odd} = \frac{\pi^{2}}{8}$

Final Answer: $\frac{\pi^{2}}{8}$

Q13:ipmat indore 2025QA › LogarithmsMediumMCQ · MCQ

If

y=a+b\log_{e}(x)

then which of the following is true?

A $\frac{1}{y-a}$ is proportional to $x^{b}$
B $y-a$ is proportional to $x^{b}$
C $e^{y}$ is proportional to $x^{b}$
D $\log_{e}(y)$ is proportional to $x^{b}$

Pick an option to attempt

The Setup: We are dealing with a classic logarithmic equation, and we need to determine the true proportionality statement. Proportionality basically means two variables level up together at a constant rate. To figure out who is vibing with $x^b$ , we need to rescue $x$ from inside that natural log function. Time to hit it with an exponential reverse-uno card.

Step 1: Isolate the log term.
First, we have to get the log term by itself. Move $a$ to the other side like you are clearing out your inventory.
$y-a = b\log_{e}(x)$

Step 2: Apply the power rule of logarithms.
In log lore, a multiplier on the outside can be sent up top as an exponent. That $b$ in the front? Yeet it up.
$y-a = \log_{e}(x^{b})$

Step 3: Convert from log form to exponential form.
Here is the main event. A log with base $e$ is a natural log. To break $x^{b}$ out of the log prison, we exponentiate both sides using base $e$ .
$e^{y-a} = e^{\log_{e}(x^{b})}$
Because $e$ and $\log_{e}$ cancel each other out, this simplifies perfectly to:
$e^{y-a} = x^{b}$

Step 4: Expose the proportionality.
Let's split that exponent on the left side using basic exponent rules ( $z^{m-n} = z^{m} \cdot z^{-n}$ ).
$e^{y} \cdot e^{-a} = x^{b}$
Now, rearrange it to isolate $e^{y}$ :
$e^{y} = e^{a} \cdot x^{b}$
Since $a$ is a constant, $e^{a}$ is also just a constant (let's call it $k$ , like a random NPC).
$e^{y} = k \cdot x^{b}$
Whenever two variables are linked purely by a constant multiplier, they are directly proportional. Therefore, $e^{y} \propto x^{b}$ .

Final Answer: $e^{y}$ is proportional to $x^{b}$

Q14:ipmat indore 2025QA › PolynomialsMediumMCQ · MCQ

If

a_{1},a_{2},...,a_{8}

are the roots of the equation

x^{8}+x^{7}+...+x+1=0,

then the value of

a_{1}^{2025}+a_{2}^{2025}+...+a_{8}^{2025}

is.

A $0$
B $2$
C $8$
D $4$

Pick an option to attempt

The Setup: This polynomial might look like an absolute unit, but it is actually just a Geometric Progression (GP) in disguise. We are dealing with the classic "roots of unity" concept. Think of it like a Pokémon evolving into its final form when you hand it the right evolution stone—in this case, multiplying by $(x-1)$ will reveal its true identity.

Step 1: Reveal the true equation.
The given equation is $x^{8}+x^{7}+...+x+1=0$ . Let's multiply both sides by $(x-1)$ to trigger the evolution:
$(x-1)(x^{8}+x^{7}+...+x+1) = 0$
Using the standard algebraic identity for the difference of powers, this massive bracket collapses beautifully into:
$x^{9}-1 = 0$
$x^{9} = 1$

Step 2: Analyze the roots.
The roots of $x^{9}=1$ are the 9th roots of unity. Because our original equation was just the polynomial *without* the $(x-1)$ factor, its roots ( $a_{1}, a_{2}, ..., a_{8}$ ) are all the 9th roots of unity *except* for $x=1$ .
The ultimate cheat code here? For any root $a_{k}$ from this set, we know for an absolute fact that:
$a_{k}^{9} = 1$

Step 3: Evaluate the massive exponent.
We need to find the value of $a_{k}^{2025}$ for each of these roots. The exponent $2025$ seems like a final boss, but let's check its vibe with the number $9$ . The sum of the digits is $2+0+2+5=9$ , meaning $2025$ is perfectly divisible by $9$ .
$2025 = 9 \times 225$
Using basic exponent rules, we can rewrite $a_{k}^{2025}$ :
$a_{k}^{2025} = (a_{k}^{9})^{225}$
Since we already established that $a_{k}^{9}=1$ , this simplifies instantly:
$a_{k}^{2025} = (1)^{225} = 1$

Step 4: Sum it all up.
The question asks for the sum of all 8 roots, each raised to the power of $2025$ .
$a_{1}^{2025}+a_{2}^{2025}+...+a_{8}^{2025}$
Substitute the $1$ we found in Step 3 for every single term:
$1 + 1 + 1 + 1 + 1 + 1 + 1 + 1$
Since there are 8 roots in total, we just add $1$ to itself 8 times. No cap, it really is that simple.
$1 \times 8 = 8$

Final Answer: $8$

Q15:ipmat indore 2025QA › Mean, Median & ModeMediumMCQ · MCQ

Suppose

a, b,

and

c

are three real numbers such that

Max(a,b,c)+Min(a,b,c)=15

, and

Median(a,b,c)-Mean(a,b,c)=2

. Then the median of

a, b,

and

c

is.

A $11$
B $10.5$
C $10$
D $9.5$

Pick an option to attempt

The Setup: We are dealing with basic statistical measures—max, min, median, and mean—across three numbers. To keep things structured and avoid getting lost in the sauce, let's sort these variables from lowest to highest.

Step 1: Establish the hierarchy.
Let's assume $a \le b \le c$ .
Therefore, the minimum $Min(a,b,c) = a$ , the maximum $Max(a,b,c) = c$ , and the middle-child $Median(a,b,c) = b$ .
The mean is just the standard average: $\frac{a+b+c}{3}$ .

Step 2: Translate the given conditions into equations.
Condition 1 gives us the sum of the extremes:
$a + c = 15$
Condition 2 gives us the relationship between the median and the mean:
$b - \frac{a+b+c}{3} = 2$

Step 3: Substitute and solve.
We know that $a+c = 15$ , so we can drop that right into the numerator of our mean equation like a perfect puzzle piece:
$b - \frac{15+b}{3} = 2$
Multiply the entire equation by $3$ to clear out the fraction (because nobody likes dealing with denominators):
$3b - (15+b) = 6$
Distribute the negative sign and simplify:
$2b - 15 = 6$
$2b = 21$
$b = 10.5$
Since $b$ represents our median, we have officially locked in the answer.

Final Answer: $10.5$

Q16:ipmat indore 2025QA › LogarithmsEasyMCQ · MCQ

If

\log_{25}[5\log_{3}(1+\log_{3}(1+2\log_{2}(x)))]=\frac{1}{2}

, then

x

is:

A $4$
B $16$
C $2$
D $8$

Pick an option to attempt

The Setup: We are looking at a heavily nested logarithm problem, which is basically the math equivalent of the movie *Inception*—a dream inside a dream inside a dream. To find $x$ , we have to work our way from the outside in, peeling back the layers one by one by converting each logarithmic equation into its exponential form.

Step 1: Break the outermost layer.
The outer boss is $\log_{25}$ . To defeat it, we convert the equation to exponential form. The base $25$ swings over to the right side and pushes the $\frac{1}{2}$ up into the exponent slot:
$5\log_{3}(1+\log_{3}(1+2\log_{2}(x))) = 25^{\frac{1}{2}}$
Since $25^{\frac{1}{2}}$ is just the square root of $25$ , this simplifies down to $5$ .
$5\log_{3}(1+\log_{3}(1+2\log_{2}(x))) = 5$

Step 2: Clear the multiplier.
Divide both sides by $5$ to clean up the front of our next log term.
$\log_{3}(1+\log_{3}(1+2\log_{2}(x))) = 1$

Step 3: Break the second layer.
Now we hit the next layer. Convert the $\log_{3}$ to exponential form. The base $3$ swings over and lifts the $1$ :
$1+\log_{3}(1+2\log_{2}(x)) = 3^{1}$
$1+\log_{3}(1+2\log_{2}(x)) = 3$

Step 4: Isolate the next log.
Subtract $1$ from both sides to keep peeling the onion.
$\log_{3}(1+2\log_{2}(x)) = 2$

Step 5: Break the third layer.
Trigger the exponential conversion one more time for the remaining $\log_{3}$ . The base $3$ yeets the $2$ up top:
$1+2\log_{2}(x) = 3^{2}$
$1+2\log_{2}(x) = 9$

Step 6: Clear the remaining constants.
Subtract $1$ from both sides, then divide by $2$ to leave the final log completely isolated.
$2\log_{2}(x) = 8$
$\log_{2}(x) = 4$

Step 7: The final unboxing.
One last exponential conversion to free $x$ . The base $2$ swings over to hold up the $4$ .
$x = 2^{4}$
$x = 16$
No cap, we just speedran that nested log.

Final Answer: $16$

Q17:ipmat indore 2025QA › MiscellaneousMediumMCQ · MCQ

A natural number

n

lies between

100

and

400

, and the sum of its digits is

10

. The probability that

n

is divisible by

4

is:

A $\frac{1}{4}$
B $\frac{7}{27}$
C $\frac{1}{3}$
D $\frac{2}{9}$

Pick an option to attempt

The Setup: We are dealing with classic probability, which means we need to find the total number of possible outcomes (the entire lobby) and divide it by the specific outcomes we actually want (the VIPs). The overarching rules: our number is a 3-digit integer between $100$ and $400$ , the digits must add up to exactly $10$ , and our winning condition is that the number is divisible by $4$ . Let our 3-digit number be represented as $abc$ .

Step 1: Define the hundreds digit.
Since $n$ lies between $100$ and $400$ , the hundreds digit ( $a$ ) is heavily restricted. It can only be $1, 2,$ or $3$ . We will map out the possibilities for each scenario.

Step 2: Scout the $100$ s tier ( $a = 1$ ).
If the first digit is $1$ , then the remaining two digits must add up to $9$ to satisfy the sum rule ( $1 + b + c = 10 \implies b + c = 9$ ).
The possible pairs for $(b, c)$ are:
$(0,9), (1,8), (2,7), (3,6), (4,5), (5,4), (6,3), (7,2), (8,1), (9,0)$ .
That gives us $10$ total numbers in this tier.
Now, apply the divisibility rule for $4$ (the last two digits must form a number divisible by $4$ ). Looking at our pairs, only $36$ and $72$ pass the vibe check.
* Total numbers: $10$
* Valid hits: $2$ (which are $136$ and $172$ )

Step 3: Scout the $200$ s tier ( $a = 2$ ).
If the first digit is $2$ , the remaining digits must sum to $8$ ( $b + c = 8$ ).
The possible pairs are:
$(0,8), (1,7), (2,6), (3,5), (4,4), (5,3), (6,2), (7,1), (8,0)$ .
That is $9$ total numbers. Checking the last two digits for multiples of $4$ , we keep $08, 44,$ and $80$ .
* Total numbers: $9$
* Valid hits: $3$ (which are $208, 244,$ and $280$ )

Step 4: Scout the $300$ s tier ( $a = 3$ ).
If the first digit is $3$ , the remaining digits must sum to $7$ ( $b + c = 7$ ).
The possible pairs are:
$(0,7), (1,6), (2,5), (3,4), (4,3), (5,2), (6,1), (7,0)$ .
That gives us $8$ total numbers. Sifting for multiples of $4$ in those last two digits, only $16$ and $52$ survive.
* Total numbers: $8$
* Valid hits: $2$ (which are $316$ and $352$ )

Step 5: Calculate the final probability.
Now we just tally up the final score.
Total sample space (every possible valid number) $= 10 + 9 + 8 = 27$
Favorable outcomes (the ones divisible by $4$ ) $= 2 + 3 + 2 = 7$

The probability $P$ is the ratio of favorable outcomes to the total sample space:
$P = \frac{7}{27}$

Final Answer: $\frac{7}{27}$

Q18:ipmat indore 2025QA › TrianglesMediumMCQ · MCQ

In triangle

ABC

,

AB=AC=x

,

\angle ABC=\theta

and the circumradius is equal to

y

. Then

\frac{x}{y}

equals

A $\sin(\theta)$
B $\cos(\theta)$
C $2\cos(\theta)$
D $2\sin(\theta)$

Pick an option to attempt

The Setup: We have an isosceles triangle stepping into the arena, and we need to link its side lengths, angles, and circumradius. The ultimate cheat code for any circle circumscribing a triangle is the Extended Law of Sines. It is basically the universal API bridging triangle sides, their opposite angles, and the circumcircle's radius.

Step 1: Decode the triangle's geometry.
We are given that $AB=AC=x$ , making triangle $ABC$ isosceles. In geometry lore, the angle opposite to side $AC$ is $\angle ABC$ , which we are told is $\theta$ .

Step 2: Deploy the Extended Law of Sines.
The law dictates that the ratio of any side length to the sine of its opposite angle is exactly equal to the circumcircle's diameter (twice the circumradius, or $2R$ ).
The general framework is:
$\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)} = 2R$

Step 3: Plug in the stats.
Let's target side $AC$ (which has a length of $x$ ) and its corresponding opposite angle, $\angle ABC$ (which is $\theta$ ). The circumradius $R$ is given to us as $y$ . Substitute these into the formula:
$\frac{x}{\sin(\theta)} = 2y$

Step 4: Isolate the target ratio.
The prompt is asking for the exact value of $\frac{x}{y}$ . We just need to execute a quick algebraic shuffle—divide both sides by $y$ and multiply both sides by $\sin(\theta)$ to get the variables aligned perfectly.
$\frac{x}{y} = 2\sin(\theta)$
The math literally solves itself.

Final Answer: $2\sin(\theta)$

Q19:ipmat indore 2025QA › PolynomialsMediumMCQ · MCQ

If

8x^2-2kx+k=0

is a quadratic equation in

x

, such that one of its roots is

p

times the other, and

p, k

are positive real numbers, then

k

equals

A $(p+\frac{1}{p})$
B $2(p+\frac{1}{p})$
C $2(\sqrt{p}+\frac{1}{\sqrt{p}})^2$
D $(\sqrt{p}+\frac{1}{\sqrt{p}})^2$

Pick an option to attempt

The Setup: We are dealing with a quadratic equation where one root is basically a scaled-up clone of the other (like a Pokémon evolution). We need to find the specific value of $k$ that makes this relationship valid. The classic strat here is to use Vieta's formulas to link the roots directly to the coefficients of the polynomial.

Step 1: Define the roots and pull the stats.
Our quadratic is $8x^2 - 2kx + k = 0$ .
Let the first root be $\alpha$ . Since the second root is $p$ times the first, it is $p\alpha$ .
Using Vieta's formulas, we grab the sum and product of the roots.
Sum of the roots ( $-\frac{b}{a}$ ):
$\alpha + p\alpha = \frac{-(-2k)}{8}$
$\alpha(1+p) = \frac{k}{4}$

Product of the roots ( $\frac{c}{a}$ ):
$(\alpha)(p\alpha) = \frac{k}{8}$
$p\alpha^2 = \frac{k}{8}$

Step 2: Isolate and substitute the base root.
From our sum equation, let's isolate $\alpha$ so we can swap it out in the product equation (a classic bait-and-switch maneuver).
$\alpha = \frac{k}{4(p+1)}$
Now, drop this $\alpha$ into the product equation:
$p\left(\frac{k}{4(p+1)}\right)^2 = \frac{k}{8}$

Step 3: Expand and simplify the arena.
Square the fraction to open up the equation:
$p\left(\frac{k^2}{16(p+1)^2}\right) = \frac{k}{8}$
Since the prompt confirms $k$ is a positive real number, $k$ definitely is not zero. That gives us the green light to divide both sides by $k$ without breaking the math.
$\frac{pk}{16(p+1)^2} = \frac{1}{8}$

Step 4: Solve for the boss, $k$ .
Cross-multiply to get $k$ on its own.
$pk = \frac{16(p+1)^2}{8}$
$pk = 2(p+1)^2$
Divide by $p$ to completely isolate $k$ :
$k = \frac{2(p+1)^2}{p}$

Step 5: Shape-shift into the final form.
Expand the numerator using the standard identity $(a+b)^2 = a^2 + 2ab + b^2$ :
$k = \frac{2(p^2 + 2p + 1)}{p}$
Split the fraction by dividing each term inside by $p$ :
$k = 2\left(p + 2 + \frac{1}{p}\right)$
This right here is the expanded form of a perfect square. If we recognize that $p = (\sqrt{p})^2$ and $\frac{1}{p} = (\frac{1}{\sqrt{p}})^2$ , we can repackage it flawlessly:
$k = 2\left(\sqrt{p} + \frac{1}{\sqrt{p}}\right)^2$
Boom. The math is mathing perfectly.

Final Answer: $2\left(\sqrt{p} + \frac{1}{\sqrt{p}}\right)^2$

Q20:ipmat indore 2025QA › Straight LinesMediumMCQ · MCQ

Let

A(1,3)

and

B(5,1)

be two points. If a line with slope

m

intersects

AB

at an angle of

45^{\circ}

, then the possible values of

m

are

A $7, \frac{1}{7}$
B $3, \frac{1}{3}$
C $-3, \frac{1}{3}$
D $5, -\frac{1}{5}$

Pick an option to attempt

The Setup: We are dealing with a coordinate geometry vibe check. We have a defined line segment $AB$ and a mystery line pulling up at exactly a 45-degree angle. We need to find the slope $m$ of this mystery line. The ultimate weapon here is the angle between two lines formula, which relies on the tangent function to bridge the slopes.

Step 1: Find the slope of the base line $AB$ .
The standard formula for slope is $m_1 = \frac{y_2 - y_1}{x_2 - x_1}$ .
Plug in the coordinates for $A(1,3)$ and $B(5,1)$ :
$m_1 = \frac{1 - 3}{5 - 1}$
$m_1 = \frac{-2}{4} = -\frac{1}{2}$

Step 2: Deploy the angle formula.
The formula for the angle $\theta$ between two intersecting lines with slopes $m$ and $m_1$ is:
$\tan(\theta) = \left|\frac{m - m_1}{1 + m \cdot m_1}\right|$
We know $\theta = 45^\circ$ , and $\tan(45^\circ)$ is straight up $1$ . Let's plug everything into the mainframe:
$1 = \left|\frac{m - \left(-\frac{1}{2}\right)}{1 + m\left(-\frac{1}{2}\right)}\right|$

Step 3: Clean up the UI and set up the absolute value split.
Simplify the fraction inside the absolute value by multiplying the numerator and the denominator by $2$ to clear out the decimals/fractions:
$1 = \left|\frac{2m + 1}{2 - m}\right|$
Because it is wrapped in an absolute value, this equation branches into two separate timelines—one positive and one negative. We have to solve both to get all possible values for $m$ .

Step 4: Solve Timeline 1 (The Positive Case).
Set the inside of the absolute value equal to $1$ :
$\frac{2m + 1}{2 - m} = 1$
Cross-multiply to flatten it out:
$2m + 1 = 2 - m$
Move the $m$ 's to the left and the constants to the right:
$3m = 1$
$m = \frac{1}{3}$

Step 5: Solve Timeline 2 (The Negative Case).
Set the inside of the absolute value equal to $-1$ :
$\frac{2m + 1}{2 - m} = -1$
Cross-multiply:
$2m + 1 = -1(2 - m)$
$2m + 1 = -2 + m$
Subtract $m$ from both sides, and subtract $1$ from both sides:
$m = -3$

Final Answer: $-3, \frac{1}{3}$

Q21:ipmat indore 2025QA › PolynomialsHardMCQ · MCQ

Let

P(x)

be a quadratic polynomial such that

\begin{vmatrix} P(0) & P(1) \\ P(0) & P(2) \end{vmatrix} = 0

Let

P(0)=2

and

P(1)+P(2)+P(3)=14

. Then

P(4)

equals

A $-14$
B $30$
C $-6$
D $16$

Pick an option to attempt

The Setup: We have a mystery quadratic polynomial acting like an NPC, and we need to figure out its exact stats (the coefficients $a, b,$ and $c$ ). We are given a $2 \times 2$ determinant acting as a vibe check, an initial value, and a sum condition. We will break the matrix first, build the polynomial's profile, and then plug in the final value.

Step 1: Crack the matrix determinant.
The determinant of a $2 \times 2$ matrix is found by cross-multiplying the diagonals and subtracting them:
$P(0)P(2) - P(0)P(1) = 0$
We are given that $P(0) = 2$ . Let's plug that straight in to nerf the equation:
$2P(2) - 2P(1) = 0$
Divide by $2$ and rearrange to reveal a massive clue:
$P(2) = P(1)$

Step 2: Build the quadratic profile.
The standard form of any quadratic polynomial is:
$P(x) = ax^2 + bx + c$
Since $P(0) = 2$ , all the $x$ terms drop to zero, immediately giving us our constant:
$c = 2$
So our working model is $P(x) = ax^2 + bx + 2$ .
Now, apply the intel from Step 1 ( $P(1) = P(2)$ ) to find how $a$ and $b$ relate:
$a(1)^2 + b(1) + 2 = a(2)^2 + b(2) + 2$
$a + b + 2 = 4a + 2b + 2$
Cancel the $2$ s and shift the variables to one side:
$3a + b = 0$
$b = -3a$

Step 3: Run the sum condition.
We are given the final boss equation: $P(1) + P(2) + P(3) = 14$ .
Let's evaluate each piece using our updated model $P(x) = ax^2 - 3ax + 2$ .
For $P(1)$ :
$P(1) = a(1)^2 - 3a(1) + 2 = -2a + 2$
For $P(2)$ (which we know is identical to $P(1)$ ):
$P(2) = -2a + 2$
For $P(3)$ :
$P(3) = a(3)^2 - 3a(3) + 2 = 9a - 9a + 2 = 2$
Now, drop these into the sum equation:
$(-2a + 2) + (-2a + 2) + 2 = 14$
$-4a + 6 = 14$
$-4a = 8$
$a = -2$
Since $b = -3a$ , we know $b = -3(-2) = 6$ .
Our fully leveled-up polynomial is:
$P(x) = -2x^2 + 6x + 2$

Step 4: Secure the W.
The prompt asks for the value of $P(4)$ . Plug $x = 4$ into our confirmed polynomial.
$P(4) = -2(4)^2 + 6(4) + 2$
$P(4) = -2(16) + 24 + 2$
$P(4) = -32 + 26$
$P(4) = -6$

Final Answer: $-6$

Q22:ipmat indore 2025QA › CirclesMediumMCQ · MCQ

A circle touches the y-axis at

(0,4)

and passes through the point

(-2,0)

. Then the radius of the circle is

A $4$
B $5$
C $6$
D $7$

Pick an option to attempt

The Setup: We are doing some coordinate geometry map reconnaissance. If a circle physically touches the y-axis at a specific point, it is basically giving away its center's y-coordinate for free. It also tells us that the radius is exactly equal to the horizontal distance from the center to the y-axis. Think of the center as the player and the radius as the exact range of its hitbox.

Step 1: Establish the center and the radius.
Since the circle touches the y-axis at $(0,4)$ , the center must lie directly on the horizontal line $y=4$ .
Let the center be $(h, 4)$ .
Because the y-axis ( $x=0$ ) is a tangent line, the radius $r$ is just the absolute value of the x-coordinate: $r = |h|$ .

Step 2: Build the circle's equation.
The standard equation of a circle is $(x - h)^2 + (y - k)^2 = r^2$ .
Substitute our known center $(h, 4)$ and radius squared $r^2 = h^2$ into the mainframe:
$(x - h)^2 + (y - 4)^2 = h^2$

Step 3: Use the waypoint to solve for $h$ .
The problem states the circle passes through $(-2,0)$ . This is a verified coordinate on the circle's path, so we can plug $x = -2$ and $y = 0$ straight into our equation to see what happens.
$(-2 - h)^2 + (0 - 4)^2 = h^2$
Expand the binomial (remember that $(-2 - h)^2$ is exactly the same as $(2 + h)^2$ ) and square the $-4$ :
$4 + 4h + h^2 + 16 = h^2$

Step 4: Simplify and secure the radius.
Subtract $h^2$ from both sides to instantly nerf the quadratic down to a linear equation:
$4h + 20 = 0$
$4h = -20$
$h = -5$
So, the center of our circle is located at $(-5, 4)$ .
Since the radius $r$ is the absolute distance $|h|$ , we just take the positive value.
$r = |-5| = 5$
The math is flawless.

Final Answer: $5$

Q23:ipmat indore 2025QA › TrianglesMediumMCQ · MCQ

Consider a triangle with side lengths

4

meters,

6

meters, and

9

meters. A dog runs around the triangle in such a way that the shortest distance of the dog from the triangle is exactly

1

meter. The total distance covered (in meters) by the dog in one round is

A $22+2\pi$
B $19+2\pi$
C $22$
D $22-2\pi$

Pick an option to attempt

The Setup: This is a locus problem disguised as a dog-walking simulator. The dog maintains a strict $1$ -meter social distance from the triangle at all times. When you trace this path, the dog's route is built from two types of geometry: straight paths parallel to the triangle's sides, and circular arcs pivoting around the corners. It is basically Mario Kart drifting mechanics.

Step 1: Analyze the straightaways.
For every side of the triangle, the dog runs a parallel straight line exactly $1$ meter away. The lengths of these straight paths match the triangle's base stats perfectly.
$\text{Straight distance} = 4 + 6 + 9 = 19$

Step 2: Calculate the corner drifts.
When the dog reaches a corner, it has to swing around the vertex to stay exactly $1$ meter away from the point. This movement creates a circular arc. The angle of this arc is exactly equal to the exterior angle of the triangle at that specific vertex.

Step 3: Piece the arcs together.
In Euclidean geometry lore, the sum of the exterior angles of *any* convex polygon is an automatic $360^\circ$ (or $2\pi$ radians). This means if you take those three corner arcs and stitch them together, they form one flawless, complete circle with a radius of $r = 1$ meter. Let's run the circumference formula:
$\text{Arc distance} = 2\pi r = 2\pi(1) = 2\pi$

Step 4: Sum the total distance.
Combine the straightaways and the corner drifts to get the final run length. The total distance covered by the dog in one round is just the sum of our two parts.
$\text{Total} = 19 + 2\pi$

Final Answer: $19+2\pi$

Q24:ipmat indore 2025QA › Simple & Compound InterestMediumMCQ · MCQ

Anindita invests a total of 1 lakh rupees distributed across three schemes A, B and C for a period of two years. These schemes offer an interest rate of 10%, 8% and 12% per annum, respectively, each compounded annually. If the initial investment amount in scheme A is 30000 rupees and the total interest earned from all the three schemes during the first year is 10600 rupees, then the total interest earned, in rupees, from all the three schemes for the second year is

A10308
B11748
C22348
D19708

Pick an option to attempt

The Setup: This is a compound interest min-maxing problem. Anindita is distributing her loot (100,000 rupees) across three different investment pools (A, B, and C) that compound annually. We are given her first-year returns and need to calculate exactly how much extra gold she farms in year two. The secret sauce here is that second-year interest is just the first-year interest *plus* the interest earned on that first-year interest (the compounding buff).

Step 1: Define the initial loadout.
Total investment is 100,000.
We already know Scheme A gets 30,000. That means the remaining 70,000 is split between B and C.
$I_B + I_C = 70000$

Step 2: Analyze the first-year interest.
First-year compound interest behaves exactly like simple interest. Let's calculate the returns based on the given rates (10% for A, 8% for B, 12% for C) which sum up to 10,600.
$0.10(30000) + 0.08(I_B) + 0.12(I_C) = 10600$
$3000 + 0.08(I_B) + 0.12(I_C) = 10600$
Subtract the guaranteed 3,000 from Scheme A:
$0.08(I_B) + 0.12(I_C) = 7600$

Step 3: Solve the system of equations.
We have a basic system here. Let's multiply the equation from Step 1 by 0.08 to set up an elimination:
$0.08(I_B) + 0.08(I_C) = 5600$
Subtract this from the interest equation in Step 2:
$(0.08 I_B + 0.12 I_C) - (0.08 I_B + 0.08 I_C) = 7600 - 5600$
$0.04(I_C) = 2000$
$I_C = \frac{2000}{0.04} = 50000$
If $I_C = 50000$ , then $I_B$ must be 20,000 to complete the 70,000 remainder.

Step 4: Calculate the first-year interest breakdown.
Now we know exactly how much interest each scheme generated in Year 1:
* Scheme A: 10% of 30,000 = 3,000
* Scheme B: 8% of 20,000 = 1,600
* Scheme C: 12% of 50,000 = 6,000
(Check: 3,000 + 1,600 + 6,000 = 10,600. The math checks out perfectly.)

Step 5: Calculate the second-year interest.
Because of compounding, the second-year interest equals the first-year interest plus the new interest generated *on* that first-year interest.
* Scheme A Year 2 Interest: 3,000 + (10% of 3,000) = 3,000 + 300 = 3,300
* Scheme B Year 2 Interest: 1,600 + (8% of 1,600) = 1,600 + 128 = 1,728
* Scheme C Year 2 Interest: 6,000 + (12% of 6,000) = 6,000 + 720 = 6,720

Sum it all up for the final score:
Total Year 2 Interest = 3,300 + 1,728 + 6,720 = 11,748

Final Answer: 11748

Q25:ipmat indore 2025QA › PolynomialsEasyMCQ · MCQ

Let

f(x) = a^2x^2 + 2bx + c

where

a \neq 0

,

b

and

c

are real numbers, and

x

is a real variable. Then:

A $f(x)$ has a maximum and a minimum
B $f(x)$ has a minimum and no maximum
C $f(x)$ has a maximum and no minimum
D $f(x)$ has no minimum and no maximum

Pick an option to attempt

The Setup: We are dealing with a quadratic function, which graphically forms a parabola (think of a skateboard halfpipe). The whole problem revolves around figuring out if this parabola opens upwards or downwards. To decode this, we just need to vibe-check the leading coefficient—the term attached to the $x^2$ .

Step 1: Analyze the leading coefficient.
Let's look at the main equation:
$f(x) = a^2x^2 + 2bx + c$
The leading coefficient here is not just some random variable; it is explicitly squared ( $a^2$ ). Because any non-zero real number squared is always strictly positive (and the prompt confirms $a \neq 0$ ), we know for an absolute fact that $a^2 > 0$ .

Step 2: Determine the parabola's trajectory.
In the polynomial meta, a positive leading coefficient dictates that the parabola opens upwards. It comes down from infinity, hits a single lowest point (the vertex), and then scales back up to positive infinity.

Step 3: Lock in the extrema.
Because the arms of the parabola shoot upwards forever, there is absolutely no ceiling to this function—meaning it has no maximum. However, it definitely hits rock bottom at the vertex. That rock-bottom point is the function's absolute minimum.

Final Answer: $f(x)$ has a minimum and no maximum

Q26:ipmat indore 2025QA › Straight LinesMediumMCQ · MCQ

The area of the triangle, formed by the straight lines

y=0

,

12x-5y=0

, and

3x+4y=7

is

A $\frac{35}{27}$
B $\frac{14}{9}$
C $\frac{28}{9}$
D $\frac{35}{54}$

Pick an option to attempt

The Setup: We are dropping into a coordinate geometry battle royale. We have three straight lines acting as boundaries, and we need to find the exact area of the safe zone (the triangle) they form. To do this, we need to find the three Points of Interest (POIs) where these lines intersect, then run the standard base-times-height formula.

Step 1: Identify the boundary lines.
Here is our loadout:
Line 1: $y=0$ (This is literally just the x-axis, the ground floor).
Line 2: $12x-5y=0$
Line 3: $3x+4y=7$

Step 2: Locate the first two vertices (the ground floor POIs).
Since Line 1 is $y=0$ , finding where it crosses the other two lines is a free loot drop. Just plug $0$ in for $y$ in their equations.
Intersection of Line 1 and Line 2:
$12x - 5(0) = 0$
$12x = 0 \implies x = 0$
First vertex locked in: $(0, 0)$ .

Intersection of Line 1 and Line 3:
$3x + 4(0) = 7$
$3x = 7 \implies x = \frac{7}{3}$
Second vertex locked in: $(\frac{7}{3}, 0)$ .

Step 3: Locate the final vertex (the apex).
Now we find where Line 2 and Line 3 cross each other (the third-party fight).
From Line 2, let's isolate $12x$ :
$12x = 5y$
Now, take Line 3 ( $3x+4y=7$ ) and multiply the entire equation by $4$ to match the $12x$ term:
$12x + 16y = 28$
Substitute $5y$ in for $12x$ :
$5y + 16y = 28$
$21y = 28 \implies y = \frac{28}{21} = \frac{4}{3}$
So the y-coordinate (which is the exact vertical height of our triangle) is $\frac{4}{3}$ .
*(We do not even need the x-coordinate because we already have the base secured flat on the x-axis, saving us precious time).*

Step 4: Calculate the final area.
Because our triangle sits perfectly flat on the x-axis, the math is super clean.
Base length ( $b$ ) = Distance from $(0, 0)$ to $(\frac{7}{3}, 0)$ = $\frac{7}{3}$ .
Height ( $h$ ) = The y-coordinate of our apex vertex = $\frac{4}{3}$ .
Plug these into the classic area formula:
$\text{Area} = \frac{1}{2} \cdot b \cdot h$
$\text{Area} = \frac{1}{2} \cdot \frac{7}{3} \cdot \frac{4}{3}$
$\text{Area} = \frac{28}{18}$
Simplify the fraction by dividing the top and bottom by $2$ for the flawless victory:
$\text{Area} = \frac{14}{9}$

Final Answer: $\frac{14}{9}$

Q27:ipmat indore 2025QA › PolygonsEasyMCQ · MCQ

Area of a regular octagon inscribed in a circle of radius

1

unit is:

A $2\sqrt{2}$
B $2+\sqrt{2}$
C $\frac{9}{2\sqrt{2}}$
D $\sqrt{10}$

Pick an option to attempt

The Setup: We are dropping into a geometry raid involving a regular octagon inscribed inside a circle. Think of it like slicing up a pizza into 8 perfectly equal pieces. To find the total area, we can calculate the area of just one of those triangular pizza slices (sectors) branching out from the center, and then multiply it by 8.

Step 1: Analyze the central angles of the octagon.
A full circle is $360^\circ$ (or $2\pi$ radians). Since a regular octagon has $8$ identical sides, it also has $8$ identical triangles radiating out from the center point to each vertex.
Let's find the central angle $\theta$ for just one of these triangles:
$\theta = \frac{360^\circ}{8} = 45^\circ$

Step 2: Pull the stats for a single triangle.
Each of these $8$ triangles shares two sides that extend from the center to the circle's boundary. Since both of those lines are radii of the circle, their lengths are both equal to $r = 1$ .
So, we have an isosceles triangle with two side lengths of $1$ and an included angle of $\theta = 45^\circ$ .

Step 3: Calculate the area of one triangle.
The trigonometry cheat code for finding the area of a triangle when you know two sides ( $a$ and $b$ ) and the included angle ( $\theta$ ) is:
$\text{Area}_{\text{triangle}} = \frac{1}{2} a b \sin(\theta)$
Plug in our stats ( $a = 1$ , $b = 1$ , and $\theta = 45^\circ$ ):
$\text{Area}_{\text{triangle}} = \frac{1}{2} (1)(1) \sin(45^\circ)$
Since $\sin(45^\circ) = \frac{\sqrt{2}}{2}$ :
$\text{Area}_{\text{triangle}} = \frac{1}{2} \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{4}$

Step 4: Scale it up to the full octagon.
An octagon consists of $8$ of these exact identical triangles. To get the total area, just multiply the area of one triangle by $8$ :
$\text{Total Area} = 8 \cdot \left(\frac{\sqrt{2}}{4}\right)$
Simplify the numbers:
$\text{Total Area} = 2\sqrt{2}$
The math is clean, fast, and flawless.

Final Answer: $2\sqrt{2}$

Q28:ipmat indore 2025QA › Time, Speed & DistanceHardMCQ · MCQ

Two swimmers, Ankit and Bipul, start swimming from the opposite ends of a swimming pool at the same time. Ankit can cover the length of the pool once in 10 minutes. Bipul can cover the length of the pool once in 15 minutes. They swim back and forth for 80 minutes without stopping. The number of times they meet each other is

A8
B6
C7
D5

Pick an option to attempt

The Setup: This is a relative speed and distance problem. Instead of tracking their exact coordinates minute by minute, the most efficient speedrun strategy is to calculate the total combined distance they cover and map that to the standard formula for opposite-end meetings.

Step 1: Define the speeds.
Let the total length of the pool be $L$ .
Ankit's speed ( $v_A$ ) = $\frac{L}{10}$ per minute.
Bipul's speed ( $v_B$ ) = $\frac{L}{15}$ per minute.

Step 2: Establish the meeting logic (combined distance).
Because they start at opposite ends, for their first meeting, they must jointly cover exactly $1L$ of distance.
After they cross paths, to meet a second time, they have to reach their respective walls and bounce back towards each other, meaning they must jointly cover an additional $2L$ of distance.
Therefore, the formula for the total combined distance required for $n$ meetings is:
$D_{\text{required}} = L + (n - 1)2L = (2n - 1)L$

Step 3: Calculate the actual distance covered.
They both swim continuously for exactly 80 minutes. Let's find out how much raw distance they put on the board:
Distance covered by Ankit = $80 \times \left(\frac{L}{10}\right) = 8L$
Distance covered by Bipul = $80 \times \left(\frac{L}{15}\right) = \frac{80L}{15} = \frac{16L}{3} \approx 5.33L$

Total combined distance actually covered by both swimmers:
$D_{\text{actual}} = 8L + 5.33L = 13.33L$

Step 4: Solve for $n$ .
We set our required meeting distance formula to be less than or equal to the actual distance they managed to swim:
$(2n - 1)L \le 13.33L$
$2n - 1 \le 13.33$
$2n \le 14.33$
$n \le 7.165$

Since the number of physical meetings ( $n$ ) must be a whole integer, we round down to the highest complete integer they achieved.
$n = 7$

Final Answer: 7

Q29:ipmat indore 2025QA › Progression & SeriesMediumMCQ · MCQ

The sum of the first 5 terms of a geometric progression is the same as the sum of the first 7 terms of the same progression. If the sum of the first 9 terms is 24, then the 4th term of the progression is

A-48
B-24
C24
D48

Pick an option to attempt

The Setup: We are dealing with a geometric progression (GP) where the sums of different numbers of terms are behaving unusually. If adding more terms doesn't change the sum, those extra terms must cancel each other out. We will use the standard GP sum formulas to lock in the common ratio and the first term.

Step 1: Analyze the equal sums.
Let the first term be $a$ and the common ratio be $r$ .
We are given that the sum of the first 5 terms ( $S_5$ ) is equal to the sum of the first 7 terms ( $S_7$ ).
$S_5 = S_7$
This means the 6th and 7th terms must sum to zero:
$S_7 - S_5 = 0$
$ar^5 + ar^6 = 0$
$ar^5(1 + r) = 0$

Step 2: Determine the common ratio.
From $ar^5(1 + r) = 0$ , we have three possibilities:
1. $a = 0$ : If the first term is $0$ , all terms are $0$ , and the sum of 9 terms would be $0$ . But we are given $S_9 = 24$ , so $a \neq 0$ .
2. $r = 0$ : If $r = 0$ , the sum of any number of terms is just $a$ . This would mean $S_9 = a = 24$ . The 4th term would be $ar^3 = 24(0)^3 = 0$ . This is not among the options, so $r \neq 0$ .
3. $1 + r = 0$ : This leaves us with $r = -1$ .
Let's lock in $r = -1$ . The sequence alternates: $a, -a, a, -a, a, -a, \dots$

Step 3: Calculate the first term.
We are given that the sum of the first 9 terms ( $S_9$ ) is 24.
For a GP with $r = -1$ , the sum of an odd number of terms is always just the first term $a$ (since all the consecutive pairs cancel out to $0$ ).
$S_9 = a = 24$
Now we have our exact stats: $a = 24$ and $r = -1$ .

Step 4: Find the 4th term.
The question asks for the 4th term of the progression.
$T_4 = ar^3$
$T_4 = 24(-1)^3$
$T_4 = 24(-1) = -24$

The math is completely flawless.

Final Answer: -24

Q30:ipmat indore 2025QA › LogarithmsMediumMCQ · MCQ

The set of all values of

x

satisfying the inequality

\log_{\left(x+\frac{1}{x}\right)}\left[\log_{2}\left(\frac{x-1}{x+2}\right)\right]>0

is

A $(2,5)$
B $(-5,-2)$
C $(5,\infty)$
DNull set

Pick an option to attempt

The Setup: We are solving a nested logarithmic inequality. We need to work from the outside in, paying strict attention to the domain of logarithmic functions, especially the base, which will be the ultimate trap in this problem.

Step 1: Analyze the base of the outer logarithm.
The outer logarithm has a variable base of $b = x + \frac{1}{x}$ .
For any logarithm to be valid, its base must be strictly positive and not equal to $1$ ( $b > 0$ and $b \neq 1$ ).
For $x + \frac{1}{x} > 0$ to hold true, $x$ must be strictly greater than $0$ .
By the AM-GM inequality, if $x > 0$ , then $x + \frac{1}{x} \ge 2$ .
Thus, our base $b$ is always greater than or equal to $2$ (meaning it is securely $> 1$ ).

Step 2: Resolve the outer inequality.
We have an inequality of the form $\log_b(y) > 0$ .
Since we established that our base $b > 1$ , the inequality sign does not flip when we convert it to exponential form.
$y > b^0$
$y > 1$
Substituting our inner expression back in for $y$ :
$\log_2\left(\frac{x-1}{x+2}\right) > 1$

Step 3: Resolve the inner inequality.
Now we solve the second layer. The base here is $2$ , which is also $> 1$ , so the sign stays locked.
$\frac{x-1}{x+2} > 2^1$
$\frac{x-1}{x+2} > 2$

Step 4: Solve the rational inequality.
Subtract $2$ from both sides (never cross-multiply variables when solving inequalities, as you might multiply by a negative and fail to flip the sign):
$\frac{x-1}{x+2} - 2 > 0$
$\frac{x-1 - 2(x+2)}{x+2} > 0$
$\frac{x-1 - 2x - 4}{x+2} > 0$
$\frac{-x-5}{x+2} > 0$
Multiply the entire inequality by $-1$ and flip the sign:
$\frac{x+5}{x+2} < 0$
The critical points are $x = -5$ and $x = -2$ . Testing the intervals gives the valid solution range for this rational expression:
$x \in (-5, -2)$

Step 5: Determine the final intersection (The Trap).
From Step 4, the inner expression requires $x$ to be strictly between $-5$ and $-2$ .
However, from Step 1, the domain of the outer logarithm's base strictly requires $x > 0$ .
Since $x$ cannot be both positive and negative at the same time, the intersection of these two conditions is completely empty.
$x \in (-5, -2) \cap (0, \infty) = \emptyset$

There are no real numbers that can satisfy all the conditions simultaneously.

Final Answer: Null set

Q31:ipmat indore 2025QA › Progression & SeriesMediumMCQ · MCQ

Let

S_1 = \{100, 105, 110, 115, \dots\}

and

S_2 = \{100, 95, 90, 85, \dots\}

be two series in arithmetic progression. If

a_k

and

b_k

are the

k

-th terms of

S_1

and

S_2

, respectively, then

\sum_{k=1}^{20} a_k b_k

equals

A137275
B138250
C137225
D135375

Pick an option to attempt

The Setup: We have two arithmetic progressions (APs). One increases by 5, and the other decreases by 5. We need to find the general algebraic expressions for their $k$ -th terms, multiply them together, and compute the sum of that quadratic sequence from $k=1$ to $20$ .

Step 1: Define the $k$ -th terms.
For the first series $S_1$ , the first term is $100$ and the common difference is $5$ :
$a_k = 100 + (k - 1)5 = 5k + 95$

For the second series $S_2$ , the first term is $100$ and the common difference is $-5$ :
$b_k = 100 + (k - 1)(-5) = -5k + 105$

Step 2: Formulate the product $a_k b_k$ .
Multiply the two general terms together:
$a_k b_k = (5k + 95)(-5k + 105)$
Expand the brackets:
$a_k b_k = -25k^2 + 525k - 475k + 9975$
$a_k b_k = -25k^2 + 50k + 9975$

Step 3: Set up the summation.
We need to find the sum of this expression from $k=1$ to $20$ . Using summation properties, we can split it into three separate parts:
$\sum_{k=1}^{20} a_k b_k = -25\sum_{k=1}^{20} k^2 + 50\sum_{k=1}^{20} k + \sum_{k=1}^{20} 9975$

Step 4: Evaluate the standard sums.
Let's calculate the values for the sum of squares and the sum of integers for $n=20$ .
* Sum of squares: $\sum k^2 = \frac{n(n+1)(2n+1)}{6}$
$\sum_{k=1}^{20} k^2 = \frac{20 \times 21 \times 41}{6} = 2870$
* Sum of integers: $\sum k = \frac{n(n+1)}{2}$
$\sum_{k=1}^{20} k = \frac{20 \times 21}{2} = 210$
* Sum of a constant: $\sum_{k=1}^{20} c = n \times c$
$\sum_{k=1}^{20} 9975 = 20 \times 9975 = 199500$

Step 5: Calculate the final value.
Substitute the calculated sums back into our master equation:
$\text{Total Sum} = -25(2870) + 50(210) + 199500$
$\text{Total Sum} = -71750 + 10500 + 199500$
$\text{Total Sum} = -61250 + 199500 = 138250$

The math is perfectly locked in.

Final Answer: 138250

Q32:ipmat indore 2025QA › ProbabilityMediumMCQ · MCQ

A and B take part in a rifle shooting match. The probability of A hitting the target is 0.4, while the probability of B hitting the target is 0.6. If A has the first shot, post which both strike alternately, then the probability that A hits the target before B hits it is

A $\frac{1}{2}$
B $\frac{10}{19}$
C $\frac{2}{3}$
D $\frac{9}{19}$

Pick an option to attempt

The Setup: This is a classic infinite geometric probability game. Since they strike alternately, A can win on the 1st shot, 3rd shot, 5th shot, and so on. We need to calculate the probability of A winning on any of these odd-numbered turns and sum them up using the infinite geometric series formula.

Step 1: Define the individual probabilities.
* Probability A hits ( $P(A)$ ) = $0.4$
* Probability A misses ( $P(A')$ ) = $1 - 0.4 = 0.6$
* Probability B hits ( $P(B)$ ) = $0.6$
* Probability B misses ( $P(B')$ ) = $1 - 0.6 = 0.4$

Step 2: Map out the winning scenarios for A.
* Scenario 1 (A wins on turn 1): A hits immediately.
$P(\text{Win 1}) = 0.4$
* Scenario 2 (A wins on turn 3): A misses, B misses, A hits.
$P(\text{Win 3}) = P(A') \times P(B') \times P(A) = 0.6 \times 0.4 \times 0.4 = 0.096$
* Scenario 3 (A wins on turn 5): A misses, B misses, A misses, B misses, A hits.
$P(\text{Win 5}) = (P(A') \times P(B'))^2 \times P(A) = (0.24)^2 \times 0.4$

Step 3: Apply the infinite geometric series formula.
The probabilities form an infinite geometric progression where:
* First term ( $a$ ) = $0.4$ (Probability of winning on the first shot)
* Common ratio ( $r$ ) = $P(A') \times P(B') = 0.6 \times 0.4 = 0.24$ (The probability of a full cycle of misses before it's A's turn again)

The sum of an infinite GP is $S = \frac{a}{1 - r}$ .

Step 4: Calculate the final probability.
$S = \frac{0.4}{1 - 0.24}$
$S = \frac{0.4}{0.76}$
Multiply the numerator and denominator by $100$ to clear the decimals:
$S = \frac{40}{76}$
Divide both by their greatest common divisor ( $4$ ):
$S = \frac{10}{19}$

The math holds up perfectly.

Final Answer: $\frac{10}{19}$

Q33:ipmat indore 2025QA › Divisibility RulesMediumMCQ · MCQ

Which of the following numbers is divisible by

3^{10}+2

?

A $3^{20}+4$
B $3^{30}+2$
C $3^{20}+8$
D $3^{30}+8$

Pick an option to attempt

The Setup: We can use algebraic identities and the Factor Theorem to solve this quickly without dealing with massive exponents. By substituting a variable for the base term, we can test the options as simple polynomials.

Step 1: Define the variable.
Let $x = 3^{10}$ .
The divisor given in the question is $x + 2$ .
We are looking for an option that is completely divisible by $(x + 2)$ . According to the Factor Theorem, if a polynomial $P(x)$ is divisible by $(x + 2)$ , then $P(-2)$ must equal $0$ .

Step 2: Translate the options into polynomials.
* (A) $3^{20} + 4 = (3^{10})^2 + 4 = x^2 + 4$
* (B) $3^{30} + 2 = (3^{10})^3 + 2 = x^3 + 2$
* (C) $3^{20} + 8 = (3^{10})^2 + 8 = x^2 + 8$
* (D) $3^{30} + 8 = (3^{10})^3 + 8 = x^3 + 8$

Step 3: Apply the Factor Theorem (Test $x = -2$ ).
Let's plug $x = -2$ into our options to see which one zeros out:
* (A) $(-2)^2 + 4 = 4 + 4 = 8 \neq 0$
* (B) $(-2)^3 + 2 = -8 + 2 = -6 \neq 0$
* (C) $(-2)^2 + 8 = 4 + 8 = 12 \neq 0$
* (D) $(-2)^3 + 8 = -8 + 8 = 0$

Since option (D) evaluates exactly to $0$ , it is perfectly divisible by $(x + 2)$ .

Step 4: The Algebraic Proof (Double Check Protocol).
We can confirm this using the sum of cubes identity: $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ .
Let $a = 3^{10}$ and $b = 2$ .
$a^3 + b^3 = (3^{10})^3 + 2^3 = 3^{30} + 8$
$3^{30} + 8 = (3^{10} + 2)((3^{10})^2 - 2(3^{10}) + 2^2)$
Since $(3^{10} + 2)$ is explicitly pulled out as a factor, $3^{30} + 8$ is strictly divisible by it. The logic is fully locked in.

Final Answer: $3^{30}+8$

Q34:ipmat indore 2025QA › Set TheoryMediumMCQ · MCQ

Let

A

and

B

be two finite sets where the values of

n(A-B)

,

n(A\cap B)

, and

n(B-A)

form an arithmetic progression. (Note:

n(X)

represents the total number of elements within a finite set

X

). Given that

n(A\cup B) = 18

, determine the exact value of

n(A) + n(B)

.

A24
B27
C30
D36

Pick an option to attempt

The Setup: We are dropping into a crossover between Set Theory and Arithmetic Progressions. By breaking the sets down into their mutually exclusive, non-overlapping regions, we can algebraicize the entire Venn diagram and solve it without needing the exact value of the common difference.

Step 1: Define the AP regions.
A Venn diagram with two sets is made of three distinct zones: strictly A ( $A-B$ ), the overlap ( $A \cap B$ ), and strictly B ( $B-A$ ).
Since these three regions form an arithmetic progression, we can represent their element counts using a standard AP format with middle term $a$ and common difference $d$ :
* $n(A-B) = a - d$
* $n(A\cap B) = a$
* $n(B-A) = a + d$

Step 2: Solve for the middle term.
We are given that the total number of elements in the union $n(A \cup B)$ is 18.
The union is simply the sum of our three distinct zones:
$n(A \cup B) = n(A-B) + n(A \cap B) + n(B-A)$
Substitute our AP variables:
$18 = (a - d) + a + (a + d)$
$18 = 3a$
$a = 6$
We now know the exact size of the intersection: $n(A \cap B) = 6$ .

Step 3: Calculate the target sum.
The question asks for the total value of $n(A) + n(B)$ . Let's build both sets from our defined zones:
* $n(A) = n(A-B) + n(A \cap B) = (a - d) + a = 2a - d$
* $n(B) = n(B-A) + n(A \cap B) = (a + d) + a = 2a + d$

Now, add them together:
$n(A) + n(B) = (2a - d) + (2a + d)$
$n(A) + n(B) = 4a$

Since we already locked in $a = 6$ , we just multiply:
$n(A) + n(B) = 4(6) = 24$

The $-d$ and $+d$ perfectly cancel each other out, making the math completely flawless.

Final Answer: 24

Q35:ipmat indore 2025QA › Permutation & CombinationEasyMCQ · MCQ

Determine the total count of integers strictly greater than 5000 and exactly divisible by 5 that can be constructed using the digits {1, 3, 5, 7, 8, 9} without any digit repetition.

A276
B180
C120
D240

Pick an option to attempt

The Setup: This is a classic combinatorics and permutation problem. We need to build numbers that satisfy three conditions simultaneously: they must be strictly greater than 5000, end in 5 (to be a multiple of 5), and feature no repeating digits. Since the length of the integer isn't artificially restricted to four digits, we must account for all possible 4-digit, 5-digit, and 6-digit combinations.

Step 1: Analyze the divisibility constraint.
For an integer to be divisible by 5, its unit digit must be 0 or 5. Since our available pool is $\{1, 3, 5, 7, 8, 9\}$ , the unit digit is strictly locked to 5. This leaves us with a remaining working pool of 5 digits: $\{1, 3, 7, 8, 9\}$ .

Step 2: Calculate valid 4-digit numbers.
Format: `_ _ _ 5`
To ensure the number is $> 5000$ , the first digit (thousands place) must be 5 or greater. Since 5 is already deployed at the unit place, the first digit can only be 7, 8, or 9.
* Thousands place: $3$ options (7, 8, or 9).
* Hundreds and Tens places: We have $4$ remaining digits to fill $2$ open spots. This gives us $P(4,2) = 4 \times 3 = 12$ ways.
* Total valid 4-digit numbers = $3 \times 12 = 36$ .

Step 3: Calculate valid 5-digit numbers.
Format: `_ _ _ _ 5`
Any 5-digit number formed from this set is automatically $> 5000$ .
* Unit place: Locked to $5$ .
* Remaining 4 spots: We have $5$ available digits to fill $4$ open spots. This is $P(5,4) = 5 \times 4 \times 3 \times 2 = 120$ ways.
* Total valid 5-digit numbers = $120$ .

Step 4: Calculate valid 6-digit numbers.
Format: `_ _ _ _ _ 5`
Any 6-digit number formed from this set is automatically $> 5000$ .
* Unit place: Locked to $5$ .
* Remaining 5 spots: We have $5$ available digits to fill $5$ open spots. This is $P(5,5) = 5! = 120$ ways.
* Total valid 6-digit numbers = $120$ .

Step 5: Final tally.
Add up all the valid configurations to find the absolute total:
$\text{Total} = 36 + 120 + 120 = 276$

The math is completely flawless.

Final Answer: 276

Q36:ipmat indore 2025QA › RemainderMediumMCQ · MCQ

Calculate the exact remainder obtained when the expression

11^{1011} + 1011^{11}

is divided by 9.

A0
B8
C9
D7

Pick an option to attempt

The Setup: We are dropping into modular arithmetic. Instead of computing astronomical numbers, we will break the expression down and apply modulo $9$ properties to the bases first, then handle the exponents using cyclicity and exponent rules.

Step 1: Analyze the first term ( $11^{1011} \pmod 9$ ).
First, reduce the base by finding the remainder of $11$ divided by $9$ :
$11 \equiv 2 \pmod 9$
So, the problem simplifies to:
$11^{1011} \equiv 2^{1011} \pmod 9$
To solve this quickly, look for a power of $2$ that is close to a multiple of $9$ . We know that $2^3 = 8$ , which is exactly $1$ less than $9$ .
$2^3 \equiv -1 \pmod 9$
Let's rewrite the exponent $1011$ as a multiple of $3$ :
$1011 = 3 \times 337$
Substitute this back into our expression:
$2^{1011} = (2^3)^{337} \equiv (-1)^{337} \pmod 9$
Since $337$ is an odd number, $-1$ raised to an odd power remains $-1$ .
$-1 \pmod 9 \equiv 8 \pmod 9$
The first term leaves a remainder of $8$ .

Step 2: Analyze the second term ( $1011^{11} \pmod 9$ ).
To find $1011 \pmod 9$ , apply the divisibility rule for $9$ by summing its digits:
$1 + 0 + 1 + 1 = 3$
So, $1011 \equiv 3 \pmod 9$ .
This means our second term simplifies to:
$1011^{11} \equiv 3^{11} \pmod 9$
Notice what happens when you square $3$ :
$3^2 = 9 \equiv 0 \pmod 9$
Since $3^2$ is a perfect multiple of $9$ , any higher power of $3$ (like $3^{11}$ ) will also contain that $9$ as a factor, instantly zeroing out the remainder.
$3^{11} \equiv 0 \pmod 9$
The second term leaves a remainder of $0$ .

Step 3: Combine the remainders.
Add the individual remainders together to get the final result:
$\text{Total Remainder} = 8 + 0 = 8$

The modular logic is fully locked in.

Final Answer: 8

IPMAT Indore 2025 — QA

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