IPMAT Indore 2025 — MCQ

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

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 $x^{b}$

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: $8$

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: $10.5$

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: $16$

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: $\frac{7}{27}$

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: $2\sin(\theta)$

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

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

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: $-6$

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:
$$(x - h)^2 + (y - 4)^2 = h^2$$