IPMAT Indore 2022 — SA

The Setup: We are given a logarithmic equation involving bases of $x^2$ and $y^2$, alongside a quadratic relation $y=x^2-30$. We must simplify the log equation to find $x$ and $y$, then compute $x^2+y^2$.

Step 1: Simplify the logarithmic equation.
Use the base power rule $\log_{(b^k)} a = \frac{1}{k} \log_b a$:

$$\frac{1}{2}\log_x y + \frac{1}{2}\log_y x = 1$$

Multiply by 2:

$$\log_x y + \log_y x = 2$$

Step 2: Solve for the relationship between $x$ and $y$.
Let $a = \log_x y$. By logarithmic inversion, $\log_y x = \frac{1}{a}$.

$$a + \frac{1}{a} = 2$$

$$a^2 - 2a + 1 = 0 \implies (a - 1)^2 = 0 \implies a = 1$$

Therefore, $\log_x y = 1$, which strictly implies $y = x$.

Step 3: Apply the quadratic relation to find the valid coordinates.
Substitute $y = x$ into the second equation $y = x^2 - 30$:

$$x = x^2 - 30 \implies x^2 - x - 30 = 0$$

Factor the quadratic:

$$(x - 6)(x + 5) = 0 \implies x = 6 \text{ or } x = -5$$

Logarithmic bases ($x^2$ and $y^2$) and arguments ($x$ and $y$) must be strictly positive, discarding $x = -5$.
Thus, $x = 6$, and consequently $y = 6$.

Step 4: Calculate the final target expression.

$$x^2 + y^2 = 6^2 + 6^2 = 36 + 36 = 72$$

Final Answer: 72

The Setup: The inequality $2|x| + 3|y| \le 6$ forms a geometric shape (a rhombus) bounded by absolute value constraints. We must calculate its total enclosed area.

Step 1: Find the intercepts of the boundary lines.
Consider the boundary equation in the first quadrant where $x \ge 0$ and $y \ge 0$:

$$2x + 3y = 6$$

Find the $x$-intercept (set $y = 0$): $2x = 6 \implies x = 3$.
Find the $y$-intercept (set $x = 0$): $3y = 6 \implies y = 2$.
The vertices of the full rhombus across all four quadrants are $(\pm 3, 0)$ and $(0, \pm 2)$.

Step 2: Calculate the area of the bounded shape.
The area of a rhombus is given by $\frac{1}{2} \times d_1 \times d_2$, where $d_1$ and $d_2$ are the lengths of the diagonals.
The horizontal diagonal connects $(-3, 0)$ to $(3, 0)$, so $d_1 = 6$.
The vertical diagonal connects $(0, -2)$ to $(0, 2)$, so $d_2 = 4$.

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

Final Answer: 12

The Setup: A speed increase from $12\text{ km/hr}$ to $20\text{ km/hr}$ results in a $1$-hour time reduction to cover a fixed distance. We must build a time-difference equation to find the distance.

Step 1: Formulate the time equation.
Let $d$ be the fixed distance between the home and office.
Time at normal speed = $\frac{d}{12}$.
Time at increased speed = $\frac{d}{20}$.
The difference between these times is exactly $1$ hour:

$$\frac{d}{12} - \frac{d}{20} = 1$$

Step 2: Solve for $d$.
Find a common denominator ($60$) to combine the fractions:

$$\frac{5d}{60} - \frac{3d}{60} = 1$$

$$\frac{2d}{60} = 1 \implies \frac{d}{30} = 1 \implies d = 30$$

Final Answer: 30

The Setup: We are choosing points to form triangles from two parallel lines containing $7$ and $5$ points respectively.

Step 1: Identify the combinatorial cases for triangle formation.
A triangle requires exactly $3$ non-collinear vertices. Since the points lie on two parallel lines, taking $3$ points from the same line is invalid. Thus, we must select either:
* Case 1: $1$ point from the first line and $2$ points from the second line.
* Case 2: $2$ points from the first line and $1$ point from the second line.

Step 2: Calculate configurations for Case 1.
Choose $1$ point from the line with $7$, and $2$ points from the line with $5$:

$$\text{Ways}_1 = \binom{7}{1} \times \binom{5}{2} = 7 \times \left(\frac{5 \times 4}{2}\right) = 7 \times 10 = 70$$

Step 3: Calculate configurations for Case 2.
Choose $2$ points from the line with $7$, and $1$ point from the line with $5$:

$$\text{Ways}_2 = \binom{7}{2} \times \binom{5}{1} = \left(\frac{7 \times 6}{2}\right) \times 5 = 21 \times 5 = 105$$

Step 4: Sum the valid triangle configurations.

$$\text{Total Triangles} = 70 + 105 = 175$$

Final Answer: 175

The Setup: We have a system of profit/loss constraints across two different scenarios for an unknown quantity of apples and mangoes with fixed unit costs. We must solve the linear equations to find the exact number of apples.

Step 1: Define variables and calculate per-item cost variations.
Let $A$ be the number of apples (Cost = $20$ each).
Let $M$ be the number of mangoes (Cost = $25$ each).
* $+10\%$ on apples = $+2$ profit per apple.
* $-20\%$ on mangoes = $-5$ loss per mango.
* $-20\%$ on apples = $-4$ loss per apple.
* $+10\%$ on mangoes = $+2.5$ profit per mango.

Step 2: Formulate Scenario 1 to find the quantity ratio.
In the first scenario, the net profit is exactly $0$:

$$2A - 5M = 0 \implies 2A = 5M \implies M = 0.4A$$

Step 3: Formulate Scenario 2 to solve for $A$.
In the second scenario, the net loss is $150$ (meaning profit is $-150$):

$$-4A + 2.5M = -150$$

Multiply by $-1$ for clarity:

$$4A - 2.5M = 150$$

Step 4: Substitute the ratio and calculate $A$.
Substitute $M = 0.4A$ into the second equation:

$$4A - 2.5(0.4A) = 150$$

$$4A - 1A = 150$$

$$3A = 150 \implies A = 50$$

Final Answer: 50

The Setup: We must find the minimum value of a sum of absolute value functions $f(x)=|x|+2|x-1|+|x-2|+|x-4|+|x-6|+2|x-10|$.

Step 1: Translate the function into a median problem.
The minimum value of a sum of absolute deviations $\sum c_i |x - a_i|$ occurs at the median of the data points $a_i$, weighted by their frequencies $c_i$.
List out the points based on their respective coefficients:
* $0$ (Frequency 1)
* $1$ (Frequency 2)
* $2$ (Frequency 1)
* $4$ (Frequency 1)
* $6$ (Frequency 1)
* $10$ (Frequency 2)
The complete ordered data set contains $8$ points: $\{0, 1, 1, 2, 4, 6, 10, 10\}$.

Step 2: Identify the median of the set.
Since there is an even number of data points ($n=8$), the minimum is achieved anywhere along the interval between the 4th and 5th values.
The 4th point is $2$ and the 5th point is $4$. Any $x \in [2, 4]$ minimizes $f(x)$.

Step 3: Calculate the minimum value $f(x)$.
Substitute any value from the valid interval (e.g., $x = 2$) into $f(x)$:

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

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

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

Final Answer: 26

The Setup: A specific permutation matrix $A$ is given. We must establish the cyclical property of its powers to evaluate the absolute determinant of a matrix polynomial $(A^9+A^6+A^3+A)$.

Step 1: Establish the power cycle of Matrix $A$.

$$A = \begin{bmatrix}1&0&0\\0&0&1\\0&1&0\end{bmatrix}$$

Calculate $A^2$:

$$A^2 = \begin{bmatrix}1&0&0\\0&0&1\\0&1&0\end{bmatrix} \begin{bmatrix}1&0&0\\0&0&1\\0&1&0\end{bmatrix} = \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix} = I$$

Since $A^2 = I$, the powers of $A$ alternate: odd powers yield $A$, and even powers yield the Identity matrix $I$.

Step 2: Simplify the matrix polynomial.
The expression is $A^9 + A^6 + A^3 + A$.
Apply the parity rules:
* $A^9 = A$ (odd)
* $A^6 = I$ (even)
* $A^3 = A$ (odd)
* $A = A$ (odd)

$$\text{Sum Matrix} = A + I + A + A = 3A + I$$

Step 3: Construct the elements of the final matrix.

$$3A = \begin{bmatrix}3&0&0\\0&0&3\\0&3&0\end{bmatrix}$$

Add the Identity matrix $I$:

$$3A + I = \begin{bmatrix}4&0&0\\0&1&3\\0&3&1\end{bmatrix}$$

Step 4: Calculate the determinant.
Expand along the top row:

$$\det(3A+I) = 4(1 \times 1 - 3 \times 3) = 4(1 - 9) = 4(-8) = -32$$

The absolute value of the determinant is $|-32| = 32$.

Final Answer: 32

The Setup: We are asked to find the arithmetic sum of the coefficients of all terms within the binomial expansion of $(5x-9)^4$.

Step 1: Apply the polynomial coefficient property.
For any polynomial $P(x)$, the sum of its fully expanded coefficients is mathematically obtained by evaluating the polynomial at $x = 1$.

Step 2: Evaluate the expression at $x = 1$.

$$P(1) = (5(1) - 9)^4$$

$$P(1) = (5 - 9)^4$$

$$P(1) = (-4)^4$$

$$P(1) = 256$$

Final Answer: 256

The Setup: A sequence of positive integers removes all perfect squares. We must determine the numerical value of the $2022^{\text{nd}}$ remaining integer in this sequence.

Step 1: Establish the formula for sequence terms.
Let $N$ be the actual value of the term. The sequence loses one position for every perfect square it passes.
The number of perfect squares less than or equal to $N$ is $\lfloor\sqrt{N}\rfloor$.
Thus, the position of $N$ in the modified sequence is given by: $N - \lfloor\sqrt{N}\rfloor = 2022$.

Step 2: Estimate $N$ to find bounds.
Because $N$ will be slightly larger than $2022$, we estimate $\sqrt{N} \approx \sqrt{2022}$.
$44^2 = 1936$ and $45^2 = 2025$. Therefore, the square root will be bounded by approximately $45$.

Step 3: Test the logical value for $N$.
Add our $45$ estimated skipped squares to the $2022^{\text{nd}}$ position:
Let's test $N = 2022 + 45 = 2067$.
Check how many perfect squares exist up to $2067$:
The largest integer square less than $2067$ is $45^2 = 2025$. Thus, exactly $45$ squares are skipped.
Subtract the skipped squares to verify the position:

$$\text{Position} = 2067 - 45 = 2022$$

The number $2067$ precisely lands on the $2022^{\text{nd}}$ position, and it is not itself a perfect square, so it genuinely survives into the new sequence.

Step 4: Confirm the answer is unique.
The counting function $N - \lfloor\sqrt{N}\rfloor$ never decreases as $N$ grows, so it takes the value $2022$ on at most one non-square $N$. Hence $2067$ is the only possibility.

Final Answer: 2067

The Setup: We are given constraint equations $\sin\alpha+\sin\beta = \frac{\sqrt{2}}{\sqrt{3}}$ and $\cos\alpha+\cos\beta = \frac{1}{\sqrt{3}}$. We need to evaluate the compound trigonometric expression $\left(20\cos\left(\frac{\alpha-\beta}{2}\right)\right)^2$.

Step 1: Square and add the two primary equations.
Equation 1: $(\sin\alpha+\sin\beta)^2 = \left(\frac{\sqrt{2}}{\sqrt{3}}\right)^2 \implies \sin^2\alpha + \sin^2\beta + 2\sin\alpha\sin\beta = \frac{2}{3}$
Equation 2: $(\cos\alpha+\cos\beta)^2 = \left(\frac{1}{\sqrt{3}}\right)^2 \implies \cos^2\alpha + \cos^2\beta + 2\cos\alpha\cos\beta = \frac{1}{3}$
Summing them together:

$$(\sin^2\alpha + \cos^2\alpha) + (\sin^2\beta + \cos^2\beta) + 2(\cos\alpha\cos\beta + \sin\alpha\sin\beta) = \frac{2}{3} + \frac{1}{3}$$

Using Pythagorean and subtraction identities:

$$1 + 1 + 2\cos(\alpha-\beta) = 1 \implies 2 + 2\cos(\alpha-\beta) = 1$$

Step 2: Relate the expression to the half-angle formula.
Factor out the $2$:

$$2(1 + \cos(\alpha-\beta)) = 1$$

Apply the power-reduction identity $1 + \cos\theta = 2\cos^2\left(\frac{\theta}{2}\right)$:

$$2 \left(2\cos^2\left(\frac{\alpha-\beta}{2}\right)\right) = 1$$

$$4\cos^2\left(\frac{\alpha-\beta}{2}\right) = 1$$

Step 3: Evaluate the target expression.
The target is $\left(20\cos\left(\frac{\alpha-\beta}{2}\right)\right)^2$.
Expand the target:

$$400\cos^2\left(\frac{\alpha-\beta}{2}\right)$$

Substitute the derived identity block:

$$100 \times \left[4\cos^2\left(\frac{\alpha-\beta}{2}\right)\right] = 100 \times 1 = 100$$

Final Answer: 100

The Setup: Three selected terms ($3^{\text{rd}}, 14^{\text{th}}, 69^{\text{th}}$) of an Arithmetic Progression (AP) form a separate Geometric Progression (GP). The 4th term of this GP corresponds to the $n^{\text{th}}$ term of the AP. We must find index $n$.

Step 1: Establish the GP relationship.
Let the AP have a first term $a$ and common difference $d$.
$T_3 = a + 2d$
$T_{14} = a + 13d$
$T_{69} = a + 68d$
Because these form a GP, the square of the middle term equals the product of the outer terms:

$$(a + 13d)^2 = (a + 2d)(a + 68d)$$

$$a^2 + 26ad + 169d^2 = a^2 + 70ad + 136d^2$$

Step 2: Relate the starting term to the difference.

$$33d^2 = 44ad$$

Assuming $d \neq 0$ (otherwise all terms are identical and limits break), divide by $11d$:

$$3d = 4a \implies a = \frac{3}{4}d$$

Step 3: Calculate the GP common ratio.
Substitute $a = 0.75d$ into the first two terms of the GP:
$T_3 = 0.75d + 2d = 2.75d = \frac{11}{4}d$
$T_{14} = 0.75d + 13d = 13.75d = \frac{55}{4}d$
The common ratio $r = \frac{T_{14}}{T_3} = \frac{55/4}{11/4} = 5$.
The third term is $T_{69} = 0.75d + 68d = 68.75d = \frac{275}{4}d$. (Verified: $\frac{55}{4}d \times 5 = \frac{275}{4}d$).

Step 4: Calculate the next GP term and map it to the AP index.
The 4th GP term is $T_{69} \times r$:

$$\text{Next GP Term} = \frac{275}{4}d \times 5 = \frac{1375}{4}d$$

Set this equal to the generic $n^{\text{th}}$ term of the AP ($a + (n-1)d$):

$$\frac{3}{4}d + (n-1)d = \frac{1375}{4}d$$

$$(n-1)d = \frac{1372}{4}d \implies n-1 = 343 \implies n = 344$$

Final Answer: 344

The Setup: We need to compute the total number of elements generated through four successive iterations of the power set operation $P(X)$, starting with a null set $A$.

Step 1: Track the set cardinality through each power set iteration.
The cardinality property dictates that if a set $X$ has $k$ elements, its power set $P(X)$ will have $2^k$ elements.
* Iteration 0: The set $A$ is a null set. $|A| = 0$.
* Iteration 1: $|P(A)| = 2^0 = 1$.
* Iteration 2: $|P(P(A))| = 2^1 = 2$.
* Iteration 3: $|P(P(P(A)))| = 2^2 = 4$.
* Iteration 4: $|P(P(P(P(A))))| = 2^4 = 16$.

Final Answer: 16

The Setup: Three expressions involving exponential powers of $2$ and constants are given as consecutive terms of an Arithmetic Progression. We must solve for the unknown exponent $x$.

Step 1: Establish the arithmetic mean equation.
For three terms $A, B, C$ in an AP, $2B = A + C$.
Given terms: $A = -16$, $B = 2^{x+3} - 2^{2x-1} - 16$, and $C = 2^{2x-1} + 16$.

$$2(2^{x+3} - 2^{2x-1} - 16) = -16 + (2^{2x-1} + 16)$$

Step 2: Simplify the algebraic constraints.

$$2 \cdot 2^{x+3} - 2 \cdot 2^{2x-1} - 32 = 2^{2x-1}$$

Combine the exponential terms:

$$2^{x+4} - 2^{2x} - 32 = 2^{2x-1}$$

Group the terms involving $2x$ on the right:

$$2^{x+4} - 32 = 2^{2x-1} + 2^{2x}$$

Factor out $2^{2x}$:

$$2^{x+4} - 32 = 2^{2x} \left(\frac{1}{2} + 1\right) = \frac{3}{2} \cdot 2^{2x}$$

Step 3: Use substitution to solve the resulting quadratic.
Let $y = 2^x$. Then $2^{x+4} = 16y$, and $2^{2x} = y^2$.

$$16y - 32 = \frac{3}{2}y^2$$

Multiply by $2$ and rearrange into standard quadratic form:

$$3y^2 - 32y + 64 = 0$$

Factor the quadratic equation:

$$(3y - 8)(y - 8) = 0$$

Step 4: Evaluate valid roots for integer $x$.
This yields $y = 8/3$ or $y = 8$.
Since $y = 2^x$, resolving for $y = 8$ yields $2^x = 8 \implies x = 3$.
*(The fractional root $y=8/3$ yields an irrational logarithmic decimal out of scope for standard numeric data entry).*

Final Answer: 3

The Setup: A combinatorics problem where 8 individuals are arranged at a round table. We apply circular permutation blocks and complementary counting to satisfy simultaneous grouping constraints (one pair together, one pair separated).

Step 1: Calculate permutations satisfying the first constraint.
Total people = 8. Constraint 1: The Sharma pair must sit together.
Bind the Sharmas into a single block. We now have 7 distinct entities to arrange in a circle (the Sharma block + 6 others).

$$\text{Circular arrangements} = (n-1)! = (7-1)! = 6!$$

The Sharmas can also swap seats internally within their block ($2!$).

$$\text{Total ways (Sharmas together)} = 6! \times 2 = 720 \times 2 = 1440$$

Step 2: Calculate overlapping violations (both pairs together).
Constraint 2: The Ahuja pair must NOT sit together. We find the restricted subset where they *do* sit together to subtract it from our total.
Bind the Sharmas into one block and the Ahujas into another block. We now have 6 distinct entities to arrange.

$$\text{Circular arrangements} = (6-1)! = 5!$$

Both the Sharma block and the Ahuja block have internal swaps ($2! \times 2!$).

$$\text{Ways (Both pairs together)} = 5! \times 2 \times 2 = 120 \times 4 = 480$$

Step 3: Apply complementary counting.
Subtract the violating cases from the initial grouping constraints:

$$\text{Valid Seating} = (\text{Sharmas together}) - (\text{Both pairs together})$$

$$\text{Valid Seating} = 1440 - 480 = 960$$

Final Answer: 960

The Setup: We are asked to maximize the total sum of $50$ distinct positive integers bounded by $100$. The average of the top $25$ integers is rigidly offset $50$ units above the average of the bottom $25$ integers.

Step 1: Translate the average constraint into sum parameters.
Let $S_L$ be the sum of the largest $25$ integers, and $S_S$ be the sum of the remaining (smallest) $25$ integers.

$$\text{Average}(S_L) - \text{Average}(S_S) = 50$$

$$\frac{S_L}{25} - \frac{S_S}{25} = 50 \implies S_L - S_S = 1250$$

The goal is to maximize the grand total sum: $S_L + S_S$.

Step 2: Express the objective strictly in terms of $S_L$.
Rearranging the constraint yields $S_S = S_L - 1250$.

$$\text{Total Sum} = S_L + (S_L - 1250) = 2S_L - 1250$$

To maximize the Total Sum, we strictly need to maximize the component $S_L$.

Step 3: Maximize $S_L$ using number properties.
The set consists of strictly distinct positive integers with a ceiling of $100$.
To maximize $S_L$ (the sum of the 25 largest items), we pick the absolute highest available discrete integers descending from $100$: $\{100, 99, 98, \dots, 76\}$.
Using the arithmetic series sum formula:

$$S_L = \frac{n}{2}(a_1 + a_n) = \frac{25}{2}(100 + 76) = 25 \times 88 = 2200$$

Step 4: Verify the lower boundary $S_S$ is physically viable.
With $S_L = 2200$, the forced offset calculates $S_S = 2200 - 1250 = 950$.
We must verify if $25$ distinct integers smaller than $76$ can legitimately sum to $950$.
The absolute minimum sum of $25$ distinct positive integers is $1 + 2 + \dots + 25 = 325$.
The largest such sum is $51 + 52 + \dots + 75 = 1575$.
Because $325 \le 950 \le 1575$, the required $S_S$ sits inside the achievable range. An explicit set works: $\{26, 27, \dots, 50\}$ consists of $25$ distinct integers, all below $76$, and sums to $\frac{25}{2}(26 + 50) = 25 \times 38 = 950$. So the configuration is genuinely realisable, not merely arithmetically consistent.

Step 5: Compute the final maximum total.

$$\text{Max Total Sum} = S_L + S_S = 2200 + 950 = 3150$$

Final Answer: 3150