IPMAT Indore 2024 — QA

The Setup: We are given the equation $4^{\log_{2}x}-4x+9^{\log_{3}y}-16y+68=0$. We need to evaluate the expression $y-x$.

Step 1: Simplify the logarithmic terms using the base-change exponent identity $a^{\log_a b} = b$.
For the first term:
$$4^{\log_{2}x} = (2^2)^{\log_{2}x} = 2^{2\log_{2}x} = 2^{\log_{2}(x^2)} = x^2$$
For the third term:
$$9^{\log_{3}y} = (3^2)^{\log_{3}y} = 3^{2\log_{3}y} = 3^{\log_{3}(y^2)} = y^2$$

Step 2: Substitute the simplified terms back into the algebraic equation.
$$x^2 - 4x + y^2 - 16y + 68 = 0$$

Step 3: Complete the square for both the $x$ and $y$ variables.
Isolate the respective variables and add the required constants:
$$(x^2 - 4x + 4) + (y^2 - 16y + 64) = 0$$
$$(x - 2)^2 + (y - 8)^2 = 0$$
Note that $4 + 64 = 68$, which perfectly balances the original constant.

Step 4: Solve for $x$ and $y$.
The sum of two real squares equals zero if and only if each independent square evaluates to zero.
$x - 2 = 0 \Rightarrow x = 2$
$y - 8 = 0 \Rightarrow y = 8$
Both values are strictly positive, satisfying the logarithmic domain restrictions.

Step 5: Calculate the final target expression $y - x$.
$$8 - 2 = 6$$

Final Answer: 6

The Setup: A fruit seller has oranges, apples, and bananas in a ratio of $3:6:7$. The quantity of oranges is a multiple of both $5$ and $6$, and we must find the minimum total number of fruits.

Step 1: Define the quantities using a common scaling factor. Let $k$ be a positive integer.
Oranges = $3k$
Apples = $6k$
Bananas = $7k$
Total Fruits = $3k + 6k + 7k = 16k$

Step 2: Apply the divisibility constraint to the number of oranges.
The quantity of oranges ($3k$) must be a multiple of both $5$ and $6$.
Calculate the Least Common Multiple (LCM) of $5$ and $6$:
$$\text{LCM}(5, 6) = 30$$
Therefore, $3k$ must be a multiple of $30$.

Step 3: Determine the minimum valid scaling factor $k$.
Let $m$ be a positive integer such that:
$$3k = 30m \Rightarrow k = 10m$$
To minimize the total number of fruits, we must minimize $k$, which occurs when $m = 1$. Thus, $k = 10$.

Step 4: Calculate the total number of fruits using the minimal scale factor.
$$\text{Total} = 16(10) = 160$$

Final Answer: 160

The Setup: We must find the number of real solutions for the exponential equation $(x^{2}-15x+55)^{x^{2}-5x+6}=1$.

Step 1: Evaluate Condition 1 where the exponent is $0$ and the base is non-zero.
$$x^2 - 5x + 6 = 0$$
$$(x - 2)(x - 3) = 0$$
This yields potential solutions $x = 2$ and $x = 3$. We must verify the base is non-zero for these values:
For $x = 2$, base = $2^2 - 15(2) + 55 = 29 \neq 0$. (Valid)
For $x = 3$, base = $3^2 - 15(3) + 55 = 19 \neq 0$. (Valid)

Step 2: Evaluate Condition 2 where the base is exactly $1$.
$$x^2 - 15x + 55 = 1$$
$$x^2 - 15x + 54 = 0$$
$$(x - 6)(x - 9) = 0$$
This yields solutions $x = 6$ and $x = 9$. Both are valid for any real exponent.

Step 3: Evaluate Condition 3 where the base is $-1$ and the exponent is an even integer.
$$x^2 - 15x + 55 = -1$$
$$x^2 - 15x + 56 = 0$$
$$(x - 7)(x - 8) = 0$$
This yields potential solutions $x = 7$ and $x = 8$. We must verify the exponent is even:
For $x = 7$, exponent = $7^2 - 5(7) + 6 = 49 - 35 + 6 = 20$. (Even $\Rightarrow$ Valid)
For $x = 8$, exponent = $8^2 - 5(8) + 6 = 64 - 40 + 6 = 30$. (Even $\Rightarrow$ Valid)

Step 4: Aggregate all valid real solutions.
The complete set of solutions is $\{2, 3, 6, 9, 7, 8\}$.
Counting these unique values yields $6$ distinct solutions.

Final Answer: 6

The Setup: Person A borrows Rs. $4000$ for $4$ years, splitting the principal between a $3\%$ simple interest rate and a $4\%$ simple interest rate. The total interest earned is Rs. $520$, and we must find the amount borrowed at $3\%$.

Step 1: Define the variables for the split principal.
Let $x$ represent the principal amount borrowed at the $3\%$ rate.
Let $4000 - x$ represent the remaining principal borrowed at the $4\%$ rate.

Step 2: Calculate the annualized interest yield.
The total interest accrued over $4$ years is $520$. Because it is simple interest, the annual interest is constant:
$$\text{Annual Interest} = \frac{520}{4} = 130$$

Step 3: Construct the linear equation for the annual interest.
$$0.03x + 0.04(4000 - x) = 130$$

Step 4: Solve for $x$.
Distribute the terms and isolate the variable:
$$0.03x + 160 - 0.04x = 130$$
$$-0.01x = -30$$
$$x = 3000$$

Final Answer: 3000

The Setup: We need to calculate the total number of triangles that can be formed with integer sides and a fixed perimeter of $15$.

Step 1: Define the basic parameters and constraints.
Let the integer sides of the triangle be $a$, $b$, and $c$, strictly ordered such that $a \le b \le c$.
The perimeter condition dictates:
$$a + b + c = 15$$

Step 2: Apply the triangle inequality theorem to bound the longest side $c$.
The sum of the two shorter sides must strictly exceed the longest side:
$$a + b > c$$
Substitute $a + b = 15 - c$ into the inequality:
$$15 - c > c \Rightarrow 2c < 15 \Rightarrow c \le 7.5$$
Since $c$ is an integer, the maximum valid dimension for $c$ is $7$.
Furthermore, $c$ is the maximum side, so it must be at least the average length of the perimeter:
$$c \ge \frac{15}{3} = 5$$
Thus, $c \in \{5, 6, 7\}$.

Step 3: Systematically evaluate integer pairs $(a, b)$ for each possible value of $c$, maintaining $a \le b \le c$.
Case 1: $c = 5$
Requires $a + b = 10$. The only integer pair satisfying $a \le b \le 5$ is $(5, 5)$. (Yields $1$ triangle)
Case 2: $c = 6$
Requires $a + b = 9$. The pairs satisfying $a \le b \le 6$ are $(3, 6)$ and $(4, 5)$. (Yields $2$ triangles)
Case 3: $c = 7$
Requires $a + b = 8$. The pairs satisfying $a \le b \le 7$ are $(1, 7)$, $(2, 6)$, $(3, 5)$, and $(4, 4)$. (Yields $4$ triangles)

Step 4: Aggregate the valid triangle formations.
$$\text{Total Triangles} = 1 + 2 + 4 = 7$$

Final Answer: 7

The Setup: A $3 \times 3$ matrix is given with elements $[x_1, x_2, 7]$, $[y_1, y_2, y_3]$, and $[z_1, 8, 3]$ such that the sums across any row, column, or diagonal are identical. We must find its determinant.

Step 1: Determine the magic sum $S$ and the center element.
Let $S$ equal the constant sum of any row, column, or diagonal.
Analyze the bottom row:
$$z_1 + 8 + 3 = S \Rightarrow z_1 = S - 11$$
Analyze the right-to-left diagonal:
$$7 + y_2 + z_1 = S$$
Substitute $z_1$ into the diagonal equation:
$$7 + y_2 + (S - 11) = S \Rightarrow y_2 - 4 = 0 \Rightarrow y_2 = 4$$
In a $3 \times 3$ magic square, the central element is always exactly $\frac{1}{3}$ of the magic sum.
$$S = 3(y_2) = 12$$

Step 2: Populate the remaining elements of matrix $A$ using $S = 12$.
From Step 1, $z_1 = 12 - 11 = 1$.
Left-to-Right Diagonal:
$$x_1 + y_2 + 3 = 12 \Rightarrow x_1 + 4 + 3 = 12 \Rightarrow x_1 = 5$$
Top Row:
$$5 + x_2 + 7 = 12 \Rightarrow x_2 = 0$$
Left Column:
$$5 + y_1 + 1 = 12 \Rightarrow y_1 = 6$$
Right Column:
$$7 + y_3 + 3 = 12 \Rightarrow y_3 = 2$$

Step 3: Construct the populated matrix $A$.
$$A = \begin{bmatrix} 5 & 0 & 7 \\ 6 & 4 & 2 \\ 1 & 8 & 3 \end{bmatrix}$$

Step 4: Calculate the determinant $|A|$ by expanding along the top row.
$$|A| = 5((4)(3) - (8)(2)) - 0 + 7((6)(8) - (1)(4))$$
$$|A| = 5(12 - 16) + 7(48 - 4)$$
$$|A| = 5(-4) + 7(44) = -20 + 308 = 288$$

Final Answer: 288

The Setup: We are asked to determine the number of factors of the integer $1800$ that are also multiples of $6$.

Step 1: Extract the prime factorization of $1800$.
$$1800 = 18 \times 100 = (2 \times 3^2) \times (2^2 \times 5^2)$$
$$1800 = 2^3 \times 3^2 \times 5^2$$
Any generic factor of $1800$ takes the structure $2^a \times 3^b \times 5^c$, where constraints are $0 \le a \le 3$, $0 \le b \le 2$, and $0 \le c \le 2$.

Step 2: Apply the multiple-of-$6$ constraint to the exponents.
Because $6 = 2^1 \times 3^1$, any factor that is a multiple of $6$ must include at least one $2$ and at least one $3$ in its prime factorization.
The restricted exponent ranges become:
$a \in \{1, 2, 3\}$ (yielding $3$ valid choices)
$b \in \{1, 2\}$ (yielding $2$ valid choices)
$c \in \{0, 1, 2\}$ (yielding $3$ valid choices, as $5$ is unconstrained)

Step 3: Calculate the combinatorics of the restricted factor set.
Multiply the independent choices together:
$$\text{Total Factors} = 3 \times 2 \times 3 = 18$$

Final Answer: 18

The Setup: We are looking for the global median of a massive, perfectly sorted dataset. Since we know the strict hierarchical order of the companies ($A < B < C \dots$), the entire server population is already sorted in ascending order. We just need to locate exactly which company houses the middle employee and extract their stats. It is an absolute 'Where's Waldo?' situation, but with array indices.

Step 1: Calculate the total server population. We sum up all employees across the companies to find our $N$:
$$32 + 28 + 43 + 39 + 35 + 29 + 23 + 16 = 245$$

Step 2: Find the index of the global median. Since the total $N$ is odd ($245$), the median is simply the exact middle value in the sorted list.
$$\frac{245 + 1}{2} = 123$$
We need to find the exact target coordinates for the $123$rd employee overall.

Step 3: Track the cumulative frequencies to locate the target's spawn zone.
* Company A: $32$ employees (Cumulative: $32$)
* Company B: $28$ employees (Cumulative: $32 + 28 = 60$)
* Company C: $43$ employees (Cumulative: $60 + 43 = 103$)
* Company D: $39$ employees (Cumulative: $103 + 39 = 142$)

Since $103 < 123 \le 142$, the $123$rd employee lives right inside Company D's roster.

Step 4: Pinpoint the exact age of this specific NPC. The $123$rd employee overall is exactly the $20$th employee within Company D (since $123 - 103 = 20$).
Company D has exactly $39$ employees. Let's find Company D's local median index:
$$\frac{39 + 1}{2} = 20$$
The $20$th employee *is* the exact median of Company D! Since the table explicitly states the median age of Company D is $45$, the $20$th employee's age is hard-locked at $45$.

Final Answer: 45

The Setup: A right-angled triangle has vertex $B$ at the right angle, with legs $AB = BC = 18$. We need to find the area of the largest inscribed rectangle sharing vertex $B$.

Step 1: Establish the coordinate system.
Place the right-angled vertex $B$ at the origin $(0,0)$.
Given leg lengths $AB = BC = 18$, set vertex $A$ at $(0,18)$ and vertex $C$ at $(18,0)$.
The hypotenuse $AC$ forms a line passing through $(0,18)$ and $(18,0)$. Its linear equation is:
$$x + y = 18 \Rightarrow y = 18 - x$$

Step 2: Define the area function of the inscribed rectangle.
A rectangle sharing vertex $B$ at $(0,0)$ and bounded by the triangle will have its opposing vertex $(x, y)$ strictly on the hypotenuse $AC$.
The Area ($A$) is the product of its length and width:
$$A_{\text{rect}} = x \times y = x(18 - x) = 18x - x^2$$

Step 3: Maximize the quadratic area function.
The function is a downward-opening parabola. We find its maximum via differentiation (or the vertex formula $x = \frac{-b}{2a}$):
$$\frac{d A_{\text{rect}}}{dx} = 18 - 2x = 0 \Rightarrow x = 9$$

Step 4: Calculate the maximum area footprint.
Substitute $x = 9$ into the bounding equation to find $y$:
$$y = 18 - 9 = 9$$
$$\text{Max Area} = 9 \times 9 = 81$$

Final Answer: 81

The Setup: We must find the number of integer coordinate pairs $(x,y)$ that satisfy the absolute value inequality $|x-5|+|y-5| \le 6$.

Step 1: Translate the bounded region to center it at the origin.
Let $u = x - 5$ and $v = y - 5$. Since $x$ and $y$ are elements of $\mathbb{Z}$, $u$ and $v$ must also be elements of $\mathbb{Z}$.
Substitute into the inequality:
$$|u| + |v| \le 6$$
This bounded region forms a solid square rotated $45^{\circ}$ on the Cartesian plane.

Step 2: Apply the lattice point summation formula for Manhattan boundaries.
The exact number of integer coordinate pairs $(u, v)$ satisfying $|u| + |v| \le k$ for an integer $k \ge 0$ is governed by the discrete sequence formula $2k(k+1) + 1$.

Step 3: Evaluate the formula for the target boundary distance $k = 6$.
$$\text{Total Pairs} = 2(6)(6+1) + 1$$
$$\text{Total Pairs} = 2(6)(7) + 1 = 84 + 1 = 85$$
Because the translation mapping $(x,y) \rightarrow (u,v)$ is a bijective $1:1$ map, the count remains identical for the original uncentered inequality.

Final Answer: 85

The Setup: This is a pure min-max puzzle. We need to minimize the total sum of an array while anchored by a fixed median and bounded by a strict lower limit from the previous dataset (Company E). We are basically min-maxing a character build for the absolute lowest possible stats to clear this challenge.

Step 1: Analyze Company F's required median. Company F has $29$ employees. The median index is:
$$\frac{29 + 1}{2} = 15$$
So, the $15$th employee's age must be exactly $54$ (from the table). To minimize the total sum, all employees from index $15$ to $29$ should be exactly $54$ years old. Going any higher is an automatic L for our minimum sum objective.

Step 2: Find the absolute minimum age for the first $14$ employees in F. Because of the strict inequality rule, the youngest person in F must be strictly older than the oldest person in E.
$$f_1 > e_{35}$$
We need to shrink E's oldest age as much as possible to give F a lower floor. Company E has $35$ employees with a median age of $49$. The median is the $18$th employee.
We can set E's entire upper half to exactly $49$:
$$e_{18} = e_{19} = \dots = e_{35} = 49$$
So, the maximum age in E can be successfully nerfed down to $49$.

Step 3: Set the lower half of F's ages. Since $f_1 > e_{35}$, and $e_{35} = 49$, the lowest possible valid integer age for any employee in Company F is $50$.
We generously assign this bare minimum age to all employees below F's median:
$$f_1 = f_2 = \dots = f_{14} = 50$$

Step 4: Calculate the final minimized sum for Company F.
We have $14$ employees at age $50$, and $15$ employees (the median and everyone above) at age $54$.
$$\text{Sum} = (14 \times 50) + (15 \times 54)$$
$$\text{Sum} = 700 + 810 = 1510$$

Final Answer: 1510

The Setup: Out of $150$ students, $52$ like tea, $48$ like juice, and $62$ like coffee. Since everyone likes at least one drink, we must maximize the number of students who like more than one drink.

Step 1: Define the cardinalities of the sets.
Let the subsets of students be $T$ (Tea), $J$ (Juice), and $C$ (Coffee).
$|T| = 52$, $|J| = 48$, and $|C| = 62$.
The sum of independent choices is $52 + 48 + 62 = 162$.
Because every student likes at least one drink, the total union encapsulates the whole group: $|T \cup J \cup C| = 150$.

Step 2: Construct the overlapping set union equation.
Let $x$ denote the exact count of students liking precisely two drinks.
Let $y$ denote the exact count of students liking precisely three drinks.
The Inclusion-Exclusion Principle formula for exact counts states:
$$|T \cup J \cup C| = (|T| + |J| + |C|) - x - 2y$$
$$150 = 162 - x - 2y \Rightarrow x + 2y = 12$$

Step 3: Formulate and execute the optimization constraint.
We must maximize the parameter of students liking *more than one* drink, which equates mathematically to maximizing the sum $(x + y)$.
Rewrite $(x + y)$ by isolating $x$ in our established equation ($x = 12 - 2y$):
$$x + y = (12 - 2y) + y = 12 - y$$
To maximize the function $(12 - y)$, we must apply the minimum valid boundary for $y$. Since cardinalities must be non-negative integers, the minimum for $y$ is $0$.

Step 4: Evaluate the maximum value.
If $y = 0$, then $x = 12$.
$$\text{Max}(x + y) = 12 + 0 = 12$$

Final Answer: 12

The Setup: A chocolate's price undergoes a sequential $x\%$ increase followed by an $x\%$ decrease. The resulting price is $96.76\%$ of the original price, and we must determine the value of $x$.

Step 1: Construct the sequential price multiplier function.
Let $P$ represent the baseline price.
An $x\%$ increase applies a multiplier of $(1 + \frac{x}{100})$.
An $x\%$ decrease applies a subsequent multiplier of $(1 - \frac{x}{100})$.
$$\text{New Price} = P \times \left(1 + \frac{x}{100}\right) \times \left(1 - \frac{x}{100}\right)$$

Step 2: Simplify via the difference of squares identity.
$$\text{New Price} = P \left(1 - \frac{x^2}{10000}\right)$$

Step 3: Equate to the provided net proportional change.
The final state is $96.76\%$ (or $0.9676$) of $P$.
$$P \left(1 - \frac{x^2}{10000}\right) = 0.9676 P$$
Divide out $P$ from both sides since initial price is arbitrary:
$$1 - \frac{x^2}{10000} = 0.9676$$

Step 4: Solve for the absolute rate $x$.
$$\frac{x^2}{10000} = 1 - 0.9676 = 0.0324$$
$$x^2 = 324$$
Because a percentage rate scaling magnitude must be positive, take the principal square root:
$$x = 18$$

Final Answer: 18

The Setup: We are given the functions $f(x)=|x+|x||$ and $g(x)=1/x$ for non-zero $x$. Based on the constraint $f(a)+g(f(a))=13/6$, we must find the maximum possible value of $f(g(a))$.

Step 1: Analyze the piecewise domains of $f(x) = |x + |x||$.
If $x \le 0$, then $|x| = -x \Rightarrow f(x) = |x - x| = 0$.
If $x > 0$, then $|x| = x \Rightarrow f(x) = |x + x| = 2x$.

Step 2: Ascertain the domain of $a$ using the given composite equation.
The given condition is $f(a) + g(f(a)) = \frac{13}{6}$.
The function $g(x) = \frac{1}{x}$ is fundamentally undefined at $0$. Consequently, $f(a)$ cannot equal $0$.
By our piecewise analysis, if $f(a) \neq 0$, then $a$ must be strictly positive ($a > 0$), locking $f(a) = 2a$.

Step 3: Formulate and solve the rational equation for $a$.
Substitute $f(a) = 2a$ into the equation:
$$2a + g(2a) = \frac{13}{6}$$
$$2a + \frac{1}{2a} = \frac{13}{6}$$
Let $u = 2a$ to clarify the quadratic structure:
$$u + \frac{1}{u} = \frac{13}{6}$$
Multiply entirely by $6u$:
$$6u^2 + 6 = 13u \Rightarrow 6u^2 - 13u + 6 = 0$$
Factor the resulting quadratic:
$$(2u - 3)(3u - 2) = 0$$
Thus, $u = \frac{3}{2}$ or $u = \frac{2}{3}$.
Because $u = 2a$, we trace back to two valid positive candidates for $a$:
$$a = \frac{3}{4} \quad \text{or} \quad a = \frac{1}{3}$$

Step 4: Evaluate the maximization query for $f(g(a))$.
Since $a > 0$, $g(a) = \frac{1}{a} > 0$. Any positive input to $f$ triggers the $2x$ piecewise condition.
$$f(g(a)) = f\left(\frac{1}{a}\right) = 2\left(\frac{1}{a}\right) = \frac{2}{a}$$
Evaluate against both derived candidates of $a$:
If $a = \frac{3}{4} \Rightarrow \frac{2}{3/4} = \frac{8}{3} \approx 2.67$
If $a = \frac{1}{3} \Rightarrow \frac{2}{1/3} = 6$
The strict mathematical maximum is $6$.

Final Answer: 6

The Setup: The terms of a geometric progression are real and positive, with the $p$-th term equal to $q$ and the $q$-th term equal to $p$. We need to find the logarithm of the first term.

Step 1: Define the sequence algebraically.
Let the first term be $a$ and the common ratio be $r$.
The general term formula gives:
$$T_p = a r^{p-1} = q$$
$$T_q = a r^{q-1} = p$$

Step 2: Apply logarithms to both equations.
Taking the log of both sides yields a system of linear equations:
$$\log(a) + (p-1)\log(r) = \log(q)$$
$$\log(a) + (q-1)\log(r) = \log(p)$$

Step 3: Eliminate $\log(r)$ to solve for $\log(a)$.
Subtract the second equation from the first:
$$(p-q)\log(r) = \log(q) - \log(p) \implies \log(r) = \frac{\log(q) - \log(p)}{p-q}$$
Substitute $\log(r)$ back into the first equation:
$$\log(a) = \log(q) - (p-1)\left(\frac{\log(q) - \log(p)}{p-q}\right)$$