<img src="https://lightyellow-stork-291267.hostingersite.com/wp-content/uploads/2025/12/4.webp">Note: This IPMAT Indore 2025 question is repeated from IPMAT Indore 2020.

Subtract $a$ from both sides: $y - a = b \log_e x$ $\frac{y - a}{b} = \log_e x$ <hr class="my-[18px]" node="[object Object]"> If $\log_e x = k$, then $x = e^k$ So: $x = e^{\frac{y - a}{b}}$ <hr class="my-[18px]" node="[object Object]"> Using exponent rules: $x = e^{\frac{y - a}{b}}$ $= e^{\frac{y}{b} - \frac{a}{b}}$ $= e^{\frac{y}{b}} \cdot e^{-\frac{a}{b}}$ Since $e^{-\frac{a}{b}}$ is a constant, let's call it $k$: $x = k \cdot e^{\frac{y}{b}}$ <hr class="my-[18px]" node="[object Object]"> Raise both sides to the power $b$: $x^b = k^b \cdot e^y$ Since $k^b$ is still a constant: $x^b = \text{constant} \times e^y$ Therefore: $e^y$ is proportional to $x^b$

Given: $a_1, a_2, ..., a_8$ are the roots of $x^8 + x^7 + ... + x + 1 = 0$ Find: $a_1^{2025} + a_2^{2025} + ... + a_8^{2025}$ <hr class="my-[18px]" node="[object Object]"> We have the equation: $x^8 + x^7 + x^6 + x^5 + x^4 + x^3 + x^2 + x + 1 = 0$ This is a geometric series. Using the formula for sum of a geometric series: $\frac{x^9 - 1}{x - 1} = 1 + x + x^2 + ... + x^8$ So our equation becomes: $\frac{x^9 - 1}{x - 1} = 0$ <hr class="my-[18px]" node="[object Object]"> For a fraction to equal zero, the numerator must be zero and denominator must be non-zero. Therefore: $x^9 - 1 = 0$ and $x \neq 1$ This gives us: $x^9 = 1$ and $x \neq 1$ So each root $a_1, a_2, ..., a_8$ satisfies: $a_k^9 = 1$ <hr class="my-[18px]" node="[object Object]"> Now we need to find $a_k^{2025}$ for each root. Notice that $2025 = 9 \times 225$ Therefore: $a_k^{2025} = a_k^{9 \times 225} = (a_k^9)^{225} = (1)^{225} = 1$ <hr class="my-[18px]" node="[object Object]"> Since each of the 8 roots gives us 1 when raised to the power 2025: $a_1^{2025} + a_2^{2025} + a_3^{2025} + ... + a_8^{2025} = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 8$ Answer: $8$

Let the maximum value $=M$ Let the minimum value $=m$ Let the median value $= d$ In a way, we've gotten the 3 numbers $(a, b, c)$. $\text{Mean} = \dfrac{M + m + d}{3} \quad \rightarrow (1)$ <hr class="my-[18px]" node="[object Object]"> From the first condition: $\Rightarrow$ Max$(a, b, c) \ +$ Min$(a, b, c) = 15$ $\Rightarrow M + m = 15 \quad \rightarrow (2)$ <hr class="my-[18px]" node="[object Object]"> From the second condition: $\Rightarrow$ Median$(a, b, c) \ -$ Mean$(a, b, c) = 2$ $\Rightarrow d - \text{Mean} = 2 \quad \rightarrow (3)$ <hr class="my-[18px]" node="[object Object]"> Putting $(1)$ into $(3)$: $\Rightarrow d - \left(\dfrac{M + m + d}{3}\right) = 2$ Putting $(2)$ in the above: $\Rightarrow d - \left(\dfrac{15 + d}{3}\right) = 2$ $\Rightarrow 3d - 15 - d = 6$ $\Rightarrow 2d - 15 = 6$ $\Rightarrow d = \dfrac{21}{2} = 10.5$ Therefore, the median of $a$, $b$, and $c$ is $10.5$.

We start by simplifying the outer logarithm: If $\log_{25}(y) = \frac{1}{2}$, then $y = 25^{\frac12} = 5$. $5 \log_3 (1+\log_3(1+2\log_2x)) = 5$ <hr class="my-[18px]" node="[object Object]"> Dividing both sides by 5: $\log_3 (1+\log_3(1+2\log_2x)) = 1$ Since $\log_3(3) = 1$, we have: $1+\log_3(1+2\log_2x) = 3$ $\therefore \log_3(1+2\log_2x) = 2$ <hr class="my-[18px]" node="[object Object]"> Since $\log_3(9) = 2$, we have: $1+2\log_2x = 9$ Therefore: $2\log_2x = 8$ So: $\log_2x = 4$ <hr class="my-[18px]" node="[object Object]"> $\log_2x = 4$ means $x = 2^4 = 16$ Therefore, $x = 16$ is our answer. <hr class="my-[18px]" node="[object Object]"> Verification: Substituting $x = 16$ back into the original equation confirms our answer.

Let's denote a three-digit number as $abc$ where: $= a$ is the hundreds digit (1, 2, or 3) $= b$ is the tens digit (0-9) $= c$ is the units digit (0-9) We need $a + b + c = 10$ <hr class="my-[18px]" node="[object Object]"> When $a = 1$, we need $b + c = 9$ Possible numbers: $109, 118, 127, 136, 145, 154, 163, 172, 181, 190$ When $a = 2$, we need $b + c = 8$ Possible numbers: $208, 217, 226, 235, 244, 253, 262, 271, 280$ When $a = 3$, we need $b + c = 7$ Possible numbers: $307, 316, 325, 334, 343, 352, 361, 370$ In total, there are $10 + 9 + 8 = 27$ numbers between 100 and 400 with the digit sum $= 10$ <hr class="my-[18px]" node="[object Object]"> A number is divisible by 4 if its last two digits form a number divisible by 4. Numbers divisible by 4: $136, 172, 208, 244, 280, 316, 352$ That gives us 7 numbers (divisible by 4) out of 27 total numbers. <hr class="my-[18px]" node="[object Object]"> $\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{7}{27}$

Calculate the slope of line $AB$: $m_{AB} = \dfrac{y_B - y_A}{x_B - x_A} = \dfrac{1 - 3}{5 - 1} = \dfrac{-2}{4} = -\dfrac{1}{2}$ <hr class="my-[18px]" node="[object Object]"> When two lines with slopes $m_1$ and $m_2$ intersect at angle $θ$, we use the formula: $\tan θ = \left|\dfrac{m_2 - m_1}{1 + m_1m_2}\right|$ In our case, $m_1 = -\dfrac{1}{2}$ and $m_2 = m$. Since the angle is $45°$, $\tan(45°) = 1$. So we have: $1 = \left|\dfrac{m - (-\frac{1}{2})}{1 + (-\frac{1}{2})m}\right| = \left|\dfrac{m + \frac{1}{2}}{1 - \frac{m}{2}}\right|$ <hr class="my-[18px]" node="[object Object]"> Removing the absolute value gives us two cases: Case 1: $\dfrac{m + \frac{1}{2}}{1 - \frac{m}{2}} = 1$ $m + \dfrac{1}{2} = 1 - \dfrac{m}{2}$ $m + \dfrac{m}{2} = 1 - \dfrac{1}{2}$ $\dfrac{3m}{2} = \dfrac{1}{2}$ $m = \dfrac{1}{3}$ <hr class="my-[18px]" node="[object Object]"> Case 2: $\dfrac{m + \frac{1}{2}}{1 - \frac{m}{2}} = -1$ $m + \frac{1}{2} = -1 + \frac{m}{2}$ $m - \frac{m}{2} = -1 - \frac{1}{2}$ $\frac{m}{2} = -\frac{3}{2}$ $m = -3$ <hr class="my-[18px]" node="[object Object]"> Therefore, the possible values of $m$ are $\dfrac{1}{3}$ and $-3$.

Let's evaluate the determinant in the first condition: $\left|\begin{array}{ll}P(0) &amp; P(1) \\P(0) &amp; P(2)\end{array}\right| = P(0)P(2) - P(0)P(1) = P(0)[P(2)-P(1)]$ Since this equals zero, and $P(0)=2 \neq 0$, we must have $P(2) = P(1)$. <hr class="my-[18px]" node="[object Object]"> Since $P(x)$ is quadratic, we can write it as: $P(x) = ax^2 + bx + c$ With $P(0) = c = 2$, our polynomial is: $P(x) = ax^2 + bx + 2$ <hr class="my-[18px]" node="[object Object]"> Using $P(2) = P(1)$: $P(1) = a + b + 2$ $P(2) = 4a + 2b + 2$ Setting them equal: $a + b + 2 = 4a + 2b + 2$ $0 = 3a + b$ $b = -3a$ So our polynomial is now: $P(x) = ax^2 - 3ax + 2$ <hr class="my-[18px]" node="[object Object]"> From $P(1) + P(2) + P(3) = 14$: $P(1) = -2a + 2$ $P(2) = -2a + 2$ $P(3) = 9a - 9a + 2 = 2$ This gives us: $(-2a + 2) + (-2a + 2) + 2 = 14$ $-4a + 6 = 14$ $-4a = 8$ $a = -2$ <hr class="my-[18px]" node="[object Object]"> With $a = -2$ and $b = -3a = 6$, our polynomial is: $P(x) = -2x^2 + 6x + 2$ Therefore: $P(4) = -2(16) + 6(4) + 2 = -32 + 24 + 2 = -6$

ipmat-indore 2025 Complete Paper Solution | MCQ

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ipmat-indore 2025 Complete Paper Solution | MCQ

Question 1.

<div><div class="latex-wrapper">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</div></div>

Question 2.

<div><div class="latex-wrapper">If $y=a+b \log _{e} x$ then which of the following is true?<br /></div></div>

Question 3.

<div><div class="latex-wrapper">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<br /></div></div>

Question 4.

<div><div class="latex-wrapper">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<br /></div></div>

Question 5.

<div><div class="latex-wrapper">If $\log_{25} [5 \log_3 (1+\log_3(1+2\log_2x))] = \frac12$ then $x$ is:<br /></div></div>

Question 6.

<div><div class="latex-wrapper">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<br /></div></div>

Question 7.

<div><div class="latex-wrapper">In triangle $ABC, AB = AC = x, ∠ABC = \theta$ and the circumradius is equal to $y$. Then $\frac{x}{y}$ equals<br /></div></div>

Question 8.

<div><div class="latex-wrapper">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<br /></div></div>

Question 9.

<div><div class="latex-wrapper">Let $A(1,3)$ and $B(5,1)$ be two points. If a line with slope $m$ intersects $AB$ at an angle of $45°$, then the possible values of $m$ are<br /></div></div>

Question 10.

<div><div class="latex-wrapper">Let $P(x)$ be a quadratic polynomial such that<br />$\left|\begin{array}{ll} P(0) &amp; P(1) \\ P(0) &amp; P(2) \end{array}\right|=0$<br />Let $P(0)=2$ and $P(1)+P(2)+P(3)=14$. Then $P(4)$ equals<br /></div></div>