Create your PRT table (simplest way to solve it). We know that both working together (A + N) can complete the work in 12 days. $\begin{array}{l|c|c|c} \text{Person: } &amp; A+N &amp; A &amp; N \\ \hline \text{Rate:} &amp; &amp; \\ \hline \text{Time:} &amp;12 &amp; \\ \hline \text{Total Work:} &amp; &amp; \\ \end{array}$ If Aprita does 40% of the work and then Nikita does 60%, they'll complete the work in 24 days. Let's assume the total work to be a bigger number that is divisible by 24 too (for ease in calculation), say 240 units. We know that Rate $\times$ Time $=$ Work Done. $\begin{array}{l|c|c|c} \text{Person: } &amp; A+N &amp; A &amp;N \\ \hline \text{Rate:} &amp;20 &amp; \\ \hline \text{Time:} &amp;12 &amp; \\ \hline \text{Total Work:} &amp; 240 &amp;0.4 \times 240 = 96&amp;0.6 \times 240 = 144 \\ \end{array}$ Let's assume the time taken by $A$ and $N$ as $x$ and $y$ respectively. $\begin{array}{l|c|c|c} \text{Person: } &amp; A+N &amp; A &amp;N \\ \hline \text{Rate:} &amp;20 &amp; \\ \hline \text{Time:} &amp;12 &amp;x &amp;y \\ \hline \text{Total Work:} &amp; 240 &amp;96&amp;144 \\ \end{array}$ Using the formula, we can get the rate: $\begin{array}{l|c|c|c} \text{Person: } &amp; A+N &amp; A &amp;N \\ \hline \text{Rate:} &amp;20 &amp; \frac{96}{x}&amp; \frac{144}{y} \\ \hline \text{Time:} &amp;12 &amp;x &amp;y \\ \hline \text{Total Work:} &amp; 240 &amp;96&amp;144 \\ \end{array}$ Now, we have two equations: $x+y = 24$ (they took 24 days to complete the job) $\frac{96}{x} + \frac{144}{y} = 20$ (because their rate while working together is 20) Since the numbers are clean, we can do hit &amp; trial (try $x=12, y=12$) and get the answer too. <hr class="my-[18px]" node="[object Object]"> Let $y = 24-x$. Taking the 2nd equation and solving using the quadratic formula: $\begin{aligned} \frac{96}{x} + \frac{144}{24 - x} &amp;= 20\\ 96(24 - x) + 144x &amp;= 20x(24 - x)\\ 2304 + 48x &amp;= 480x - 20x^2\\ 20x^2 - 432x + 2304 &amp;= 0\\ 5x^2 - 108x + 576 &amp;= 0\\ \Delta = 144,\quad x &amp;= \frac{108 \pm 12}{10} = \{12,\,9.6\}\\ y = 24 - x &amp;= \{12,\,14.4\} \end{aligned}$ <hr class="my-[18px]" node="[object Object]"> Hence, one of the answers is $x=12, y=12$. $\begin{array}{l|c|c|c} \text{Person: } &amp; A+N &amp; A &amp;N \\ \hline \text{Rate:} &amp;20 &amp; \\ \hline \text{Time:} &amp;12 &amp;x &amp;y \\ \hline \text{Total Work:} &amp; 240 &amp;96&amp;144 \\ \end{array}$ Let's plug the same: $\begin{array}{l|c|c|c} \text{Person: } &amp; A+N &amp; A &amp;N \\ \hline \text{Rate:} &amp;20 &amp; 8 &amp; 12 \\ \hline \text{Time:} &amp;12 &amp;12 &amp;12 \\ \hline \text{Total Work:} &amp; 240 &amp;96&amp;144 \\ \end{array}$ So, Nikita's rate of work is $12$. If she had to do the full work (of 240 units), she'll take $\dfrac{240}{12} = 20$ days. <hr class="my-[18px]" node="[object Object]"> You can also solve with $x= 9.6, y = 14.4$. You'll get a different answer (but also valid). IPMAT Indore 2025 gave marks to both the answers.

5 teams $×$ 15 members (in each team) $=$ 75 total members. Equal distribution across skill sets $=$ 25 biologists, 25 geologists, 25 explorers. <hr class="my-[18px]" node="[object Object]"> Step 1. Create your data-structure (table). If this is properly done, half of the headache is solved: <table class="question-table"><tbody><tr><th>Team</th><th>Biologists</th><th>Geologists</th><th>Explorers</th><th>Total</th></tr><tr><td node="[object Object]">A</td><td node="[object Object]"></td><td node="[object Object]"></td><td node="[object Object]"></td><td node="[object Object]">15</td></tr><tr><td node="[object Object]">B</td><td node="[object Object]"></td><td node="[object Object]"></td><td node="[object Object]"></td><td node="[object Object]">15</td></tr><tr><td node="[object Object]">C</td><td node="[object Object]"></td><td node="[object Object]"></td><td node="[object Object]"></td><td node="[object Object]">15</td></tr><tr><td node="[object Object]">D</td><td node="[object Object]"></td><td node="[object Object]"></td><td node="[object Object]"></td><td node="[object Object]">15</td></tr><tr><td node="[object Object]">E</td><td node="[object Object]"></td><td node="[object Object]"></td><td node="[object Object]"></td><td node="[object Object]">15</td></tr><tr><td node="[object Object]">Total</td><td node="[object Object]">25</td><td node="[object Object]">25</td><td node="[object Object]">25</td><td node="[object Object]">75</td></tr></tbody></table>

To find the maximum value of $b$ where $a$, $b$, $c$ are distinct natural numbers less than 100 satisfying $|a - b| + |b - c| = |c - a|$. <hr class="my-[18px]" node="[object Object]"> First, let's analyze the equation $|a - b| + |b - c| = |c - a|$ This equation is true if and only if $b$ lies between $a$ and $c$ (either $a &lt; b &lt; c$ or $c &lt; b &lt; a$). <hr class="my-[18px]" node="[object Object]"> To maximize $b$, we need to make it as close to 100 as possible while ensuring it satisfies our constraint. Let's try $b = 98$ with two possible arrangements: <hr class="my-[18px]" node="[object Object]"> Case 1: If $a &lt; b &lt; c$ $b = 98$ $c = 99$ (largest possible value less than 100) $a = 1$ (can be any number less than 98) Verifying: $|a - b| + |b - c| = |1 - 98| + |98 - 99| = 97 + 1 = 98$ $|c - a| = |99 - 1| = 98$ <hr class="my-[18px]" node="[object Object]"> Case 2: If $c &lt; b &lt; a$ $b = 98$ $a = 99$ (largest possible value less than 100) $c = 1$ (can be any number less than 98) Verifying: $|a - b| + |b - c| = |99 - 98| + |98 - 1| = 1 + 97 = 98$ $|c - a| = |1 - 99| = 98$ <hr class="my-[18px]" node="[object Object]"> Since both cases work with $b = 98$ and we cannot choose a larger value for $b$ (as we need distinct natural numbers less than 100), the maximum possible value of $b$ is 98.

If a tournament has $n$ teams and they play against every other team once, the number of matches $=\boxed{\dfrac{n(n-1)}{2}}$ Total matches with 8 teams $= \dfrac{8 \times 7}{2} = 28$ matches <hr class="my-[18px]" node="[object Object]"> When one team got suspended after playing 3 matches: <ul node="[object Object]"> <li node="[object Object]">3 matches were already played involving the suspended team <ul node="[object Object]"> <li node="[object Object]">The suspended team would have played 7 other teams in total</li> <li node="[object Object]">Since it played 3 matches before suspension, it missed playing 4 matches</li> </ul> </li> </ul> <hr class="my-[18px]" node="[object Object]"> Total matches played $=$ Complete tournament matches $-$ Matches not played due to suspension Total matches played $= 28 - 4 = 24$

$\ln \dfrac{a}{b}, \quad \ln \dfrac{a}{b\sqrt{b}}, \quad \ln \dfrac{a}{b^{2}}, \quad \ln \dfrac{a}{b^{2}\sqrt{b}}, \dots$ The power of $b$ in the denominator follows the pattern: $1, 1.5, 2, 2.5, \dots$ This means the exponent increases by $0.5$ each time. <hr class="my-[18px]" node="[object Object]"> For the $n$th term, the exponent of $b$ in the denominator would be: $1 + (n-1) \times 0.5 = 0.5n + 0.5$ So the $n$th term is: $\ln \dfrac{a}{b^{0.5n+0.5}} = \ln \dfrac{a}{b^{0.5(n+1)}}$ <hr class="my-[18px]" node="[object Object]"> To find the sum of the first 21 terms, we have: $\sum_{i=1}^{21} \ln \dfrac{a}{b^{0.5(i+1)}}$ Using logarithm properties: $\ln \dfrac{a}{b^{0.5(i+1)}} = \ln a - \ln b^{0.5(i+1)} = \ln a - 0.5(i+1)\ln b$ <hr class="my-[18px]" node="[object Object]"> Now, we can calculate the sum: $\sum_{i=1}^{21} \ln \dfrac{a}{b^{0.5(i+1)}} = \sum_{i=1}^{21} (\ln a - 0.5(i+1)\ln b)$ $= 21\ln a - 0.5\ln b \sum_{i=1}^{21} (i+1)$ $= 21\ln a - 0.5\ln b \sum_{i=2}^{22} i$ <hr class="my-[18px]" node="[object Object]"> The sum of integers from 2 to 22 is: $\dfrac{(2+22)(22-2+1)}{2} = \dfrac{24 \cdot 21}{2} = 252$ <hr class="my-[18px]" node="[object Object]"> Therefore: $\sum_{i=1}^{21} \ln \dfrac{a}{b^{0.5(i+1)}} = 21\ln a - 0.5\ln b \cdot 252$ $= 21\ln a - 126\ln b$ $= \ln a^{21} - \ln b^{126}$ $= \ln \dfrac{a^{21}}{b^{126}}$ <hr class="my-[18px]" node="[object Object]"> Comparing with $\ln \dfrac{a^m}{b^n}$, we get $m = 21$ and $n = 126$ Therefore, $m + n = 21 + 126 = 147$

<ul node="[object Object]"> <li node="[object Object]">Students who didn't appear for Math: $x$ <ul node="[object Object]"> <li node="[object Object]">Students who didn't appear for English: $2x$ (as given)</li> <li node="[object Object]">Students who appeared but failed English: $y$ (this is what we need to find)</li> <li node="[object Object]">Students who appeared but failed Math: $z$</li> <li node="[object Object]">Students who passed Math: $2y$ (given)</li> <li node="[object Object]">Students who passed English: $2z$ (given)</li> </ul> </li> </ul> <hr class="my-[18px]" node="[object Object]"> For each exam, we can calculate the number of students who appeared: <ul node="[object Object]"> <li node="[object Object]">Students who appeared for Math: $120 - x$ <ul node="[object Object]"> <li node="[object Object]">Students who appeared for English: $120 - 2x$</li> </ul> </li> </ul> <hr class="my-[18px]" node="[object Object]"> The students who appeared can be split into those who passed and those who failed: <ul node="[object Object]"> <li node="[object Object]">For Math: $(120 - x) = 2y + z$ <ul node="[object Object]"> <li node="[object Object]">For English: $(120 - 2x) = 2z + y$</li> </ul> </li> </ul> <hr class="my-[18px]" node="[object Object]"> From the second equation: $y = 120 - 2x - 2z$ Substitute this into the first equation: $(120 - x) = 2(120 - 2x - 2z) + z$ $(120 - x) = 240 - 4x - 4z + z$ $(120 - x) = 240 - 4x - 3z$ $-x + 4x = 240 - 120 - 3z$ $3x = 120 - 3z$ $x = 40 - z$ <hr class="my-[18px]" node="[object Object]"> Substitute back to find $y$: $y = 120 - 2x - 2z$ $y = 120 - 2(40 - z) - 2z$ $y = 120 - 80 + 2z - 2z$ $y = 40$ <hr class="my-[18px]" node="[object Object]"> Therefore, the number of students who appeared but failed the English exam is $40$.

$A^2 = \begin{bmatrix} 2 &amp; n \\ 4 &amp; 1 \end{bmatrix} \begin{bmatrix} 2 &amp; n \\ 4 &amp; 1 \end{bmatrix} = \begin{bmatrix} 4 + 4n &amp; 3n \\ 12 &amp; 4n + 1 \end{bmatrix}$ $A^3 = \begin{bmatrix} 4 + 4n &amp; 3n \\ 12 &amp; 4n + 1 \end{bmatrix} \begin{bmatrix} 2 &amp; n \\ 4 &amp; 1 \end{bmatrix} = \begin{bmatrix} 8 + 20n &amp; 7n + 4n^2 \\ 28 + 16n &amp; 16n + 1 \end{bmatrix}$ <hr class="my-[18px]" node="[object Object]"> Comparing the above $A^3$ value with the given data: $27\begin{bmatrix} 4 &amp; q \\ p &amp; r \end{bmatrix} = \begin{bmatrix} 108 &amp; 27q \\ 27p &amp; 27r \end{bmatrix}=\boxed{\begin{bmatrix} 8 + 20n &amp; 7n + 4n^2 \\ 28 + 16n &amp; 16n + 1 \end{bmatrix}}$ $8 + 20n = 108 \implies n = 5$ $7n + 4n^2 = 27q \implies 35 + 100 = 27q \implies q = 5$ $28 + 16n = 27p \implies 28 + 80 = 27p \implies p = 4$ $16n + 1 = 27r \implies 80 + 1 = 27r \implies r = 3$ <hr class="my-[18px]" node="[object Object]"> Therefore, $p + q + r = 4 + 5 + 3 = 12$

In an arithmetic progression, the difference between consecutive terms is constant. Let's call this common difference $d$. $d = \log_3(2x^2 + 1) - \log_3(x^2 - 1) = \log_3\left(\dfrac{2x^2 + 1}{x^2 - 1}\right)$ For the second and third terms: $\log_3(6x^2 + 3) - \log_3(2x^2 + 1) = \log_3\left(\dfrac{6x^2 + 3}{2x^2 + 1}\right)$ For these to form an arithmetic progression, these differences must be equal: $\log_3\left(\dfrac{2x^2 + 1}{x^2 - 1}\right) = \log_3\left(\dfrac{6x^2 + 3}{2x^2 + 1}\right)$ Since the logarithms are equal, their arguments must be equal: $\dfrac{2x^2 + 1}{x^2 - 1} = \dfrac{6x^2 + 3}{2x^2 + 1}$ <hr class="my-[18px]" node="[object Object]"> Cross-multiplying: $(2x^2 + 1)(2x^2 + 1) = (x^2 - 1)(6x^2 + 3)$ Expanding and rearranging: $4x^4 + 4x^2 + 1 = 6x^4 - 6x^2 + 3x^2 - 3$ $-2x^4 + 7x^2 + 4 = 0$ Let $y = x^2$ to simplify: $-2y^2 + 7y + 4 = 0$ $2y^2 - 7y - 4 = 0$ Using the quadratic formula: $y = \dfrac{7 \pm \sqrt{49 + 32}}{4} = \dfrac{7 \pm \sqrt{81}}{4} = \dfrac{7 \pm 9}{4}$ So $y = 4$ or $y = -\dfrac{1}{2}$ Since $y = x^2$, and $x^2$ cannot be negative, $y = 4$ is our only valid solution. Therefore, $x^2 = 4$, so $x = \pm 2$. <hr class="my-[18px]" node="[object Object]"> Calculating the common difference using $x^2 = 4$: $d = \log_3\left(\dfrac{2x^2 + 1}{x^2 - 1}\right) = \log_3\left(\dfrac{2(4) + 1}{4 - 1}\right) = \log_3\left(\dfrac{9}{3}\right) = \log_3(3) = 1$ Calculating the first three terms: $a_1 = \log_3(x^2 - 1) = \log_3(4 - 1) = \log_3(3) = 1$ $a_2 = \log_3(2x^2 + 1) = \log_3(2(4) + 1) = \log_3(9) = 2$ $a_3 = \log_3(6x^2 + 3) = \log_3(6(4) + 3) = \log_3(27) = 3$ We see the pattern: $a_n = n$ Therefore: $a_4 = 4$ $a_5 = 5$ $a_6 = 6$ The sum of the next three terms = $a_4 + a_5 + a_6 = 4 + 5 + 6 = 15$ <hr class="my-[18px]" node="[object Object]"> Alternate Solution Given: $\log _{3}\left(x^{2}-1\right), \log _{3}\left(2 x^{2}+1\right), \log _{3}\left(6 x^{2}+3\right)$ are in AP. For AP: 2(middle term) $=$ first term + third term $\begin{aligned} &amp; 2 \log _{3}\left(2 x^{2}+1\right)=\log _{3}\left(x^{2}-1\right)+\log _{3}\left(6 x^{2}+3\right) \\ &amp; \log _{3}\left(2 x^{2}+1\right)^{2}=\log _{3}\left[\left(x^{2}-1\right)\left(6 x^{2}+3\right)\right] \\ &amp; \left(2 x^{2}+1\right)^{2}=\left(x^{2}-1\right)\left(6 x^{2}+3\right) \\ &amp; 4 x^{4}+4 x^{2}+1=6 x^{4}-3 x^{2}-3 \\ &amp; 2 x^{4}-7 x^{2}-4=0 \\ &amp; x^{2}=4\text { or } x^2=-0.5(\text {not possible}) \\ &amp; x^{2}=4, \text { so } x=2 \\ &amp; \text { Terms: } \log _{3}(3)=1, \log _{3}(9)=2, \log _{3}(27)=3 \\ &amp; \text { Common difference }=1 \\ &amp; \text { Next three terms: } 4,5,6 \\ &amp; \text { Sum }=\mathbf{1 5} \end{aligned}$

We need to find the area of triangle PMD in a square ABCD where a circle of radius $13$ cm touches sides AB and BC at points M and N respectively, and intersects sides CD and DA at points P and Q. Let's place the square in a coordinate system with A at origin $(0,0)$, B at $(18,0)$, C at $(18,18)$, and D at $(0,18)$. <hr class="my-[18px]" node="[object Object]"> The center of the circle must be at a distance of $13$ cm from both sides AB and BC. Since the circle touches side AB (x-axis) and side BC (line $x = 18$), the center is at: $(18-13, 13) = (5, 13)$ <hr class="my-[18px]" node="[object Object]"> Point M is where the circle touches side AB. Since AB is along the x-axis, M is directly below the center: $M = (5, 0)$ <hr class="my-[18px]" node="[object Object]"> To find point P on side CD, we need to find where the circle intersects the line $y = 18$. The equation of the circle is: $(x-5)^2 + (y-13)^2 = 13^2$ Substituting $y = 18$: $(x-5)^2 + (18-13)^2 = 13^2$ $(x-5)^2 + 25 = 169$ $(x-5)^2 = 144$ $x-5 = \pm 12$ So $x = 17$ or $x = -7$ Since P must be on CD, $P = (17, 18)$ <hr class="my-[18px]" node="[object Object]"> Now we have the coordinates: $M = (5, 0)$ $P = (17, 18)$ $D = (0, 18)$ Using the formula for area of a triangle with coordinates: Area = $\dfrac{1}{2}|x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$ Area = $\dfrac{1}{2}|5(18-18) + 17(18-0) + 0(0-18)|$ = $\dfrac{1}{2}|0 + 17(18) + 0|$ = $\dfrac{1}{2}(306)$ = $153$ sq. cm Therefore, the area of triangle PMD is $153$ sq. cm.

ipmat-indore 2025 Complete Paper Solution | SA

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ipmat-indore 2025 Complete Paper Solution | SA

Question 1.

Instructions

Question 2.

<div><div class="latex-wrapper">The number of biologists in team E is _____<br /></div></div>

Question 3.

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

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Question 5.

<div><div class="latex-wrapper">If the sum of the first $21$ terms of the sequence: $\ln \frac{a}{b}, \ln \frac{a}{b \sqrt{b}}, \ln \frac{a}{b^{2}}, \ln \frac{a}{b^{2} \sqrt{b}}, \ldots$ is $\ln \frac{a^{m}}{b^{n}}$, then the value of $m+n$ is $\qquad$<br /></div></div>

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Question 7.

<div><div class="latex-wrapper">If $A = \begin{bmatrix} 2 &amp; n \\ 4 &amp; 1 \end{bmatrix}$ such that $A^3 = 27 \begin{bmatrix} 4 &amp; q \\ p &amp; r \end{bmatrix}$, then $p + q + r$ equals _________<br /></div></div>

Instructions

Question 8.

<div><div class="latex-wrapper">The number of teams having more geologists than biologists is ______<br /></div></div>

Question 9.

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

Question 10.

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