Given a series inGP which are positive real numbers ..Let the first term be "a" and common ratio be "r". $p^{th}$ term =$a\times r^{p-1}$ = q $q^{th}$ term =$a\times r^{q-1}$ = p By dividing both, we get $r=\ \left(\dfrac{\ q}{p}\right)^{\ \dfrac{\ 1}{p-q}}$ By substituting this in 2nd equation , we get : $a\left(\ \dfrac{\ q}{p}\right)^{\ \dfrac{q-1\ }{p-q}}=p$ Now applying log on both sides : Log a + $\dfrac{\ q-1}{p-1}\left(\log\dfrac{\ q}{p}\right)$ = log p Therefore , log a = log p $-$ $\ \dfrac{\ q-1}{p-q}\left(\log\ q-\log\ p\right)$ Hence, log a = $\ \dfrac{\ \left(p-1\right)\log\ p\ +\ \left(q-1\right)\log\ q}{p-q}$

Lets assume the distance b/w point and centre of circle as "d" . let radius of circle be R . There are two cases , one case is in which point is inside the circle and the other one is point lies totally outside circle : Case I : Point lying outside the circle<img src="https://lightyellow-stork-291267.hostingersite.com/wp-content/uploads/2025/12/figure-1.png">The shortest distance is PB = PO - BO = d - r = 4The longest distance is PA = PO + OA = d + r = 9Solving this we get r = 2.5 . Case II : Point lying inside the circle<img src="https://lightyellow-stork-291267.hostingersite.com/wp-content/uploads/2025/12/figure-2.png">The shortest distance is PB = OB - OP = r - d = 4The longest distance is PA = PO+ OA = r + d = 9Solving this we get r = 6.5 Therefore the radius of circle can be either 2.5 cm or 6.5 cm

If $\mid x + 1 \mid + (y + 2)^2 = 0$
That means x = -1 and y = -2
Then, only LHS can be equal to zero.
And,$ax - 3ay = 1$
$-a +6a = 1$
$5a=1$
Thus,$a=\dfrac{1}{5}$

The number of real solutions of the equation $x^2 - 10 \mid x \mid - 56 = 0$ isWe need to find the solutions to the equation -$x^2 - 10 \mid x \mid - 56 = 0$Taking two casesCase 1: x 0$x^2-10x-56=0$$x^2-14x+4x-56=0$$x\left(x-14\right)+4\left(x-14\right)=0$$\left(x+4\right)\left(x-14\right)=0$x = -4 and x = 14Since, x 0x = 14 is the only possible value.Case 2: x 0$x^2+10x-56=0$
$x^2+14x-4x-56=0$=$x\left(x+14\right)-4\left(x+14\right)=0$=$\left(x-4\right)\left(x+14\right)=0$x = 4 and x = -14Since, x 0x = -14 is the only possible value.Therefore, there are 2 possible real solutions.

$4^{100}$ can be written as $2^{200}$ which is less than $2^{300}$ . $2^{100}+3^{100}$ $2(3^{100})$ $3^{200}$ Now we have to see which is greatest among $2^{300}$ $3^{200}$ . $2^{300}$ = $2^{3^{100}}$ = $8^{100}$ $3^{200}$ = $3^{2^{100}}$ = $9^{100}$ Therefore,$3^{200}$is greatest among all.

Let 'a' be first term, 'r' be common ratio of an infinite GP : $\ \dfrac{\ a}{1-r}=80$ ..... $1^{st}$ Equation . sum of first two terms = a + ar = a(1 + r) = 35.... Equation 2. Substituting equation 1 in $2^{nd}$ equation : 80(1 - r)(1 + r) = 35 (1 - r)(1 + r) = $1-r^2$ = $\dfrac{\ 7}{16}$ hence,$r=\pm\ \dfrac{\ 3}{4}$ The possible values for (a,r) = (20, 3/4) or (140, -3/4) Sum of first 'n' terms of GP = a($\dfrac{\ 1-r^n}{1-r}$) = 100 If we check for (a,r) = (20, 3/4): It gives us that : $1-\left(\dfrac{\ 3^n}{4^n}\right)$ = $\dfrac{\ 5}{4}$ . But 1 minus some positive value can never be$\dfrac{\ 5}{4}$ which is greater than 1. Hence, this possibility is ignored. Now, the only possibility is (a,r) =(140, -3/4) : It gives us that :$1-\dfrac{\ \left(-1\right)^n3^n}{4^n}=\dfrac{\ 5}{4}$ This implies , $\dfrac{-1}{4}=\dfrac{\left(-1\right)^n3^n}{4^n}$ Therefore, closest integral value of n is 5.

Let's start by taking cases for the number of balloons a boy can get. We'll consider the following cases:
<ul><li>The maximum balloons any boy gets is 1: In this case, all boys will have to have at least one balloon (1 way)and there will be $17$ balloons that will be distributed among girls in $^{17-1}C_{5-1} = ^{16}C_4 = 1820$ ways. </li><li>The maximum balloons any boy gets is 2: In this case, we can have three sub-cases, the first where only one boy gets two balloons (3 ways), the second where two boys get two balloons and the other gets one (3 ways), and the third where all boys get 2 balloons (1 way). Keeping in mind that no girl gets fewer balloons than any boy,we get the total number of cases as $3*{}^{6+5-1}C_{5-1} + 3* {}^{5+5-1}C_{5-1} + 1* {}^{4+5-1}C_{5-1} = 630+378+70= 1078$</li><li>The maximum balloons any boy gets is 3: In this case, there is only one possibility. One of the three boys should get 3 balloons, other two should get 1 balloon each, and all the 5 girls get 3 balloons each as well. This amounts to a total of 3 possibilities.</li></ul>
The total number of possibilities, therefore, are 1820+1078+3=2901 The correct answer is $n7000$
<ul></ul>
<ul></ul>

$a = \dfrac{(\log_7 4)(\log_7 5 - \log_7 2)}{\log_7 25(\log_7 8 - \log_7 4)}$ $a=\dfrac{\ 2\left(\log_72\right)\left(\log_72.5\right)}{2\left(\log_75\right)\left(\log_72\right)}$ $a=\dfrac{\ \log_72.5}{\log_75}=\log_52.5$ Therefore ,$5^a=2.5$

Let, the number of girls be $N$.Now, in the first option:a.) $\dfrac{48}{100}\times\ 27=12.96$ and $\dfrac{50}{100}\times\ 27=13.5$Now, $12.96$So, we can have an integer value of $N$, $N=13$So, option(a) can be a possible answer choiceSo, automatically option (b) and (c) are eliminated as they are greater than 27So, we need to check only option (d)Now,$\dfrac{48}{100}\times\ 25=12$ and $\dfrac{50}{100}\times\ 25=12.5$Now,$12$In this range we can't have an integer value for $N$So, option(d) is also eliminatedSo the smallest value of $N$ is 27.

Based on the information given in the question, a diagram is constructed where BM is the median of AC. Since the question deals with areas, we construct BM as the altitude for the triangle.$\angle\ BDA=\theta$ (given) and$\angle\ BAD=\alpha$ is assumed. The following figure is obtained.<img src="https://lightyellow-stork-291267.hostingersite.com/wp-content/uploads/2025/12/Screenshot-2025-07-17-2.57.42-PM.png">BM is the height, and so$\angle\ BMA=\angle\ BMD=90^{\circ\ }$So, in $\triangle\ BAM$,$\sin\alpha\ =\frac{BM}{AB}=\ BM=c\ \sin\alpha$$\cos\ \alpha\ =\frac{AM}{AB}=AM=c\cos\alpha$Again, in $\triangle\ BDM$,$\sin\theta\ =\frac{BM}{BD}=\ BM=k\ \sin\theta$$\cos\ \theta\ =\frac{DM}{BD}=DM=k\cos\theta$Now, area of$\triangle\ ABD$ can be calculated as$\dfrac{1}{2}\times\ AD\times\ BM=\frac{1}{2}\times\ \left(AM+BM\right)\times\ BM$Putting the values obtained we have,$ar\triangle\ ABD=\frac{1}{2}\times\ \left(c\ \cos\alpha\ +k\ \cos\theta\ \right)\times\ k\ \sin\theta$ ________(A)Now,$c\ \sin a=\ k\ \sin\theta\ =\sin\ \alpha\ =\dfrac{k\sin\theta\ }{c}$So,$\cos\ \alpha\ =\sqrt{\ 1-\left(\frac{k\sin\theta\ }{c}\ \right)^2}=\dfrac{\sqrt{\ c^2-k^2\sin^2\theta\ }}{c}$Substituting this in (A) we have$ar\triangle\ ABD=\frac{1}{2}\times\ \left(\sqrt{\ c^2-k^2\sin^2\theta\ }\ +k\ \cos\theta\ \right)\times\ k\ \sin\theta$or, $ar\triangle\ ABD=\frac{1}{2}\left(\ k\sin\theta\ \sqrt{c^2-k^2\sin^2\theta\ }\ +k^2\ \cos\theta\sin\theta\ \right)$or, $ar\triangle\ ABD=\frac{1}{2}\left(\ k\sin\theta\ \sqrt{c^2-k^2\sin^2\theta\ }\ +\frac{k^2}{2}\times\ \ 2\cos\theta\sin\theta\ \right)$or, $ar\triangle\ ABD=\dfrac{1}{2}\left(\ k\sin\theta\ \sqrt{c^2-k^2\sin^2\theta\ }\ +\dfrac{k^2}{2}\sin2\theta\ \right)$Now, since BD is the median, this means the area of$ar\triangle\ ABC=2\ ar\triangle\ ABD$so ,$ar\triangle\ ABC=\ k\sin\theta\ \sqrt{c^2-k^2\sin^2\theta\ }\ +\dfrac{k^2}{2}\sin2\theta$ ,which is option B.

ipmat-indore 2024 Complete Paper Solution | All

Question 1.

Question 2.

Question 3.

Question 4.

Question 5.

Question 6.

Question 7.

Question 8.

Question 9.

Question 10.

ipmat-indore 2024 Complete Paper Solution | All

Question 1.

<div><p>The terms of a geometric progression are real and positive. If the p-th term of the progression is q and the q-th term is p, then the logarithm of the first term is</p></div>

Question 2.

<div><p>If the shortest distance of a given point to a given circle is 4 cm and the longest distance is 9 cm, then the radius of the circle is</p></div>

Question 3.

<div><p>If <span>$\mid x + 1 \mid + (y + 2)^2 = 0$</span> and <span>$ax - 3ay = 1$</span>, Then the value of a is</p></div>

Question 4.

<div><p>The number of real solutions of the equation <span>$x^2 - 10 \mid x \mid - 56 = 0$</span> is</p></div>

Question 5.

<div><p>The greatest number among <span>$2^{300}, 3^{200}, 4^{100}, 2^{100} + 3^{100}$</span> is</p></div>

Question 6.

<div><p>The sum of a given infinite geometric progression is 80 and the sum of its first two terms is 35. Then the value of n for which the sum of its first n terms is closest to 100, is</p></div>

Question 7.

<div><p>Let <span>$n$</span> be the number of ways in which <span>$20$</span> identical balloons can be distributed among <span>$5$</span> girls and <span>$3$</span> boys such that everyone gets at least one balloon and no girl gets fewer balloons than a boy does. Then</p></div>

Question 8.

<div><p>Let <span>$a = \dfrac{(\log_7 4)(\log_7 5 - \log_7 2)}{\log_7 25(\log_7 8 - \log_7 4)}$</span>. Then the value of <span>$5^a$</span> is</p></div>

Question 9.

<div><p>The smallest possible number of students in a class if the girls in the class are less than 50% but more than 48% is</p></div>

Question 10.

<div><p>The side AB of a triangle ABC is c. The median BD is of length k. If $\angle BDA = \theta$ and $\theta 90^\circ$, then the area of triangle ABC is</p></div>