Let the first term of A.P. be $a$, common difference be $d$
So, sum of first n term is $S_n=\dfrac{n}{2}\left[2a+\left(n-1\right)d\right]$
Now, using the information given in question,
$\dfrac{15}{2}\left[2a+14d\right]=200$
or,$2a+14d=200\times\ \dfrac{2}{15}=\dfrac{80}{3}$ ------(1)
Also, sum of next 15 terms is 350
So, sum of first 30 terms =200+350=550
So,$\dfrac{30}{2}\left[2a+29d\right]=550$
or,$2a+29d=550\times\ \dfrac{2}{30}=\dfrac{110}{3}$ ------(2)
Subtracting equation (1) from equation (2),
$15d=\dfrac{110}{3}-\dfrac{80}{3}$
or,$15d=\dfrac{110-80}{3}=\dfrac{30}{3}=10$
or,$d=\dfrac{10}{15}=\dfrac{2}{3}$

We know when $0 \lt \theta \lt \frac{\pi}{4}$, $\cos\theta \gt \sin\theta$Also, all the terms $a,b,c,d$ are multiplication of an exponential term (which is always positive) and a logarithmic term either $\log_2\cos\theta$ or $\log_2\sin\theta$Now, both $\sin\theta$ and $\cos\theta$ are fractional values in $0 \lt \theta \lt \dfrac{\pi}{4}$ rangeSo, both$\log_2\cos\theta$ and$\log_2\sin\theta$ will definitely be negativeSo, $a,b,c,d$ all are negative termsNow, let's compare $a,b,c,d$i.) $a$ and $d$:$\dfrac{a}{d}=\dfrac{((\sin\theta)^{\sin\theta})(\log_2\cos\theta)}{((\sin\theta)^{\sin\theta})(\log_2\sin\theta)}$or,$\dfrac{a}{d}=\dfrac{(\log_2\cos\theta)}{(\log_2\sin\theta)}$Since,$\cos\theta \gt \sin\theta$or,$\log_2\cos\theta \gt \log_2\sin\theta$ (as $\log_ax$ is an increasing function for $a \gt 1$)or, $\dfrac{\log_2\cos\theta}{\log_2\sin\theta} \lt 1$ (as both $\log_2\cos\theta$ and $\log_2\sin\theta$ are negative)or, $\dfrac{a}{d} \lt 1$or, $a \gt d$ (as $a$ and $d$ both are negative) ------(1)ii.) $b$ and $d$:$\dfrac{b}{d}=\dfrac{((\cos\theta)^{\sin\theta})(\log_2\sin\theta)}{((\sin\theta)^{\sin\theta})(\log_2\sin\theta)}$or,$\dfrac{b}{d}=\dfrac{((\cos\theta)^{\sin\theta})}{((\sin\theta)^{\sin\theta})}=\left(\cot\ \theta\ \right)^{\sin\ \theta\ }$Now for$0 \lt \theta \lt \frac{\pi}{4}$,$\cot\theta$ greater than 1 alwaysSo,$\dfrac{b}{d} \gt 1$or, $d \gt b$(as $b$ and $d$ both are negative)-------(2)iii.) $a$ and $c$:$\dfrac{a}{c}=\dfrac{((\sin\theta)^{\sin\theta})(\log_2\cos\theta)}{((\sin\theta)^{\cos\theta})(\log_2\cos\theta)}$or,$\dfrac{a}{c}=\left(\sin\theta\ \right)^{\sin\ \theta\ -\cos\ \theta }$Now,$\sin\theta\ -\cos\theta$ is a negative term and$\sin\theta$ has a fractional valueSo, $\left(\sin\theta\right)^{\sin\theta\ -\cos\theta\ }$ will definitely be greater than 1 (as fractional term having negative power is always greater than 1 )So, $\dfrac{a}{c} \gt 1$or, $c \gt a$ (as both a and c are negative) --------(3)So, from (1),(2) and (3),$c \gt a \gt d \gt b$So $a,b,c,d$ are now arranged in descending order.Also, there are 4 terms in total.So, median = average of $3^{rd}$ term and $4^{th}$ term =$\dfrac{a+d}{2}$

We can draw a figure like this,<img src="https://lightyellow-stork-291267.hostingersite.com/wp-content/uploads/2025/12/Screenshot-2025-06-17-152611_3qyCFwY.png">D and E are mid points of AB and AC respectivelyBy, midpoint theorem we can say,$DE=\frac{1}{2}BC$So,$BC=2\cdot DE$or,$BC=2\times\ 6=12\ cm$

Initially,there are $n$ persons whose average height is 160 cm.
So, sum of all their heights =$160n$
Also, $m$ more persons enter the room with average height 172 cm.
Now, sum of their heights =$172m$
So, total height of all members =$160n+172m$
And total no. of persons are $m+n$
The final average height is given as 164
So,$\dfrac{160n+172m}{m+n}=164$
or,$160n+172m=164\left(m+n\right)$
or,$160n+172m=164m+164n$
or,$8m=4n$
or,$m:n=1:2$

There are 9000 four digit integers.
Now if the integers are divisible by 2 and 3, then theymust be divisible by 6 also.
Number of four digit integers divisible by 6 =$\dfrac{9000}{6}=1500$
But, we also need to eliminate the integers which are divisible by 5
Now, integers divisible by 6 and 5, means it is divisible by 30 also
So, number of four digit integers divisible by 30=$\dfrac{9000}{30}=300$
So, required number of integers =$1500-300=1200$

MANAGEMENT has 2 M's, 2 A's, 2 N's, 1 G, 2 E's, 1 TVowels are A and E.No two vowels are together. So, between every vowel there has to be atleast a consonant.So, if we write the 6 consonants like:$\times M\times M\times N\times N\times G\times T\times$There are 7 gaps, out of which we need to place 4 vowels.No. of ways of selecting 4 places out of 7 and then placing the 4 vowels are $^7C_4\times \dfrac{4!}{2!\cdot2!}$Number of ways in which the 6 consonants can be arranged =$\dfrac{6!}{2!\cdot2!}$So, total number of permutations =$^7C_4\times \dfrac{4!}{2!\cdot2!}\times \dfrac{6!}{2!\cdot2!}$=$\dfrac{7!}{4!\cdot3!}\times \dfrac{4!}{2!\cdot2!}\times \dfrac{6!}{2!\cdot2!}=\dfrac{7!}{3!}\times \dfrac{6!}{\left(2!\right)^4}=\dfrac{5040}{6}\times \dfrac{720}{16}=315\times 120=37800$

Given,$81^x+81^{f\left(x\right)}=3$Now,$81^{f\left(x\right)}=3-81^x$or,$81^{f\left(x\right)}=3-3^{4x}$Now,$81^{f\left(x\right)}$ is an exponential term which is always greater than zeroSo,$3-3^{4x}0$or,$33^{4x}$or,$4x1$or,$x0.25$So, x will lie in the interval $(-\infty, 0.25)$

The lines are concurrent, so area of triangle formed by the lines will be zero.So,$\begin{vmatrix}1ㅤ-1ㅤ-1 \\ 2ㅤ3ㅤ-12\\ 2ㅤ-3ㅤk \end{vmatrix}$=0Applying transformation, C2-&gt;C1+C2 and C3--C3+C1,$\begin{vmatrix}1ㅤ0ㅤ0 \\ 2ㅤ5ㅤ-10\\ 2ㅤ-1ㅤ(k+2) \end{vmatrix}$=0Expanding along R1,$1\cdot$$\begin{vmatrix}5ㅤ-10 \\ -1ㅤ(k+2)\end{vmatrix}$=0So, $5\left(k+2\right)-\left(-1\right)\cdot\left(-10\right)=0$or, $5k+10-10=0$or, $5k=0$or, $k=0$

The standard ellipse is in the form of $\ \dfrac{X^2}{\left(A\right)^2}+\dfrac{Y^2}{\left(B\right)^2}=1$, where the major axis is known by comparing the values of A and B. For instance, If AB , then major axis is X-axis and focus lie on the X axis, else if BA, then major axis is Y axis and focus lie on Y axis.Now, first lets see if a and b , both positive or not :a = $Sin\sqrt{2}-Sin\sqrt{3}$, b =$Cos\sqrt{2}-Cos\sqrt{3}$$\sqrt{2}\ =\ 1.414\ \ ,\ \sqrt{3}\ =1.732, \dfrac{\pi\ }{2} =1.5714$Now, Lets use Sin A - Sin B = 2$Cos\dfrac{A+B}{2}Sin\dfrac{A-B}{2}$$Sin\sqrt{2}-Sin\sqrt{3} = 2Cos\dfrac{\sqrt{2}+\sqrt{3}}{2}Sin\dfrac{\sqrt{2} -\sqrt{3}}{2}$= $2Cos\left(1.573\right)Sin\left(-0.159\right)$Now, as 1.573 $\frac{\pi}{2}$ , this implies Cos(1.573) is negative value Sin(-0.159) is also negative ..Therefore (-Ve)x(-Ve) is positive . Hence we can conclude 'a' is positive ..Therefore it can be written in the form of $A^2$ ..Similarly, we can check for b :b =$Cos\sqrt{2}-Cos\sqrt{3}$Now, lets use$CosA-CosB=-2 Sin\dfrac{A+B}{2}Sin\dfrac{A-B}{2}$$Cos\sqrt{2}-Cos\sqrt{3}$ =$-2Sin\left(1.573\right)Sin\left(-0.159\right)$As both the values 1.573 and 0.159 are less than $\pi$ , both sin values are positive , but there are two negative signs , hence again positive number. Therefore b 0.. So this can also be written in the format of $B^2$.Therefore, we can conclude that its a ellipse. Now lets find the major axis by comparing a b.Now lets check for a b :$Sin\sqrt{2}-Sin\sqrt{3}$ $\gt$ $Cos\sqrt{2}-Cos\sqrt{3}$,$Sin\sqrt{2}-Cos\sqrt{2}$ $\gt$ $Sin\sqrt{3}-Cos\sqrt{3}$,now, squaring on both sides : { $Sin^2\left(\theta\ \right)+Cos^2\left(\theta\ \right)\ = 1$, $\left(Sin(2\theta\right)\ = 2Sin\left(\theta\ \right)Cos\left(\theta\ \right)$ }1 - 2($Sin\sqrt{\ 2}Cos\sqrt{\ 2}$) $\gt$ 1 - 2($Sin\sqrt{\ 3}Cos\sqrt{\ 3}$)-$\left(Sin(2\sqrt{2}\right)$ $\gt$ -$\left(Cos(2\sqrt{3}\right)$This implies ,$Sin2\sqrt{2} \lt Sin2\sqrt{3}$$2\sqrt{2} = 2.828 , which is less\ than \pi$$2\sqrt{3} = 3.464 , which is greater than \pi$Hence, we can say that the $Sin2\sqrt{2} \gt 0,Sin2\sqrt{\ 3} \lt 0$Therefore , we can say our assumption of a b is wrong, and the b a is true relation.Hence, we can say Its an ellipse with major axis as Y - axis , therefore the focus lies on Y axis .

Set A has 3 elements and set B has 2 elements.
Number of relations from set A to B = $2^{\left(3\times\ 2\right)}=2^6$
(Number of relations from a set having $m$ elements to a set having $n$ elements is given by the formula $2^{mn}$)
Number of functions from set A to set B =$2^3$
(Number of functions from a set having $m$ elements to a set having $n$ elements is given by the formula $n^m$)
So, probability that a relation is also a function = $\dfrac{2^3}{2^6}=\dfrac{1}{2^3}=\dfrac{1}{8}$

ipmat-indore 2022 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 2022 Complete Paper Solution | All

Question 1.

<div><p>The sum of the first 15 terms in an arithmetic progression is 200, while the sum of the next 15 terms is 350, then the common difference is</p></div>

Question 2.

Question 3.

<div><p>In a right-angled triangle ABC, the hypotenuse AC is of length 13 cm. A line drawn connecting the midpoints D and E of sides AB and AC is found to be 6 cm in length. The length of BC is</p></div>

Question 4.

<div><p>In a room, there are n persons whose average height is 160 cm. If m more persons, whose average height is 172 cm, enter the room, then the average height of all persons in the room becomes 164 cm. Then m : n is</p></div>

Question 5.

<div><p>The number of four-digit integers which are greater than 1000 and divisible by both 2 and 3, but not by 5, is</p></div>

Question 6.

<div><p>In how many ways can the letters of the word MANAGEMENT be arranged such that no two vowels appear together?</p></div>

Question 7.

<div><p>The set of all possible values of f(x) for which <span>$(81)^{x} + (81)^{(f(x)} = 3$</span> is</p></div>

Question 8.

<div><p>The value of k for which the following lines<br>x-y-1=0<br>2x+3y-12=0<br>2x-3y+k=0<br>are concurrent is</p></div>

Question 9.

<div><p>The curve represented by the equation</p> <p>$\frac{x^{2}}{\sin \sqrt{2} - \sin \sqrt{3}} + \frac{y^{2}}{\cos \sqrt{2} - \cos \sqrt{3}} = 1$</p></div>

Question 10.

<div><p>Let A = {1, 2, 3} and B = {a, b}. Assuming all relations from set A to set B are equally likely, what is the probability that a relation from A to B is also a function?</p></div>