IPM Indore 2022 Question Paper

Let, the line having 7 points be denoted as L1 and the line having 5 points be denoted as L2

A triangle is formed by 3 points (not all collinear)

In this case, in 2 ways a traingle can be formed.

i.) One point is at L1 and the other two points are at L2:

Number of ways of choosing one point in L1 is $$^7C_1$$ and, number of ways of choosing two points in L2 is $$^5C_2$$

So, number of triangles in this case = $$^7C_1\cdot^5C_2=7\times\ \dfrac{5\times\ 4}{2}=7\times\ 5\times\ 2=70$$

ii.) One point is at L2 and the other two points are at L1:

Number of ways of choosing one point in L2 is $$^5C_1$$ and, number of ways of choosing two points in L1 is $$^7C_2$$

So, number of triangles in this case = $$^5C_1\cdot^7C_2=5\times\ \dfrac{7\times\ 6}{2}=5\times\ 7\times\ 3=105$$

So, total number of triangles = $$70+105=175$$

We know. $$45^2=2025$$

Now, if the series was 1,2,3,4,.....,2026,..... then 2026 number would have been $$2026^{th}$$ term only

But since we are eliminating all perfect squares, so the number 2026 has shifted 45 places forward.

(It shifts 45 places forward because before 2026, there are 45 perfect squares, as $$45^2=2025$$)

So, now, 2026 is the $$(2026-45)^{th}$$ term = $$1981^{st}$$ term

So, $$2022^{nd}$$ term =$$(1981+41)^{th}$$term = 41 numbers after $$1981^{th}$$ term = 2026+41=2067

Let the time taken by Geeta usually be $$t$$ hours.

Now, when Geeta is walking at a speed of 12 km/hr, time taken = $$t$$ hours.

So, Distance covered =$$12t$$ km

Also, when Geeta is walking at a speed of 20 km/hr, time taken = $$(t-1)$$ hours.

So, Distance covered = $$20\cdot(t-1)$$ km

Now, distance covered in both the cases is same.

So, $$12\cdot t=20\cdot(t-1)$$

or, $$8\cdot t=20$$

or, $$t=2.5$$ hours

So, distance between her home and office is $$12\times\ 2.5=30$$ km

Let the first term of arithmetic progression be a and common difference be d.

So, $$3^{rd}$$ term = $$a+2d$$, $$14^{th}$$ term = $$a+13d$$, $$69^{th}$$ term = $$a+68d$$.

These terms are in G.P.

So, $$\left(a+13d\right)^2=\left(a+2d\right)\cdot\left(a+68d\right)$$

or, $$a^2+26ad+169d^2=a^2+2ad+68ad+136d^2$$

or, $$169d^2-136d^2=70ad-26ad$$

or, $$33d^2=44ad$$

or, $$3d=4a$$

or, $$a=3d/4$$

So, $$3^{rd}$$ term =$$a+2d=\dfrac{3d}{4}+2d=\dfrac{11d}{4}$$

$$14^{th}$$ term =$$a+13d=\dfrac{3d}{4}+13d=\dfrac{55d}{4}$$

$$69^{th}$$ term =$$a+68d=\dfrac{3d}{4}+68d=\dfrac{275d}{4}$$

We can see that the common ratio of this GP is 5

So, next term of this G.P. will be $$\dfrac{275d}{4}\cdot5=\dfrac{1375d}{4}$$

And, $$\dfrac{1375d}{4}=a+\left(n-1\right)d=\dfrac{3d}{4}+\left(n-1\right)d$$

or, $$\dfrac{1375d}{4}-\dfrac{3d}{4}=\left(n-1\right)d$$

or, $$\dfrac{1372d}{4}=\left(n-1\right)d$$

or, $$n-1=343$$

or, $$n=344$$

The critical points for $$f(x)$$ are $$x=0,1,2,4,6,10.$$

When $$x=0$$, $$f(x)=0+2\times\ 1+2+4+6+2\times\ 10=0+2+2+4+6+20=34$$

when $$x=1$$, $$f(x)=1+2\times\ 0+1+3+5+2\times\ 9=28$$

when $$x=2$$, $$f(x)=2+2\times1+0+2+4+2\times\ 8=26$$

when $$x=4$$, $$f(x)=4+2\times3+2+0+2+2\times\ 6=26$$

when $$x=6$$, $$f(x)=6+2\times5+4+2+0+2\times\ 4=30$$

when $$x=10$$, $$f(x)=10+2\times9+8+6+4+0=46$$

For, $$x<0$$, value of $$f(x)$$ clearly will be greater than f(0)=34

Similarly, for $$x>0$$, value of $$f(x)$$ will be greater than f(10)=46

So, the minimum value of $$f\left(x\right)=26$$ when $$x$$ lies in the interval [2,4]

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 $$0 < \theta < \frac{\pi}{4}$$, $$\cos\theta\ >\sin\theta\ $$

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 $$0 < \theta < \dfrac{\pi}{4}$$ range

So, both $$\log_2\cos\theta\ $$ and $$\log_2\sin\theta\ $$ will definitely be negative

So, $$a,b,c,d$$ all are negative terms

Now, 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, $$\cos\theta\ >\sin\theta\ $$

or, $$\log_2\cos\theta\ >\log_2\sin\theta\ $$  (as $$\log_ax$$ is an increasing function for $$a>1$$)

or, $$\dfrac{\log_2\cos\theta}{\log_2\sin\theta}<1$$ (as both $$\log_2\cos\theta\ $$ and $$\log_2\sin\theta\ $$ are negative)

or, $$\dfrac{a}{d}<1$$
or, $$a>d$$ (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 $$0 < \theta < \frac{\pi}{4}$$, $$\cot\theta\ $$ greater than 1 always

So, $$\dfrac{b}{d}>1$$ 

or, $$d>b$$ (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 value

So, $$\left(\sin\theta\right)^{\sin\theta\ -\cos\theta\ }$$ will definitely be greater than 1 (as fractional term having negative power is always greater than 1)

So, $$\dfrac{a}{c}>1$$

or, $$c>a$$ (as both a and c are negative) -------->(3)

So, from (1),(2) and (3),

$$c>a>d>b$$

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,

D and E are mid points of AB and AC respectively

By, 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 they must 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 =$$1500-300=1200$$