IPM Indore 2023 Question Paper

$$10^{2023}=2^{2023}\cdot5^{2023}$$

$$10^{2001}=2^{2001}\cdot5^{2001}$$

$$10^{2023}$$ can be written as $$10^{2001}*2^{22}\cdot5^{22}$$

The number of factors of $$10^{2023}$$ that are multiple of $$10^{2001}$$ will be the number of factors of  $$2^{22}\cdot5^{22}$$

Number of factors of $$2^{a}\cdot5^{b} = (a+1)(b+1)$$

Number of factors of  $$2^{22}\cdot5^{22}$$ = 23*23 = 529

And the total number of factors for $$10^{2023}=2^{2023}\cdot5^{2023}$$ is (2023+1)*(2023+1) = $$2024^2$$

Probability = $$\frac{529}{2024^2}$$

We can draw a figure like this:

Let us consider the lengths of AB,BC and CA be 2 units, 4 units and 3 units respectively.

As D is the mid point of BC, AD is the median. Also, BD=CD= 2 units.

Using Apollonius theorem, 

$$AB^2+AC^2=2\left(AD^2+BD^2\right)$$

or, $$2^2+3^2=2\left(AD^2+2^2\right)$$

or, $$AD^2=\dfrac{5}{2}$$

or, $$AD=\dfrac{\sqrt{\ 10}}{2}$$ ------>(1)

Now, $$AM$$ is perpendicular to $$BC$$.

So, $$\triangle\ AMB$$ and $$\triangle\ AMC$$ both are right angled triangles.

Using Pythagoras theorem in $$\triangle\ AMB$$, we can write,

$$AM^2=AB^2-BM^2$$

or, $$AM^2=2^2-BM^2$$ ----->(2)

Similarly, using Pythagoras theorem in $$\triangle\ AMC$$, we can write,

$$AM^2=AC^2-CM^2$$

or, $$AM^2=3^2-\left(4-BM\right)^2$$ ------->(3)

From equation (2) and (3),

$$3^2-\left(4-BM\right)^2=2^2-BM^2$$

or, $$3^2-2^2=\left(4-BM\right)^2-BM^2$$

or, $$5=\left(4-2BM\right)\cdot4$$

or, $$BM=\dfrac{11}{8}$$ ----->(4)

From equation (1) and (4),

$$\dfrac{BM}{AD}=\dfrac{\dfrac{11}{8}}{\dfrac{\sqrt{\ 10}}{2}}=\dfrac{11}{4\sqrt{\ 10}}$$

So, $$BM:AD=11:4\sqrt{\ 2}$$

Given, det $$(A^{3} − 3A^{2} − 5A) =0$$

So, det $$(A^3-3A^2-5A)$$ = det $$\left((A)\left(A^2-3A-5I\right)\right)$$=det $$A$$ $$\times\ $$ det $$(A^2-3A-5I)$$ =0

So, det $$A$$=0

or, $$\begin{vmatrix}1&2 \\ 3&a\end{vmatrix}$$=0

or, $$1\times\ a-2\times\ 3=0$$

or, $$a=6$$

The equation of circle is $$x^{2} + y^{2} − 6x + 4y − 12 = 0$$

We can compare this with standard equation of circle, $$x^2+y^2+2gx+2fy+c=0$$

so, $$g=-3\ ,\ f=2,\ c=-12$$

Now, centre of circle is $$\left(-g,-f\right)$$. So, centre will be $$\left(3,-2\right)$$

Radius of circle will be r = $$\sqrt{\ g^2+f^2-c}=\sqrt{\ 3^2+\left(-2\right)^2-\left(-12\right)}=\ \sqrt{\ 25\ }=5$$ units

Now for a line to be tangent to a circle, perpendicular distance from the centre of the circle to the line, should be equal to its radius.

For the line $$4x+3y+19=0$$, perpendicular distance from the centre = $$\dfrac{\left|4\left(3\right)+3\left(-2\right)+19\right|}{\sqrt{\ \left(4^2+3^2\right)}}=\dfrac{25}{5}=5\ $$ units = radius

So, $$4x+3y+19=0$$ is a tangent to the circle.

For the line $$4x+3y-31=0$$, perpendicular distance from the centre = $$\dfrac{\left|4\left(3\right)+3\left(-2\right)-31\right|}{\sqrt{\ \left(4^2+3^2\right)}}=\dfrac{25}{5}=5\ $$ units = radius

So, $$4x+3y-31=0$$ is also a tangent to the circle.

Centre of circle is $$\left(3,-2\right)$$

Purple colour line is $$4x+3y-31=0$$

Green colour line is $$4x+3y+19=0$$

OA and OB are radii perpendicular to the respective tangents.

Given, $$a_{1} a_{2}, a_{3}$$ are in G.P.

Let the common ratio of the G.P. be $$r$$

So, $$a_2=a_1r,\ a_3=a_1r^2$$

Putting the values in equation, $$a_{1}x^{2} + 2a_{2} x + a_{3} = 0$$

So, $$a_1x^2+2a_1xr+a_1r^2=0$$

or, $$a_1\left(x^2+2xr+r^2\right)=0$$

or, $$a_1\left(x+r\right)^2=0$$

Now, $$a_1\ne\ 0$$

So, $$\left(x+r\right)^2=0$$

or, $$x=-r$$ is the only root

So the common root is also $$x=-r$$

Putting this in the second equation, $$b_{1}x^{2} + 2b_{2}x + b_{3} = 0$$

or, $$b_{1}r^{2} - 2b_{2}r + b_{3} = 0$$

Now we can replace $$r^2=\dfrac{a_3}{a_1},\ r=\dfrac{a_2}{a_1}$$

So, $$b_1\cdot\dfrac{a_3}{a_1}+b_3=2b_2\cdot\dfrac{a_2}{a_1}$$

Dividing all terms by $$a_3$$,

$$\dfrac{b_1}{a_1}+\dfrac{b_3}{a_3}=2\cdot\dfrac{b_2a_2}{a_1a_3}=\dfrac{2b_2a_2}{a_2^2}=\dfrac{2b_2}{a_2}$$

(Since, $$a_1,a_2,a_3$$ are in G.P., $$a_1\cdot a_3=a_2^2$$)

so, $$\dfrac{b_1}{a_1}+\dfrac{b_3}{a_3}=\dfrac{2b_2}{a_2}$$

or, $$\dfrac{b_1}{a_1},\dfrac{b_2}{a_2},\dfrac{b_3}{a_3}$$ are in A.P.

Let the rabbit take $$a$$ times one step and $$b$$ times two steps.

So, $$a+2b=10$$

The possible values of $$\left(a,b\right)$$ can be $$\left(0,5\right),\left(2,4\right),\left(4,3\right),\left(6,2\right),\left(8,1\right),\left(10,0\right)$$

Now, when $$a=0$$, $$b=5$$, the person basically is taking all two steps at a time, 5 times.

So, number of ways of doing this = $$\dfrac{5!}{5!}$$

Similarly, when $$a=2$$, $$b=4$$, the person basically is taking 2 times one step at a time and 4 times two step at a time.

So, number of ways of doing this = $$\dfrac{6!}{4!2!}$$

So, overall total number of ways to reach the top

=$$\dfrac{5!}{5!}+\dfrac{6!}{4!2!}+\dfrac{7!}{4!3!}+\dfrac{8!}{6!2!}+\dfrac{9!}{8!1!}+\dfrac{10!}{10!}$$

=$$1+15+35+28+9+1=89$$ ways

Probability of getting at least one head = 1 - Probability of getting no heads = 1 - Probability of getting all tails.

Let the number of times the coin be tossed be n.

Probability of getting one tail =$$\dfrac{1}{2}$$

So, probability of getting all tails in n trials = $$\left(\dfrac{1}{2}\right)^n$$

So, probability of getting at least one head = $$1-\left(\dfrac{1}{2}\right)^n$$

According to question,

$$1-\left(\dfrac{1}{2}\right)^n>0.8$$

So, $$1-0.8>\left(\dfrac{1}{2}\right)^n$$

or, $$0.2>0.5^n$$

for n = 2, $$0.5^2=0.25$$ which is greater than $$0.2$$

for n = 3, $$0.5^3=0.125$$ which is lesser than 0.2

So, minimum value of n is 3

For each row, a rook can be placed in one of the eight cells. Once a rook is placed in a column, in no other row can a rook be placed in the same column. 

So, in other rows to place the second rook, there will be seven ways. Once the second rook is placed, there will only be 6 for the next row.

This way, the total ways = 8*7*6*5*4*3*2*1 = 8!

Given,

$$\log_{n}(\log_{2}a) = 1$$

So, $$\log_2a\ =n^1= n$$

or, $$a=2^n$$

So, $$a^n=\left(2^n\right)^n=2^{n^2}$$

Now, $$\log_{n}(\log_{2}b) = 2$$

So, $$\log_2b\ =n^2$$

or, $$b=2^{n^2}$$

Now, $$a^n+b=2^{n^2}+2^{n^2}=2^{n^2}\times\ 2=2^{n^2+1}$$

So, $$\left(a^n+b\right)^n=2^{\left(n^2+1\right)\cdot n}=2^{n^3+n}$$ ------->(1)

Now, $$\log_{n}(\log_{2}c) = 3$$

So, $$\log_2c\ =n^3$$

or, $$c=2^{n^3}$$

From equation (1), $$\left(a^n+b\right)^n=2^{n^3+n}=2^{n^3}\times\ 2^n=c\times\ a=ac$$

So, option B is the correct answer.

Let the amount invested be P.

Now, compound interest for 3 years = Amount after three years - Principal

or, C.I. = $$P\left(1+\dfrac{10}{100}\right)^3-P=P\left(1.1\right)^3-P=1.331P-P=0.331P$$

Also, simple interest for 3 years = S.I. = $$\dfrac{P\times\ 10\times\ 3}{100}=0.3P$$

Given, C.I. - S.I. =527

or, $$0.331P-0.3P=527$$

or, $$0.031P=527$$

or, $$P=\dfrac{527}{0.031}=17000$$

So, amount invested is 17000 INR.