IPM Indore 2025 Question Paper

The figure below represents a regular octagon ABCDEFGH inscribed in a circle with center at O and radius 1 unit.

Since the octagon is a regular octagon, on drawing diagonals AE, BF, CG, and DH, all of them will pass through the center. We can therefore divide the octagon into 8 congruent isosceles triangles.

Hence, OA=OBV=OC=OD=OE=OF=OG=OH=1 unit(since all of them are radii of the same circle)

And, AB=BC=CD=DE=EF=FG=GH=HA ( since all of them are sides of the same regular octagon)

So we can say that $$\triangle AOB,\ \triangle BOC,\ \triangle COD,\ \triangle DOE,\ \triangle EOF,\ \triangle FOG,\ \triangle GOH\ and\ \triangle HOA\ $$ are all congruent triangles, which would mean that all the angles at the center are also equal. 

So, $$\angle\ AOB=\frac{360}{8}=45^{\circ\ }$$

Hence, area of$$\triangle AOB=\frac{1}{2}\times\ AO\times\ BO\times\ \sin\ \left(\angle\ AOB\right)=\frac{1}{2}\times\ 1\times\ 1\times\ \sin\ 45=\frac{1}{2\sqrt{\ 2}}\ sq\ units$$

As the octagon is divided into 8 congruent triangles, so the area of the octagon is $$\frac{1}{2\sqrt{\ 2}}\times\ 8=2\sqrt{\ 2}\ sq\ units$$

Consider the speeds of Ankit and Bipul as $$A$$ and $$B$$ respectively. The total distance of the swimming pool is covered by Ankit in 10 minutes and by Bipul in 15 minutes. This gives us the following relation:

$$10A=15B$$  or  $$A:B=3:2$$. We can therefore assume the value of $$A$$ as $$3k$$ and the value of $$B$$ as $$2k$$. This makes the total length of the swimming pool $$3k*10 = 2k*15 = 30k$$

Since they are running in the same direction, the first meeting will be at $$\dfrac{30k}{3k+2k} = 6$$ minutes. $$80-6=74$$ minutes of their journey is still remaining.

Since they are swimming back and forth, the entire back and forth journey can be taken as a circular track the length of which is double the length of the swimming pool, or $$30k+30k=60k$$. 

On this circular journey, Ankit and Bipul will meet every $$\dfrac{60k}{3k+2k}=12$$ minutes. In remaining $$74$$ minutes of journey, therefore, they will meet a total of $$\lfloor \frac{74}{12} \rfloor = 6$$ times. 

Along with the first meeting after $$6$$ minutes, there will thus be a total of $$6+1=7$$ meetings in total in a span of $$80$$ minutes.

Sum of first $$n$$ terms of a G.P. is given by the formula, $$S$$=$$\dfrac{a\cdot\left(r^n-1\right)}{r-1}$$ where $$a$$ is the first term and $$r$$ is the common ratio.

So according to question,

$$\dfrac{a.\left(r^5-1\right)}{r-1}=a\cdot\dfrac{\left(r^7-1\right)}{r-1}$$

or, $$r^5=r^7$$

or, $$r^5(1-r^2)=0$$

or, $$r=0,1,-1$$

But $$r$$ can't be 1 as then sum of terms will be not defined (in denominator $$r-1$$ is there so for $$r=1$$, sum will be not defined)

Also, $$r$$ can't be 0 as then all the other terms except the first term will be zero.

So, $$r=-1$$

Now, sum of first 9 terms=24

or, $$\dfrac{a.\left(r^9-1\right)}{r-1}=\dfrac{a\left(-1-1\right)}{\left(-1-1\right)}=24$$

or, $$a=24$$

So, fourth term = $$a\cdot r^{4-1}=a\cdot r^3=24\left(-1\right)^3=-24$$

Given, $$\log_{\left(x+\dfrac{1}{x}\right)}\left[\log_2\left(\dfrac{x-1}{x+2}\right)\right]>0$$

The first constraint for this inequality is:

i.)$$x+\dfrac{1}{x}>0$$

or, $$\dfrac{x^2+1}{x}>0$$

Now, $$x^2+1$$ is always positive

So, for  $$\dfrac{x^2+1}{x}$$ to be positive, $$x$$ has to be positive i.e. $$x>0$$ ------>(1)

ii.)$$\log_2\left(\dfrac{x-1}{x+2}\right)>1$$

or, $$\dfrac{x-1}{x+2}>2^1$$

or, $$\dfrac{x-1}{x+2}-2>0$$

or, $$\dfrac{x-1-2x-4}{x+2}>0$$

or, $$\dfrac{-x-5}{x+2}>0$$

or, $$\left(x+5\right)\left(x+2\right)<0$$

or, $$-5<x<-2$$ ----->(2)

(1) and (2) are contradictory to each other

So, there is no value of $$x$$ possible satisfying this inequality

So, solution set is null set

Since $$a_k$$ is the $$k^{th}$$term of $$S_1$$

So, $$a_k=100+5\left(k-1\right)=95+5k$$

Similarly, $$b_k=100-5\left(k-1\right)=105-5k$$

So, $$\sum_{k=1}^{20}a_k b_k$$ = $$\sum_{k=1}^{20}\left(105-5k\right)\left(95+5k\right)$$

=$$5\times\ 5\sum_{k=1}^{20}\left(21-k\right)\left(19+k\right)$$

=$$25\sum_{k=1}^{20}\left(399+21k-19k-k^2\right)$$

=$$25\left(\sum_{k=1}^{20}\left(399+2k-k^2\right)\right)$$

=$$25\left(\sum_{k=1}^{20}399+2\sum_{k=1}^{20}k-\sum_{k=1}^{20}k^2\right)$$

Now, we know $$\sum_{k=1}^nk=\dfrac{n\left(n+1\right)}{2}$$ and $$\sum_{k=1}^nk^2=\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}$$ and here $$n=20$$

=$$25\left(\ 399\times\ 20+2\times\ 20\times\ \dfrac{21}{2}-20\times\ 21\times\ \dfrac{41}{6}\right)$$

=$$25\left(399\times\ 20+420-2870\right)$$

=$$25\times\ 5530$$

=$$138250$$

According to question, probability of A hitting the target=$$P(A)=0.4$$

probability of B hitting the target=$$P(B)=0.6$$

So, probability of A missing the target=$$P(A^c)=0.6$$

probability of B missing the target=$$P(B^c)=0.4$$

Let, E denote the event of A hitting the target first.

The event E can occur in the following ways:

i.) A hits the target in the first shot

ii.) A and B both miss the target in first shot and after that A hits in the second shot

iii.) A and B both miss the target in first and second shot and after that A hits in the third shot and so on....

So basically, $$P\left(E\right)=P\left(A\right)+P\left(A^c∩\ B^c∩\ A\right)+P\left(A^c∩\ B^c∩\ A^c\ ∩\ B^c∩\ A\right)+......$$

or, $$P\left(E\right)=P\left(A\right)+P\left(A^c\right)\cdot P\ \left(B^c\right)\cdot P\left(A\right)+P\left(A^c\right)\cdot P\left(B^c\right)\cdot P\left(A^c\right)\cdot P\left(B^c\right)\cdot P\ \left(A\right)+......$$

or, $$P\left(E\right)=0.4+0.6\times\ 0.4\times\ 0.4+0.6\times\ 0.4\times\ 0.6\times\ 0.4\times\ 0.4+.....$$

or, $$P\left(E\right)=0.4\left(1+0.6\times\ 0.4+0.6\times\ 0.4\times\ 0.6\times\ 0.4+.....\right)$$

or, $$P\left(E\right)=0.4\left(1+0.24+0.24^2+.....\right)$$

Now, the term in bracket is an infinite G.P. series with common ratio 0.24

or, $$P\left(E\right)=\dfrac{0.4}{1-0.24}=\dfrac{0.4}{0.76}=\dfrac{40}{76}=\dfrac{10}{19}$$

The given expression, $$3^{10}+2$$, when cubed, gives us $$(3^{10}+2)^3= 3^{30} + 8 + (6\times 3^{20}) + (12\times 3^{10})$$

This expansion can be rewritten as $$(\underline{3^{10}+2)^3} = (3^{30}+8) + \underline{(6\times 3^{10})(3^{10}+2)}$$

Since the two underlined parts are both divisible by $$(3^{10}+2)$$, we can conclude that the remaining non-underlined part, $$(3^{30}+8)$$ should also be divisible by $$(3^{10}+2)$$. Therefore option D, is correct. We can do the same for other options.

We can try $$(3^{10}+2)^2$$ to verify option A, the expansion will be $$(3^{20}+4+4*3^{10})$$. Thus, after dividing this by $$(3^{10}+2)$$ the remainder will clearly be $$4*3^{10}$$, since this is definitely not divisible by $$(3^{10}+2)$$, as $$3^{10}$$ is not divisible by $$(3^{10}+2)$$, which is an odd number, we can conclude option A is incorrect.

Option B is also incorrect because if $$3^{30}+8$$ is divisible after the cubic expansion, a number $$6$$ less than it cannot be.

Finally, if we assume option C to be divisible, then the difference between option C and option D should also be divisible by $$(3^{10}+2)$$, this difference is nothing but $$(3^{30}-3^{20}) = 3^{20}(3^{10}-1)$$, since none of the two parts in this difference are divisible, the difference itself is not divisible, and option C will not be divisible either.

Alternate Explanation:

Let $$3^{10}=a$$ and $$2=b$$

Then the given expression is of form $$a+b$$

So, option A is of form $$a^2+b^2$$

option B is of form $$a^3+b$$

option C is of form $$a^2+b^3$$

option D is of form $$a^3+b^3$$

Among all these polynomials, we know only $$a^3+b^3$$ has a factor $$a+b$$

So, option D is the correct answer.

Given, $$n(A - B), n(A \cap B), n(B - A)$$ are in A.P.

Now, $$n\left(A-B\right)=n\left(A\right)-n\left(A∩B\right)$$

$$n\left(B-A\right)=n\left(B\right)-n\left(A∩B\right)$$

Now, adding these two equations, $$n\left(A-B\right)+n\left(B-A\right)=n\left(A\right)+n\left(B\right)-2\cdot n\left(A∩B\right)$$--->(1)

Since, $$n(A - B), n(A \cap B), n(B - A)$$ are in A.P.

So, $$n\left(A-B\right)+n\left(B-A\right)=2\cdot n\left(A∩B\right)$$---->(2)

From (1) and (2), $$n\left(A\right)+n\left(B\right)-2\cdot n\left(A∩B\right)=2\cdot n\left(A∩B\right)$$

So, $$n\left(A\right)+n\left(B\right)=4\cdot n\left(A∩B\right)$$ ------>(3)

Now we know, $$n\ \left(A∪B\right)=n\left(A\right)+n\left(B\right)-n\left(A∩B\right)$$

or, $$18=4n\left(A∩B\right)-n\left(A∩B\right)=3.n\left(A∩B\right)$$  (From (3))

So, $$n\left(A∩B\right)=\dfrac{18}{3}=6$$

From (3), $$n\left(A\right)+n\left(B\right)=4\times\ 6=24$$

We are being given with six digits- 1,3,5,7,8,9

So for a number formed by these digits to be greater than 5000, the number can be either 4-digit,5-digit or 6-digit

i.)4-digit number:

Since the number is divisible by 5, so last digit has to be 5

Also, the number is greater than 5000 so first digit has to be 7,8 or 9

So, the first digit can be selected in $$^3C_1$$ ways

For the remaining 2-places, we can select 2 digits from the remaining 4 digits in $$^4C_2$$ ways and arrange them in 2! ways

So, number of ways of forming the 4-digit number=$$^3C_1\times^4C_2\times\ 2!=36$$ ways

ii.)5-digit number:

The last digit has to be 5 as the number is divisible by 5

Now, for the remaining 4 places we have to select 4 digits from 5 digits which can be done in $$^5C_4$$ ways and rearrange them in $$4!$$ ways

So, number of ways of forming the 5-digit number=$$^5C_4\times\ 4!=120$$ ways

iii.)6-digit number:
The last digit has to be 5 as the number is divisible by 5

So, the remaining 5 places can be filled by remaining 5 digits in $$5!=120$$ ways

So, number of ways of forming the 6-digit number=$$5!=120$$ ways

So, number of integers=$$36+120+120=276$$

11 when divided by 9 remainder is 2

1011 when divided by 9 remainder is 3

Now, $$11≡2\ mod(9)$$

so, $$11^{1011}≡2^{1011\ }mod(9)$$ ----->(1)

Now $$2^{1011}=\left(2^3\right)^{337}=8^{337}$$

Also, $$8≡-1mod(9)$$

So, $$8^{337}≡-1mod(9)$$

So, we can say $$2^{1011}≡8^{337}mod(9)≡-1\ mod(9)$$

So, $$11^{1011}≡-1\ mod(9)$$ (From (1))

Now, $$-1$$ is a negative remainder, so positive remainder will be 8 --------->(2)

Similarly, $$1011≡3\ mod(9)$$

So, $$1011^{11}≡3^{11\ }mod(9)$$

Now, $$3^{11}=3^2\cdot3^9=9\cdot3^9$$ ,so it is definitely divisible by 9

So, in this part remainder will be $$0$$ --------->(3)

From (2) and (3) we can say, remainder=$$8+0=8$$