IPM Indore 2025 Question Paper

Let the total work done be 1 unit.

Now, let us assume the total number of days taken by Arpita is A, then her efficiency will be = $$\dfrac{1}{A}$$

And the total number of days taken by Nikita is N, then her efficiency will be = $$\dfrac{1}{N}$$

It is given that they complete the work together in 12 days. Thus => $$\dfrac{1}{A}+\dfrac{1}{N}=\dfrac{1}{12}\rightarrow1$$

Now, it is given that Arpita works initially to complete 40% of the job, which means she does 0.4 units of work. For 1 unit, she was taking A days; for 0.4 units of work, she will take 0.4 A days. 

Then, Nikita completes the remaining job alone. Thus, Nikita will do 0.6 units of work, which will take 0.6N days. It is given that in this way, they take 24 days to complete the job.

=> $$0.4A+0.6N=24\rightarrow2$$

=> $$2A+3N=120$$

=> $$A=\dfrac{\left(120-3N\right)}{2}\rightarrow3$$

From eq. 1, we can get => $$\dfrac{\left(A+N\right)}{AN}=\dfrac{1}{12}$$

=> $$12A+12N=AN$$

Substituting the value of A from eq. 3 -

=> $$12\left(\dfrac{120-3N}{2}\right)+12N=\left(\dfrac{120-3N}{2}\right)N$$

=> $$1440-36N+24N=120N-3N^2$$             (Dividing the entire equation by 3)

=> $$480-4N=40N-N^2$$

=> $$N^2-44N+480=0$$

=> $$N^2-24N-20N+480=0$$          (Splitting the middle term)

=> $$\left(N-24\right)\left(N-20\right)=0$$

=> $$N=20$$ or $$N=24$$ 

Thus, Nikita alone takes to complete that work is either 20 or 24 days. 

There are five teams, each with 15 members, for a total of 75 members.

Also, the number of biologists, geologists and explorers is equal, thus this will be equal to 25.

It is given that Teams C and D have 6 biologists each, and Team A has 6 geologists.

Now, each team has at least 2 members from each skill set, and the number of explorers in each team is distinct and decreases in the order A, B, C, D, and E. Thus, if we assume there are 2 explorers in team E, then team D, C, B and A will have 3, 4, 5, and 6 explorers respectively. Thus, the total number of explorers will be (2 + 3 + 4 + 5 + 6 = 20), less than 25. Now, if we assume there are 3 explorers in team E, then team D, C, B and A will have 4, 5, 6, and 7 explorers respectively. Thus, the total number of explorers will be (3 + 4 + 5 + 6 + 7 = 25), equal to 25. This is the only case valid; in all the other cases, it will be either more than 25 or less than 25.

In the diagram, we can see that team A has 2 biologists, and teams C and D will have 4 and 5 geologists, respectively.

Now, each team has more biologists than explorers except team A. Thus, the number of biologists in team B will be 7 (If it is 8, then the number of geologists will be 1, which is not possible, because there are 2 members of each kind in each team). Thus, our final table will look like this.

Thus, the number of biologists in team E is 4. 

$$\mid a - b \mid + \mid b - c \mid = \mid c - a \mid$$. This equation is only valid when b lies between a and c. Thus, there are two cases possible - $$a<b<c$$ or $$c<b<a$$

Case - 1 :- $$a<b<c$$

Now, we need to find the maximum value of b. All, a, b, and c are natural numbers is less than 100, and we $$c>b$$. Thus, if we assume the value of c = 99 and b = 98 (which are the two maximum possible values), then -

$$\mid a-98\mid+\mid98-99\mid=\mid99-a\mid$$, and we assumed that a < b, thus a < 98.

=> $$98-a+1=99-a$$ => 0 = 0

Thus, any value of a < 98 is possible for this equation.

Therefore, c = 99, b = 98 and a < 98

Case - 2 :- $$c<b<a$$

Now, we need to find the maximum value of b. All, a, b, and c are natural numbers is less than 100, and we $$a>b$$. Thus, if we assume the value of a = 99 and b = 98 (which are the two maximum possible values), then -

$$\mid99-98\mid+\mid98-c\mid=\mid c-99\mid$$, and we assumed that c < b, thus c < 98.

=>$$1+98-c=99-c$$ => 0 = 0

Thus, any value of c < 98 is possible for this equation.

Therefore, a = 99, b = 98 and c < 98

Thus, in both cases, the maximum value of b = 98. 

The total number of matches played by all the teams will be = $$^8C_2=28$$ matches.

Now, one of the team members has been suspended after playing three matches. If they had not been suspended, they would have played seven games in total. Thus, the four matches played by this team won't be happening in the tournament further.

Thus, if the team is suspended after three matches, the total number of matches will be = $$28-4=24$$ matches.

$$a_1=\ln\left(\dfrac{a}{b}\right),\ a_2=\ln\left(\dfrac{a}{b^{\frac{3}{2}}}\right)$$. Thus, the common difference will be = $$a_2-a_1$$

=> $$d=\ln\left(\dfrac{a}{b^{\frac{3}{2}}}\right)-\ln\left(\dfrac{a}{b}\right)=\ln\left(\dfrac{a\times b}{a\times b^{\frac{3}{2}}}\right)=-\dfrac{1}{2}\ln\left(b\right)$$

The, sum of first n terms is given by = $$\dfrac{n}{2}\left[2a+\left(n-1\right)d\right]$$

Here, n = 21; thus, the sum of the first 21 terms = $$\dfrac{21}{2}\left[2a+20d\right]=21\left[a+10d\right]$$

=> $$S_{21}=21\left[\ln\left(\dfrac{a}{b}\right)+10\left(-\dfrac{1}{2}\right)\ln b\right]$$

=> $$S_{21}=21\left[\ln a-\ln b-5\ln b\right]$$

=> $$S_{21}=21\left[\ln a-6\ln b\right]$$

=> $$S_{21}=21\ln\left(\dfrac{a}{b^6}\right)$$

=> $$S_{21}=\ln\left(\dfrac{a^{21}}{b^{126}}\right)$$

It is given that the sum of first 21 terms = $$\ln\left(\dfrac{a^m}{b^n}\right)$$, therefore m = 21 and n = 126.

Thus, $$m+n=21+126=147$$

The number of students who did not appear for the English exam is twice that of those who did not appear for the Math exam. Let us assume that the number of students who did not appear for Math = X, then the number of students who did not appear for English = 2X

The number of students who passed the Math exam is twice that of those who appeared but failed the English exam. Let us assume the number of students who appeared but failed the English exam is Y, then the number of students who passed the Math exam = 2Y

The number of students who passed the English exam is twice that of those who appeared but failed the Math exam. Let us assume that the number of students who appeared but failed the Math exam = Z, then the number of students who passed the English exam = 2Z.

$$2X+2Z+Y=120\rightarrow1$$

$$X+2Y+Z=120\rightarrow2$$

Multiply eq. 2 by 2, and subtract eq. 1 from it.

=> $$(2X+4Y+2Z)-(2X+Y+2Z)=240-120$$

=> $$3Y=120$$

=> $$Y=40$$

$$A = \begin{bmatrix}2 & n \\4 & 1 \end{bmatrix}$$

=> $$A^2 = \begin{bmatrix}2 & n \\4 & 1 \end{bmatrix}\begin{bmatrix}2 & n \\4 & 1 \end{bmatrix}$$

=> $$A^2=\begin{bmatrix}4+4n & 3n \\12 & 4n+1 \end{bmatrix}$$

And, $$A^3 = \begin{bmatrix}4+4n & 3n \\12 & 4n+1 \end{bmatrix}\begin{bmatrix}2 & n \\4 & 1 \end{bmatrix}$$

=> $$A^3=\begin{bmatrix}8+20n & 4n^2+7n \\16n+28 & 16n+1 \end{bmatrix}\rightarrow1$$

And, it is given that $$A^3 = 27\begin{bmatrix}4 & q \\p & r \end{bmatrix}$$ 

=> $$A^3 = \begin{bmatrix}108 & 27q \\27p & 27r \end{bmatrix}\rightarrow2$$

Comparing each cell value in eq. 1 and eq. 2 - 

=> $$8+20n=108$$ => $$n=5$$

=> $$4n^2+7n=27q$$ => $$135=27q$$ => $$q=5$$

=> $$16n+28=27p$$ => $$108=27p$$ => $$p=4$$

=> $$16n+1=27r$$ => $$81=27r$$ => $$r=3$$

Thus, the value of $$p+q+r=4+5+3=12$$

There are five teams, each with 15 members, for a total of 75 members.

Also, the number of biologists, geologists and explorers is equal, thus this will be equal to 25.

It is given that Teams C and D have 6 biologists each, and Team A has 6 geologists.

Now, each team has at least 2 members from each skill set, and the number of explorers in each team is distinct and decreases in the order A, B, C, D, and E. Thus, if we assume there are 2 explorers in team E, then team D, C, B and A will have 3, 4, 5, and 6 explorers respectively. Thus, the total number of explorers will be (2 + 3 + 4 + 5 + 6 = 20), less than 25. Now, if we assume there are 3 explorers in team E, then team D, C, B and A will have 4, 5, 6, and 7 explorers respectively. Thus, the total number of explorers will be (3 + 4 + 5 + 6 + 7 = 25), equal to 25. This is the only case valid; in all the other cases, it will be either more than 25 or less than 25.

In the diagram, we can see that team A has 2 biologists, and teams C and D will have 4 and 5 geologists, respectively.

Now, each team has more biologists than explorers except team A. Thus, the number of biologists in team B will be 7 (If it is 8, then the number of geologists will be 1, which is not possible, because there are 2 members of each kind in each team). Thus, our final table will look like this.

Thus, the number of teams having more geologists than biologists is 2 {Team A and Team E}

$$\log_3(x^2 - 1), \log_3(2x^2 + 1)$$ and $$\log_3(6x^2 + 3)$$ are the first three terms of an arithmetic progression.

=> $$\log_3(2x^2 + 1)-\log_3(x^2 - 1)=\log_3(6x^2 + 3)-\log_3(2x^2 + 1)$$

=> $$\log_3\left[\dfrac{2x^2+1}{x^2-1}\right]=\log_3\left[\dfrac{6x^2+3}{2x^2+1}\right]$$

=> $$\log_3\left[\dfrac{2x^2+1}{x^2-1}\right]=\log_33$$

=> $$\dfrac{2x^2+1}{x^2-1}=3$$

=> $$2x^2+1=3x^2-3$$

=> $$x^2=4$$

Thus, the first three terms of the AP will be $$\log_3(x^2 - 1)=1$$, $$\log_3(2x^2 + 1)=2$$, $$\log_3(6x^2 + 3)=3$$.

Therefore, the next three terms of the AP will be 4, 5, and 6, and their sum will be = 4 + 5 + 6 = 15.