IPM Indore 2025 Question Paper

Age of Monica = 18

Age of Monica's father = 3 * 18 = 54

Let us assume that after x years, Monica's age will be half that of her father's age.

=> $$18+x=\dfrac{1}{2}\left(54+x\right)$$

=> $$36+2x=54+x$$

=> $$x=18$$

Thus, after 18 years, Monica's age will be half that of her father's age. At that time, Monica will be = 18 + 18 = 36.

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 arranged in ascending order in each team is = (2, 4, 6, 6, 7). Thus, the median value is 6. 

$$7m + 11n = 200$$

It is given that both m and n are positive integers. We need to find one integer pairs of (m,n) that satisfy the equation for these questions. 

If we put the value of n = 1, then the value of m = 27. Thus, one solution will be (27,1).

Now, to find the next solution of n, we need to add the coefficient of m in the original value of n, and to find the next value of m, we need to subtract the coefficient of n in the original value of m. (One should be added, the other should be subtracted. Since we need the positive values of m and n, thus we are adding in n, and subtracting from m). We will continue this process till we get any value of m or n as negative. 

Thus, the other solutions of these equations will be - (m,n) = (16,8), (5,15)

If (m,n) = (27,1) => m+n = 28.

If (m,n) = (16,8) => m+n = 24.

If (m,n) = (5,15) => m+n = 20.

Thus, the minimum value of m + n = 20. 

If $$N=p^a\times q^b\times r^c$$, then the number of factors is given by $$(a+1)(b+1)(c+1)$$

Now, we are given $$N=3^5 \times 5^8 \times 7^2$$, and we need to find the factors which are perfect squares.

A number is a perfect square if all exponents in its prime factorization are even. So, we need to find the even exponent possible values for each prime number in N

Exponent of 3 in N is 5, thus, the even values which are possible = 0, 2, 4 {3 possibilities}

Exponent of 5 in N is 8, thus, the even values which are possible = 0, 2, 4, 6, 8 {5 possibilities}

Exponent of 7 in N is 2, thus, the even values which are possible = 0, 2 {2 possibilities}

Thus, the total number of even factors for N will be $$=3\times5\times2=30$$

$$f(x)=ax^2 + bx + 5$$

Polynomial $$ax^2 + bx + 5$$ leaves a remainder 3 when divided by $$x - 1$$. Thus, as per the remainder theorem, when x = 1, the remainder is 3. 

f(1) = 3 => $$a+b+5=3$$ => $$a+b=-2\rightarrow1$$

Polynomial $$ax^2 + bx + 5$$ leaves a remainder 2 when divided by $$x + 1$$. Thus, as per the remainder theorem, when x = -1, the remainder is 2.

f(-1) = 3 => $$a-b+5=2$$ => $$a-b=-3\rightarrow2$$

Adding eq. 1 and eq. 2 - 

=> $$2a=-5$$ => $$a=-2.5$$

Subtracting eq. 2 from eq. 1 -

=> $$2b=1$$ => $$b=0.5$$

Therefore the value of $$2b-4a=2(0.5)-4(-2.5)=1+10=11$$

Given,

$$1 + \dfrac{1}{2^2} + \dfrac{1}{3^2} + \dfrac{1}{4^2} + ...... = \dfrac{\pi^2}{6}$$ ------>(1)

Multiplying all the terms by $$\dfrac{1}{4}$$ we get,

$$\dfrac{1}{4}+\dfrac{1}{2^2\cdot4}+\dfrac{1}{3^2\cdot4}+\dfrac{1}{4^2\cdot4}+......=\dfrac{\pi^2}{6\cdot4}=\dfrac{\pi^2}{24}$$

or, $$\dfrac{1}{2^2}+\dfrac{1}{2^2\cdot2^2}+\dfrac{1}{3^2\cdot2^2}+\dfrac{1}{4^2\cdot2^2}+......=\dfrac{\pi^2}{6\cdot4}=\dfrac{\pi^2}{24}$$

or, $$\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+\dfrac{1}{8^2}+......=\dfrac{\pi^2}{6\cdot4}=\dfrac{\pi^2}{24}$$ ------->(2)

Subtracting (2) from (1),

$$1+\dfrac{1}{3^2}+\dfrac{1}{5^2}+\dfrac{1}{7^2}+......=\dfrac{\pi^2}{6}-\dfrac{\pi^2}{24}=\dfrac{3\pi^2}{24}=\dfrac{\pi^2}{8}\ $$

$$y = a + b \log_e x$$

=> $$b\log_ex=y-a$$

=> $$\log_ex^b=y-a$$

=> $$x^b=e^{\left(y-a\right)}$$

=> $$x^b=\dfrac{e^y}{e^a}$$

=> $$e^y=e^ax^b$$

Thus, we can clearly see that $$e^y$$ is directly proportional to $$x^b$$

$$x^8 + x^7 + ..... + x + 1 = 0$$

Since $$1,x,x^2,x^3,......$$ are in GP, thus we will apply the formula of sum of GP.

=> $$1\left[\dfrac{x^9-1}{x-1}\right]=0$$

=> $$x^9-1=0$$

=> $$x^9=1$$

Now, $$a_1, a_2, ......., a_8$$ are the roots of the equation thus - 

$$\left(a_1\right)^9=\left(a_2\right)^9=\left(a_3\right)^9=....=\left(a_8\right)^9=1$$

We need to find the value of $$a^{2025}_1 + a^{2025}_2 + .... + a^{2025}_8$$.

$$\left(a_1^9\right)^{225}+\left(a_2^9\right)^{225}+....+\left(a_8^9\right)^{225}$$

$$\left(1\right)^{225}+\left(1\right)^{225}+....+\left(1\right)^{225}=1+1+.....+1=8$$

Let us assume $$a<b<c$$

Max(a, b, c) = c

Min(a, b, c) = a

Median(a, b, c) = b

Mean(a, b, c) = (a+b+c)/3

Now, Max(a, b, c) + Min(a, b, c) = 15

=> $$c+a=15\rightarrow1$$

Also, Median(a, b, c) - Mean(a, b, c) = 2

=> $$b-\dfrac{\left(a+b+c\right)}{3}=2$$

=> $$2b-(a+c)=6\rightarrow2$$

Adding eq. 1 and eq. 2 -

=> $$2b=21$$ 

=> $$b=10.5$$

Thus, the median of a, b, and c = b = 10.5

$$\log_{25}\left[5 \log_3 (1 + \log_3(1 + 2 \log_2 x))\right] = \dfrac{1}{2}$$

$$\left[5\log_3(1+\log_3(1+2\log_2x))\right]=(25)^{\frac{1}{2}}$$

$$\left[5\log_3(1+\log_3(1+2\log_2x))\right]=5$$

$$\log_3(1+\log_3(1+2\log_2x))=1$$

$$1+\log_3(1+2\log_2x)=3$$

$$\log_3(1+2\log_2x)=2$$

$$(1+2\log_2x)=9$$

$$2\log_2x=8$$

$$\log_2x=4$$

$$x=16$$