IPM Indore 2025 Question Paper

Let us assume the number as $$xyz$$

Total outcomes = All the numbers between 100 and 400 for which the sum of digits = 10

Favourable outcomes = All the numbers between 100 and 400 for which the sum of digits = 10, and the number is divisible by 4.

$$P=\dfrac{Favourable\ Outcomes}{Total\ Outcomes}$$

Now, for the total outcomes, we have 3 cases as per the hundredth place digit.

Case-1: When the hundredth place digit = 1 (x = 1)

Since the sum of digits = 10, (y + z) = 9. Thus, there are 10 possible pairs for (y,z) = (0,9), (1,8), (2,7),......,(9,0)

=> 10 possibilities

Case-2: When the hundredth place digit = 2 (x = 2)

Since the sum of digits = 10, (y + z) = 8. Thus, there are 9 possible pairs for (y,z) = (0,8), (1,7), (2,6),......,(8,0)

=> 9 possibilities

Case-3: When the hundredth place digit = 3 (x = 3)

Since the sum of digits = 10, (y + z) = 7. Thus, there are 8 possible pairs for (y,z) = (0,7), (1,6), (2,5),......,(7,0)

=> 8 possibilities

For 400, the sum of the digits is 4; thus, we don't need to worry about 400.

Thus, the total outcomes = 10 + 9 + 8 = 27

For the favourable outcomes, we need to check whether the number's last two digits are divisible by 4, that is, yz is divisible by 4 or not.

For Case-1 => the only possible values for yz = 36 and 72. Thus, 2 possibilities.

For Case-2 => the only possible values for yz = 08, 44, and 80. Thus, 3 possibilities.

For Case-3 => the only possible values for yz = 16 and 52. Thus, 2 possibilities.

So, number of favourable outcomes = 2 + 3 + 2 = 7 

=> $$P=\dfrac{Favourable\ Outcomes}{Total\ Outcomes}=\dfrac{7}{27}$$

In triangle ABC, AB = AC = x, therefore $$\angle ABC=\angle ACB=\theta\ $$, since angles opposite to equal sides are equal. Let AD be the perpendicular to BC.

$$\cos\theta=\dfrac{base}{hypoteneuse}=\dfrac{BD}{AB}=\dfrac{BD}{x}$$ => $$BD=x\cos\theta$$

$$\cos\theta=\dfrac{base}{hypoteneuse}=\dfrac{CD}{AB}=\dfrac{CD}{x}$$ => $$CD=x\cos\theta$$

=> $$BC=BD+DC=2x\cos\theta$$

$$\sin\theta=\dfrac{perpendicular}{hypoteneuse}=\dfrac{AD}{AB}=\dfrac{AD}{x}$$ => $$AD=x\sin\theta$$

It is given that the circumradius of the triangle is y.

We also know that => $$Area\ of\ \triangle=\dfrac{abc}{4R}$$, where R is the circumradius, and a, b, and c are the sides of the triangle.

=> $$\dfrac{1}{2}\times2x\cos\theta\times x\sin\theta\ =\dfrac{x\times x\times2x\cos\theta\ }{4y}$$

=> $$\dfrac{1}{2}\times2x^2\cos\theta\sin\theta\ =\dfrac{2x^3\cos\theta\ }{4y}$$

=> $$\dfrac{1}{2}\times\sin\theta\ =\dfrac{x\ }{4y}$$

=> $$\dfrac{x\ }{y}=2\sin\theta\ $$

$$8x^2 - 2kx + k = 0$$. Let us assume the two roots are $$a$$ and $$ap$$

Now, $$a+ap=\dfrac{2kx}{8}$$ => $$a\left(1+p\right)=\dfrac{kx}{4}$$ => $$a=\dfrac{kx}{4\left(1+p\right)}\rightarrow1$$

And, $$a^2p=\dfrac{k}{8}$$

Substituting the value of a from eq. 1 -

=> $$p\left[\dfrac{k}{4\left(1+p\right)}\right]^2=\dfrac{k}{8}$$

=> $$p\times\dfrac{k^2}{16\left(1+p\right)^2}=\dfrac{k}{8}$$

=> $$k=\dfrac{2\left(1+p\right)^2}{p}$$

=> $$k=2\left[\dfrac{1+p}{\sqrt{p}}\right]^2$$

=> $$k=2\left[\sqrt{p}+\dfrac{1}{\sqrt{p}}\right]^2$$

Slope of line AB = $$m_{AB}=\dfrac{1-3}{5-1}=-\dfrac{2}{4}=-\dfrac{1}{2}$$. And, the slope of the other line is m. The angle between the two lines is 45 degrees, and we know the formula -

$$\tan\theta=\left|\dfrac{m_2-m_1}{1+m_1m_2}\right|$$

$$\tan\left(45\right)=\left|\dfrac{m-\left(-\frac{1}{2}\right)}{1+m\left(-\frac{1}{2}\right)}\right|$$

$$1=\left|\dfrac{2m+1}{2-m}\right|$$

$$\left|2m+1\right|=\left|m-2\right|$$

There are two possible cases for this.

Case-1: $$2m+1=m-2$$ => $$m=-3$$

Case-2: $$2m+1=-m+2$$ => $$3m=1$$ => $$m=\dfrac{1}{3}$$

Thus, the two possible values for m = -3 and m = 1/3.

Let $$P(x)=ax^2+bx+c$$, and it is given that P(0) = 2.

=> $$a(0)+b(0)+c=2$$ => $$c=2$$

$$\begin{vmatrix}P(0) & P(1)\\P(0) & P(2)\end{vmatrix} = 0$$

=> $$P\left(0\right)P\left(2\right)-P\left(0\right)P\left(1\right)=0$$

=> $$P(1)=P(2)$$

=> $$a+b+2=4a+2b+2$$

=> $$b=-3a$$

It is also given that -

$$P(1) + P(2) + P(3) = 14$$

$$a+b+2+4a+2b+2+9a+3b+2=14$$

$$14a+6b=8$$                (Substituting the value of b = -3a)

$$14a-18a=8$$ => $$a=-2$$

Thus, $$b=-3(-2)=6$$

Therefore, $$P(x)=-2x^2+6x+2$$

=> $$P(4)=-2*4^2+6*4+2$$

=> $$P(4)=-32+24+2$$

=> $$P(4)=-6$$

A circle touches the y-axis at (0, 4). Therefore, the centre of the circle is on the horizontal line whose y coordinate = 4. Thus, the coordinates of the center of the circle will be (a,4).

Now, the circle also passes through the point (-2,0)

The radius of the circle remains constant. 

$$\left(a-0\right)^2+\left(4-4\right)^2=\left(a-\left(-2\right)\right)^2+\left(4-0\right)^2$$

$$a^2=a^2+4a+4+16$$

=> $$a=-5$$

Therefore, the radius of the circle = $$\sqrt{\left(4-4\right)^2+\left(0-\left(-5\right)\right)^2}=5\ cm$$