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 = $$\sqrt{\left(4-4\right)^2+\left(0-\left(-5\right)\right)^2}=5\ cm$$

The amount invested in Scheme A is 30,000. Let the amount invested in scheme B be X, and the amount invested in scheme C be 70000 - X. The interest rates in scheme A, scheme B, and scheme C are 10%, 8%, and 12%, compounded annually.

Now, since in the first year, the amount of simple interest is the same as the amount of compound interest, we will use the formulas of simple interest for the first year for calculations. 

It is given that the total interest earned from all three schemes during the first year is 10600.

=> $$\dfrac{30000\times10\times1}{100}+\dfrac{X\times8\times1}{100}+\dfrac{\left(70000-X\right)\times12\times1}{100}=10600$$

=> $$3000+\dfrac{8X}{100}+8400-\dfrac{12X}{100}=10600$$

=> $$\dfrac{4X}{100}=800$$

=> $$X=20000$$

Thus, the amount invested in scheme B is 20,000, and the amount invested in scheme C is 50,000.

The total amount at the end of 2 years will be = $$30000\left(1.1\right)^2+20000\left(1.08\right)^2+50000\left(1.12\right)^2$$

The total amount at the end of 2 years will be = $$36300+23328+62720=122348$$

Thus, the interest earned in the 2 years will be = $$122348-100000=22348$$

Now, the interest earned in the first year was 10600. Thus, the interest earned in the second year will be equal to the total interest earned over the two years minus the interest earned in the first year.

Thus, the interest earned in the second year will be = $$22348-10600=11748$$ Rupees. 

$$f(x) = a^2x^2 + 2bx + c$$

$$a^2$$ is a positive number. This means the coefficient of $$x^2$$ is a positive number. Whenever the coefficient of $$x^2$$ is positive, the graph will look something like this - 

Thus, we can see that this graph will have a definite minimum, but no definite maximum value. 

We are given 3 lines : y = 0, $$12x - 5y = 0$$ , and $$3x + 4y = 7$$ . The triangle formed by these three lines will have its vertices at the points of intersection of the lines with each other. These points can be found by solving the equations simultaneously. 

The lines y=0 and 12x-5y=0 intersect at x=0 ie (0,0)
The lines y=0 and 3x+4y=0 intersect at x= $$\frac{7}{3}$$ ie $$\left(\frac{7}{3},0\right)$$
And the intersection point of the lines 12x-5y=0 and 3x+4y=0 is found to be $$\left(\frac{5}{9},\frac{4}{3}\right)$$

So the area of the triangle formed by these 3 vertices is given by the formula $$\triangle\ =\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|$$

Taking (0,0), $$\left(\frac{7}{3},0\right)$$, and $$\left(\frac{5}{9},\frac{4}{3}\right)$$ as (x1,y1),(x2,y2) and (x3,y3) we have ,

$$\triangle\ =\frac{1}{2}\left|0\left(0-\frac{4}{3}\right)+\frac{7}{3}\left(\frac{4}{3}-0\right)+\frac{5}{9}\left(0-0\right)\right|$$

or, $$\triangle\ =\frac{1}{2}\left|0+\frac{28}{9}+0\right|=\frac{14}{9}$$ sq units.