IPM Indore 2024 Question Paper

Let there be four values $$a$$, $$b$$, $$c$$, and $$d$$, all greater than or equal to zero, such that $$x_1=a+1$$, satisfying $$x_1\ge 1$$ and, $$x_2= b+2$$, satisfying $$x_2\ge 2$$, and $$x_3=c$$ and $$x_4=d$$.

We have, $$a+1+b+2+c+d=50$$ or $$a+b+c+d=47$$

Since all of the variables are greater than or equal to zero, the number of solutions to this equation will simply be $${}^{47+4-1}C_{4-1} = \dfrac{50*49*48}{6} = 19600$$

The number of digits in a number $$a^b$$ is given by $$\lfloor \log_{10}(a^b)  + 1\rfloor$$

Therefore, the number of digits in $$2^{2024}$$ is $$\lfloor \log_{10} 2^{2024} +1 \rfloor = \lfloor 2024(\log_{10} 2)+1 \rfloor = \lfloor 2024*0.301 + 1 \rfloor = 610$$

And the number of digits in $$5^{2024}$$ is $$\lfloor \log_{10}{5^{2024}} +1\rfloor = \lfloor 2024(\log_{10} 5) + 1 \rfloor =\lfloor 2024*0.698 + 1\rfloor = 1415$$

Therefore, the total number of digits on the page are $$610+1415= 2025$$

The equation of the circle reduces to $$(x-7)^2+(y+8)^2=5^2$$ when we add $$49+64$$ to both sides and subtract $$88$$ from both sides to complete the squares. This means that the centre of the circle is at $$(7,-8)$$ and its radius is $$5$$ units.

Using the Distance Formula, we get the distance between the centre and the point where tangents originate (say $$P$$) as:

$$\sqrt{(7-0)^2 + (-8-3.5)^2} = \dfrac{5\sqrt{29}}{2}$$

Along with the radius of the circle in question, the tangent and this distance between the centre and the point $$P$$ form a right angled triangle. Therefore, the length of each of the tangents will be $$\sqrt{\left(\dfrac{5\sqrt{29}}{2}\right)^2 - 5^2} = \dfrac{25}{2}$$.

$$\text{tan }\dfrac{\theta}{2}$$ is therefore equal to $$\dfrac{5}{\frac{25}{2}} = \dfrac{2}{5}$$

Using the formula $$\text{tan }2x = \dfrac{2\text{ tan }x}{1-\text{tan }^2 x}$$, we obtain $$\text{tan }\theta$$ as:

$$\dfrac{2*\frac{2}{5}}{1-\frac{4}{25}} = \dfrac{20}{21}$$

The  given equation is $$(x^2 - 5)^4 + (x^2 - 7)^4 = 16$$

Let, $$x^2-5=a$$ then $$x^2-7=a-2$$

The equation, therefore, becomes, $$a^4+\left(a-2\right)^4=16$$. We simplify this equation to find real values of a , and then consequently find real values of x.

Now, $$a^4+\left(a-2\right)^4=16$$

or, $$\left(a^2\right)^2+\left(\left(a-2\right)^2\right)^2=16$$

or, $$\left(a^2-\left(a-2\right)^2\right)^2+2\times\ a^2\times\ \left(a-2\right)^2=16$$

or, $$\left(\left(a+a-2\right)\times\ \left(a-a+2\right)\right)^2+2\times\ a^2\times\ \left(a-2\right)^2=16$$

or, $$\left(2\times\ 2\left(a-1\right)\right)^2+2\times\ a^2\times\ \left(a-2\right)^2=16$$

or, $$16\times\ \left(a-1\right)^2+2\times\ a^2\times\ \left(a-2\right)^2=16$$

or, $$8\times\ \left(a-1\right)^2+\ a^2\times\ \left(a-2\right)^2=8$$

or, $$a^2\times\ \left(a-2\right)^2=8-\ 8\times\ \left(a-1\right)^2=8\left\{1-\left(a-1\right)^2\right\}=8\left(1-a+1\right)\left(1+a-1\right)=8a\left(2-a\right)$$

$$\therefore\ a^2\times\ \left(a-2\right)^2=8a\left(2-a\right)$$

or, $$a^2\times\ \left(a-2\right)^2-8a\left(2-a\right)=0$$

or, $$a^2\times\ \left(a-2\right)^2+8a\left(a-2\right)=0$$

or, $$a\times\ \left(a-2\right)\left\{a\left(a-2\right)+8\right\}=0$$

$$\therefore\ a=0$$ or $$a-2=0$$ or $$a\left(a-2\right)+8=0$$

From the first eqn, a=0, $$x^2-5=0\ =>x=\pm\ \sqrt{\ 5}$$

From the second eqn, a-2=0,  $$x^2-7=0\ =>x=\pm\ \sqrt{\ 7}$$

ANd from the third eqn, $$a\left(a-2\right)+8=0$$ $$a\left(a-2\right)+8=0\ =>\ a^2-2a+8=0$$ 

Discriminant, $$D=2^2-4\times\ 8\times\ 4<0$$

This has no real solutions for a , and then there will be no real solutions for x as well. 

So x has 4 real solutions, the maximum among them is $$\sqrt{7}$$ and the minimum is $$-\sqrt{7}$$

So their difference is $$\sqrt{\ 7}-\left(-\sqrt{7}\right)=2\sqrt{\ 7}$$ which is option D. 

Since $$\triangle ACD$$ has $$\angle ADC =\angle CAD =45^\circ$$, it is an isosceles right-angled triangle, and the lengths $$DC$$ and $$AD$$ will both be the same. $$DC$$ being the length of the pole, we find $$DC=AD= p$$ meters.

Since $$AB$$ is the length walked, it is $$20$$ meters, and the total length $$BC$$ is $$p+20$$ meters.

In $$\triangle BCD$$, since the pole is upright, $$\angle DCB$$ is $$90^\circ$$, and the other two angles are $$30^\circ$$ and $$60^\circ$$. We therefore get it as a $$30-60-90$$ triangle and the ratio of sides $$1:\sqrt{3}:2$$ will give $$DC:BC = 1:\sqrt{3}$$.

We finally have $$\dfrac{p}{p+20}=\dfrac{1}{\sqrt{3}}$$ which gives $$p=\dfrac{20}{\sqrt{3}-1} = 10(\sqrt{3}+1)$$

$$C\left(\begin{array}{c}17-x\\ 3x + 1\end{array}\right)$$ ; here the C refers to combination , ie nCr where n=17-x and r=3x+1

Now, we know nCr is defined as an integer when $$n\ge0,\ r\ge0\ and\ n\ge r$$ , where n and r are integers.

In the given expression, n=17-x, and r=3x+1

So, $$17-x\ge0;\ \ 3x+1\ge0\ and\ 17-x\ge3x+1$$

From 1st eqn, $$x\le17$$

From 2nd eqn, $$x\ge-\frac{1}{3}$$

And , from 3rd eqn, $$17-x\ge3x+1\ =>16\ge4x\ \ \ or,\ x\le4$$

Combining all these , we get one relation that satisfies all three inequalities, which is $$-\frac{1}{3}\le x\le4$$

And since 17-x and 3x+1 need to be integers, so x has to be an integer. The integers in this range are 0,1,2,3,4. 

So the number of values of x is 5. 

ABC is an equilateral triangle with side length k , and a circle is drawn with AB as the diameter. So the radius of the circle is $$\dfrac{k}{2}$$

Let D and E be the points where the circle and triangle intersect. O is the centre of the circle. The following figure represents the given situation.

$$\angle\ ABC=\angle\ ACB=\angle\ BAC=60^{\circ\ }$$

Also, OA=OB=OD=OE as all of them are radii of the same circle. So, this means, for $$\triangle\ OAD,\ \angle\ OAD=\angle\ ODA=60^{\circ\ }$$, as the sides are equal. By angle sum, we can say that $$\angle\ DOA=180-60-60=60^{\circ\ }$$. So, AOD is also equilateral with side length $$\dfrac{k}{2}$$ 

Similarly we can also conclude that BOE is also equilateral with side length $$\frac{k}{2}$$

Again, AOB is a straight line, so $$\angle\ AOD+\angle\ DOE+\angle\ EOB=180^{\circ\ }$$, which gives us $$\angle\ DOE=180-60-60=60^{\circ\ }$$

Now, we are asked to find the area of the portion of the triangle lying inside the circle, which is the sum of the areas of the 2 triangles AOD and COB and the sector DOE of the circle. 

Now, 

$$ar\triangle\ AOD=\dfrac{\sqrt{\ 3}}{4}\left(side\right)^2=\dfrac{\sqrt{\ 3}}{4}\left(\dfrac{k}{2}\right)^2=\dfrac{\sqrt{\ 3}k^2}{16}sq\ units$$

which is also the area of $$\triangle\ BOE$$

And fo the sector DOE, area = $$\dfrac{60}{360}\times\ \pi\ r^2=\dfrac{1}{6}\times\ \pi\ \left(\dfrac{k}{2}\right)^2=\dfrac{\pi k^2}{24}\ sq\ units$$

So the required total area = $$\dfrac{\sqrt{\ 3}k^2}{16}+\dfrac{\pi k^2}{24}+\dfrac{\sqrt{\ 3}k^2}{16}=2\dfrac{\sqrt{\ 3}k^2}{16}+\dfrac{\pi k^2}{24}=\dfrac{\left(3\sqrt{\ 3}+\pi\ \right)}{24}k^2$$ , which is option A

Let Sagarika divide the 10,000 rupees into x and y, such that she invests x in scheme A and y in scheme B.

So x+y=10,000 ; or x=10,000-y _______(1)

Scheme A is compound interest at 10% compounded half-yearly. 

So, athe mount after 1 year will be $$A=x\left(1+\frac{5}{200}\right)^{2.1}=x\left(1.05\right)^2=1.1025x$$,

And, interest is 1.1025x-x=0.1025x

Scheme B is simple interest at 12% per annum,

So interest is $$SI=\frac{y\times\ 12\times\ 1}{100}=0.12y$$

And the amount is $$SI+P=0.12y+y=1.12y$$

Now, it is given that the value of her investment in scheme B exceeds the value of her investment in scheme A by 2310 rupees,

So, 1.12y-1.1025x=2310

or, 1.12y-1.1025(10,000-y)=2310

or, 1.12y+1.1025y-11025=2310

or,2.222y=13335

or, y=6000. so x=4000

Now, total interest = 0.12y+0.1025x=1130, which is the answer. 

The situation is best represented using a Venn diagram. Here, let A denote the set of people who own a 4-wheeler, and B denote the set of people who own a 2-wheeler. Then the number of people who own a 4-wheeler but not a 2-wheeler is A-B, which is igvwn to be 250. And the number of people who own a two-wheeler but not a 4-wheeler is B-A=100 , the number of people who own none is 100, and the total number of people in the survey is 500. We have to find the number of people who own both a 2-wheeler and a 4-wheeler., which is $$A\cap B$$

Total number of people who own at least one of either a 2-wheeler or a 4-wheeler is $$A\cup B$$ which is total - number of people who own none =500-100=400.

Now, $$n\left(A\cup B\right)=n\left(A-B\right)+n\left(A\cap B\right)+n\left(B-A\right)$$

So, $$400=250+n\left(A\cap B\right)+100$$

or, $$n\left(A\cap B\right)=400-250-100=50$$ 

Hence, the number of people who own both is 50.

Since the product $$ac$$ is greater than zero, we get that either both $$a$$ and $$c$$ are positive, or both are negative. $$b$$ will have the opposite sign as $$a$$ and $$c$$ because the fraction $$\frac{b}{c}$$ is negative. This gives us the following cases.

1. Case 1: Both $$a$$ and $$c$$ are positive, and $$b$$ is negative.
   
   We will conclude $$\frac{c}{a}=4$$, $$\frac{a}{b}=\frac{-1}{3}$$, and $$\frac{b}{c}=\frac{-3}{4}$$
   Therefore $$\frac{b}{a}=-3$$ and we get $$\frac{b+c}{a} = \frac{c}{a}+\frac{b}{a} = 4-3=1$$
2. Case 2: Both $$a$$ and $$c$$ are negative, and $$b$$ is positive
   
   The calculations will be the same.

Therefore, the correct answer is $$1$$.