IPM Indore 2024 Question Paper

Lets assume |X-5| = a & |Y-5| = b :
$$\left|a\right|\ +\ \left|b\right|\ \le\ 6$$
The graph with horizontal axis as "a" & vertical axis "b" will look like :

The points which satisfy this inequality will be (a,b) :

Case I : Points on the line/boundary : 
(0,$$\pm6$$) (1,$$\pm5$$) (2,$$\pm4$$) (3,$$\pm3$$) (4,$$\pm2$$) (5,$$\pm1$$) (6,0) 
(-1,$$\pm5$$) (-2,$$\pm4$$) (-3,$$\pm3$$) (-4,$$\pm2$$) (-5,$$\pm1$$) (-6,0)
Therefore , on the boundary  = 24

Case II : Points which lie inside the boundary :
(0,0)
($$\pm1$$,0) ($$\pm2$$,0) ($$\pm3$$,0), ($$\pm4$$,0) ,($$\pm5$$,0) 
(0,$$\pm1$$) (0,$$\pm2$$) (0,$$\pm3$$), (0,$$\pm4$$) ,(0,$$\pm5$$)
For X = $$\pm1$$ : 2x2x4 = 16
For X = $$\pm2$$ : 2x2x3 = 12
For X = $$\pm3$$ : 2x2x2 = 8
For X = $$\pm4$$ : 2x2x1 = 4
Therefore , inside no of points = 1+10+10+16+12+8+4 = 61

Hence total no of integral points which satisfy the inequality will be = 61+24 = 85.

F has 29 employees. So the median will be the age of the 15th employee, which is 54. E has 35 employees, so the median will be the age of the 18th employee, which is 49.

We need to find the minimum possible sum of the ages of all the employees of F.

For that, we also need to minimise the highest possible age in E, as no employee in E is older than the employees in F. 

Now, the median age of E is 49, which would mean that if at least 18 people in E are aged 49, the remaining aged below 49, then too, the median will be 49. This would mean that the minimum possible age in Company F is 50. 

Also, since the median age in F is 54, at least 15 employees should be aged 54 for the median to hold true. 

So the minimum age will be when 15 employees are aged 54 and the remaining 14 employees are aged 50, the minimum sum of ages being 50x14+54x15 = 1510. 

Given total no of students 150. 
52 (liked tea) = only tea + (tea, coffee) + (tea, juice) + (tea, coffee, juice) .
48 (like juice) = only juice + (juice, tea) + (juice, coffee) + (coffee, juice, tea)
62 (like coffee) = only coffee + (coffee, juice) + (coffee, tea) + (coffee, juice, tea).

adding all three we get :
162 = Exactly 1 drink + 2(Exactly 2 drinks) + 3(Exactly all 3 drinks) . 
but given total 150 =  Exactly 1 drink + Exactly 2 drinks + Exactly all 3 drinks .
Substituting equation 2 in equation 1: 
162 = 150 + Exactly 2 drinks + 2(Exactly all 3 drinks) .

12 = {Exactly 2 drinks + Exactly 3 drinks} + {Exactly 3 drinks} .
12 = {More than 1 drink} + {Exactly 3 drinks}.

Therefore, the maximum of More than 1 drink occurs at exactly 3 drinks is 0. Hence, maximum of More than 1 drink is 12. 

Let the price of the chocolate be P.

It is given that the price of a chocolate is increased by x% and then reduced by x%..

The new price is 96.76% of the original price.

So, the equation will be

$$P\left(1+\dfrac{x}{100}\right)\left(1-\dfrac{x}{100}\right)=\dfrac{96.76}{100}P$$

=> $$1-\dfrac{x^2}{10000}=\dfrac{9676}{10000}$$

=> $$10000-x^2=9676$$

=> $$x^2=324$$

=> $$x\ =18$$

The value of x is equal to 18.

Given that f(a) + g(f(a)) = $$\dfrac{\ 13}{6}$$

f(a) + $$\dfrac{\ 1}{f\left(a\right)}$$ = $$\dfrac{\ 13}{6}$$ 

$$6(f(a))^2-13\left(f\left(a\right)\right)+6\ =\ 0$$

f(a) = $$\dfrac{13\pm\sqrt{13^2-24\left(6\right)\ }\ }{2\times6}$$ 

this implies , f(a) = $$\ \dfrac{\ 3}{2}\ (or)\ \dfrac{\ 2}{3}$$ 

Given that f(a) = |a+|a||
if $$a\le0\ $$, f(a) = |a-a| = 0
else if $$a\ge\ 0$$ , then f(a) = 2a.

Therefore , 2a = $$\ \dfrac{\ 3}{2}\ (or)\ \dfrac{\ 2}{3}$$ 
that is, a = $$\ \dfrac{\ 3}{4}\ (or)\ \dfrac{\ 1}{3}$$

g($$\dfrac{\ 3}{4}$$) = $$\dfrac{\ 4}{3}$$

g($$\dfrac{\ 1}{3}$$) = $$\dfrac{\ 3}{1}$$

Therefore f(g(a)) = $$\dfrac{\ 8}{3}$$ or 6.

Maximum value is 6.

Given a series in GP which are positive real numbers ..Let the first term be "a" and common ratio be "r".
$$p^{th}$$ term =$$a\times r^{p-1}$$ = q 
$$q^{th}$$ term =$$a\times r^{q-1}$$ = p
By dividing both, we get $$r=\ \left(\dfrac{\ q}{p}\right)^{\ \dfrac{\ 1}{p-q}}$$
By substituting this in 2nd equation , we get : $$a\left(\ \dfrac{\ q}{p}\right)^{\ \dfrac{q-1\ }{p-q}}=p$$
Now applying log on both sides :
Log a + $$\dfrac{\ q-1}{p-1}\left(\log\dfrac{\ q}{p}\right)$$ = log p
Therefore , log a = log p $$-$$ $$\ \dfrac{\ q-1}{p-q}\left(\log\ q-\log\ p\right)$$

Hence, log a = $$\ \dfrac{\ \left(p-1\right)\log\ p\ +\ \left(q-1\right)\log\ q}{p-q}$$

Lets assume the distance b/w point and centre of circle as "d" . let radius of circle be R .

There are two cases , one case is in which point is inside the circle and the other one is point lies totally outside circle :

Case I : Point lying outside the circle 
  

The shortest distance is PB = PO - BO = d - r = 4
The longest distance is PA = PO + OA = d + r = 9
Solving this we get r = 2.5 .

Case II : Point lying inside the circle 

The shortest distance is PB = OB - OP = r - d = 4
The longest distance is PA = PO + OA = r + d  = 9
Solving this we get r = 6.5

Therefore the radius of circle can be either 2.5 cm or 6.5 cm

If $$\mid x + 1 \mid + (y + 2)^2 = 0$$

That means x = -1 and y = -2

Then, only LHS can be equal to zero.

And, $$ax - 3ay = 1$$

$$-a +6a = 1$$

$$5a=1$$

Thus, $$a=\dfrac{1}{5}$$

The number of real solutions of the equation $$x^2 - 10 \mid x \mid - 56 = 0$$ is

We need to find the solutions to the equation - $$x^2 - 10 \mid x \mid - 56 = 0$$

Taking two cases

Case 1: x > 0

$$x^2-10x-56=0$$

$$x^2-14x+4x-56=0$$

$$x\left(x-14\right)+4\left(x-14\right)=0$$

$$\left(x+4\right)\left(x-14\right)=0$$

x = -4 and x = 14

Since, x > 0

x = 14 is the only possible value.

Case 2: x < 0

$$x^2+10x-56=0$$

$$x^2+14x-4x-56=0$$

=> $$x\left(x+14\right)-4\left(x+14\right)=0$$

=> $$\left(x-4\right)\left(x+14\right)=0$$

x = 4 and x = -14

Since, x < 0

x = -14 is the only possible value.

Therefore, there are 2 possible real solutions.

$$4^{100}$$ can be written as $$2^{200}$$ which is less than $$2^{300}$$ .

$$2^{100}+3^{100}$$ < $$2(3^{100})$$ < $$3^{200}$$

Now we have to see which is greatest among $$2^{300}$$ & $$3^{200}$$ .
$$2^{300}$$ = $$2^{3^{100}}$$ = $$8^{100}$$
$$3^{200}$$ = $$3^{2^{100}}$$ = $$9^{100}$$

Therefore, $$3^{200}$$ is greatest among all.