IPM Indore 2024 Question Paper

$$4^{\log_2 x} - 4x + 9^{\log_3 y} - 16y + 68 = 0$$

Using the property of logarithms$$a^{\log_xy}=y^{\log_xa}$$ and $$\log_xx^2=2$$

We can rewrite $$4^{\log_2 x} - 4x + 9^{\log_3 y} - 16y + 68 = 0$$

=> $$x^{\log_24}-4x+y^{\log_39}-16y+68=0$$

=> $$x^2-4x+y^2-16y+68=0$$

We can rewrite the equation as sum of two perfect squares

=> $$\left(x-2\right)^2+\left(y-8\right)^2=0$$

We know that RHS is equal to zero so both perfect squares should be equal to zero.

Thus, x = 2 and y =8

Therefore, y-x = 8-2 = 6

A fruit seller has oranges, apples, and bananas in the ratio 3 : 6 : 7.

Let the quantities of oranges, apples, and bananas equal 3k, 6k, and 7k, respectively.

It is given that the number of oranges is a multiple of both 5 and 6.

That means oranges = 5*6*N = 30N

So, if oranges = 30N

Then, apples = 60N

bananas = 70N

The minimum number of fruits the seller has is = 30N+60N+70N = 160N

It depends on the value of N, and the minimum value of N can be 1.

So, minimum number if fruits that seller has = 160.

We need to find out the number of real solutions of the equation $$(x^2 -15x + 55)^{x^2 - 5x + 6} = 1$$

We can assume that $$x^2-15x+55$$ = $$P$$ and $$x^2-5x+6$$ = $$Q$$

$$P^Q=1$$

There are three possible cases

Case 1: $$P\ =\ 1$$ and $$Q\in\ R$$

$$x^2-15x+55$$ = 1

$$x^2-15x+54$$ = 0

$$\left(x-9\right)\left(x-6\right)=0$$

x = 9 and 6

Case 2: $$P\ \in\ R$$ and $$Q\ =0$$

$$x^2-5x+6$$ = 0

x = 2 and 3

Case 3: $$P\ =-1$$ and $$Q\ =$$ even positive power

$$x^2-15x+55$$ = -1

$$x^2-15x+56$$ = 0

x = 8 and 7

When x = 8 and 7 , Q = 30 and 20

Therefore, the number of possible values of x is 6. (9,6,2,3,8,7)

Person A borrows Rs. 4,000 from Person B for a period of 4 years.

It is given that he borrows a portion of it at 3% simple interest per annum, while the rest is borrowed at 4% simple interest per annum.

Let the portion borrowed at 3% is P, therefore, the portion borrowed at 4% is 4000-P

Let us calculate the interest earned by them in 4 years for both portions.

For P, rate of interest = 3%, time = 4 years

Interest earned = $$\dfrac{P\times3\times\ 4}{100}$$    ------(1)

Similarly, for 4000-P, rate of interest = 4% and time = 4 years

Interest earned =$$\dfrac{\left(4000-P\right)\times4\times\ 4}{100}$$    -------(2)

Therefore, total interest earned =  (1) + (2)

It is also given that B gets Rs. 520 as total interest.

(1) + (2) = 520

12P+64000-16P = 52000

4P = 12000

P = 3000

Therefore, the amount A borrowed at 3% per annum in Rs. is P = 3000

The number of triangles with integral sides possible with perimeter 'x', if x is odd = $$\left\{\dfrac{\left(x+3^2\right)}{48}\right\}\ $$

The number of triangles with integral sides possible with perimeter 'x', if x is even = $$\left\{\dfrac{\left(x^2\right)}{48}\right\}\ $$

{n}->represents the integer nearest to 'n'

The number of triangles with integer sides and with a perimeter of 15.

15 is an odd number.

The number of triangles with integral sides possible with perimeter 15 = $$\left\{\dfrac{\left(15+3^2\right)}{48}\right\}\ $$

=> $$\dfrac{324}{48}$$ = 6.75 $$\approx\ 7$$

Given ,
$$A = \begin{bmatrix}x_1 & x_2 & 7 \\y_1 & y_2 & y_3 \\z_1 & 8 & 3 \end{bmatrix}$$
Now lets assume the sum of any row or column or diagonals to be "K".

We can use the property of determinants C1 to C1+C2+C3 :
$$A = \begin{bmatrix}K & x_2 & 7 \\K & y_2 & y_3 \\K & 8 & 3 \end{bmatrix}$$

Now apply R1 to R1+R2+R3 :
$$A = \begin{bmatrix}3K & K & K \\K & y_2 & y_3 \\K & 8 & 3 \end{bmatrix}$$

Now as we assumed : $$K=7+3+Y_3$$ , this implies $$Y_3=K-10$$ .... Equation 1

Now by equating Diagonal sum to C2 sum :$$X_2+Y_2+8=X_1+Y_2+3$$
This implies $$X_1=X_2+5$$
And we also know that sum of R1 to be K : $$X_1+X_2+7=K$$
Now by substituting $$X_1$$ in terms of $$X_2$$ : we get $$X_2$$ = $$\dfrac{\ K-12}{2}$$
And we also know that Sum of C2 to be K : $$X_2+Y_2+8=K$$
By substituting $$X_2$$ we get  $$Y_2=\dfrac{\ K-4}{2}$$... Equation 2 

Now we can substitute Equation 1 & Equation 2 in the matrix :
$$A = \begin{bmatrix}3K & K & K \\K & K/2-2 & K-10 \\K & 8 & 3 \end{bmatrix}$$

Now to find K :
We already know that Sum of R2 = Sum of C3 = K
$$Y_1+Y_2+Y_3=10+Y_3$$
Hence,$$Y_1+Y_2=10$$
We already deduced $$Y_2$$ = $$\dfrac{\ K-4}{2}$$
Hence, $$Y_1=12-\dfrac{\ K}{2}$$

Now , we also know that Sum of all elements should be 3(K) , and
 This 3K should be equal to {Sum of two diagonals - $$Y_2$$ + 8 +$$Y_1+X_2+Y_3$$} 
3K = 2K - $$\dfrac{\ K-4}{2}$$ + 8 + ($$12-\dfrac{\ K}{2}$$) + ($$\dfrac{\ K-12}{2}$$) + (K-10)
This implies K = 12.

Therefore the determinant of :
$$A = \begin{bmatrix}3K & K & K \\K & K/2-2 & K-10 \\K & 8 & 3 \end{bmatrix}$$

$$A = \begin{bmatrix}36 & 12 & 12 \\12 & 4 & 2 \\12 & 8 & 3 \end{bmatrix}$$

= $$36(-4)-12(12)+12(12(4))$$
= $$12\times12\times2$$
= 288.

The number of factors of 1800 that are multiples of 6 is:

Prime factorisation of 1800 = $$2^3\times3^2\times\ 5^2$$

We know that the total number of factors of N = $$p^a\ \times q^b\times\ r^c$$ will be $$(a+1)(b+1)(c+1)$$

And, the number of factors that are multiple of 6 will be

N =$$6\times\ \left(2^2\times3^1\times\ 5^2\right)$$

Total factors will be (2+1)*(1+1)*(2+1) = 18

ABC is a right angled triangle at B, and said that rectangle has to inscribed inside with B as one of its vertices : 

As triangle ABC and EFC are similar , we can say that : $$\dfrac{\ AB}{BC}=\ \dfrac{\ EF}{FC}$$
$$\dfrac{\ 18}{18}=\ \dfrac{\ X}{18-Y}$$ 

This implies X+Y = 18.
Area of rectangle is X(Y) = X(18-X) = $$18X-X^2$$
This will be maximum at X = 18/2 = 9 

Therefore largest area is 9x9 = 81.