CAT 2024 Slot 2 Question Paper

We must bring the right-hand side in the form so that everything has the same power. 

25 has factors 1, 5 and 25
The only common factor 40 and 25 have is 5 (other than 1 of course, which does not work)

So the right-hand side can be rewritten as $$\left(2^5\right)^5\times\ \left(3^8\right)^5$$
$$\left(32\times\ 81\times\ 81\right)^5$$
$$\left(209952\right)^5$$

Giving the value of m - n as 209952 - 5 = 209947

Therefore, Option D is the correct answer. 

There are multiple ways of solving these sorts of questions. One method is to look for powers of the term in the numerator that leave a remainder of 1 or -1 when divided by the denominator. 

Noting down the powers of 3, 3, 9, 27, 81, 243

243 is one such number, 242 is multiple of 11 (11 times 22), hence 243 will leave a remainder of 1 when divided by 11. 

243 is 3 raised to power 5; we can rewrite the given term as $$\dfrac{3^{330}\times\ 3^3}{11}$$
The overall remainder will be $$\left[\dfrac{3^{330}}{11}\right]_R\times\ \left[\dfrac{3^3}{11}\right]_R$$

$$\left[\dfrac{3^{5\times\ 66}}{11}\right]_R\times\ \left[\dfrac{3^3}{11}\right]_R$$
$$\left[\dfrac{243^{66}}{11}\right]_R\times\ \left[\dfrac{3^3}{11}\right]_R$$
$$1^{66}\times\ \left[\dfrac{27}{11}\right]_R$$
$$1\times\ 5$$
$$5$$

Therefore, Option A is the correct answer. 

In such questions, we should be trying to complete the squares. 
We see a $$xy$$ term; we need to accommodate that in a square that has both x and y terms. 

Since there is only one other term with x, we also need to have it entirely in the square. 
$$\left(2x-y\right)^{^2}=4x^2+y^2-4xy$$
Using this in the given equation, we are left with $$\left(2x-y\right)^{^2}+3y^2+3-6y$$

This can be written as $$\left(2x-y\right)^{^2}+3\left(y^2+1-2y\right)$$
$$\left(2x-y\right)^{^2}+3\left(y-1\right)^2=0$$

Since both the squares add up to 0, this is only possible when the squares themselves are 0
This would give us y=1 from the second term, and using that, we get x= 1/2 from the first term. 

Therefore the value of 4x+5y will be 2+5 = 7

Hence, 7 is the correct answer. 

Squaring on both sides, we get:

$$x+6\sqrt{\ 2}+x-6\sqrt{\ 2}-2\left(x^2-72\right)^{\frac{1}{2}}=8$$
$$x-\left(x^2-72\right)^{\frac{1}{2}}=4$$

Bringing x to the other side, we get:
$$-\left(x^2-72\right)^{\frac{1}{2}}=4-x$$
Squaring on both sides again, we get:

$$x^2-72=16+x^2-8x$$
$$8x=88$$
$$x=11$$

Therefore, 11 is the correct answer. 

Let's take the number of paths between Q and R to be b and the number of paths between R and S to be a

We are given the paths from P to S through R (which would be 4a), the paths from P to S through Q (which would be 12) and the paths from P to Q to R to S, which would be 3ab) is equal to 62
Giving the relation 4a+12+3ab = 62
Or 4a+3ab = 50

The paths from Q to R directly (which would be b), through P( which would be 12) and through S (which would be 4a) are 27
Giving the relation b+12+4a = 27
Or 4a+b = 15

Subtracting this equation from the first one we got, we get 3ab-b=35, or b(3a-1)=35
b can be 1, 3, 5 or 7
Substituting these values in the second equation, we see that it can not be 1 or 5, leaving only 3 or 7 as the possible values. 

Substituting b as 3 in the first equation would give 13a=50, which is not true. 
Substituting bas 7 in the first equation would give 25a= 50, which would give a=2

We are asked the number of paths from Q to R, which is b=7

Therefore, 7 is the correct answer. 

We can consider the quadrants of a graph:

First quadrant: Both x and y are positive
This would change the equation to 2x+y=15 and x=20, giving a negative value of y; hence, this is not the case. 

Second quadrant: x is negative, but y is positive
This would change the equations to y=15 and x=20, giving a positive value of x, which hence can not be the case.  

Third quadrant: Both x and y are negative
This would change the equation to y=15 and x-2y=20; this gives a positive value of y and hence can not be the case. 

Fourth quadrant: x is positive, but y is negative
This would change the equations to 2x+y=15 and x-2y=20; this gives the value of x as 10 and y as -5, which would lie in the fourth quadrant. 

The value of x-y would be 10-(-5)=15

Therefore, Option B is the correct answer. 

Let's start from the step when there was 50% concentration. 
Let's take there to ee 2T solution: T acid and T water.

Adding 15 litres of acid increases the acid concentration to  80%, giving the equation $$\frac{T+15}{2T+15}=\frac{4}{5}$$
Solving this would give us T=5

This means that there were 5 litres of acid and 5 litres of water after mixing 2 litres of water. 

Therefore, there would be 5 litres of acid and 3 litres of water before adding the water. 
We are asked the ratio of water to acid, which would be 3:5

Therefore, Option B is the correct answer. 

Opening the brackets, we get the series as: $$\left(\dfrac{1}{5}\right)^2-\left(\dfrac{1}{5}\times\ \dfrac{1}{7}\right)+\left(\dfrac{1}{5}\right)^4-\left(\dfrac{1}{5}\times\ \dfrac{1}{7}\right)^2+\left(\dfrac{1}{5}\right)^6-\left(\dfrac{1}{5}\times\ \dfrac{1}{7}\right)^6+...$$

These are two infinite GPs when rearranged:
$$\left(\dfrac{1}{5}\right)^2+\left(\dfrac{1}{5}\right)^4+\left(\dfrac{1}{5}\right)^6+...-\left(\dfrac{1}{5}\times\ \dfrac{1}{7}\right)-\left(\dfrac{1}{5}\times\ \dfrac{1}{7}\right)^6-\left(\dfrac{1}{5}\times\ \dfrac{1}{7}\right)^2-...$$

The sum of the first series would be $$\dfrac{\dfrac{1}{25}}{1-\dfrac{1}{25}}=\dfrac{1}{24}$$

The sum of the second series would be $$\frac{\dfrac{1}{35}}{1-\dfrac{1}{35}}=\dfrac{1}{34}$$

The answer to the given series would then be $$\dfrac{1}{24}-\dfrac{1}{34}=\dfrac{10}{816}=\dfrac{5}{408}$$

Therefore, Option B is the correct answer.