CAT 2024 Slot 3 Question Paper

This is the figure in the question, 

We are given that each side is 6cm long, 
To find the side AC, we can use cosine rule, since we know each interior angle of the octagon(which is 135 degrees)

$$\cos\left(\angle ABC\right)=\dfrac{\left(AB^2+BC^2-AC^2\right)}{2\left(AB\right)\left(AC\right)}$$

$$\cos\left(135\right)=\dfrac{\left(36+36-AC^2\right)}{2\left(36\right)}$$

$$-\dfrac{1}{\sqrt{2}}=\dfrac{\left(72-AC^2\right)}{72}$$

$$AC^2=72+36\sqrt{2}$$

$$AC^2=36\left(2+\sqrt{2}\right)$$

Since AC is the side of the square, and the area of a square is square of the side. 

Answer is $$36\left(2+\sqrt{2}\right)$$

The moduli will give out only non-negative outputs, and since we are to consider only integer values of x and y, this drastically reduces the possible cases. 

We can get 2 from either 2+0 or 1+1

We get a 2+0 form when either the first term or the second term is 0
The second term is 0; this is when x=y, in this case,|2x|=2, where x can be 1 or -1; therefore, the two cases are (1,1) and (-1,-1)
The first term is 0; this is the case when x = -y, in this case, |x- (-x)|=2, giving x=1 or -1 yet again, here the two cases are (1,-1) and (-1,1)

The other way we can get 2 is through 1+1

This is possible when one of the terms is 0; if y=0, |x|+|x|=2, where x can be 1 or -1, giving two cases (1,0) and (-1,0)
Similarly,y for x=0, we get two cases, (0,1) and (0,-1)

Therefore, there are 8 pairs of (x,y) that satisfy the given equation. 

We are given, $$f(x) + 2f \left(\cfrac{1}{x}\right) = 3x$$
Substituting $$\frac{1}{x}\ for\ x$$

$$f\left(\dfrac{1}{x}\right)+2f\left(x\right)=\dfrac{3}{x}$$
Multiplying the second equation by 2 we will have 

$$2f\left(\dfrac{1}{x}\right)+4f\left(x\right)=\dfrac{6}{x}$$

Subtracting the first equation from the second equation we have, 

$$3f\left(x\right)=\frac{6}{x}-3x$$

$$f\left(x\right)=\frac{2}{x}-x$$
We want the sum of values when this function equals 3, 

$$\frac{2}{x}-x=3$$
$$x^2+3x-2=0$$
Since the discriminant is greater than zero, both values of x will be real, and we can directly take the sum of values of $$x$$ to be, 
$$-\frac{3}{1}$$

Answer is -3. 

Arranging the equation, we know that there are no solutions when the lines are parallel:
for that the condition had to $$\frac{p}{3}=-\frac{4}{k}\ne\ \frac{2}{a}$$

Checking through options: 
Option A: using the first and last terms in our relation, we see that ap must not be equal to 6 for the lines to be parallel. This option puts no conditions on that and thus is not relevant. 

Option C: This question is the opposite of what we want; if this is true, the lines can never be parallel. 

Option D: Using the first and second terms of the relation, we see that we want kp = -12, or kp-12 = 0. Hence, this statement is not what we want. 

Option B: Using the second and third terms, we see that we do not want k equals to -2a, or we do not want k+2a to be equal to 0

Therefore, B is a condition that is necessary for the lines to be parallel and have no solution. 

Therefore, Option B is the correct answer. 

We are told that Rajesh manages 20 hectares and Vimal manages 30 hectares

For Vimal, we know the distribution of the land between Wheat and Mustard, 5:3
So, wheat area will be, $$\frac{5}{8}\left(30\right)$$
Mustard area will be, $$\frac{3}{8}\left(30\right)$$

Similarly, let us assume that the distribution of crops between Wheat and Mustard to be k:1
Wheat will be, $$\frac{k}{k+1}\left(20\right)$$
Mustard will be, $$\frac{1}{k+1}\left(20\right)$$

We are told that total area of Wheat and Mustard is in the ratio 11:9

Adding them up we get, 
$$\dfrac{\left(\frac{150}{8}+\frac{20k}{k+1}\right)}{\left(\frac{90}{8}+\frac{20}{k+1}\right)}=\dfrac{11}{9}$$

$$\dfrac{\left(\frac{15}{8}+\frac{2k}{k+1}\right)}{\left(\frac{9}{8}+\frac{2}{k+1}\right)}=\dfrac{11}{9}$$

$$\frac{135}{8}+\frac{18k}{k+1}=\frac{99}{8}+\frac{22}{k+1}$$

$$\frac{36}{8}=\frac{22-18k}{k+1}$$

$$44-36k=9k+9$$

$$45k=35$$

$$k=\frac{7}{9}$$

Hence the ratio of distribution of area between Wheat and Mustard for Rajesh is $$\dfrac{7}{9}$$

The question describes a figure of the following nature, 

We are told that the area of the triangle ABC is 1440. 
By proportionality theorem, the area of MPN should be one fourth of that. 
Since we know that, $$\frac{\left(Area\ \triangle_1\right)}{Area\ \triangle_2}=\frac{\left(side\ of\ \triangle_1\right)^2}{\left(side\ of\ \triangle_2\right)^2}$$

And since it is given that MPN are midpoints of the respective sides, area of triangle MPN is $$\triangle\ \frac{ABC}{4}=\frac{1440}{4}=360$$

Now, the medians from each side intersect, MP MN and NP are X Y and Z respectively, since these sides are proportional to the main triangle, the point of intersection of the medians with these sides will divide the side MP, MN and NP in half. 

Then, Area of Triangle XYZ will be one fourth the area MPN, 
$$\triangle\ \frac{MPN}{4}=\frac{360}{4}=90$$

The answer is 90. 

We need to consider single-digit, two-digit and three-digit numbers

Single digit: There are only nine numbers for a single-digit number (not including zero since we are looking for positive integers only)

Double digits: The ten's digit can be chosen in 9 ways (not including 0), and the unit digit can be chosen in 9 ways (including 0 but excluding the digit used in the ten's place). Hence, a total of 81 numbers. 

Three-digit: The hundred's digit can be 1, 2, 3 or 4; hence, there are four options for the hundred's digit; for the ten's digit, there will be 9 options (numbers from 0 to 9 except the one chosen in the hundred's place); for unit's digit there would be 8 options. Hence,e a total of 288 numbers

Giving the total numbers as 288+81+9 = 378

Therefore, 378 is the correct answer. 

We are told that there was two successive increments in the salary, with the second increment percentage twice the first one. Total Increment was 187.5%. 

Drawing up the equation
$$\left(1+z\right)\left(1+2z\right)=1.875$$
$$1+3z+2z^2=1.875$$
$$2z^2+3z-0.875$$
$$z=\frac{\left(-3\pm\sqrt{9+8\left(0.875\right)}\right)}{4}$$
$$z=\frac{\left(-3\pm\ 4\right)}{4}$$
$$z=0.25$$

Answer is 25%.