CAT 2023 Slot 1 Question Paper

Given equation of circle = $$x^{2} + y^{2} + 4x - 6y - 3 = 0$$

Center of the circle is (-2,3) and radius of the circle = $$\sqrt{\ g^2+f^2-c}=\sqrt{\ 4+9+3}=4$$

Let us assume the point of the intersection of the tangents is ( h,k)

The angle made by the line joining (h,k) to the centre makes an angle of 30 degrees with the tangent, and sin(30) will be the ratio of the radius and the distance between the center and (h,k)

=> $$\sin\left(30\right)=\dfrac{4}{\sqrt{\left(\ h+2\right)^2+\left(k-3\right)^2}}$$

Squaring on both sides:

$$\dfrac{1}{4}=\dfrac{16}{\left(h+2\right)^2+\left(k-3\right)^2}$$

=> $$\left(h+2\right)^2+\left(k-3\right)^2=64$$

When x = 6 => h = 6 => $$64+\left(k-3\right)^2=64$$ => k = 3.

=> required point is (6,3)

Given ABCD is a cyclic quadrilateral.

Angle ADB = Angle ACB (Angle subtended by chord on the same side of arc)

Angle DAC = Angle DBC (Angle subtended by chord on the same side of arc)

=> Triangles AED and BEC are similar triangles

Similarly triangles AEB and DEC are also similar using AA similarity property.

Now, given that AB : CD = 2 : 1 and BC : AD = 5 : 4

AE/BE = AD/BC = 4/5 (Similar Triangles AED and BEC)

BE/CE = AB/CD = 2/1 (Similar Triangles AEB and DEC)

Multiplying both, we get AE/CE = 8/5.

Given that ABC is a right-angled triangle with AB = 5 and BC = 12 => Area of the triangle = 0.5 * 5 * 12 = 30.

Let us assume BP = p, BQ = q

=> Area of ABP = 0.5 * 5 * p = 2.5p

=> Area of ABQ = 0.5 * 5 * q = 2.5q

Given the area of ABC is 1.5 times that of ABP

=> 30 = 1.5 * 2.5p => 20 = 2.5p => p = 8.

Given Areas of ABP, ABQ and ABC are in A.P. => 2 * 2.5q = 2.5 * 8 + 30 => 5q = 50 => q = 10.

PQ = BQ - BP = q - p = 10 - 8 = 2.

Given that $$\dfrac{1}{\sqrt{y}+\sqrt{z}}$$ is the arithmetic mean of $$\dfrac{1}{\sqrt{x}+\sqrt{z}}$$ and $$\dfrac{1}{\sqrt{x}+\sqrt{y}}$$

=> $$\dfrac{2}{\sqrt{\ y}+\sqrt{\ z}}=\dfrac{1}{\sqrt{\ x}+\sqrt{\ z}}+\dfrac{1}{\sqrt{\ x}+\sqrt{\ y}}$$

=> $$2\left(\sqrt{\ x}+\sqrt{\ z}\right)\left(\sqrt{\ x}+\sqrt{\ y}\right)=\left(\sqrt{\ y}+\sqrt{\ z}\right)\left(\sqrt{\ x}+\sqrt{\ y}+\sqrt{\ x}+\sqrt{\ z}\right)$$

=> $$2\left(x+\sqrt{\ xy}+\sqrt{\ xz}+\sqrt{\ yz}\right)=2\sqrt{xy}+y+\sqrt{\ yz}+2\sqrt{\ xz}+\sqrt{\ yz}+z$$

=> $$2x=y+z$$

=> y, x, z are in A.P. as x is the arithmetic mean of y and z.

1-digit numbers => We have 1 to 9 => 9

2-digit numbers => x y, we have 9 ways to choose x from 1 to 9 => 9 ways and 9 ways to choose y (0 to 9 except x) => 9*9 = 81

3-digit numbers => x y z, we have 9 ways to choose x, 9 ways to choose y and 8 ways to choose z => 9*9*8 = 648.

Total numbers till 1000 without digits repeated in them is 9 + 81 + 648 = 738.

Given on day-1, there are 2 organisms.

On day-2, there are 2*2 + 3 = 7 and on day-3, there are 2*7 + 3 = 17..

Let us try to form a pattern:

2 = 2 + 0 (n = 1)

7 = 4 + 3 (n = 2)

17 = 8 + 9 [8 + 3*3] (n = 3)

37 = 16 + 21[16 + 3*7] n = 4

T(n) = $$2^n+3\left(2^{n-1}-1\right)$$

We know that $$2^{20}=2^{10}\times\ 2^{10}=1024\times\ 1024$$, which is more than 1 million.

Let us check for n = 19

$$2^{19}+3\left(2^{18}-1\right)=2^{19}\ +\ 3\cdot2^{18}-3=2\cdot2^{19}+2^{18}-3=2^{20}+2^{18}-3$$, which is more than 1 million.

Let us check for n = 18

=> $$2^{18}+3\left(2^{17}-1\right)=2^{18}\ +\ 3\cdot2^{17}-3=2\cdot2^{18}+2^{17}-3=2^{19}+2^{17}-3$$ which is not more than a million.

=> n = 19.