CAT 2023 Slot 1 Question Paper - QA

Prime Factorising 1134, we get 1134 = $$2\times\ 3^4\times\ 7$$ and 168 = $$2^3\times\ 3\times\ 7$$

$$1134^n$$ is a factor of 168 => the factor of 2 should be atleast 3, for 168 to be a factor => n = 3.

Now, $$1134^n$$ = $$1134^3=2^3\times\ 3^{12}\times\ 7^3$$ is a factor of $$168^m=\left(2^3\times\ 3\times\ 7\right)^m$$ => m = 12 as power of 3 should be atleast 12.

=> So, m + n = 15.

Given, $$\log_{x}(x^2 + 12) = 4$$

=> $$x^2+12=x^4$$

=> $$x^4-x^2-12=0$$

=> $$x^4-4x^2+3x^2-12=0$$

=> $$x^2\left(x^2-4\right)+3\left(x^2-4\right)=0$$

=> $$\left(x^2-4\right)\left(x^2+3\right)=0$$ => since, x is a positive real number (given) => x = 2.

Now, Given $$3 \log_{y} x = 1$$

=> $$\log_yx=\frac{1}{3}$$

=> $$x=y^{\frac{1}{3}}$$

=> $$y=x^3$$ => y = 8.

=> x + y = 2 + 8 = 10.

Given, $$\sqrt{5x+9} + \sqrt{5x - 9} = 3(2 + \sqrt{2})$$

=> $$\sqrt{\ 5x+9}+\sqrt{\ 5x-9}=6+3\sqrt{\ 2}$$

=> $$\sqrt{\ 5x+9}+\sqrt{\ 5x-9}=\sqrt{\ 36}+\sqrt{\ 18}$$

Comparing the L.H.S. and R.H.S.

=> $$5x+9=36\ $$ => $$5x=27$$ => $$x=\dfrac{27}{5}$$ (can be verified using the second term as well).

=> $$\sqrt{10x+9}$$ = $$\sqrt{\left(10\times\dfrac{27}{5}\right)+9}$$ = $$\sqrt{\ 63}=3\sqrt{\ 7}$$

Given, $$x^{2} + (x - 2y - 1)^{2} = -4y(x + y)$$

=> $$x^2+4xy+4y^2+\left(x-2y-1\right)^2=0$$

=> $$\left(x+2y\right)^2+\left(x-2y-1\right)^2=0$$

For the L.H.S. of the equation to be 0, each of the square terms should be 0 (as squares cannot be negative)

=> x - 2y - 1 = 0 => x - 2y = 1

Let us consider 3 cases:

1) x = 0, This is a solution, as both L.H.S and R.H.S will be equal (0) when x = 0. (1 solution)

2) x > 0

=> $$2x\left(x^2+1\right)=5x^2$$

=> $$2\left(x^2+1\right)=5x$$

=> $$2x^2-5x+2=0$$ => $$2x^2-4x-x-2=0$$

=> $$2x\left(x-2\right)-1\left(x-2\right)=0$$

=> $$\left(x-2\right)\left(2x-1\right)=0$$ => x = 2 or 1/2 => (1 integer solution)

3) x < 0

=> $$-2x\left(x^2+1\right)=5x^2$$

=> $$2x^2+5x+2=0$$

=> $$2x^2+4x+x+2=0$$

=> $$2x\left(x+2\right)+1\left(x+2\right)=0$$

=> $$\left(x+2\right)\left(2x+1\right)=0$$ => x = -2 or -1/2 => (1 integer solution)

So, the total number of integer solutions are 0, 2, -2 => 3.

Given that -2 is a root of the given cubic equation.

=> Dividing the given equation by (x + 2), Using the Horners method of synthetic division:

coefficient of $$x^2$$ is 1, and coefficient of x is (2r+1)-2 = 2r-1 and the constant term = (4r-1)-2(2r-1) = 1.

=> The quadratic obtained by dividing the cubic = $$x^2+\left(2r-1\right)x+1=0$$, Since, this equation has 2 real roots => Discriminant should be greater than 0

=> $$\left(2r-1\right)^2>4$$ => 2r-1 > 2 or 2r-1 < -2 => r > 3/2 or r < -1/2.

=> Minimum possible non-negative integer value of r is 2.

Given a and b are the distinct roots of the equation $$2x^{2} - 6x + k = 0$$

=> a + b = -(-6/2) = 3 (Sum of the roots)

=> ab = k/2 (Product of the roots)

Now, (a+b) and ab are the roots of the quadratic equation $$x^{2} + px + p = 0$$

=> a + b + ab = -p => 3 + k/2 = -p ---(1)

=> (a + b)(ab) = p => 3(k/2) = p ---(2)

$$3+\dfrac{k}{2}=-\dfrac{3k}{2}$$ => 2k = -3 => k = $$-\dfrac{3}{2}$$

p = $$\dfrac{3k}{2}=\dfrac{3}{2}\left(-\dfrac{3}{2}\right)=-\dfrac{9}{4}$$

=> 8(k-p) = $$8\left(-\frac{3}{2}+\frac{9}{4}\right)=-12+18=6$$

Given that in the final mixture, the ratio of coffee and cocoa is 16:9

Let us assume coffee is 16 units and cocoa is 9 units.

=> Initially, there are 25 units of coffee and 0 units of cocoa

Let's say x units of the mixture is removed and replaced with cocoa

=> Now, we have (25-x) coffee and x units of cocoa. => Mixture P

Now, if x units of the mixture is removed:

Amount of coffee present = (25-x) - $$\dfrac{\left(25-x\right)}{25}\times\ x$$

=> $$\left(25-x\right)\left(1-\dfrac{x}{25}\right)=16$$

=> $$\left(25-x\right)^2=16\times\ 25$$

=> 25 - x = 20 => x = 5.

In mixture P, cocoa = x = 5

In mixture Q, cocoa = 9 units.

=> Required ratio = 5:9

The given time is 8:48 AM.

Angle made by hours hand w.r.t 12 is 8 * 30 (30 degrees in 1 hour) + 0.5 * 48 (0.5 degree in 1 minute) = 240 + 24 = 264 degrees.

Angle made by minutes hands w.r.t 12 is 48 * 6 = 288 degrees.

=> The angle between them is 288 - 264 = 24 degrees.

This should further increase by 12 degrees (50% of 24)

After m minutes, the further increase in angle = (6 - 0.5)*m = $$\dfrac{11}{2}m=12$$ => m = $$\dfrac{24}{11}$$

Let us assume the initial selling prices of A and B is p.

Given, she made profit of 20% on A => 1.2 * c = p => c = 5p/6 => cost of A is $$\dfrac{5}{6}p$$

Given, she made a loss of 10% on B => 0.9 * c = p => c = 10p/9 => cost of B is $$\dfrac{10}{9}p$$

Now, she sold them at a price such that a 10% profit is made on B

=> Selling price = s = 11/10 * 10/9 p => $$\dfrac{11}{9}p$$

=> Profit % on A = $$\dfrac{\left(\dfrac{11}{9}-\dfrac{5}{6}\right)}{\left(\dfrac{5}{6}\right)}\times\ 100$$ = 46.66% = nearly 47%

Given the total number of kilometres travelled, including the trip = is 26862 Km, and the duration of the trip is 8 hrs.

If avg. speed of the car during the trip is 's' => the km travelled till just before the trip is 26862 - 8s, which should also be a palindrome.

=> From the options if s = 110 => The reading will be 26862 - 110*8 = 25982 (Not a palindrome)

=> If s = 100 => The reading will be 26862 - 100*8 = 26062 => It is a palindrome.

=> s = 100 is the correct option.

Given that the average marks of 4 girls and 6 boys is 24.

Let us assume 'b' is the marks scored by a boy and 'g' is the marks scored by a girl.

=> 4g + 6b = 10*24 = 240 ---(1)

Given that, $$b\le g\le2b$$

We need to find the distinct possible values of 2g + 6b = 2g + 240 - 4g = 240 - 2g.

From (1)

when b = g => 10g = 240 => g = 24

when b = g/2 => 7g = 240 => g = 240/7

=> 240 - 2g varies from 240 - 2*24 to 240 - 2*240/7

=> 171.42 to 192 => Integer values of 172 to 192 => 21 values.

Given, the salaries of Sita, Gita and Mita are initially in the ratio 5 : 6 : 7, respectively, Let us assume their salaries are 5p, 6p and 7p.

They get salary hikes of 20%, 25% and 20%, respectively.

=> Their salaries are 6/5 * 5p, 5/4 * 6p and 6/5 * 7p => 6p, 7.5p, 8.4p

Now, Sita and Mita get salary hikes of 40% and 25%, respectively

=> Sita's salary = 1.4 * 6p = 8.4p and Mita's salary = 1.25 * 8.4p = 10.5p

Let Gita's salary be 'g' after hike

=> 3g = 8.4p + g + 10.5p => 2g = 18.9p => g = 9.45p

=> Hike % = $$\dfrac{\left(9.45-7.5\right)}{7.5}\times\ 100$$ = 26%

Let us assume the speeds of Arvind and Surbhi are 'a' and 's', respectively.

Let us say they meet after 't' hours

=> Arvind travelled s*t distance in 6 hrs and Surbhi travelled a*t in 24 hrs

=> s*t = a*6 and a*t = s*24 => $$t^2=6\times\ 24$$ => t = 12

Given a = 54 => s*12 = 54*6 => s = 27.

=> Total distance between A and B is (s+a)*t = (54+27)*12 = 81*12 = 972 Kms.

Anil invested 22000 for 6 years at 4% interest compounded half-yearly

=> Amount = $$22000\left(1.02\right)^6$$

Let Sunil invest 'S' rupees for 5 years at 4% C.I. half-yearly and 10% S.I. for 1 additional year

=> Amount = $$S\left(1.02\right)^{10}\left(1.1\right)$$

Given that the both amounts are equal

=> $$22000\left(1.02\right)^{12}=S\left(1.02\right)^{10}\left(1.1\right)$$

=> $$S=\dfrac{22000\left(1.02\right)^2}{1.1}=20808$$

Let us assume the efficiencies of Amal, Sunil, and Kamal are a, s, and k, respectively.

Given that they are in H.P.

=> $$\dfrac{2}{s}=\dfrac{1}{a}+\dfrac{1}{k}$$ ---(1)

Also, given that Kamal takes twice as much time as Amal to do the same amount of job

=> a = 2k

Given that when Amal and Sunil work for 4 days and 9 days, respectively, Kamal needs to work for 16 days to finish the remaining job.

=> If W is the total work => 4a + 9s + 16k = W.

from (1)$$\dfrac{2}{s}=\dfrac{1}{a}+\dfrac{2}{a}$$ => $$a=\dfrac{3}{2}s$$ and $$k=\dfrac{3}{4}s$$

=> $$4\left(\dfrac{3s}{2}\right)+9s+16\left(\dfrac{3s}{4}\right)=W$$

=> $$6s+9s+12s=W$$

=> $$27s=W\ =>\ s\ =\ \dfrac{W}{27}$$

=> Sunil will take 27 days to finish the work when working alone.

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.