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(x^2 - 4) + 3(x^2 - 4) = 0 $$ (x^2 - 4)(x^2 + 3) = 0 \implies$ since $x$ is a positive real number, $x = 2$Now, given $ 3 \log_{y} x = 1 $$ \log_{y} x = \frac{1}{3} $$ x = y^{\frac{1}{3}} $$ y = x^3 $$ y = 8 $$ x + y = 2 + 8 = 10 $

Given,$ x^{2} + (x - 2y - 1)^{2} = -4y(x + y) $$ x^2 + 4xy + 4y^2 + (x-2y-1)^2 = 0 $$ (x+2y)^2 + (x-2y-1)^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 $ =&gt; $ x - 2y = 1 $

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 $ =&gt; $ 5x=27 $ =&gt; $ 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} $

Prime factorising 1134, we get&nbsp;&nbsp;$1134 = 2 \times 3^4 \times 7$&nbsp;&nbsp;and&nbsp;&nbsp;$168 = 2^3 \times 3 \times 7$$1134^n$ is a factor of 168&nbsp;&nbsp;$\Rightarrow$ the factor of 2 should be at least 3, for 168 to be a factor&nbsp;&nbsp;$\Rightarrow n = 3$Now,&nbsp;&nbsp;$1134^n = 1134^3 = 2^3 \times 3^{12} \times 7^3$&nbsp;&nbsp;is a factor of&nbsp;&nbsp;$168^m = \left(2^3 \times 3 \times 7\right)^m$&nbsp;&nbsp;$\Rightarrow m = 12$ as power of 3 should be at least 12.$\Rightarrow$ So, $m + n = 15$

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) $x &gt; 0$&nbsp;&nbsp;$\Rightarrow\ 2x\left(x^2+1\right)=5x^2$&nbsp;&nbsp;$\Rightarrow\ 2\left(x^2+1\right)=5x$&nbsp;&nbsp;$\Rightarrow\ 2x^2-5x+2=0\ \Rightarrow\ 2x^2-4x-x-2=0$&nbsp;&nbsp;$\Rightarrow\ 2x(x-2)-1(x-2)=0$&nbsp;&nbsp;$\Rightarrow\ (x-2)(2x-1)=0\ \Rightarrow\ x=2\ \text{or}\ \frac{1}{2}\ \Rightarrow$ (1 integer solution) 3) $x &lt; 0$&nbsp;&nbsp;$\Rightarrow\ -2x\left(x^2+1\right)=5x^2$&nbsp;&nbsp;$\Rightarrow\ 2x^2+5x+2=0$&nbsp;&nbsp;$\Rightarrow\ 2x^2+4x+x+2=0$&nbsp;&nbsp;$\Rightarrow\ 2x(x+2)+1(x+2)=0$&nbsp;&nbsp;$\Rightarrow\ (x+2)(2x+1)=0\ \Rightarrow\ x=-2\ \text{or}\ -\frac{1}{2}\ \Rightarrow$ (1 integer solution) So, the total number of integer solutions are $0,\,2,\,-2\ \Rightarrow\ 3$.

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 -2 is a root of the given cubic equation.=&gt; Dividing the given equation by (x + 2), using the Horner's 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.=&gt; The quadratic obtained by dividing the cubic = $x^2 + (2r - 1)x + 1 = 0$, Since, this equation has 2 real roots =&gt; Discriminant should be greater than 0=&gt; $(2r - 1)^2 \gt 4 $ =&gt; 2r-1 &gt; 2 or 2r-1 &lt; -2 =&gt; r &gt; 3/2 or r &lt; -1/2.=&gt; Minimum possible non-negative integer value of r is 2.

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' =&gt; the km travelled till just before the trip is 26862 - 8s, which should also be a palindrome.=&gt; From the options if s = 110 =&gt; The reading will be 26862 - 110*8 = 25982 (Not a palindrome)=&gt; If s = 100 =&gt; The reading will be 26862 - 100*8 = 26062 =&gt; It is a palindrome.=&gt; s = 100 is the correct option.

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.=&gt; 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 $ =&gt; m = $ \dfrac{24}{11} $

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.=&gt; 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 =&gt; 10g = 240 =&gt; g = 24when b = g/2 =&gt; 7g = 240 =&gt; g = 240/7=&gt; 240 - 2g varies from 240 - 2*24 to 240 - 2*240/7=&gt; 171.42 to 192 =&gt; Integer values of 172 to 192 =&gt; 21 values.

Given that in the final mixture, the ratio of coffee and cocoa is 16:9Let us assume coffee is 16 units and cocoa is 9 units.=&gt; Initially, there are 25 units of coffee and 0 units of cocoaLet's say x units of the mixture is removed and replaced with cocoa=&gt; Now, we have (25-x) coffee and x units of cocoa. =&gt; Mixture PNow, if x units of the mixture is removed:Amount of coffee present = (25-x) - $ \dfrac{\left(25-x\right)}{25}\times\ x $=&gt; $ \left(25-x\right)\left(1-\dfrac{x}{25}\right)=16 $=&gt; $ \left(25-x\right)^2=16\times\ 25 $=&gt; 25 - x = 20 =&gt; x = 5.In mixture P, cocoa = x = 5In mixture Q, cocoa = 9 units.=&gt; Required ratio = 5:9

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.=&gt; Their salaries are 6/5 * 5p, 5/4 * 6p and 6/5 * 7p =&gt; 6p, 7.5p, 8.4pNow, Sita and Mita get salary hikes of 40% and 25%, respectively=&gt; Sita's salary = 1.4 * 6p = 8.4p and Mita's salary = 1.25 * 8.4p = 10.5pLet Gita's salary be 'g' after hike=&gt; 3g = 8.4p + g + 10.5p =&gt; 2g = 18.9p =&gt; g = 9.45p=&gt; Hike % = $\dfrac{(9.45-7.5)}{7.5}\times 100$ = 26%

Let us assume the initial selling prices of A and B is p.Given, she made profit of 20% on A =&gt; 1.2 * c = p =&gt; c = 5p/6 =&gt; cost of A is $\dfrac{5}{6}p$Given, she made a loss of 10% on B =&gt; 0.9 * c = p =&gt; c = 10p/9 =&gt; cost of B is $\dfrac{10}{9}p$Now, she sold them at a price such that a 10% profit is made on B=&gt; Selling price = s = 11/10 * 10/9 p =&gt; $\dfrac{11}{9}p$=&gt; Profit % on A = $\dfrac{\left(\dfrac{11}{9}-\dfrac{5}{6}\right)}{\left(\dfrac{5}{6}\right)}\times 100$ = 46.66% = nearly 47%

Let us assume the speeds of Arvind and Surbhi are 'a' and 's', respectively.Let us say they meet after 't' hours=&gt; Arvind travelled s*t distance in 6 hrs and Surbhi travelled a*t in 24 hrs=&gt; s*t = a*6 and a*t = s*24 =&gt; $t^2=6\times 24$ =&gt; t = 12Given a = 54 =&gt; s*12 = 54*6 =&gt; s = 27.=&gt; Total distance between A and B is (s+a)*t = (54+27)*12 = 81*12 = 972 Kms.

Let us assume the efficiencies of Amal, Sunil, and Kamal are a, s, and k, respectively.Given that they are in H.P.=&gt; $ \frac{2}{s} = \frac{1}{a} + \frac{1}{k} $ ---(1)Also, given that Kamal takes twice as much time as Amal to do the same amount of job=&gt; a = 2kGiven 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.=&gt; If W is the total work =&gt; 4a + 9s + 16k = W.from (1) $ \frac{2}{s} = \frac{1}{a} + \frac{2}{a} $ =&gt; $ a = \frac{3}{2}s $ and $ k = \frac{3}{4}s $=&gt; $ 4\left(\frac{3s}{2}\right) + 9s + 16\left(\frac{3s}{4}\right) = W $=&gt; 6s + 9s + 12s = W=&gt; 27s = W =&gt; $ s = \frac{W}{27} $=&gt; Sunil will take 27 days to finish the work when working alone.

Anil invested 22000 for 6 years at 4% interest compounded half-yearly=&gt; Amount = 22000 $ (1.02)^{12} $Let Sunil invest 'S' rupees for 5 years at 4% C.I. half-yearly and 10% S.I. for 1 additional year=&gt; Amount = $ S (1.02)^{10} (1.1) $Given that the both amounts are equal=&gt; $ 22000 (1.02)^{12} = S (1.02)^{10} (1.1) $=&gt; $ S = \frac{22000 (1.02)^2}{1.1} = 20808 $

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)=&gt; $ \sin(30) = \frac{4}{\sqrt{( h+2)^2+(k-3)^2}} $Squaring on both sides:$ \frac{1}{4} = \frac{16}{(h+2)^2+(k-3)^2} $=&gt; $ (h+2)^2+(k-3)^2 = 64 $When x = 6 =&gt; h = 6 =&gt; $ 64 + (k-3)^2 = 64 $ =&gt; k = 3.=&gt; required point is (6,3)

<img src="https://lightyellow-stork-291267.hostingersite.com/wp-content/uploads/2025/11/211.image113.png"> Given that ABC is a right-angled triangle with AB = 5 and BC = 12 =&gt; Area of the triangle = 0.5 * 5 * 12 = 30.Let us assume BP = p, BQ = q=&gt; Area of ABP = 0.5 * 5 * p = 2.5p=&gt; Area of ABQ = 0.5 * 5 * q = 2.5qGiven the area of ABC is 1.5 times that of ABP=&gt; 30 = 1.5 * 2.5p =&gt; 20 = 2.5p =&gt; p = 8.Given Areas of ABP, ABQ and ABC are in A.P. =&gt; 2 * 2.5q = 2.5 * 8 + 30 =&gt; 5q = 50 =&gt; q = 10.PQ = BQ - BP = q - p = 10 - 8 = 2.

Given that $ \frac{1}{\sqrt{y} + \sqrt{z}} $ is the arithmetic mean of $ \frac{1}{\sqrt{x} + \sqrt{z}} $ and $ \frac{1}{\sqrt{x} + \sqrt{y}} $=&gt; $ \frac{2}{\sqrt{y} + \sqrt{z}} = \frac{1}{\sqrt{x} + \sqrt{z}} + \frac{1}{\sqrt{x} + \sqrt{y}} $=&gt; $ 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) $=&gt; $ 2 \left(x + \sqrt{xy} + \sqrt{xz} + \sqrt{yz}\right) = 2\sqrt{xy} + y + \sqrt{yz} + 2\sqrt{xz} + \sqrt{yz} + z $=&gt; $ 2x = y + z $=&gt; y, x, z are in A.P. as x is the arithmetic mean of y and z.

1-digit numbers =&gt; We have 1 to 9 =&gt; 92-digit numbers =&gt; x y, we have 9 ways to choose x from 1 to 9 =&gt; 9 ways and 9 ways to choose y (0 to 9 except x) =&gt; 9*9 = 813-digit numbers =&gt; x y z, we have 9 ways to choose x, 9 ways to choose y and 8 ways to choose z =&gt; 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 = 4T(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 \cdot 2^{18} - 3 = 2 \cdot 2^{19} + 2^{18} - 3 = 2^{20} + 2^{18} - 3 $, which is more than 1 million.Let us check for n = 18=&gt; $ 2^{18} + 3 \left(2^{17} - 1\right) = 2^{18} + 3 \cdot 2^{17} - 3 = 2 \cdot 2^{18} + 2^{17} - 3 = 2^{19} + 2^{17} - 3 $ which is not more than a million.=&gt; n = 19.

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cat 2023 Complete Paper Solution | Quant Slot 1

Question 1.

<div><p>If x and y are positive real numbers such that $\log _x\left(x^2+12\right)=4$ and $3 \log _y x=1$ , then x+y equals</p></div>

Question 2.

<div><p>If x and y are real numbers such that $x^2+(x-2 y-1)^2=-4 y(x+y)$, then the value x&minus;2y is</p></div>

Question 3.

<div><p>If $\sqrt{5 x+9}+\sqrt{5 x-9}=3(2+\sqrt{2})$, then $\sqrt{10 x+9}$ is equal to</p></div>

Question 4.

<div><p>Let n be the least positive integer such that 168 is a factor of $1134^n$. If m is the least positive integer such that $1134^n$ is a factor of $168^m$, then m+n equals</p></div>

Question 5.

<div><p>The number of integer solutions of equation $2|x|\left(x^2+1\right)=5 x^2$ is</p></div>