It is given that $ a^m \cdot b^n = 144^{145} $, where $ a &gt; 1 $ and $ b &gt; 1 $.144 can be written as $ 144 = 2^4 \times 3^2 $Hence, $ a^m \cdot b^n = 144^{145} $ can be written as $ a^m \cdot b^n = \left(2^4 \times 3^2\right)^{145} = 2^{580} \times 3^{290} $We know that $ 3^{290} $ is a natural number, which implies it can be written as $ a^1 $, where $ a &gt; 1 $Hence, the least possible value of m is 1. Similarly, the largest value of n is 580.Hence, the largest value of (n-m) is (580-1) = 579The correct option is&nbsp;C

It is given that $ 2^{4x^2} - 2^{2x^2+x+16} + 2^{2x+30} = 0 $, which can be written as:=&gt; $ \left(2^{2x^2}\right)^2 - 2^{2x^2} \cdot 2^{x+15} \cdot 2^1 + \left(2^{x+15}\right)^2 = 0 $=&gt; $ \left(2^{2x^2} - 2^{x+15}\right)^2 = 0 $=&gt; $ 2^{2x^2} - 2^{x+15} = 0 $ (Since $ (a-b)^2 = 0 \Rightarrow a-b = 0 $)=&gt; $ 2x^2 = x + 15 $=&gt; $ 2x^2 - x - 15 = 0 $=&gt; $ 2x^2 - 6x + 5x - 15 = 0 $=&gt; $ 2x(x-3) + 5(x-3) = 0 $=&gt; $ (2x+5)(x-3) = 0 $Hence, the possible values of x are $ -\frac{5}{2} $, and 3, respectively.Therefore, the sum of the possible values is $ \left(3 - \frac{5}{2}\right) = \frac{1}{2} $The correct option is C

It is given that k divides m+2n and 3m+4n.Since k divides (m+2n), it implies k will also divide 3(m+2n). Therefore, k divides 3m+6n.Similarly, we know that k divides 3m+4n.We know that if two numbers a, and b both are divisible by c, then their difference (a-b) is also divisible by c.By the same logic, we can say that {(3m+6n)-(3m+4n)} is divisible by k. Hence, 2n is also divisible by k.Now, (m+2n) is divisible by k, it implies 2(m+2n) =2m+4n&nbsp;is also divisible by k.Hence, {(3m+4n)-(2m+4n)} = m is also divisible by k.Therefore, m, and 2n are also divisible by k.The correct option is B

It is given that $ \frac{x}{y} $ &lt; $ \frac{x+3}{y-3} $, which can be written as $ \frac{x}{y} - \frac{x+3}{y-3} $ &lt; $ 0 $=&gt; $ \frac{x(y-3) - y(x+3)}{y(y-3)} $ &lt; $ 0 $=&gt; $ \frac{xy - 3x - xy - 3y}{y(y-3)} $ &lt; $ 0 $=&gt; $ \frac{-3(x+y)}{y(y-3)} $ &lt; $ 0 $=&gt; $ \frac{3(x+y)}{y(y-3)} $ &gt; $ 0 $From this inequality, we can say that, when $ y $ &lt; $ 0 \Rightarrow y(y-3) $ &gt; $ 0 $. Now to satisfy the given equation $ \frac{3(x+y)}{y(y-3)} $ &gt; $ 0 $,$ (x+y) $ must be greater than zero Hence, $ x $ &gt; $ 0 $ and $ |x| $ &gt; $ |y| $Therefore, the magnitude of $ x $ is greater than the magnitude of $ y $.Hence, $ x $ &gt; $ y $, and $ |x| $ &gt; $ |y| \Rightarrow -x $ &lt; $ y $ (Since the magnitude of $ x $ is greater than the magnitude of $ y $.)The correct option is C.

It is given that $ \log_{\sqrt{3}}(x) + \frac{\log_3(25)}{\log_3(0.008)} = \frac{16}{3} $, which can be written as:=&gt; $ 2 \log_3 x + \log_{0.008} 25 = \frac{16}{3} $=&gt; $ 2 \log_3 x + \log_{\frac{8}{1000}} 25 = \frac{16}{3} $=&gt; $ 2 \log_3 x + \log_{\frac{1}{125}} 25 = \frac{16}{3} $=&gt; $ 2 \log_3 x + \log_{5^{-3}} (5)^2 = \frac{16}{3} $=&gt; $ 2 \log_3 x - \frac{2}{3} = \frac{16}{3} $=&gt; $ 2 \log_3 x = \frac{16}{3} + \frac{2}{3} $=&gt; $ 2 \log_3 x = 6 $=&gt; $ \log_3 x^2 = 6 \Rightarrow x^2 = 3^6 $Hence, $ \log_3 \left(3 \cdot x^2\right) = \log_3 \left(3 \cdot 3^6\right) = \log_3 3^7 = 7 $

Since there are two distinct factors other than 1, and itself, which implies the total number of factors of N is 4.It can be done in two ways.&nbsp;First case:&nbsp;N = $p^3$&nbsp;(where p is a prime number)Second case: N&nbsp;=&nbsp;$p_1$ *&nbsp;$p_2 $&nbsp;(Where&nbsp;&nbsp;$p_1$,&nbsp;$p_2 $&nbsp;are the prime numbers)From case 1, we can see that the numbers which is a cube of prime and less than 50 are 8, and 27 (2 numbers).From case 2, we will get the numbers in the form (2*3), (2*5), (2*7), (2*11), (2*13), (2*17), (2*19), (2*23), (3*5), (3*7), (3*11), (3*13), (5*7) {(13 numbers)}Hence, the total number of numbers having two distinct factors is (13+2) = 15.

It is given that $ (x-1)^2 + 2kx + 11 = 0 $ has no real roots. (Where k is the largest integer)$ (x-1)^2 + 2kx + 11 = 0 $, which can be written as:=&gt; $ x^2 - 2x + 1 + 2kx + 11 = 0 $=&gt; $ x^2 - 2(k-1)x + 12 = 0 $We know that for no real roots, $ D $ &lt; $ 0 $ =&gt; $ b^2 - 4ac $ &lt; $ 0 $Hence, $ {(2(k-1))}^2 - 4 \cdot 1 \cdot 12$ &lt; $ 0 $=&gt; $ 4(k-1)^2 $ &lt; $ 48 $=&gt; $ (k-1)^2 $ &lt; $ 12 $Since k is an integer, it implies (k-1) is also an integer.Therefore, from the above inequality, we can say that the largest possible value of (k-1) = 3=&gt; The largest possible value of k is 4.Now we need to calculate the least possible value of $ \frac{k}{4y} + 9y $.$ \frac{k}{4y} + 9y $ can be written as $ \frac{4}{4y} + 9y = \frac{1}{y} + 9y $The least possible value of $ 9y + \frac{1}{y} $ can be calculated using A.M-G.M inequality.Using A.M-G.M inequality, we get:$ \frac{9y + \frac{1}{y}}{2} \geq \sqrt{9y \times \frac{1}{y}} $=&gt; $ \frac{9y + \frac{1}{y}}{2} \geq \sqrt{9} $=&gt; $ \frac{9y + \frac{1}{y}}{2} \geq 3 $=&gt; $ 9y + \frac{1}{y} \geq 6 $Hence, the least possible value is 6

Let the number of total employees in the company be 100x, and the total salary of all the employees be 100y.It is given that&nbsp;20% of the employees work in the manufacturing department, and&nbsp;the total salary obtained by all the manufacturing employees is one-sixth of the total salary obtained by all the employees in the company.Hence, the total number of employees in the manufacturing department is 20x, and the total salary received by them is (100y/6)Average salary in the manufacturing department = (100y/6*20x) = 5y/6xSimilarly, the total number of employees in the nonmanufacturing department is 80x, and the total salary received by them is (500y/6)Hence, the average salary in the nonmanufacturing department = (500y/6*80x) = 25y/24xHence, the ratio is:- (5y/6x): (25y/24x)&nbsp;=&gt; 120: 150 = 4:5The correct option is A

The cost price of the sunglass for Meenu when he purchased it for the first time was 1000 rupees, and he sold it to Kanu at 20% profit. Hence, the selling price of the sunglass is 1200 rupees, which Kanu purchased. Hence, the profit made by Meenu is (1200-1000) = 200 rupees.Hence, the cost price of the same sunglass for Kanu is 1200 rupees, and now he sold it to Meenu at a 20% loss. Hence, the selling price of the sunglass now is (1200*0.8) = 960 rupees.The cost price of the same sunglass for Meenu when he purchased it for the second time was 960 rupees. Now Meenu sold it Tanu, at a certain price such that the total profit of Meenu becomes 500 rupees.Hence, on the second transaction (selling it to Tanu), Meenu made a profit of (500-200) = 300 rupees.Hence, the profit made by Minu in the second transaction is (300/960)*100% = 31.25%The correct option is B

Let the time taken by A to fill the tank alone be x hours, which implies the time taken by B to empty the tank alone is (x-1) hours (B is the drainage pipe), and the time taken by C to fill the tank is y hours.It is given that when pipes A, B, and C are turned on together, the empty tank is filled in two hours.Hence, $ \frac{1}{x} - \frac{1}{x-1} + \frac{1}{y} = \frac{1}{2} $ .... Eq(1)It is given that if pipes B and C are turned on together when the tank is empty and Pipe B is turned off after one hour, then Pipe C takes another one hour and 15 minutes to fill the remaining tank.Hence, B worked for 1 hour, and C worked for 2 hours 15 minutes, which is equal to $ \frac{9}{4} $ hours.In 1 hour, B worked $ -\frac{1}{x-1} $ units, and in $ \frac{9}{4} $ hours, C worked $ \frac{9}{4y} $ units.Hence, $ \frac{9}{4y} - \frac{1}{x-1} = 1 $ .... Eq(2)Solving both equations, we get $ y = \frac{3}{2} $, and $ x = 3 $Hence, the time taken by C is $ \frac{3}{2} $ hours, which is equal to 90 minutes.The correct option is A

cat 2023 Complete Paper Solution | Slot 2

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cat 2023 Complete Paper Solution | Slot 2

Question 1.

<div><p>Let a,b,m and n be natural numbers such that a&gt;1 and b&gt;1 . If $a^m b^n=144^{145}$, then the largest possible value of n&minus;m is</p></div>

Question 2.

<div><p>The sum of all possible values of x satisfying the equation $2^{4 x^2}-2^{2 x^2+x+16}+2^{2 x+30}=0$, is</p></div>

Question 3.

<div><p>For any natural numbers m,n, and k, such that k divides both m+2n and 3m+4n,k must be a common divisor of</p></div>

Question 4.

<div><p>Any non-zero real numbers x,y such that $y &ne; 3$ and $\frac{x}{y} \lt \frac{x+3}{y-3}$, will satisfy the condition</p></div>

Question 5.

<div><p>For some positive real number x , if $\log _{\sqrt{3}}(x)+\frac{\log _x(25)}{\log _x(0.008)}=\frac{16}{3}$, then the value of $\log _3\left(3 x^2\right)$ is</p></div>

Question 6.

<div><p>The number of positive integers less than 50, having exactly two distinct factors other than 1 and itself, is</p></div>

Question 7.

<div><p>Let k be the largest integer such that the equation $(x-1)^2+2 k x+11=0$ has no real roots. If y is a positive real number, then the least possible value of $\frac{k}{4 y}+9 y$ is</p></div>

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