It is given that $ \frac{1}{2} $, $ \frac{\log_3(2^x - 9)}{\log_3 4} $, and $ \frac{\log_5(2^x + \frac{17}{2})}{\log_5 4} $ are in an arithmetic progression.$ \frac{\log_3(2^x - 9)}{\log_3 4} $ can be written as $ \log_4(2^x - 9) $, and $ \frac{\log_5(2^x + \frac{17}{2})}{\log_5 4} $ can be written as $ \log_4(2^x + \frac{17}{2}) $Hence, $ 2 \log_4(2^x - 9) = \frac{1}{2} + \log_4(2^x + \frac{17}{2}) $$ \frac{1}{2} $ can be written as $ \log_4 2 $.Therefore,=&gt; $ 2 \log_4(2^x - 9) = \log_4 2 + \log_4(2^x + \frac{17}{2}) $=&gt; $ \log_4(2^x - 9)^2 = \log_4 2(2^x + \frac{17}{2}) $=&gt; $ (2^x - 9)^2 = 2(2^x + \frac{17}{2}) $=&gt; $ 2^{2x} - 18 \cdot 2^x + 81 = 2 \cdot 2^x + 17 $=&gt; $ 2^{2x} - 20 \cdot 2^x + 64 = 0 $=&gt; $ 2^{2x} - 16 \cdot 2^x - 4 \cdot 2^x + 64 = 0 $=&gt; $ 2^x(2^x - 16) - 4(2^x - 16) = 0 $=&gt; $ (2^x - 4)(2^x - 16) = 0 $The values of $ 2^x $ can't be 4 (log will be undefined), which implies The value of $ 2^x $ is 16.Therefore, the common difference is $ \log_4(2^x - 9) - \log_4 2 $=&gt; $ \log_4 7 - \log_4 2 = \log_4\left(\frac{7}{2}\right) $

It is given that there are exactly 41 numbers, which can be expressed as the power of two, and exist&nbsp;between&nbsp;$8^m$&nbsp;and&nbsp;$8^n$, (where m, and n are positive integers, and m &lt; n)Hence,&nbsp;$2^{3m}$&nbsp;&lt;&nbsp;41&nbsp;numbers&nbsp;&lt;&nbsp;$2^{3n}$Since, m is a positive integer, the least value of m is 1. Therefore,&nbsp;$2^{3m}$&nbsp;= $2^3$, hence, the 41 numbers between them are&nbsp;$2^4$,&nbsp;$2^5$,&nbsp;$2^6$, ..., $2^{44}$Then the lowest possible value of&nbsp;$8^n$&nbsp;is&nbsp;$2^{45}$. Hence, the smallest value of n is&nbsp;$2^{45}$&nbsp;=&nbsp;$8^n$&nbsp;=&gt;&nbsp;$2^{3n}$&nbsp;= $2^{45}$&nbsp;=&gt;&nbsp;n&nbsp;=&nbsp;15Hence, the smallest value of m+n is (15+1) = 16

It is given that&nbsp;for some real numbers a and b, the system of equations&nbsp;x+y=4 and&nbsp;(a + 5)x + ($b^2$ − 15)y= 8b has infinitely many solutions for x and y.Hence, we can say that&nbsp;=&gt;&nbsp;&nbsp;$\frac{a + 5}{1}$ = $\frac{b^2 - 15}{1}$ = $\frac{8b}{4}$​This equation can be used to find the value of a, and b.Firstly,&nbsp;we will determine the value of b.&nbsp;=&gt;&nbsp;&nbsp;&nbsp;$\frac{b^2 - 15}{1}$ = $\frac{8b}{4}$​ =&gt; $b^2 - 2b - 15$ = 0=&gt; (b - 5)(b + 3) = 0Hence, the values of b are 5, and -3, respectively.&nbsp;The value of a can be expressed in terms of b, which is&nbsp;a + 5 = $b^2 − 15$&nbsp;=&gt;&nbsp;a&nbsp;= $b^2 − 20$When&nbsp;b=5, a = $5^2$ − 20 = 5When&nbsp;b=−3, a = $3^2$ − 20 = −11The maximum value of&nbsp;ab = (−3)⋅(−11) = 33

It is given that $ x^8 + \left(\frac{1}{x}\right)^8 = 47 $, which can be written as:=&gt; $ \left(x^4\right)^2 + \left(\frac{1}{x^4}\right)^2 = 47 $=&gt; $ \left(x^4 + \frac{1}{x^4}\right)^2 - 2 \cdot x^4 \cdot \frac{1}{x^4} = 47 $=&gt; $ \left(x^4 + \frac{1}{x^4}\right)^2 = 49 $=&gt; $ x^4 + \frac{1}{x^4} = 7 $Similarly, $ x^4 + \frac{1}{x^4} = 7 $ can be expressed as:=&gt; $ \left(x^2\right)^2 + \left(\frac{1}{x^2}\right)^2 = 7 $=&gt; $ \left(x^2 + \frac{1}{x^2}\right)^2 - 2 \cdot x^2 \cdot \frac{1}{x^2} = 7 $=&gt; $ \left(x^2 + \frac{1}{x^2}\right)^2 = 9 $=&gt; $ x^2 + \frac{1}{x^2} = 3 $By the same logic, we get $ x + \frac{1}{x} = \sqrt{5} $Now, $ x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x}\right)^3 - 3 \cdot x \cdot \frac{1}{x}\left(x + \frac{1}{x}\right) $=&gt; $ x^3 + \frac{1}{x^3} = \left(\sqrt{5}\right)^3 - 3\sqrt{5} = 2\sqrt{5} $By the same logic, we can say that=&gt; $ x^9 + \frac{1}{x^9} = \left(x^3 + \frac{1}{x^3}\right)^3 - 3 \cdot x^3 \cdot \frac{1}{x^3}\left(x^3 + \frac{1}{x^3}\right) $=&gt; $ x^9 + \frac{1}{x^9} = \left(2\sqrt{5}\right)^3 - 3\left(2\sqrt{5}\right) $=&gt; $ x^9 + \frac{1}{x^9} = 40\sqrt{5} - 6\sqrt{5} = 34\sqrt{5} $

It is given that $ x^2 + bx + c = 0 $ has two real roots. Let the roots of the equation be $ \alpha, \beta $. ($ \alpha$ &gt; $\beta $)Then, we can say that $ \frac{1}{\alpha} - \frac{1}{\beta} = \frac{1}{3} $ .... Eq(1)Similarly, $ \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{5}{9} $ .... Eq(2)Eq(2) can be written as $ \left(\frac{1}{\alpha} - \frac{1}{\beta}\right)^2 + 2 \cdot \frac{1}{\alpha} \cdot \frac{1}{\beta} = \frac{5}{9} $=&gt; $ \left(\frac{1}{3}\right)^2 + 2 \cdot \frac{1}{\alpha \cdot \beta} = \frac{5}{9} $=&gt; $ \frac{2}{\alpha \cdot \beta} = \frac{4}{9} \Rightarrow \frac{1}{\alpha \cdot \beta} = \frac{2}{9} $=&gt; $ \alpha \cdot \beta = \frac{9}{2} $We know that the product of the roots is equal to c, which implies $ c = \frac{9}{2} $We also know that the sum of the roots is equal to -b.=&gt; $ \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \left(\frac{1}{\alpha} + \frac{1}{\beta}\right)^2 - \frac{2}{\alpha \beta} = \frac{5}{9} $=&gt; $ \left(\frac{\alpha + \beta}{\alpha \beta}\right)^2 - \frac{4}{9} = \frac{5}{9} $=&gt; $ \left(\frac{\alpha + \beta}{\alpha \beta}\right)^2 = (1)^2 $=&gt; $ \alpha + \beta = \pm \alpha \beta $Hence, the maximum value of b is $ \frac{9}{2} $.Hence, the maximum value of (b+c) is 9

We know that the number of factors of these two numbers is 15. We know that the factors of 15 are 1, 3, 5, and 15.The number of factors of N is&nbsp;(p+1)⋅(q+1)(Where,&nbsp;N = $a^p ⋅ b^q$,&nbsp;and a, b are prime numbers).Hence, the value of N will be least when (p+1) and (q+1) are as close as possible and a, and b are the least distinct prime numbers.Hence, p + 1 = 3 =&gt; p = 2, and q + 1 = 5 =&gt; q = 4, and the prime numbers a, and b are 2, and 3, respectively.Hence, the lowest value of N is&nbsp;N = $2^4$ ×&nbsp;$3^2$&nbsp;= $144$, and the second lowest value of N is&nbsp;N = $2^2$×&nbsp;$3^4$&nbsp;= 324Hence, the sum is (144 + 324) = 468

It is given that&nbsp;$5^{n−1}$ &lt; $3^{n+1}$, where n is a natural number. By inspection, we can say that the inequality holds when n = 1, 2, 3, 4, and 5.Now, we need to find the least integer value of m that satisfies&nbsp;$3^{n+1}$ &lt; $2^{n+m}$For, n =1, the least integer value of m is 3.For, n = 2, the least integer value of m is 3For, n = 3, the least integer value of m is 4.For, n = 4, the least integer value of m is 4.For, n = 5, the least integer value of m is 5.Hence, the least integer value of m such that for all the values of n, the equation holds is 5.3

Let the work done by&nbsp;Rahul, Rakshita, and Gurmeet be a, b, and c units per day, respectively, and the total units of work are W.Hence, we can say that 7(a+b+c) &lt; W ( Rahul, Rakshita, and Gurmeet, working together, would have taken more than 7 days to finish a job).Similarly, we can say that 15(a+c) &gt; W ( Rahul and Gurmeet, working together would have taken less than 15 days to finish the job)Now, comparing these two inequalities, we get: 7(a+b+c) &lt; W &lt; 15(a+c)It is also known that&nbsp;they all worked together for 6 days, followed by Rakshita, who worked alone for 3 more days to finish the job. Therefore, the total units of work done is: W = 6(a+b+c)+3bHence, we can say that&nbsp;7(a+b+c) &lt;&nbsp;6(a+b+c)+3b&nbsp;&lt; 15(a+c)Therefore, (a+b+c) &lt; 3b =&gt; a+c &lt; 2b, and 9b &lt; 9(a+c) =&gt; b &lt; a+c=&gt; a+b+c &lt; 3b =&gt; 7(a+b+c) &lt; 21b , and 15b &lt; 15(a+c)Hence, the number of days required for b must be in between 15 and 21 (both exclusive).Hence, the correct option is B

Let the volume of mixture A be 200 ml, which implies the quantity of cocoa in the mixture is 120 ml, and the quantity of sugar In the mixture 80 ml.Similarly, let the volume of the mixture be 300 ml, which implies the quantity of coffee, and sugar in the mixture is 210, and 90 ml, respectively.Now we combine mixture A, and B in the ratio of 2:3 (if 200 ml mixture A, then 300 ml of mixture B).Hence, the volume of the mixture C is (200+300) = 500 ml, and the quantity of the sugar is (90+80) = 170 ml.&nbsp;Now he mixes&nbsp;C with an equal amount of milk to make a drink, which implies the quantity of the final mixture is (500+500) = 1000 ml.The quantity of sugar in the final mixture is 170 ml.Hence, the percentage is 17%&nbsp;The correct option is A

It is given that the population of the town in 2020 was 100000. The population decreased by y% from the year 2020 to 2021 and increased by x% from the year 2021 to 2022, where x and y are two natural numbers.Hence, the population in 2021 is $ 100000 \left(\frac{100-y}{100}\right) $.The population in 2022 is $ 100000 \left(\frac{100-y}{100}\right)\left(\frac{100+x}{100}\right) $It is also given that the population in 2022 was greater than the population in 2020, and the difference between x and y is 10.Hence,$ 100000 \left(\frac{100-y}{100}\right)\left(\frac{100+x}{100}\right) \gt 100000 $, and (x-y) = 10=&gt; $ 100000 \left(\frac{100-y}{100}\right)\left(\frac{110+y}{100}\right) \gt 100000 $=&gt; $ \frac{100-y}{100} \left(\frac{110+y}{100}\right) \gt 1 $To get the minimum possible value of 2021, we need to increase the value of y as much as possible.Hence, $ (100-y)\{(100+y) + 10\} \gt 10000 $=&gt; $ 10000 - y^2 + 1000 - 10y \gt 10000 $=&gt; $ y^2 + 10y \lt 1000 $=&gt; $ y^2 + 10y + 25 \lt 1025 $=&gt; $ (y + 5)^2 = 1024 \lt 1025 $=&gt; $ (y + 5)^2 = 32^2 $=&gt; $ y = 27 $Hence, the population in 2021 is 100000*(100-27) = 73000

cat 2023 Complete Paper Solution | Slot 3

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cat 2023 Complete Paper Solution | Slot 3

Question 1.

<div><p>For a real number x , if $\frac{1}{2}, \frac{\log _3\left(2^x-9\right)}{\log _3 4}$, and $\frac{\log _5\left(2^x+\frac{17}{2}\right)}{\log _5 4}$ are in an arithmetic progression, then the common difference is</p></div>

Question 2.

<div><p>Let n and m be two positive integers such that there are exactly 41 integers greater than $8^m$ and less then $8^n$, which can be expressed as powers of 2. Then, the smallest possible value of n + m is</p></div>

Question 3.

<div><p>For some real numbers a and b, the system of equations x + y = 4 and $(a+5) x+\left(b^2-15\right) y=8 b$ has infinitely many solutions for x and y. Then, the maximum possible value of ab is</p></div>

Question 4.

<div><p>If x is a positive real number such that $x^8+\left(\frac{1}{x}\right)^8=47$, then the value of $x^9+\left(\frac{1}{x}\right)^9$ is</p></div>

Question 5.

<div><p>A quadratic equation $x^2+b x+c=0$ has two real roots. If the difference between the reciprocals of the roots is $\frac{1}{3}$, and the sum of the reciprocals of the squares of the roots is $\frac{5}{9}$, then the largest possible value of (b + c) is</p></div>

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<div><p>The sum of the first two natural numbers, each having 15 factors (including 1 and the number itself), is</p></div>

Question 7.

<div><p>Let n be any natural number such that $5^{n-1} \lt 3^{n+1}$. Then, the least integer value of m that satisfies $3^{n+1} \lt 2^{n+m}$ for each such n, is</p></div>

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