CAT 2023 Slot 3 Quant Question Paper

It is given that $$x^8 + \left(\frac{1}{x}\right)^8 = 47$$, which can be written as:

=> $$\left(x^4\right)^{^2}+\left(\ \frac{\ 1}{x^4}\right)^{^2}=47$$

=> $$\left(x^4+\frac{\ 1}{x^4}\right)^{^2}-2\cdot x^4\cdot\frac{1}{x^4}=47$$

=> $$\left(x^4+\frac{\ 1}{x^4}\right)^{^2}=49$$

=> $$x^4+\frac{\ 1}{x^4}=7$$

Similarly, $$x^4+\frac{\ 1}{x^4}=7$$ can be expressed as:

=> $$\left(x^2\right)^{^2}+\left(\frac{\ 1}{x^2}\right)^{^2}=7$$

=> $$\left(x^2+\frac{\ 1}{x^2}\right)^{^2}-2\cdot x^2\cdot\frac{1}{x^2}=7$$

=> $$\left(x^2+\frac{\ 1}{x^2}\right)^{^2}=9$$

=> $$x^2+\frac{\ 1}{x^2}=3$$

By the same logic, we get $$x+\frac{1}{x}=\sqrt{\ 5}$$

Now, $$x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)^{^3}-3\cdot x\cdot\frac{1}{x}\left(x+\frac{1}{x}\right)$$

=> $$x^3+\frac{1}{x^3}=\left(\sqrt{\ 5}\right)^{^3}-3\sqrt{\ 5}=2\sqrt{\ 5}$$

By the same logic, we can say that 

=> $$x^9+\frac{1}{x^9}=\left(x^3+\frac{1}{x^3}\right)^{^3}-3\cdot x^3\cdot\frac{1}{x^3}\left(x^3+\frac{1}{x^3}\right)$$

=> $$x^9+\frac{1}{x^9}=\left(2\sqrt{\ 5}\right)^{^3}-3\left(2\sqrt{\ 5}\right)$$

=> $$x^9+\frac{1}{x^9}=40\sqrt{\ 5}-6\sqrt{\ 5}=34\sqrt{\ 5}$$

The correct option is D

It is given that there are exactly 41 numbers, which can be expressed as the power of two, and exist between $$8^m$$ and $$8^n$$, (where m, and n are positive integers, and m < n)

Hence, $$2^{3m}\ <\ \text{41 numbers }<\ 2^{3n}$$

Since, m is a positive integer, the least value of m is 1. Therefore, $$2^{3m\ }=2^3$$, hence, the 41 numbers between them are $$2^4,\ 2^5,\ 2^6,...,2^{44}$$.

Then the lowest possible value of $$8^n$$ is $$2^{45}$$. Hence, the smallest value of n is $$2^{45}\ =\ 8^n\ =>\ 2^{3n\ }=2^{45\ }=>\ n\ =\ 15$$

Hence, the smallest value of m+n is (15+1) = 16

The correct option is D

It is given that for some real numbers a and b, the system of equations $$x + y = 4$$ and $$(a+5)x+(b^2-15)y=8b$$ has infinitely many solutions for x and y.

Hence, we can say that 

=> $$\ \frac{\ a+5}{1}=\ \frac{\ b^2-15}{1}=\ \frac{\ 8b}{4}$$

This equation can be used to find the value of a, and b.

Firstly, we will determine the value of b. 

=> $$\ \frac{\ b^2-15}{1}=\ \frac{\ 8b}{4}\ =>\ b^2-2b-15=0$$

=> $$\left(b-5\right)\left(b+3\right)=0$$

Hence, the values of b are 5, and -3, respectively. 

The value of a can be expressed in terms of b, which is $$a+5=b^2-15\ =>\ a\ =b^2-20$$

When $$b=5,a=5^2-20=5$$

When $$b=-3,a=3^2-20=-11$$

The maximum value of $$ab=(-3)\cdot(-11)=33$$

The correct option is A

It is given that $$\frac{1}{2}, \frac{\log_3(2^x - 9)}{\log_3 4}$$, and $$\frac{\log_5\left(2^x + \frac{17}{2}\right)}{\log_5 4}$$ are in an arithmetic progression.

$$\frac{\log_3(2^x-9)}{\log_34}$$ can be written as $$\log_4\left(2^x-9\right)$$, and $$\frac{\log_5\left(2^x + \frac{17}{2}\right)}{\log_5 4}$$ can be written as $$\log_4\left(2^x+\frac{17}{2}\right)$$ 

Hence, $$2\log_4\left(2^x-9\right)=\frac{1}{2}+\log_4\left(2^x+\frac{17}{2}\right)$$

$$\frac{1}{2}$$ can be written as $$\log_42$$.

Therefore, 

=> $$2\log_4\left(2^x-9\right)=\log_42+\log_4\left(2^x+\frac{17}{2}\right)$$

=> $$\log_4\left(2^x-9\right)^{^2}=\log_42\left(2^x+\frac{17}{2}\right)$$

=> $$\left(2^x-9\right)^{^2}=2\left(2^x+\frac{17}{2}\right)$$

=> $$2^{2x}-18\cdot2^x+81=2\cdot2^x+17$$

=> $$2^{2x}-20\cdot2^x+64=0$$

=> $$2^{2x}-16\cdot2^x-4\cdot2^x+64=0$$

=> $$2^x\left(2^x-16\right)-4\left(2^x-16\right)=0$$

=> $$\left(2^x-4\right)\left(2^x-16\right)=0$$

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 $$\log_4\left(2^x-9\right)-\log_42$$

=> $$\log_47-\log_42\ =\ \log_4\left(\frac{7}{2}\right)$$

The correct option is D

It is given that $$5^{n-1} < 3^{n + 1}$$, 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 $$3^{n+1} < 2^{n+m}$$

For, n =1, the least integer value of m is 3.

For, n = 2, the least integer value of m is 3

For, 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.

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 $$(p+1)\cdot(q+1)$$(Where, $$N=a^p\cdot b^q,$$ 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 => p = 2, and q+1 = 5 => q = 4, and the prime numbers a, and b are 2, and 3, respectively.

Hence, the lowest value of N is $$N=2^4\times\ 3^2\ =144$$, and the second lowest value of N is $$N=2^2\times\ 3^4\ =324$$.

Hence, the sum is (144+324) = 468

It is given that $$x^2 + bx + c = 0$$ has two real roots. Let the roots of the equation be $$\alpha\ ,\beta\ $$. ($$\alpha\ \ >\ \beta\ $$)

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 $$\left(\frac{1}{\alpha\ }-\frac{1}{\beta\ }\right)^{^2}+2\cdot\frac{1}{\alpha\ }\cdot\frac{1}{\beta\ }=\frac{5}{9}$$

=> $$\left(\frac{1}{3}\right)^{^2}+2\cdot\frac{1}{\alpha\ }\cdot\frac{1}{\beta\ }=\frac{5}{9}$$

=> $$\frac{2}{\alpha\ \cdot\beta\ }=\frac{4}{9}=>\ \frac{1}{\alpha\ \cdot\beta\ }=\frac{2}{9}$$

=> $$\alpha\ \cdot\beta\ =\frac{9}{2}$$

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. 

=> $$\frac{1}{\alpha\ ^2}+\frac{1}{\beta\ ^2}=\left(\frac{1}{\alpha\ }+\frac{1}{\beta\ }\right)^{^2}-\frac{2}{\alpha\ \beta\ }=\frac{5}{9}$$

=> $$\left(\ \frac{\ \alpha\ +\beta\ }{\alpha\ \beta\ }\right)^{^2}-\frac{4}{9}=\frac{5}{9}$$

=> $$\left(\ \frac{\ \alpha\ +\beta\ }{\alpha\ \beta\ }\right)^{^2}=\left(1\right)^2$$

=> $$\alpha\ +\beta\ \ =\ \pm\ \alpha\ \beta\ $$

Hence, the maximum value of b is $$\frac{9}{2}$$.

Hence, the maximum value of (b+c) is 9

It is given that a merchant purchases a cloth at a rate of Rs.100 per meter and receives 5 cm length of cloth free for every 100 cm length of cloth purchased by him.

Hence, the cost price of 105 cm clothes is 100 rupees. 

It is also known that he marked the price of 100 cm clothes as 110 rupees, and gave a 5% discount, and he cheated his customers by giving 95 cm length of cloth for every 100 cm length of cloth purchased by the customers.

Hence, the selling price of 95 cm clothes is 110*(19/20) rupees.

Therefore, the selling price of 105 cm clothes is 115.5 rupees.

Hence, the profit is 15.5%

The correct option is C

Let the work done by 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) < 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) > 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) < W < 15(a+c)

It is also known that 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)+3b

Hence, we can say that 7(a+b+c) < 6(a+b+c)+3b < 15(a+c)

Therefore, (a+b+c) < 3b => a+c < 2b, and 9b < 9(a+c) => b < a+c

=> a+b+c < 3b => 7(a+b+c) < 21b , and 15b < 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 D

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, 

$$100000\left(\ \frac{\ 100-y}{100}\right)\left(\ \frac{\ 100+x}{100}\right)>\ 100000$$, and (x-y) = 10

=> $$100000\left(\ \frac{\ 100-y}{100}\right)\left(\ \frac{\ 110+y}{100}\right)>\ 100000$$

=> $$\ \frac{\ 100-y}{100}\left(\ \frac{\ 110+y}{100}\right)>\ 1$$

To get the minimum possible value of 2021, we need to increase the value of y as much as possible.

Hence, $$\left(\ \ 100-y\right)\left\{\left(\ \ 100+y\right)+10\right\}>\ 10000$$

=> $$10000-y^2+1000-10y>\ 10000$$

=> $$y^2+10y<1000$$

=> $$y^2+10y+25<1025$$

=> $$\left(y+5\right)^2=1024\ <\ 1025$$

=> $$\left(y+5\right)^2=32^2$$

=> $$y=27$$

Hence, the population in 2021 is 100000*(100-27) = 73000

The correct option is C

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. 

Now he mixes 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% 

The correct option is A

Let us assume that A, B, C, D, and E weights are a, b, c, d, and e.

1st condition

$$\frac{\left(a+b+c\right)}{3}-\frac{\left(a+b+c+d\right)}{4}=x$$

2nd condition

$$\frac{\left(a+b+c+e\right)}{4}-\frac{\left(a+b+c\right)}{3}=2x$$

Adding both the equations, we get:

$$\frac{\left(e-d\right)}{4}=3x$$

=> $$\frac{\left(e-d\right)}{4}=3x$$ => e - d = 12x

Given that 12x = 12 => x = 1.

Let us assume the speed of the 1st boat is b, the 2nd boat is s, and the river's speed is r.

Let 'd' be the distance between A and B.

=> d = 2(b+r) and d = 3(b-r)

=> b + r = d/2 and b - r = d/3 => r = d/12 (subtracting both equations).

Now, it is given that

$$\dfrac{d}{s+r}+\dfrac{d}{s-r}=6$$

=> $$\dfrac{d}{s+\dfrac{d}{12}}+\dfrac{d}{s-\dfrac{d}{12}}=6$$

=> $$2ds=6\left(s^2-\dfrac{d^2}{144}\right)$$

=> $$144s^2-48ds-d^2=0$$

Solving the quadratic equation, we get:

$$s=d\left(\dfrac{\left(48+\sqrt{\ 48^2+4\left(144\right)}\right)}{2\times\ 144}\right)$$

$$s=d\left(\dfrac{1}{6}+\dfrac{\sqrt{\ 5}}{12}\right)$$

=> Required value of $$\dfrac{d}{s+r}$$

= $$\dfrac{d}{\dfrac{d}{6}+\dfrac{\sqrt{5}d}{12}+\dfrac{d}{12}}$$

= $$\dfrac{12}{3+\sqrt{\ 5}}=\dfrac{\left(12\right)\left(3-\sqrt{\ 5}\right)}{4}$$

= $$3\left(3-\sqrt{\ 5}\right)$$

Given that the number of coins collected per week by two coin-collectors, A and B, are in the ratio 3: 4

Let us assume A collects 3c coins per week and B collects 4c coins per week.

Total number of coins collected by A in 5 weeks = 5*3c = 15c, which should be multiple of 7 => c should be multiple of 7.

Total number of coins collected by B in 3 weeks = 3*4c = 12c, which should be a multiple of 24 => c should be a multiple of 2.

So, the least possible value of c is lcm(2,7) = 14.

Coins sold by A in a week = 3c = 3*14 = 42.

Let us assume the initial stock of all the fruits is S.

Let us take we have 'b' and 'a' mangoes initially.

Stock of Mangoes = 40% of S = 2S/5

The total number of fruits sold are Mangoes Sold + Apples Sold + Bananas Sold

= 2S/10 + 96 + 4a/10 = S/2 (Given)

=> S/5 + 96 + 2a/5 = S/2

=> S = $$\dfrac{\left(4a+960\right)}{3}$$

=> $$\dfrac{4a}{3}+320$$

'a' has to be a multiple of 3 for the above term to be an integer.

But 'a' has to be a multiple of 5 for 4a/10 to be an integer.

=> The smallest value of 'a' satisfying both conditions is 15.

=> $$\dfrac{4a}{3}+320=\dfrac{4\left(15\right)}{3}+320$$ = 340

Let 'g' and 's' be the efficiencies of Gautam and Suhani. Let W is the total amount of work.

=> g + s = W/20 (1 day work) ----(1)

Also Gautam doing only 60% => 3g/5 and Suhani doing 150% => 3s/2

=> 3g/5 + 3s/2 = W/20 (1 day work)

=> $$g+s=\dfrac{3g}{5}+\dfrac{3s}{2}$$

=> $$\dfrac{s}{g}=\dfrac{4}{5}$$ => Gautam is the more efficient person.

Now, from the 1st equation

=> $$g+\dfrac{4g}{5}=\dfrac{W}{20}$$

=> $$\dfrac{9}{5}g=\dfrac{W}{20}$$

=> $$g=\dfrac{W}{36}$$

=> Gautam takes 36 days to finish the complete work.

Given that AB = AC => Angle C = Angle B (1)

AD and BE are altitudes => they make 90 degrees with the sides

Angle AOB = 105 => Angle EOD = 105 (Vertically Opposite Angles)

In quadrilateral DOEC

Angle C = 360 - 105 - 90 - 90 = 75 => Angle B = 75 (from 1)

We know that from the area of the triangle AD * BC = BE * AC

=> $$\dfrac{AD}{BE}=\dfrac{AC}{BC}=\dfrac{2R\sin B}{2R\sin A}=\dfrac{\sin\left(75\right)}{\sin\left(30\right)}=2\sin\left(75\right)$$ = 2cos(15)

[sin(x) = cos(90-x)]

Let us assume the length of the rectangle is 'l' and breadth of the rectangle is 'b'.

The radius, l/2 and b in the above diagram form a right-angled triangle.

=> $$\left(\frac{l}{2}\right)^2+b^2=2^2$$

We know that the area of the rectangle is l*b, which can be obtained by considering 2 times the geometric mean of $$\left(\frac{l}{2}\right)^2$$ and $$b^2$$.

Therefore, for the maximum area, the equality condition of AM-GM inequality should be satisfied

=> $$\left(\frac{l}{2}\right)^2=b^2$$ => l = 2b.
=> l/b = 2/1.

The sum of the interior angles of a polygon of 'n' sides is given by $$\left(2n-4\right)\times\ 90$$, and the sum of the exterior angles of a polygon is 360 degrees.

So, the difference between them will be 120 * n

=> $$\left(2n-4\right)90-360=120n$$

=> 60n = 720 => n = 12.

We know that the number of diagonals of a regular polygon is nC2 - n = 12C2 - 12 = 66 - 12 = 54.

The given sequence can be written as:

$$1\left(1+\ \frac{1}{4}+\frac{1}{16}+\frac{1}{64}+...\right)+\frac{1}{3}\left(\frac{1}{4}+\frac{1}{16}+...\right)+\frac{1}{9}\left(\frac{1}{16}+\frac{1}{64}+...\right)+..$$

We know that the sum of an infinite G.P. is $$\dfrac{a}{1-r}$$, where a is the first term and r is the common ratio.

=> The first term = $$\frac{1}{1-\frac{1}{4}}=\dfrac{4}{3}$$

=> The second term = $$\frac{1}{3}\left(\frac{\left(\frac{1}{4}\right)}{1-\left(\frac{1}{4}\right)}\right)=\dfrac{1}{9}$$

=> The third term = $$\frac{1}{9}\left(\frac{\left(\frac{1}{16}\right)}{1-\left(\frac{1}{4}\right)}\right)=\dfrac{1}{108}$$

Observing these three terms, we see that they are in G.P. with a common ratio of $$\dfrac{1}{12}$$

=> Sum of this infinite G.P. = $$\dfrac{\left(\dfrac{4}{3}\right)}{1-\left(\dfrac{1}{12}\right)}=\dfrac{16}{11}$$

The first series goes as follows:

54, 62, 70, 78, 86, 94, 102,....

The second series goes as follows:

102, 106, 110,...

The first common term is 102 (first term of the common terms) and the common difference between them will be LCM(4,8) = 8

=> The required sequence is 102, 110, 118,..... (last term should be less than 468 (100th term of second series))

=> 102 + (n-1)(8) $$\le$$ 498

=> n is less than or equal to 50.5  =>  n = 50

Using the summation of A.P. formula:

Required sum = $$\dfrac{n}{2}\left(2a+\left(n-1\right)d\right)=\dfrac{50}{2}\left(2\times\ 102+49\times\ 8\right)=14900$$

Given that f(3x + 2y, 2x - 5y) = 19x.

Let us assume the function f(a,b) is a linear combination of a and b.

=> f(3x+2y, 2x-5y) = m(3x+2y) + n(2x-5y) = 19x

=> 3m + 2n = 19 and 2m - 5n = 0

Solving we get m = 5 and n = 2

=> f(a,b) = 5a+2b

=> f(x,2x) = 5x + 2(2x) = 9x = 27 => x = 3.