CAT 2024 Slot 2 Question Paper

Looking at the additional information about the prime numbers should make one realise that they are the key to solving the question. 
f(16000) can be written as $$f\left(2^8\times\ 5^4\right)$$

Now, we can try to find these individual values:

For any prime p: f(p)=1
$$f\left(p^2\right)=f\left(p\right)f\left(p\right)+f\left(p\right)+f\left(p\right)=1+1+1=3$$
$$f\left(p^3\right)=f\left(p^2\right)f\left(p\right)+f\left(p^2\right)+f\left(p\right)=3+3+1=7$$

This way, we can find the function output for any prime number raised to a power. 

We can see that each new exponent is twice the previous output +1, solving this way till prime raised to power 8
$$f\left(p^4\right)=7+7+1=15$$
$$f\left(p^5\right)=15+15+1=31$$
$$f\left(p^6\right)=31+31+1=63$$
$$f\left(p^7\right)=63+63+1=127$$
$$f\left(p^8\right)=127+127+1=255$$

Using these values in the original expression of $$f\left(2^8\times\ 5^4\right)=f\left(2^8\right)f\left(5^4\right)+f\left(2^8\right)+f\left(5^4\right)$$ we get
$$f\left(2^8\times\ 5^4\right)=\left(255\times\ 15\right)+255+15=4095$$

Therefore, Option A is the correct answer. 

From the sum and product of roots, we get: $$\alpha\ +\beta\ =-\frac{\lambda}{3}$$ and $$\alpha\ \beta\ =-\frac{1}{3}$$

Simplifying the expression given in the question, we get: $$\frac{\alpha^2+\beta^2\ }{\alpha^2\beta^2\ }=15$$
and substituting the denominator's value as 1/9, we get:$$\alpha^2+\beta^2\ =\frac{15}{9}$$

We want the expression $$\alpha^3+\beta^3\ $$, so multiplying both sides by $$\alpha+\beta$$, we get:
$$\alpha^3+\beta^3+\alpha\beta\left(a+\beta\ \right)=\frac{15}{9}\left(\alpha\ +\beta\ \right)$$
$$\alpha^3+\beta^3+\frac{\lambda}{9}\ =\frac{15}{9}\left(-\frac{\lambda}{3}\ \right)$$
$$\alpha^3+\beta^3+\frac{\lambda}{9}\ =-\frac{5\lambda}{9}-\frac{\lambda}{9}=-\frac{2\lambda\ }{3}\ \ $$

We would still need to find the value of $$\lambda$$
This we can do from the initial relation we had:

$$\alpha^2+\beta^2\ =\frac{15}{9}$$
$$\alpha^2+\beta^2\ =\left(\alpha+\beta\right)^2-2\alpha\ \beta\ \ \ =\frac{15}{9}$$
$$\frac{\lambda^2}{9}+\frac{2}{3}\ \ \ =\frac{15}{9}$$
$$\frac{\lambda^2}{9}\ =\frac{15-6}{9}=\frac{9}{9}=1$$
This would finally give us $$\lambda^2=9$$

Using this in our required expression, we get:

$$\left(\alpha^3+\beta^3\right)^2=\left(-\frac{2\lambda}{3}\ \ \right)^2=\frac{4\times\ 9}{9}=4$$

Therefore, Option B is the correct answer. 

Let's take Rajesh and Garima's ages to be R and G, respectively
From the given ratio, we can see that Rajesh is older than Garima, so let's take R=G+x

When Rajesh was of age G, which was x years ago, Garima was of G-x years old
Giving the ratio as $$\frac{G}{G-x}=\frac{3}{2}$$
This gives us G as 3x, which in turn gives R as 4x

We are asked the ratio when Gramia becomes 4x years old. 
By that time, Rajesh will be 5x years old. 

Giinv their ratio as $$\frac{5x}{4x}=5:4$$

Therefore, Option C is the correct answer. 

Let's take the radius of the original circles to be R and that of the circle in between the three circles to be r. 

Joining the centres of the three circles, we will get an equilateral triangle or length 2R. 
The distance between the circle's centre and the original circle's centre would be R+r.

Using this right angle triangle, we can get the relation: $$\frac{R}{R+r}=\frac{\sqrt{\ 3}}{2}$$
We can take $$R=\sqrt{\ 3}a$$ and R+r as 2a, this would give us r as $$R=\left(2-\sqrt{\ 3}\right)a$$

The outer circle will have a radius of 2R+r
We need to find the ratio of $$\frac{2R+r}{r}$$

This will be equal to $$\frac{\left(2\sqrt{\ 3}+2-\sqrt{\ 3}\right)a}{\left(2-\sqrt{\ 3}\right)a}=\frac{2+\sqrt{\ 3}}{2-\sqrt{\ 3}}=4+3+4\sqrt{\ 3}=7+4\sqrt{\ 3}:1$$

Therefore, Option A is the correct answer. 

The simplest way to visualise this would be with symmetry. 

We can take AD to be x and CB to be 3-x; since the perimeter of ABCD is 6, and AB and CD are given as 2 and 1 cm, respectively, that leaves AD+BC as three only. 

We can see that triangles AEB and DEC are similar, with the lengths of AB being double of CD, essentially making D the mid-point of AE and C the mid-point of EB. 

Through this, we get the length of DE and CE to be x and 3-x. 
Give us the AE and BE lengths as 2x and 6-2x, respectively. 

 Giving the perimeter of AEB as 2x+6-2x+2 = 8 cm

Therefore, Option A is the correct answer. 

We can divide the list into 4 elements: the first 10 as a, the next 10 as b, the next 10 as c, and the last 10 as d 

From the relations we are given, we can form the equations: $$\frac{a+b+c}{3}=40,000$$
$$\frac{b+c+d}{3}=60,000$$ and $$\frac{a+d}{2}=50,000$$

Adding the first two equations, we get $$a+2\left(b+c\right)+d=300,000$$
We can substitute the value of a+d as 100,000 to get b+c as 100,000

Using this value in the first and second equation would give a and d as 20,000 and 80,000, respectively. 

We are told that the average of the first 10 employees increases by 100%, that is, it changes from 20,000 to 40,000
The average of the last 10 increases by 200%; that is, it changes from 80,000 to 240,000

The total of all the four elements would be 40,000+100,000+240.000 = 380,000

Giving the average to be $$\frac{380,000}{4}=95,000$$

Therefore, Option A is the correct answer. 

We can take the work done by Amal, Vimal and Sunil to be A, V and S, respectively. 

Let's take the total work they did to be T.

We are given the equations:
150A + 150V = T      ...(1)
100V + 100S = T      ...(2)

75A + 135V + 45S = T    ...(3)

Adding (1) and (2), we get: 150A + 250V + 100S = 2T   ...(4)

And multiplying (3) with 2 we get: 150A + 270V + 100S = 2T  ...(5)

Subtracting (5) from (4), we get 10S = 20V or simply S = 2V

Using this in (2), we get the total work T to be 300V, and using that result in (1), we get A=V

Therefore, the work done by A, V and S equals V, V, and 2V units per day. 

Now, in the question, we are given work cycles. To simplify the calculations, we should consider a time duration that is the LCM of the period taken by the three agents, which in this case would be 6

In 6 days, Amal will work for 6 days, doing 6V units of work.
Vimal will work for 3 days, doing 3V units of work. 
Sunil will work for 2 days, doing 4V units of work. 

So, in one 6-day cycle, 13V units of work will be done. 
Dividing the total work (300V) by 13V we can see that in 23 cycles 299V units of work will be done. 

These 23 cycles will be $$23\times\ 6\ =\ 138$$ days
The remaining 1V units of work will be done the next day. 

Therefore, a total of 139 days will be required. 

There are two critical points for the inequality to consider: x=-5 and x=3/2

Region I: x is greater than 3/2
In this scenario, both the terms would be positive; cross-multiplying, we get the relation $$2x-3\le x+5$$
Giving the boundary $$x\le8$$, hence giving us the valid range as $$\frac{3}{2}<x\le8$$

Region II: $$-5<x<\frac{3}{2}$$
In this case, the right-hand side will be a negative value, and hence, the sign would change when multiplying, giving the inequality
$$2x-3\ge x+5$$
Which will give x>8, which is out of bounds for this region 

Another way is to put a value in the region to check for the validity of the inequality; by putting x=0, we could see that the inequality does not hold in this region 

Region III: x less than -5
In this scenario, both the terms are negative, essentially giving us the same boundary as region 1; we take the lower bounds, giving us that x has to be less than 5

Therefore, for the given inequality to hold true x<-5 or $$\frac{3}{2}<x\le8$$

Hence, Option A is the correct answer. 

Let's take the amount invested by Sunil to be X. 

The amount received by Anil at the end of 6 years would be $$22000\left(1+\frac{4}{2\times\ 100}\right)^{6\times\ 2}=22000\left(1.02\right)^{12}$$

The amount received by Sunil at the end of 5 years would be $$X\left(1.02\right)^{10}$$
In the 6th year, Sunil invests this at a simple interest of 10%, giving him an interest of $$X\left(1.02\right)^{10}\times\ 0.1$$
Giving the total amount with him at the end of 6 years to be $$X\left(1.02\right)^{10}\times\ \left(1+0.1\right)$$

Equating the final amount with Sunil and Anil, we get:
$$X\left(1.02\right)^{10}\times\ \left(1.1\right)=22000\left(1.02\right)^{12}$$
$$X=\frac{22000\left(1.02\right)^2}{1.1}=20808$$

Therefore, Option D is the correct answer. 

Let's take the scheduled time taken by the bus to be t
From the first statement (bus travelling at 60 kmph), we can get the total distance travelled by bus to 60(t+3.5)

The second scenario gives us that the bus covered two-thirds of the distance in one-third of the time, meaning that the remaining one-third distance was covered in two-thirds of the time, giving us the relation $$\frac{1}{3}st$$ covered in $$\frac{2}{3}t$$ giving the speed to be $$\frac{s}{2}$$ which is given as 40 km/h, thereby giving the usual speed of the bus to be 80 km/hr

Now the first relation we get 60(t+3.5)=80t
Giving us t=10.5 hours

Thus, the bus usually takes 10.5 hours on its journey. 
Staring at 9:00, it will complete the journey at 7:30 pm

Therefore, Option A is the correct answer.