CAT 2024 Slot 1 Question Paper

We are looking for the minimum possible number of votes that Ramya can get when she runs an attacking campaign. 

To minimise the number of votes, we can have Ramya run a staid campaign to minimise the votes, so minimum intensity, which will get her 20% of the votes if she ran with issues. Now that she is running with attacking, she will loose 20% of the votes to Amiya and 5% of the votes will not vote anymore.

That is a total 25% loss. Remaining votes she will get is 75% of the 20% which will leave her with 15% of the votes. 

We are looking for the minimum possible number of votes that Ramya can get and maximise the number of votes that Amiya can get. 

We can borrow the scenario from the previous question where Ramya runs an attacking campaign, and we minimised the number of votes she can get. 

To minimise the number of votes, we can have Ramya run a staid campaign to minimise the votes, so minimum intensity, which will get her 20% of the votes if she ran with issues. Now that she is running with attacking, she will loose 20% of the votes to Amiya and 5% of the votes will not vote anymore.

That is a total 25% loss. Remaining votes she will get is 75% of the 20% which will leave her with 15% of the votes.

And to maximise the number of votes Amiya can get, we will have her run an vigorous issues campaign, which will give her 2x20% of the votes, that is 40% of the votes. And since Ramya has been running an attacking campaign, 20% of her votes are transferred to Amiya. 20% of the 20% of the votes which is 4% that were going to Ramya will now go to Amiya. That will bring up Amiya's tally up to 44% leaving Ramya's tally at 15%. 

The difference in the votes will be 44-15=29%. 

This is the maximum possible vote difference between the two candidates that is possible. 

Writing down the data with us in a table form. 

But this includes countries which have been visited by one person, two people and three people. Essentially, there is an overlap of countries. 
It is given to us that 32 distinct countries were visited among the three people, 
Using that we can write down the equations, 
$$I+II+III=32$$...(1)
$$I+2II+3III=43$$...(2)
Where I, II, and III denote the number of countries visited exactly by 1 of them, 2 of them and all of them. 
We are told that USA(ROW) is the only country which was visited by all three people, that means $$$III=1$$

Subtracting Equation 1 from 2 we get, $$II+2III=11$$
And since III=1 we get $$II=9$$

So, we are looking for 9 countries visited by exactly 2 people, and 22 countries visited by exactly 1 person. 
Using the information from the problem set, we can note down the following, 

We can see that Dheeraj visited only 1 country in ROW so every other value in that column has to be zero for wherever Dinesh is present. And Samantha has not visited any country in Asia so every value with Samantha present in Asia should be zero as well. 

We have the following distribution, we are looking for 9 countries visited by exactly 2 people. And we have identified 2 countries, that means, 7 were countries were visited by both Samantha and Nitesh only. And given that half the countries visited by both of them are in Europe, and total number of countries visited by both of them is 7(only two of them)+1(USA)=8, they visited 4 countries in Europe together and 3 countries in ROW together. And using that information, we can fill in the rest of the table as well. Since we know the number of countries visited by each of them in a particular continent. 

How many countries in Asia were visited by at least one of Dheeraj, Samantha and Nitesh is 3. 

Writing down the data with us in a table form.

But this includes countries which have been visited by one person, two people and three people. Essentially, there is an overlap of countries.
It is given to us that 32 distinct countries were visited among the three people,
Using that we can write down the equations,
$$I+II+III=32$$...(1)
$$I+2II+3III=43$$...(2)
Where I, II, and III denote the number of countries visited exactly by 1 of them, 2 of them and all of them.
We are told that USA(ROW) is the only country which was visited by all three people, that means $$$III=1$$

Subtracting Equation 1 from 2 we get, $$II+2III=11$$
And since III=1 we get $$II=9$$

So, we are looking for 9 countries visited by exactly 2 people, and 22 countries visited by exactly 1 person.
Using the information from the problem set, we can note down the following,

We can see that Dheeraj visited only 1 country in ROW so every other value in that column has to be zero for wherever Dinesh is present. And Samantha has not visited any country in Asia so every value with Samantha present in Asia should be zero as well.

We have the following distribution, we are looking for 9 countries visited by exactly 2 people. And we have identified 2 countries, that means, 7 were countries were visited by both Samantha and Nitesh only. And given that half the countries visited by both of them are in Europe, and total number of countries visited by both of them is 7(only two of them)+1(USA)=8, they visited 4 countries in Europe together and 3 countries in ROW together. And using that information, we can fill in the rest of the table as well. Since we know the number of countries visited by each of them in a particular continent.

Number of countries visited by only Nitesh is 2. 

Writing down the data with us in a table form.

But this includes countries which have been visited by one person, two people and three people. Essentially, there is an overlap of countries.
It is given to us that 32 distinct countries were visited among the three people,
Using that we can write down the equations,
$$I+II+III=32$$...(1)
$$I+2II+3III=43$$...(2)
Where I, II, and III denote the number of countries visited exactly by 1 of them, 2 of them and all of them.
We are told that USA(ROW) is the only country which was visited by all three people, that means $$$III=1$$

Subtracting Equation 1 from 2 we get, $$II+2III=11$$
And since III=1 we get $$II=9$$

So, we are looking for 9 countries visited by exactly 2 people, and 22 countries visited by exactly 1 person.
Using the information from the problem set, we can note down the following,

We can see that Dheeraj visited only 1 country in ROW so every other value in that column has to be zero for wherever Dinesh is present. And Samantha has not visited any country in Asia so every value with Samantha present in Asia should be zero as well.

We have the following distribution, we are looking for 9 countries visited by exactly 2 people. And we have identified 2 countries, that means, 7 were countries were visited by both Samantha and Nitesh only. And given that half the countries visited by both of them are in Europe, and total number of countries visited by both of them is 7(only two of them)+1(USA)=8, they visited 4 countries in Europe together and 3 countries in ROW together. And using that information, we can fill in the rest of the table as well. Since we know the number of countries visited by each of them in a particular continent.

Number of countries in ROW visited by both Nitesh and Samantha is 3+1=4. 

Writing down the data with us in a table form.

But this includes countries which have been visited by one person, two people and three people. Essentially, there is an overlap of countries.
It is given to us that 32 distinct countries were visited among the three people,
Using that we can write down the equations,
$$I+II+III=32$$...(1)
$$I+2II+3III=43$$...(2)
Where I, II, and III denote the number of countries visited exactly by 1 of them, 2 of them and all of them.
We are told that USA(ROW) is the only country which was visited by all three people, that means $$$III=1$$

Subtracting Equation 1 from 2 we get, $$II+2III=11$$
And since III=1 we get $$II=9$$

So, we are looking for 9 countries visited by exactly 2 people, and 22 countries visited by exactly 1 person.
Using the information from the problem set, we can note down the following,

We can see that Dheeraj visited only 1 country in ROW so every other value in that column has to be zero for wherever Dinesh is present. And Samantha has not visited any country in Asia so every value with Samantha present in Asia should be zero as well.

We have the following distribution, we are looking for 9 countries visited by exactly 2 people. And we have identified 2 countries, that means, 7 were countries were visited by both Samantha and Nitesh only. And given that half the countries visited by both of them are in Europe, and total number of countries visited by both of them is 7(only two of them)+1(USA)=8, they visited 4 countries in Europe together and 3 countries in ROW together. And using that information, we can fill in the rest of the table as well. Since we know the number of countries visited by each of them in a particular continent.

Number of countries in EU visited by exactly 1 person is 12. 

Set A={2,3,5,7,11,13} so |A|=6
Set B={1, 8, 27} so |B|=3

Without any restrictions, each element in A can map to any of the 3 elements in B. Thus, the total number of functions is: $$3^6=729$$

Excluding Functions That Miss One Element in B: If a function does not map to an element in B, there are 2 elements in B left for mapping. The total number of such functions (for each specific element not mapped) is: $$2^6=64$$

Since there are 3 elements in B, the total number of such functions is:3x64=192

Adding Back Functions That Miss Two Elements in B: If a function misses two elements in B, there is only 1 element left for mapping. The total number of such functions is: 1^6=1. 

Since there are $$^3C_2$$ ways to choose which two elements are missed, the total number of such functions is: 3

Using the inclusion-exclusion principle, the number of functions where all elements of B are mapped by at least one element of A is:
729-192+3=540. 

We have two equations,

$$4(x^{2}+y^{2}+z^{2}) = a$$ ---(1)

$$4(x - y - z) = 3 + a$$ ---(2)

Substituting the value of a from equation 1 in equation 2, we get,

$$4\left(x\ -\ y\ -\ z\right)\ =\ 3\ +\ 4(x^2\ +\ y^2\ +\ z^2)$$

$$\ 3\ +\ 4(x^2\ +\ y^2\ +\ z^2)\ -4\left(x\ -\ y\ -\ z\right)\ =\ 0$$

$$\ 3\ +\ 4x^2\ +\ 4y^2\ +\ 4z^2\ -4x\ \ +\ 4y\ +\ 4z\ =\ 0$$

It can be written as,

$$4x^2\ -4x\ +\ 1+\ 4y^2\ +\ 4y\ +\ 1\ +\ 4z^2\ +\ 4z\ +\ 1\ =\ 0$$

$$\left(2x\ -\ 1\right)^2\ +\ \left(2y\ +\ 1\right)^2\ +\ \left(2z\ +\ 1\right)^2\ \ =0$$

We know that if the sum of the squares of terms is 0, then all the terms must be equal to 0

2x - 1 = 0

x = $$\dfrac{1}{2}$$

2y + 1 = 0

y = $$-\dfrac{1}{2}$$

2z + 1 = 0

z = $$-\dfrac{1}{2}$$

Substituting the values in equation 2, we get,

$$4\left(\dfrac{1}{2}\ -\ \left(-\dfrac{1}{2}\right)\ -\ \left(-\dfrac{1}{2}\right)\right)\ =\ 3\ +\ a$$

$$4\left(\dfrac{3}{2}\right)\ =\ 3\ +\ a$$

$$6\ =\ 3\ +\ a$$

$$\ a\ =\ 3$$

Therefore, the correct answer is option A.

When given more than one equations, stating the fact that there is a common root, 
We need to equate the two equations to get discernible values for $$x$$

Here, we are given three equations with the values of $$m$$, $$n$$

$$x^2+mx+9=x^2+\left(m+n\right)x+35$$
$$mx+9=mx+nx+35$$
$$nx=-26$$

Similarly, we can do it for the other equation as well, 
$$x^2+nx+17=x^2+\left(m+n\right)x+35$$
$$mx=-18$$

Substituting the value of either $$mx$$ or $$nx$$ in the original equations, we get 
$$x^2-18+9=0$$
$$x^2=9$$
$$x=\pm\ 3$$
Since we are given that the root is negative, $$x=-3$$

$$n=-\frac{26}{-3}$$
$$m=-\frac{18}{-3}$$

$$3n=26$$
$$2m=12$$

$$2m+3n=38$$

Using the arithmetic progression formula for the nth term, where
$$x_n=a+\left(n-1\right)d$$
Substituting the value for n and using that in the equation that is given, 
$$2x_{6}+2x_{9}=x_{11}+x_{13}$$, Then,$$x_{100}$$ equals

We get, $$2\left(a+5d\right)+2\left(a+8d\right)=a+10d+a+12d$$
$$4a+26d=2a+22d$$
$$2a=-4d$$
$$a=-2d$$

We are given, $$x_5=-4$$
$$a+4d=-4$$
Substituting the value for a in terms of d, 
$$2d=-4$$
$$d=-2$$
$$a=4$$

$$x_{100}=a+99d$$
$$x_{100}=4-198=-194$$