CAT 2024 Slot 1 Question Paper

Let us assign the amount of work done by Renu in one hour is R
And the amount of work done by Seema in one hour is S

We are told that a certain task with different time durations for Seema and Renu
Renu=15 days with 4 hours each day, that is total work done by Renu is $$15\times\ 4\times\ R=60R$$
Similarly for Seema, 8 days and 5 hours each day, total work done by Seema is $$40S$$
We know that $$60R=40S$$ or $$S=1.5R$$

For this task, let us assume that the amount of days that Seema decides to work is X and the hours per day that Renu decides to work is Y
We are then told that, Seema works for 2Y hours per day and Renu works for 2X days, 
Work done by them will be $$2XYR$$ and $$2XYS$$

We are told that Y=2, Making this $$4XR$$ and $$4XS$$
$$S=1.5R$$
Total work will be, $$4XR$$+$$6XR$$=$$60R$$
We get the value of $$X=6$$

X is the number of days Seema will work, which is 6. 

To find the value of $$10^{100}mod\left(7\right)$$
When 10 is divided by 7, it leaves a remainder 3, so the above equation can be written as, 
$$3^{100}mod\left(7\right)$$

Now looking at the cyclicality of powers of 3 when divided by 7, 
$$3^1mod  7=3$$

$$3^2mod  7=2$$

$$3^3mod  7=6$$

$$3^4mod  7=4$$

$$3^5mod  7=5$$

$$3^6mod  7=1$$

From this calculation, it is evident that the powers of 3 modulo 7 repeat every 6 steps. This forms a cycle: 3, 2, 6, 4, 5, 1

$$3^{100}=\left(3^6\right)^{16}\times\ \left(3^4\right)$$

Since $$3^6mod 7=1$$

We just need to consider $$3^4mod 7$$ which equals 4

Hence the answer is 4. 

To solve this question, we need to immediately recognise the fact that, $$32768=8^5$$
Substituting this in the above given equation, 

$$\left(\dfrac{1}{8}\right)^k\times\ \left(\dfrac{1}{8}\right)^{5\times\ \dfrac{1}{3}}=\left(\dfrac{1}{8}\right)\times\ \left(\dfrac{1}{8}\right)^{5\times\ \dfrac{1}{k}}$$

Since the bases are equal, we can equate the powers on either side of the equation, 

$$k+\frac{5}{3}=1+\frac{5}{k}$$

$$\frac{\left(3k+5\right)}{3}=\frac{\left(k+5\right)}{k}$$

$$3k^2+5k=3k+15$$

$$3k^2+2k-15=0$$

Here in the given quadratic equation, the Discriminant is greater than 0, $$2^2-\left(4\right)\left(3\right)\left(-15\right)>0$$
That means both the roots are real, hence we can simply take the sum of the roots of the quadratic equation in k, 
Which in a standard quadratic equation of the form $$ax^2+bx+c$$ is $$-\frac{b}{a}$$

Here, the sum of the real values of k is $$-\frac{2}{3}$$

We are told that, for any natural number $$n_{1}$$ let $$a_{n}$$ be the largest integer not exceeding $$\sqrt{n}$$

So for n=1, the largest integer not exceeding $$\sqrt{1}$$ will be 1
For n=2, the largest integer not exceeding $$\sqrt{2}$$ will be 1
For n=3, the largest integer not exceeding $$\sqrt{3}$$ will be 1
For n=4, the largest integer not exceeding $$\sqrt{4}$$ will be 2

We see a pattern here regarding the squares of the numbers, 
Listing down all the perfect squares, 
1, 4, 9, 16, 25, 36, 49, 64, ...
We see that the difference between 4 and 1 is 3 and there were three natural numbers in the given pattern with the value as 1, 
So we can write for the rest of the numbers as well, 
3 numbers will have value 1, giving a total value of 3
5 numbers will have value 2, giving a total value of 10
7 numbers will have value 3, giving a total value of 21
9 numbers will have value 4, giving a total value of 36
11 numbers will have value 5, giving a total value of 55
13 numbers will have value 6, giving a total value of 78
Now, only the values of $$a_{49},\ a_{50}$$ will have the value of 7, total value of 14. 

Adding these values, we get the total sum as 217, which is the answer. 

We are told that, the incomes of Kamal, Amal and Vimal are in the ratio 8 ∶ 6 ∶ 5
Lets assume them to be 80X, 60X, 50X respectively.

Money that each one of them pays towards rent, 
Kamal 15% which comes to 12X
Amal 12% which comes to 7.2X
Vimal 18% which comes to 9X
Total Rental expenditure will be, 28.2X

Their incomes increase by 10%, 12% and 15% respectively, 
Kamal 10% which comes to 88X
Amal 12% which comes to 67.2X
Vimal 15% which comes to 57.5X
Total income will be 212.7X

We are told the rent is the same, so it will be 28.2X

Percentage of income going towards rent in October will be, $$\dfrac{28.2X}{212.7X}=0.13258$$

Hence, the answer is 13.26%

When gives n distinct numbers, and asked to find the sum of the possible n distinct numbers that can be formed, 
We use the formula, $$\left(10^{n-1}+10^{n-2}+10^{n-3}+...\right)\left(\left(n-1\right)!\right)\left(Sum\ of\ numbers\right)$$

Inserting n=4, $$\left(1000+100+10+1\right)\left(3!\right)\left(a+b+c+d\right)$$
$$\left(6666\right)\left(a+b+c+d\right)$$
We are told that this value equals, $$153310+n$$
Since we are told that n is a single digit natural number, the total value cannot be that much greater and 6666 should perfectly divide $$153310+n$$

Upon dividing 153310 by 6666 we get the quotient as 22.99
Nearest value being 23, We take 6666x23 giving us the value 153318

Hence the value of n=8 and the value of a+b+c+d=23

Value of a+b+c+d+n=31. 

Drawing the figure described in the question, we get the following representation, 

Since E is the midpoint of CD, the length of each half will be 28cm. 

We need to find the incircle of the triangle ADE, which is a right angled triangle, we can do that with the formula, 
$$\dfrac{\left(a+b-h\right)}{2}$$

$$h^2=28^2+45^2$$
$$h^2=784+2025=2809$$
$$h=53$$

$$r=\dfrac{\left(28+45-53\right)}{2}=10$$

We have two inequalities,

$$y\ \ge\ x\ +\ 4$$

$$-4\ \le\ x^2\ +\ y^2\ +\ 4\left(x\ -\ y\right)\ \le\ 0$$

The second inequality can be written as two separate inequalities,

$$-4\ \le\ x^2\ +\ y^2\ +\ 4\left(x\ -\ y\right)$$ and $$\ x^2\ +\ y^2\ +\ 4\left(x\ -\ y\right)\ \le\ 0$$

The first inequality can be written as,

$$\ x^2\ +\ y^2\ +\ 4x\ -\ 4y\ +4\ \ge\ 0$$

$$\ x^2+\ 4x\ +4\ +\ y^2-\ 4y\ +4\ -4\ge\ 0$$

$$\left(x\ +\ 2\right)^2\ +\ \left(y\ -\ 2\right)^2\ \ge\ 4$$

The second inequality can be written as,

$$\ x^2\ +\ y^2\ +\ 4x\ -\ 4y\ \ \le\ \ 0$$

$$\ x^2+\ 4x\ +4\ +\ y^2-\ 4y\ +4\ -8\ \le\ \ 0$$

$$\left(x\ +\ 2\right)^2\ +\ \left(y\ -\ 2\right)^2\ \le\ \ 8$$

Representing all three inequalities in the graph, we get the graph as shown above, and we must calculate the intersection of all three inequalities,

We can see that the line passes through the centre of both circles (-2, 2), and the area obtained from the second inequality is the area between the two circles. So, the area of intersection of all three graphs is half of the area between both the circles as the line divides the circle in half, and we must only consider the area above the line as per the given inequality. We know that the area of the bigger circle is $$\sqrt{\ 8}$$ and the area of the smaller circle is $$\sqrt{\ 4}$$ from the equations of the circles as we know that equation of circle as $$\left(x\ -\ a\right)^2\ +\ \left(y\ -\ b\right)^2\ =\ \left(radius\right)^2$$ where (a,b) is the centre of the circle.

The area of intersection 

= $$\dfrac{1}{2}\left(Area\ of\ bigger\ circle\ -\ Area\ of\ smaller\ circle\right)$$

= $$\dfrac{1}{2}\left(\pi\ \left(\sqrt{\ 8}\right)^2\ -\pi\ \left(\sqrt{\ 4}\right)^2\right)$$

= $$\dfrac{1}{2}\left(8\pi\ \ -4\pi\right)$$

= $$\dfrac{1}{2}\left(4\pi\ \right)$$

= $$2\pi\ $$

Therefore, the correct answer is option A.

Using the logarithmic property that $$\log_{a^p}b=\frac{1}{p}\log_ab$$

$$4 \log_{10} x + 4 \log_{100} x + 8 \log_{1000} x = 13$$
Can be written as 
$$4\log_{10}x+2\log_{10}x+\frac{8}{3}\log_{10}x=13$$
$$\frac{26}{3}\log_{10}x=13$$
$$\log_{10}x=1.5$$
$$x=10^{1.5}$$
$$x=\sqrt{1000}$$
$$\left[\sqrt{1000}\right]=31$$
Where [.] is Greatest Integer Function since that is what is asked in the question, 
31 is the greatest integer that does not exceed x. 

Let us fix the Cost Price of the product to be X, and the Selling Price of the product to be 1.4X, since it is given that it is fixed to have a profit of 40%. 
If the CP has been 40% less, making the CP 0.6X, 
And the selling price is 5 rupees less, making it 1.4X-5
Profit will be 50%, 
So, $$1.5\left(0.6X\right)=1.4X-5$$
$$0.9X=1.4X-5$$
$$0.5X=5$$
$$X=10$$

Original selling price will be 14.