XAT 2025 Question Paper

So, there are 25 rooms in total and 55 people need to be occupied in the these 25 rooms. We need to maximize the revenue.

The cost of a single, double and triple occupancy room is Rs. 2,000, Rs. 3,000 and Rs. 3,500 respectively. 

Now, if we look at the per person cost from a single, double and triple occupancy room, it will be Rs. 2,000, Rs. 1,500 and Rs. 1,166.67

Now, we clearly see that the per person cost is maximum for single occupancy room but we know that there are only 25 rooms which are not sufficient. Hence, we will aim to adjust all the 55 people to the maximum possible single occupancy then double occupancy and then triple occupancy room.

Let the number of single, double and triple occupancy rooms used are $$x$$, $$y$$ and $$z$$ respectively. 

We know $$x+y+z=25$$ as the total number of rooms are 25. 

Further, $$x+2y+3z=55$$ I.e. total number of people.

If z = 1, y = 28 which is not possible as the number of rooms are limited to 25. 

If z = 2, y = 26 which is again not possible. 

If z = 3, y = 24 which is again not possible. 

If z = 4, y = 22 which is again not possible. 

If z = 5, y = 20 which is possible. 

Hence, to maximize the revenue, we have to use 20 room of double occupancy and 5 rooms of triple occupancy. 

Hence, the total revenue will be $$20\times3,000\ +5\times3.500=77,500$$

Given that the beam got broken into two pieces, with length in a ratio of 4 : 9.

Let the lengths of the new beams be $$4x$$ and $$9x$$ respectively.

So, the length of the original beam is $$13x$$.

Now, given the value is proportional to the square of its length.

Value of the original beam = $$k\left(13x\right)^2=169kx^2$$, where $$k$$ is the constant of proportionality.

The value of new beams is $$k\left(4x\right)^2+k\left(9x\right)^2=16kx^2+81kx^2=97kx^2$$

Hence, the gain/loss with respect to the original beam is $$169kx^2-97kx^2=72kx^2$$

In percentage terms, Loss % = $$\dfrac{72kx^2}{169kx^2}\times100=42.60\%$$

Hence, the answer is 42.60% loss

CD is on of the side of the rectangle. So, two of the remaining 3 sides will be perpendicular to CD.

 Length of CD is 4. We are told that area is 24.

We know area of rectangle is lb. 4*l = 24

l = 6

So, the other two points should be such that when jined with C and D respectively, should be perpendicular to CD and lie at a distance of 6.

There will be two pair of such points. One pair is (-2,6), (2,6) and the other pair is (-2,-6), (2,-6)

So, there are two possible equations of AB. One is y = 6 and the other is y = -6.

It is given that the perimeters of areas are equal for both of them. Let the equal perimeter be P.

Adu's piece is a square, and the perimeter of the square =  4s = P

Side of the square $$s=\ \dfrac{P}{4}$$

Amu's piece is a circle, and the perimeter of the circle = $$2\pi\ r=\ P$$

The radius of the circle $$r=\ \dfrac{P}{2\pi\ }$$

The ratio of areas is,

$$s^2\ :\ \pi\ r^2$$  $$=\left(\dfrac{P}{4}\right)^{^2}:\ \pi\ \left(\dfrac{P}{2\pi\ }\right)^{^2}$$ $$=\ \dfrac{P^2}{16}\ :\ \pi\times\dfrac{P^2}{4\pi^2\ }$$  $$=\ \dfrac{1}{4}\ :\ \dfrac{1}{\ \pi\ }\ =\ $$  $$\pi\ \ :\ \ 4$$

Hence, the correct answer is option D.

Let the total price of the mobile phone is Rs. $$60X$$.

Ramesh paid 1/6 of the price via UPI i.e. $$60X\times\ \dfrac{1}{6}=10X$$

Further, he paid 1/3 of the price via cash i.e. $$60X\times\ \dfrac{1}{3}=20X$$

Remaining amount to be paid by Ramesh = $$60X-10X-20X=30X$$

Further, 10% interest charged on the balance amount = Rs. 6,000

Or, we can say, $$10\%$$ of $$30X=6,000$$

i.e. $$3X=6,000$$

Or, $$X=2,000$$

Since the total price of the mobile phone was 60X, the actual cost is $$60\times\ 2,000\ =\ Rs.\ 1,20,000$$

Given that the equation of line $$L_1$$ is $$y = k(x - 1)$$ and it intersects line $$L_2$$ at (5, 8).

So, the point (5, 8) must satisfy the equation of line $$L_1$$

We get $$8=k\left(5-1\right)$$

or $$k=2$$

Hence, the equation of line $$L_1$$ is $$y = 2(x - 1)$$ . . . (1)

Now, $$L_2$$ has a slope of 1 

Using slope form, the equation of line is $$y=mx+c$$, where $$m$$ is the slope and $$c$$ is the y-intercept

So, the equation of line $$L_2$$ is $$y=x+c$$

Now, as it passes through the point (5, 8), the equation must satisfy

Hence, $$8=5+c$$

or, $$c=3$$

Hence, the equation of line $$L_2$$ is $$y=x+3$$ . . . (2)

Plotting both lines on the graph, we get

From the graph we can clearly see that the distance between the y-intercepts of the two lines is 5 not 6. 

Hence, the statement the distance between the y-intercepts of the two lines is 6 is definitely false.

In a 4*4 gird, If all 4 diagonal cells are ingored and the remaining cells are marked, no row or column will be filled. So, we can mark 12 cells without finishing the game. When we mark the 13th one, the game gets finished.

The given expression is,  $$\sqrt[3]{(3)^a(21)^{(3a - b)}(49)^{(2b + c)}}$$ and it can also be written as,

$$\sqrt[3]{(3)^a\ \left(3\right)^{3a\ -\ b}\ \ \left(7\right)^{3a\ -\ b}\ (7)^{4b+2c}}$$ $$=\ \sqrt[3]{\ \left(3\right)^{4a\ -\ b}\ \ (7)^{3a\ +\ 3b+2c}}$$

For the above expression to be a positive integer, the power of the expression must be an integer after applying the cube root for the expression inside. This means that the power of the expression inside must be a multiple of 3.

The expression is,

$$\sqrt[3]{\ \left(3\right)^{4a\ -\ b}\ \ (7)^{3a\ +\ 3b+2c}}\ =\ 3^{\frac{\left(4a\ -\ b\right)}{3}}\ \times\ 7^{\frac{\left(3a\ +\ 3b\ +\ 2c\right)}{3}}$$

The value   $$\dfrac{4a\ -\ b}{3}=\ \dfrac{3a\ +\ a\ -\ b}{3}\ =\ a\ +\dfrac{a\ -\ b}{3}$$   must be an integer which means that a - b must be a multiple of 3.

Similarly, the value   $$\dfrac{3a\ +\ 3b\ +\ 2c}{3}=\ a\ +\ b\ +\ \dfrac{2c}{3}$$   must also be an integer which gives us the condition that c must be a multiple of 3.

So, the necessary and sufficient conditions are a - b and c must be a multiple of 3.

Hence, the correct answer is option E.

According to the question, we have the following figure:

According to the question, we have a frustrum and a semisphere with flat surface and we need to find the curved surface area of the frustrum along with the curved surface area of the hemisphere and the bottom flat area of the frustrum. 

Curved surface area of the Frustrum = $$\pi\ \times\left(r_1+r_2\right)\times l$$

Here, r1 and r2 refers to the two different radii and l refers to the slant height. 

$$l=\sqrt{h^2+\left(r_1-r_2\right)^2}$$

$$l=\sqrt{1600+100}=\sqrt{1700}=10\sqrt{17}$$

Hence, curved surface area of the frustrum = $$\pi\ \times\left(20+30\right)\times10\sqrt{17}$$

$$\pi\ \times50\times10\sqrt{17}$$

$$500\sqrt{17}\pi\ $$

Now, curved surface area of the hemisphere = $$2\pi r^2=2\times\pi\ \times\ \left(20\right)^2\ =\ 800\pi\ $$

And the flat bottom area of the frustrum = $$\pi r^2=\pi\ \times\left(30\right)^2=900\pi\ $$

Hence, the total surface area of the trophy =$$500\sqrt{17}\pi\ +\ 800\pi\ +\ 900\pi\ =500\sqrt{17}\pi\ +\ 1700\pi\ $$

The cost of gold coating is Rs. 40 per square cm

Hence, the total cost of coating = $$40\times\ \left(500\sqrt{17}\pi\ +\ 1700\pi\ \right)$$

$$40\times\ \dfrac{22}{7}\times\ \left(500\sqrt{17}\ +\ 1700\ \right)$$

$$40\times\ \dfrac{22}{7}\times\ \left(2061.55\ +\ 1700\ \right)$$

$$40\times\ \dfrac{22}{7}\times\ 3761.55=Rs.\ 4,72,880.571\ \approx\ Rs.\ 4,73,000$$

According to the question, we can draw the following table:

Here, EC refers to the shelf which is 7 feet and AB refers to the height of the robot. Here, AE reflects the hand of the robot. Since we need to find the minimum possible height, we need to maximize the length of the hands of the robot. The hands of the robot can be tilted to the maximum angle of 60, hence, we will assume that the hands of the robot tilt to exactly $$60^{\circ\ }$$.

Let the height of the robot i.e. AB be $$x$$ feet.

Now, we need to find the value of x. 

Since, BC = 1 foot, AD is also 1 foot. 

In triangle ADE, $$\tan\ 60^{\circ\ }=\dfrac{ED}{AD}$$

$$\tan\ 60^{\circ\ }=\dfrac{7-x}{1}$$

$$\sqrt{3}=\dfrac{7-x}{1}$$

$$x=7-\sqrt{3}$$