XAT 2025 Quantitative Aptitude & Data Interpretation
Q1–10 of 28There are 25 rooms in a hotel. Each room can accommodate at most three people. For each room, the single occupancy charge is Rs. 2000 per day, the double occupancy charge is Rs. 3000 per day, and the triple occupancy charge is Rs. 3500 per day.
Q.If there are 55 people staying in the hotel today, what is the maximum possible revenue from room occupancy charges today?
To maximize revenue with 55 people, the hotel should prioritize room configurations with the highest total return per person/room mix. A triple occupancy room earns Rs. 3500, double earns Rs. 3000, and single Rs. 2000. Using all 25 rooms is optimal. The strategy is to fill all 25 rooms as doubles first (50 people) for Rs. 75,000 (25 x 3000). The remaining 5 people are added to 5 of these rooms to make them triples. Each upgrade adds Rs. 500 ($3500-3000$). Total revenue = 75,000 + 2,500 = Rs.77,500.
Ramesh bought a mobile from a local store. He paid 1/6 of the price via UPI and 1/3 of the price via cash. He agreed to pay the balance amount a year later. While paying back the balance amount, Ramesh paid 10% interest on the balance amount.
Q.If the interest paid was Rs. 6000, what was the original price of the mobile?
Ramesh paid 1/6 + 1/3 = 1/2 of the price upfront, leaving a balance of 1/2 of the original price. He paid 10% interest on this balance, which was Rs. 6000. Since 10\% of the balance is 6000, the balance must be Rs. 60,000. Since this balance represents half the total price, the original price was 60,000 X 2 = Rs.1,20,000.
Adu and Amu have bought two pieces of land on the Moon from an e-store. Both the pieces of land have the same perimeters, but Adu’s piece of land is in the shape of a square, while Amu’s piece of land is in the shape of a circle.
Q.The ratio of the areas of Adu’s piece of land to Amu’s piece of land is:
For a square and circle with equal perimeters P:
Square side s = P/4,
Area As = P²/16.
Circle radius r = P/2π,
Area Ac = π(P/2π)² = P²/4π.
The ratio As : Ac is (P²/16) : (P²/4π) = 1/4 : 1/π = π : 4.
The market value of beams, made of a rare metal, has a unique property: the market value of any such beam is proportional to the square of its length. Due to an accident, one such beam got broken into two pieces having lengths in the ratio 4:9.
Q.Considering each broken piece as a separate beam, how much gain or loss, with respect to the market value of the original beam before the accident, is incurred?
The percentage loss is $(72/169) \times 100 \approx 42.60\%$.The value V is proportional to L². Let the original length be 13 (4+9). Original value ∝ 13² = 169. The broken pieces have values ∝ 4² = 16 and 9² = 81. Total new value is 16 + 81 = 97. The loss is 169 - 97 = 72. The percentage loss is (72/169) × 100 ≈ 42.60%.
ABCD is a rectangle, where the coordinates of C and D are (−2,0) and (2,0), respectively.
Q.If the area of the rectangle is 24, which of the following is a possible equation representing the line AB?
Points C(-2,0) and D(2,0) lie on the x-axis with a distance of 4 units. For the rectangle area to be 24, the height must be 24/4 = 6. The line AB is parallel to CD at a distance of 6 units, so its equation is y=6.
Consider the quadratic function f(x) = a
The probability is 7/15.For f(x) = ax² + bx + a to have irrational roots, the discriminant D = b² - 4a² must be positive and non-square. This implies b > 2a. Counting valid pairs (a,b) from integers 1 to 9 yields 15 pairs. The condition a + b > 9 is met by 7 of these pairs (e.g., 1,9; 2,8; 3,7; etc.).
A farmer has a quadrilateral parcel of land with a perimeter of 700 feet. Two opposite angles of that parcel of land are right angles, while the remaining two are not. The farmer wants to do organic farming on that parcel of land. The cost of organic farming is Rs. 400 per square foot.
Consider the following two additional pieces of information:
The length of one of the sides of that parcel of land is 110 feet.
The distance between the two corner points where the non-perpendicular sides of that parcel of land intersect is 265 feet.
Q.To determine the amount of money the farmer needs to spend to do organic farming on the entire parcel of land, which of the above additional pieces of information are MINIMALLY SUFFICIENT?
The problem involves a quadrilateral with a fixed perimeter and two right angles. Neither the length of one side (I) nor the intersection distance (II) alone defines the shape's area uniquely. However, combining both constraints allows for the unique determination of the sides and area, solving the problem.
A straight line L₁ has the equation y = k(x - 1), where k is some real number. The straight line L₁ intersects another straight line L₂ at the point (5, 8).
Q.If L₂ has a slope of 1, which of the following is definitely FALSE?
Line L1 passes through (1,0) and (5,8), giving the equation y = 2x - 2 (y-intercept is -2). Line L2 has a slope of 1 through (5,8), giving the equation y = x + 3 (y-intercept is 3). The distance between the y-intercepts is |3 - (-2)| = 5. Option C claims it is 6, which is definitely false.
For how many distinct real values of x does the equation below hold true? (Consider a > 0)
Simplifying the terms (e.g., log_a(16) / log_a(32) = 4/5) leads to the quadratic equation 4x² - 5x - 6 = 0. This yields roots x = 2 and x = -0.75. Since x appears only in polynomial positions or squared inside the log, both distinct real roots are valid.
In a computer game, each move requires pressing a button. When the button is pressed for the first time, as a move, the computer randomly chooses a cell from a 4x4 grid of sixteen cells and puts an "X" mark on that cell. When the button is pressed subsequently, the computer randomly chooses a cell from the remaining unmarked cells and puts an "X" mark on that cell. This goes on till the end of the game. The game ends when either all the cells in any one row, or all the cells in any one column, are marked with "X".
Q.What is the maximum possible number of times a player has to press the button to finish the game?
To prolong the game, a player avoids completing any row or column. In a 4×4 grid, one can mark at most 3 cells in every row and every column (12 marks total). The 13th mark is forced to land in an empty spot that completes both a row and a column, ending the game.