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Quant XAT 2024 Question paper with solution

XAT 2024 Question Paper – QA & DI

XAT 2024 – QA & DI

  1. Q.1 College Specialization

    In a small college, students are allowed to take only one specialization. Traditionally, only two specializations are offered: Science and Arts. Students enrolled to specialize in Science must take Physics and Mathematics subjects, while students enrolled to specialize in Arts must take Economics and Political Science subjects. Students enrolled in Science are not allowed to take either Economics or Political Science, while students enrolled in Arts are not allowed to take either Physics or Mathematics.

    Recently, the college has started a third specialization called MatEco that requires students to take Economics and Mathematics. However, MatEco students would not be allowed to take either Physics or Political Science. When the college opens this new specialization for enrolment, it allows students, originally enrolled in Science or Arts, to switch to MatEco.

    From among the students originally enrolled in Arts, 20 students switch to MatEco. This makes the number of Science students twice the number of Arts students. After this, from among the students who originally enrolled in Science, 45 students switch to MatEco. This makes the number of Arts students twice the number of Science students.

    In total, how many students, from among those originally enrolled in Science or Arts, are now taking Economics?

    1. None of the remaining options is correct.
    2. 95
    3. 45
    4. 65
    5. 80
    Difficulty: Moderate
    Correct Option: Choice B (95)
    Rationale: Let the number of Arts students be A and Science students be S.
    From the first switch (20 Arts to MatEco): \(S = 2(A – 20) \implies S = 2A – 40\).
    From the second switch (45 Science to MatEco): \((A – 20) = 2(S – 45) \implies A – 20 = 2S – 90 \implies A = 2S – 70\).
    Substitute (1) into (2): \(A = 2(2A – 40) – 70 \implies A = 4A – 80 – 70 \implies 3A = 150 \implies A = 50\).
    Then \(S = 2(50) – 40 = 60\).
    Total MatEco = 20 (Arts) + 45 (Science) = 65.
    Remaining Arts = 50 – 20 = 30.
    Students taking Economics = MatEco + Arts = 65 + 30 = 95.

  2. Q.2 Movie Theatre Profit

    The cost of running a movie theatre is Rs. 10,000 per day, plus additional Rs. 5000 per show. The theatre has 200 seats. A new movie released on Friday. There were three shows, where the ticket price was Rs. 250 each for the first two shows and Rs. 200 for the late-night show.

    For all shows together, total occupancy was 80%. What was the maximum amount of profit possible?

    1. Rs. 95,000
    2. Rs. 87,000
    3. Rs. 1,16,000
    4. Rs. 91,000
    5. Rs. 1,20,000
    Difficulty: Easy
    Correct Option: Choice D (Rs. 91,000)
    Rationale: Total Capacity = 200 * 3 = 600. Occupancy = 80% of 600 = 480 tickets.
    Maximize Revenue by selling higher priced tickets first.
    Max @ Rs. 250 = 400 (Show 1 & 2 full). Remaining @ Rs. 200 = 80.
    Revenue = (400 * 250) + (80 * 200) = 1,00,000 + 16,000 = 1,16,000.
    Cost = 10,000 (Fixed) + 3 * 5,000 (Variable) = 25,000.
    Profit = 1,16,000 – 25,000 = Rs. 91,000.

  3. Q.3 Flight Delay

    A flight, traveling to a destination 11,200 kms away, was supposed to take off at 6:30 AM. Due to bad weather, the departure of the flight got delayed by three hours. The pilot increased the average speed of the airplane by 100 km/hr from the initially planned average speed, to reduce the overall delay to one hour.

    Had the pilot increased the average speed by 350 km/hr from the initially planned average speed, when would have the flight reached its destination?

    1. 11:30 PM
    2. 5:10 PM
    3. 8:10 PM
    4. 10:36 PM
    5. 7:50 PM
    Difficulty: Moderate
    Correct Option: Choice C (8:10 PM)
    Rationale: Solving for planned Time (T) and Speed (S):
    (S+100)(T-2) = 11200. Solving leads to T=16 hrs, S=700 km/hr.
    New Speed = 700 + 350 = 1050 km/hr.
    New Time = 11200 / 1050 = 10 hr 40 min.
    Departure (3 hrs late) = 9:30 AM.
    Arrival = 9:30 AM + 10h 40m = 8:10 PM.

  4. Q.4 Log Equation

    Consider the equation \(\log _5(x-2)=2 \log _{25}(2 x-4)\), where x is a real number.

    For how many different values of x does the given equation hold?

    1. 2
    2. 4
    3. 0
    4. 1
    5. Infinitely many
    Difficulty: Moderate
    Correct Option: Choice C (0)
    Rationale: \(\log_{25}(A) = \frac{1}{2}\log_5(A)\).
    Equation becomes: \(\log_5(x-2) = \log_5(2x-4)\).
    \(x – 2 = 2x – 4 \implies x = 2\).
    Check validity: For \(\log(x-2)\) to be defined, \(x-2 > 0\).
    If \(x=2\), \(x-2=0\), undefined. No solution.

  5. Q.5 AP Students

    In a school, the number of students in each class, from Class I to X, in that order, are in an arithmetic progression. The total number of students from Class I to V is twice the total number of students from Class VI to X.

    If the total number of students from Class I to IV is 462, how many students are there in Class VI?

    1. 77
    2. 83
    3. 93
    4. None of the remaining options is correct.
    5. 88
    Difficulty: Moderate
    Correct Option: Choice A (77)
    Rationale: Using Sum formulas for AP:
    Sum(1-5) = 2 * Sum(6-10) leads to relation \(a = -12d\).
    Sum(1-4) = 462 leads to \(2a + 3d = 231\).
    Solving: \(d = -11, a = 132\).
    Class VI term = \(a + 5d = 132 – 55 = 77\).

  6. Q.6 Football Ground

    A group of boys is practising football… Raju and Ratan are standing at the two opposite mid-points of the two shorter sides… Consider the following two additional pieces of information:

    I. The dimension of the ground is 80 metres × 50 metres.
    II. The area of the triangle formed by Raju, Rivu and Ratan is 1000 square metres.

    Consider the problem of computing the following: how many seconds does it take for Raju to get the ball back since he passed it to Rivu? Choose the correct option.

    1. I alone is sufficient to solve the problem.
    2. The problem cannot be solved even with both I and II.
    3. I and II both are required to solve the problem.
    4. II alone is sufficient to solve the problem.
    5. Either of I or II, by itself, is sufficient to solve the problem.
    Difficulty: Moderate
    Correct Option: Choice A
    Rationale: Statement I provides the coordinates of Raju and Ratan and the y-coordinate of Rivu. Using the property that the paths are perpendicular (product of slopes = -1), we can solve for Rivu’s position and thus the distance.
    Statement II provides Area = 1000. Area = 0.5 * Base * Height = 0.5 * 80 * 25 = 1000. This is always true for the mid-point configuration and provides no new information to locate Rivu.
    Thus, I alone is sufficient.

  7. Q.7 Biryani Stall

    FS food stall sells only chicken biryani… If FS fixes a selling price of Rs. 160 per plate, 300 plates of biriyani are sold. For each increase in the selling price by Rs. 10 per plate, 10 fewer plates are sold… Cost is Rs. 120… Max capacity 400.

    If the selling price… can only be a multiple of Rs. 10, then what is the maximum profit that FS can achieve in a day?

    1. Rs. 41,400
    2. None of the remaining options is correct.
    3. Rs. 28,900
    4. Rs. 52,900
    5. Rs. 25,300
    Difficulty: Easy
    Correct Option: Choice C (Rs. 28,900)
    Rationale: Profit Function \(P(x) = (160 + 10x – 120)(300 – 10x) = 100(4+x)(30-x)\).
    Max at \(x = (-4+30)/2 = 13\).
    Price = 290, Quantity = 170.
    Max Profit = \((290-120) * 170 = 170 * 170 = 28,900\).

  8. Q.8 Coinciding Lines

    Consider the system of two linear equations as follows: 3x + 21y + p = 0; and qx + ry – 7 = 0…

    Which of the following statements DEFINITELY CONTRADICTS the fact that the lines represented by the two equations are coinciding?

    1. p cannot be 0
    2. p and q must have opposite signs
    3. The largest among p, q, and r is q
    4. r and q must have same signs
    5. The smallest among p, q, and r is r
    Difficulty: Moderate
    Correct Option: Choice C
    Rationale: For coinciding lines: \(3/q = 21/r = p/(-7) = k\).
    \(r = 7q\). Thus \(|r| > |q|\).
    If \(q\) and \(r\) are positive, \(r\) is largest. If negative, \(q\) is greater (closer to 0) but \(p\) would be positive. Testing values shows \(q\) can never be the largest value among p, q, r.

  9. Q.9 King’s Jewels

    A king has distributed all his rare jewels in three boxes. The first box contains 1/3 of the rare jewels, while the second box contains k/5 of the rare jewels… The third box contains 66 rare jewels.

    How many rare jewels does the king have?

    1. 240
    2. 1080
    3. Cannot be determined uniquely from the given information.
    4. 990
    5. 660
    Difficulty: Easy
    Correct Option: Choice D (990)
    Rationale: \(N = N/3 + kN/5 + 66 \implies N(10 – 3k) = 990\).
    Since \(N\) must be an integer and \(k\) is integer, testing \(k=1, 2, 3\).
    Only \(k=3\) yields an integer \(N = 990\).

  10. Q.10 Menu Combinations

    A local restaurant has 16 vegetarian items and 9 non-vegetarian items… Rohit and his friends… Bela and her friends… The number of item combinations that Rohit… was 12 times the number… Bela…

    How many menu items contain gluten?

    1. 5
    2. 1
    3. 4
    4. 2
    5. 3
    Difficulty: Difficult
    Correct Option: Choice D (2)
    Rationale: Let \(v_2\) and \(n_2\) be gluten-free items.
    Rohit: \(^{16}C_2 \times ^9C_3 = 10080\).
    Bela: \(^{v_2}C_2 \times ^{n_2}C_1\).
    Equation: \(10080 = 12 \times (^{v_2}C_2 \times n_2)\).
    Solving for integers yields \(v_2 = 16\) and \(n_2 = 7\).
    Total Gluten = (16-16) + (9-7) = 2.

  11. Q.11 Number Divisibility

    Consider a 4-digit number of the form abbb…

    Which of the following conditions is both NECESSARY and SUFFICIENT to ensure that the 4-digit number is divisible by a?

    1. b is equal to 0
    2. b is divisible by a
    3. 9b is divisible by a
    4. 21b is divisible by a
    5. 3b is divisible by a
    Difficulty: Moderate
    Correct Option: Choice E
    Rationale: \(N = 1000a + 111b\).
    \(N/a = 1000 + (111b)/a\).
    \(111b\) must be div by \(a\). \(111 = 3 \times 37\).
    Since \(a\) is single digit, \(\gcd(a, 37)=1\).
    Thus, \(3b\) must be divisible by \(a\).

  12. Q.12 LCM and GCD

    The least common multiple of a number and 990 is 6930. The greatest common divisor of that number and 550 is 110.

    What is the sum of the digits of the least possible value of that number?

    1. 14
    2. None of the remaining options is correct.
    3. 9
    4. 6
    5. 18
    Difficulty: Moderate
    Correct Option: Choice A (14)
    Rationale: Comparing prime factorizations of 990, 6930, 550, and 110.
    The number \(n\) must be 770.
    Sum of digits = \(7+7+0 = 14\).

  13. Q.13 Polynomial Roots

    The roots of the polynomial \(P(x)=2 x^3-11 x^2+17 x-6\) are the radii of three concentric circles.

    The ratio of their area, when arranged from the largest to the smallest, is:

    1. None of the remaining options is correct.
    2. 6:2:1
    3. 16:6:3
    4. 36:16:1
    5. 9:4:1
    Difficulty: Easy
    Correct Option: Choice D (36:16:1)
    Rationale: Roots are found to be 3, 2, and 0.5.
    Ratio of areas (squares of radii) = \(3^2 : 2^2 : 0.5^2 = 9 : 4 : 0.25\).
    Multiply by 4: \(36 : 16 : 1\).

  14. Q.14 Farmer’s Pond

    A farmer has a triangular plot of land…

    If the market rate per square feet of land is Rs. 1400, how much does the farmer must pay to buy the land from his neighbour for the pond?

    1. Rs. 3,16,80,000
    2. Rs. 6,33,60,000
    3. Rs. 2,98,20,000
    4. Rs. 7,42,80,000
    5. Rs. 4,25,60,000
    Difficulty: Moderate
    Correct Option: Choice A
    Rationale: Calculating the radius of the inscribed semi-circle on the base gives \(r = 120\) ft.
    Area to buy (outside triangle) = Area of semi-circle = \(0.5 \pi r^2 \approx 22,619.5\).
    Cost = Area * 1400 ≈ 3,16,67,250.

  15. Q.15 Circles in Triangle

    Consider a right-angled triangle ABC…
    If AB = 18 cm and BC = 24 cm, then find the value of r.

    1. 3.5 cm
    2. 4.5 cm
    3. 4 cm
    4. 3 cm
    5. None of the remaining options is correct.
    Difficulty: Moderate
    Correct Option: Choice C (4 cm)
    Rationale: Using coordinate geometry or similarity, we determine the centers of the two circles and set the distance from the second center to the hypotenuse equal to \(r\). Solving the equation gives \(r=4\).

  16. Q.16 Unused Coupon

    Four customers used four different discount coupons… total discount of Rs. 710.

    Which discount coupon was not used?

    1. Coupon D
    2. Coupon A
    3. Coupon E
    4. Coupon B
    5. Coupon C
    Difficulty: Moderate
    Correct Option: Choice C (Coupon E)
    Rationale: Sum of max discounts A+B+C+D = 250+300+100+100 = 750.
    With D providing partial discount (60), sum matches 710.
    Combinations involving E do not reach 710 perfectly.

  17. Q.17 Minimum Spend

    Four customers used four different discount coupons… nobody used any discount coupon sub-optimally.

    What was the minimum combined spend (before application of any discount)?

    1. Rs. 1550
    2. Rs. 2300
    3. Rs. 2250
    4. Rs. 2350
    5. Rs. 2500
    Difficulty: Moderate
    Correct Option: Choice E (Rs. 2500)
    Rationale: Adding the minimum threshold spends where each coupon becomes the optimal choice: A(1200) + B(500) + C(600) + E(200) = 2500.

  18. Q.18 Family Purchase

    A family wanted to purchase four products worth Rs. 1000 each, and another product worth Rs. 300…

    What was the maximum discount that they could obtain for their purchase?

    1. Rs. 700
    2. None of the remaining options is correct.
    3. Rs. 650
    4. Rs. 645
    5. Rs. 600
    Difficulty: Moderate
    Correct Option: Choice A (Rs. 700)
    Rationale: Strategic grouping of items:
    1. (1000+1000) using B (Max 300) = 300 off.
    2. (1000+300) using A (Flat 250) = 250 off.
    3. (1000) using B (15%) = 150 off.
    Total = 700.

  19. Q.19 Vegetable Purchase

    Aman has come to the market with Rs. 100… Aman decides to buy only onion…

    How much money will he be left with after the purchase?

    1. Rs. 7
    2. Rs. 12
    3. Re. 1
    4. Rs. 5
    5. Rs. 9
    Difficulty: Easy
    Correct Option: Choice C (Re. 1)
    Rationale: Solving the linear equations gives prices: C=12, O=9, P=5.
    buying max Onion with 100: \(100 = 9 \times 11 + 1\).
    Balance is 1.

  20. Q.20 Probability

    Aman decides to buy only onion and potato… money left… insufficient to buy a full kilogram…

    If all such permissible combinations of purchases are equally likely, what is the probability that Aman buys more onion than potato?

    1. 5/6
    2. 2/9
    3. 4/10
    4. 3/10
    5. 7/20
    Difficulty: Moderate
    Correct Option: Choice D (3/10)
    Rationale: Checking all pairs \((O, P)\) such that cost is between 96 and 100. There are 10 valid combinations. Only 3 of them have \(O > P\).

  21. Q.21 Run Rate N

    The following table indicates the run rate…
    What is the value of N?

    1. 14
    2. 7
    3. 13
    4. 9
    5. 12
    Difficulty: Moderate
    Correct Option: Choice B (7)
    Rationale: Based on data analysis of the run rates and integer score constraints, \(N=7\) is the only value fitting the sequence.

  22. Q.22 Score 22

    In which of these pairs of over numbers, the team could have scored 22 runs in total?

    1. 8 and 9
    2. 6 and 7
    3. 7 and 8
    4. 9 and 10
    5. 10 and 11
    Difficulty: Moderate
    Correct Option: Choice D (9 and 10)
    Rationale: With \(N=7\), calculating the scores for each over reveals that the sum of scores for overs 9 and 10 allows for 22 runs.

  23. Q.23 Least Runs

    In which of the following over numbers, the team MUST have scored the least number of runs?

    1. 10
    2. 8
    3. 7
    4. 11
    5. 9
    Difficulty: Moderate
    Correct Option: Choice C (7)
    Rationale: Calculations show that over 7 corresponds to the dip in performance required by the run rate data.

  24. Q.24 Value of x

    41 applicants…
    What BEST can be concluded about the value of x?

    1. 0, 1 or 2
    2. 2 only
    3. 0 or 1 only
    4. 1 only
    5. 1 or 2 only
    Difficulty: Difficult
    Correct Option: Choice B (2 only)
    Rationale: Solving the set theory constraints provided by the table restricts the intersection of all three sets (\(x\)) to exactly 2.

  25. Q.25 No Expertise

    How many applicants DID NOT have advanced expertise in any of the three given fields?

    1. 27
    2. 28
    3. Cannot be determined uniquely from the given information.
    4. 26
    5. 25
    Difficulty: Difficult
    Correct Option: Choice D (26)
    Rationale: Total – Union of the three sets = 41 – 15 = 26.

  26. Q.26 Hinge Data B

    A student has surveyed thirteen of her teachers… Minimum: 2, Lower Hinge: 6.5, Median: 12, Upper Hinge: 21, Maximum: 29. Two values smudged A and B (A < B).
    Which of the following is a possible value of B?

    1. 13
    2. 8
    3. 29
    4. 6
    5. 2
    Difficulty: Moderate
    Correct Option: Choice B (8)
    Rationale: To satisfy the Lower Hinge (6.5) and Median (12), B must fall in the range [8, 12]. Only 8 is in the options.

  27. Q.27 Average Experience

    Based on the information recorded, which of the following can be the average work experience of the thirteen teachers?

    1. 13.5
    2. 12.5
    3. 12
    4. 14
    5. 13
    Difficulty: Easy
    Correct Option: Choice D (14)
    Rationale: Using the valid values for A and B, the sum is 181. Average = 181/13 ≈ 13.9. Closest is 14.

  28. Q.28 Correct B Value

    The student found that one of the eleven recorded work experience values… was recorded wrongly as half of its correct value… Recalculated average was 15.
    What is the value of B?

    1. 7
    2. 9
    3. 12
    4. Cannot be determined from the given information.
    5. 10
    Difficulty: Difficult
    Correct Option: Choice E (10)
    Rationale: The new average requires a sum of 195 (increase of 14). Testing the error correction logic against the constraints for B reveals B=10 is the only consistent solution.

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