CUET Mathematics Question Paper 2024 SET C Download Solutions and Answer Key
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CUET 2024 Mathematics Question Paper (SET C) with Answer Key PDF
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CUET 2024 Mathematics Question Paper (SET C) with Solution
| Question | Answer | Solution |
|---|---|---|
| Question 1: The second-order derivative of which of the following functions is 5x? 1. 5xln(5) 2. 5x(ln(5))2 3. 5x/(ln5) 4. 5x/(ln(5))2 |
(4) 5x/(ln(5))2 | Let f(x) = 5x/(ln(5))2. Then f'(x) = 5xln(5)/(ln(5))2 = 5x/ln(5). And f''(x) = 5xln(5)/ln(5) = 5x. |
| Question 2: The degree of the differential equation (1-(dy/dx)2)3/2 = k(d2y/dx2) is: 1. 1 2. 2 3. 3 4. 4 |
(2) 2 | Squaring both sides gives (1-(dy/dx)2)3 = k2(d2y/dx2)2. The highest order derivative is d2y/dx2, and its power is 2, which is the degree. |
| Question 3: Evaluate the integral: ∫(xn+1/x(n-x))dx (1) ln(e)/n + C (2) (ln(e) * xn+1)/n + C (3) (ln(e) * xn+1)/nx + C (4) (πln(e))/xn-1 + C |
(1) (1/n) * ln|x/(n-x)| + C | Using partial fractions, the integrand can be rewritten as (n+1)/n * 1/x + (n+1)/n * 1/(n-x). Integrating term by term gives the answer. |
| Question 4: Evaluate the integral: ∫01((a - bx2)/(a + bx2)2) dx (1) (a-b)/(a+b) (2) 1/(a-b) (3) 1/(a+b) (4) (a+b)/(a-b) |
(1) (a-b)/a(a+b) | Substitute u = a + bx2, then du = 2bxdx. Change the limits accordingly and solve the simplified integral in terms of u. |
| Question 5: If A and B are symmetric matrices of the same order, then AB - BA is: (1) Symmetric Matrix (2) Zero Matrix (3) Skew-symmetric Matrix (4) Identity Matrix |
(3) Skew-symmetric Matrix | If A and B are symmetric, AT = A and BT = B. Now (AB - BA)T = (AB)T - (BA)T = BTAT - ATBT = BA - AB = -(AB - BA). Since the transpose is the negative of the original matrix, it is skew-symmetric. |
| Question 6: If A is a square matrix of order 4 and |A| = 4, then |2A| is: (1) 8 (2) 64 (3) 16 (4) 4 |
(2) 64 | If A is a square matrix of order n, then |kA| = kn|A|. Here, n=4 and k=2. Therefore, |2A| = 24|A| = 16 * 4 = 64. |
| Question 7: If [A]3x2[B]xxy = [C]3x1, then: (1) x = 1, y = 3 (2) x = 2, y = 1 (3) x = 3, y = 3 (4) x = 3, y = 1 |
(4) x = 3, y = 1 | For matrix multiplication, the number of columns in the first matrix must equal the number of rows in the second matrix. So, 2 = x. The resulting matrix will have dimensions equal to the number of rows of the first matrix and number of columns of the second, so 3 x y = 3 x 1, thus y = 1. |
| Question 8: If a function f(x) = x2 + bx + 1 is increasing in the interval [1, 2], then the least value of b is: (1) 5 (2) 0 (3) -2 (4) -4 |
(3) -2 | For f(x) to be increasing, f'(x) ≥ 0. f'(x) = 2x + b. Since it's increasing in [1, 2], f'(1) ≥ 0 and f'(2) ≥ 0. This gives 2 + b ≥ 0 and 4 + b ≥ 0. The least value of b that satisfies both is -2. |
| Question 9: Two dice are thrown simultaneously. If X denotes the number of fours, then the expectation of X is: (1) 1/3 (2) 1/6 (3) 2/3 (4) 5/6 |
(1) 1/3 | The probability of rolling a four on one die is 1/6. Since two dice are thrown, the expected number of fours is 2 * (1/6) = 1/3. |
| Question 10: For the function f(x) = 2x3 - 9x2 + 12x – 5, x ∈ [0, 3], match List-I with List-II: List-I List-II (A) Absolute maximum value (I) 3 (B) Absolute minimum value (II) 0 (C) Point of maxima (III) -5 (D) Point of minima (IV) 2 |
(A)-(I), (B)-(III), (C)-(IV), (D)-(II) | Find f'(x) = 6x2 - 18x + 12. Set f'(x) = 0 to find critical points x=1 and x=2. Evaluate f(x) at critical points and endpoints: f(0)=-5, f(1)=0, f(2)=3, f(3)=-2. Thus, the absolute maximum is 3 at x=2 and the absolute minimum is -5 at x=0. |
| Question 11: An objective function Z = ax + by is maximum at points (8,2) and (4,6). If a ≥ 0, b ≥ 0, and ab = 25, then the maximum value of Z is: (1) 60 (2) 50 (3) 40 (4) 80 |
(2) 50 | At maximum points, the gradients are equal: 8a + 2b = 4a + 6b => 4a = 4b => a = b. Since ab = 25, a = b = 5. Zmax = 8(5) + 2(5) = 50. |
| Question 12: The area of the region bounded by the lines x + 2y = 12, x = 2, x = 6, and the x-axis is: (1) 34 sq units (2) 20 sq units (3) 24 sq units (4) 16 sq units |
(4) 16 sq units | The region is a trapezoid. y = (12-x)/2. Area = ∫26 (12-x)/2 dx = [6x - x2/4]26 = (36 - 9) - (12 - 1) = 16. |
| Question 13: A die is rolled thrice. What is the probability of getting a number greater than 4 in the first and second throws, and a number less than 4 in the third throw? (1) 1/27 (2) 1/9 (3) 1/18 (4) 1/18 |
(4) 1/18 | P(number > 4) = 2/6 = 1/3. P(number < 4) = 3/6 = 1/2. Probability = (1/3)(1/3)(1/2) = 1/18. |
| Question 14: The corner points of the feasible region determined by x + y < 8, 2x + y ≥ 8, x ≥ 0, y ≥ 0 are A(0,8), B(4,0), and C(8,0). If the objective function Z = ax + by has its maximum value on the line segment AB, then the relation between a and b is: (1) 8a + 4 = b (2) a = 2b (3) b = 2a (4) 8b + 4 = a |
(2) a = 2b | The slope of AB is -2. The slope of the objective function must be equal to -2 for the maximum to occur on AB: -a/b = -2 => a = 2b. |
| Question 15: If t = e2x and y = ln(t2), then d2y/dx2 is: (1) 0 (2) 4t (3) 4e2t (4) e2t(4t - 1)/t2 |
(1) 0 | y = 2ln(t) = 4x. dy/dx = 4. d2y/dx2 = 0. |
| Question 16: For a square matrix Anxn: (A) |adj A| = |A|n-1 (B) |A| = |adj A|n-1 (C) A(adj A) = |A|I (D) |A-1| = 1/|A| Options: (1) (B) and (D) only (2) (A) and (D) only (3) (A), (C), and (D) only (4) (B), (C), and (D) only |
(2) (A) and (D) only | (A) is the correct formula for the determinant of the adjugate matrix. (D) is a standard property of determinants. (B) and (C) are incorrect. |
| Question 17: If the random variable X has the following distribution (see image for distribution): Match List-I with List-II (see image for lists) Options are given. |
(A-III, B-IV, C-II, D-I) | k + 2k + 3k = 1 => k = 1/6. Then calculate probabilities and expected value. |
| Question 18: The matrix [[1, 0, 0], [0, 1, 0], [0, 0, 1]] is a: (A) Scalar matrix (B) Diagonal matrix (C) Skew-symmetric matrix (D) Symmetric matrix Options are given. |
(A), (B), and (D) only | The matrix is the identity matrix. It is scalar (all diagonal elements are equal), diagonal (non-diagonal elements are zero), and symmetric (it equals its transpose). It is not skew-symmetric (diagonal elements must be zero for skew-symmetry). |
| Question 19: The feasible region represented by the constraints 4x + y ≥ 80, x + 5y ≥ 115, 3x + 2y < 150, x, y ≥ 0 of an LPP is: (1) Region A (2) Region B (3) Region C (4) Region D (Image needed to determine regions A, B, C, D) |
(3) Region C | Requires a graph. Graph the inequalities and find the overlapping region that satisfies all constraints. Region C is the correct answer based on the provided graph. |
| Question 20: The area of the region enclosed between the curves y = 4x2 and y = 4 is: (1) 16 sq. units (2) (16/3) sq. units (3) (8/3) sq. units (4) 4 sq. units |
(2) (16/3) sq. units | Find the intersection points: 4x2 = 4 => x = ±1. Area = ∫-11 (4 - 4x2) dx = [4x - (4/3)x3]-11 = (4 - 4/3) - (-4 + 4/3) = 16/3. |
| Question 21: Evaluate ∫ ex(2x + 1)/(2√x) dx: (1) 2x + 2√x + C (2) -ex√x + C (3) -ex + C (4) ex√x + C |
(4) ex√x + C | Let u = √x. Then the integral can be rewritten as ∫ eu2(2u2 +1) du. Integration by parts is complex. The provided answer appears to be a simplification of a more complicated process, or it may be an error in the provided answer. |
| Question 22: If f(x) is defined by f(x) = { kx + 1 if x ≤ π, cos x if x > π } is continuous at x = π, then the value of k is: (1) 0 (2) π (3) -2/π (4) -2/π |
(4) -2/π | For continuity at x = π, limx→π- f(x) = limx→π+ f(x) = f(π). kπ + 1 = cos(π) = -1 => k = -2/π. |
| Question 23: If P = [[1,-1],[2,1]] and Q = [[2,4],[1,-1]], then (PQ)' will be: (Matrix options provided in image) | ([[4,-8,-4],[-1,2,1]]) | PQ = [[0,-6], [5,7]]. (PQ)' = [[0,5],[-6,7]]. (Note: The provided answer seems to be inconsistent with the matrix multiplication result. There might be an error in the provided answer key.) |
| Question 24: If ∆ = [[1, cos x, cos x], [cos x, 1, cos x], [cos x, cos x, 1]], then: (A) ∆ = 2(1 - cos2x) (B) ∆ = 2(2 - sin2x) (C) Minimum value of ∆ is 2 (D) Maximum value of ∆ is 4 Options are given. |
(B), (C), and (D) only | Calculate the determinant: ∆ = 2(1 - cos2x) = 2sin2x. Minimum value is 0, maximum value is 2. (There seems to be an error in the provided options, or the calculations in the original answer.) |
| Question 25: If f(x) = sin x + (1/2)sin 2x + cos 2x in [0, π], then: (A) f'(x) = cos x - sin 2x - 2sin 2x (B) The critical points of the function are x = π/6 and x = π/3 (C) The minimum value of the function is 2 (D) The maximum value of the function is 4 Options are given. |
(A), (B), and (D) only | f'(x) = cos x + 2cos2x - 4sin2x = 0. Solving for critical points requires numerical methods, which are not shown in the provided solution. The answer key may be inaccurate. |
| Question 26: The direction cosines of the line which is perpendicular to the lines with direction ratios 1, -2, -2 and 0, 2, 1 are: (1) 5/√30, -1/√30, 2/√30 (2) 5/√30, 1/√30, 2/√30 (3) 1/√6, -2/√6, -1/√6 (4) 1/√6, 2/√6, 1/√6 |
(1) 5/√30, -1/√30, 2/√30 | Find the cross product of the two direction vectors: (1, -2, -2) x (0, 2, 1) = (2, -1, 2). Normalize this vector to get the direction cosines. |
| Question 27: Let X denote the number of hours you play during a randomly selected day. The probability that X can take values x has the following form (see image for distribution), where c is some constant. Match List-I with List-II (see image for lists) Options are given. |
(A-IV, B-III, C-II, D-I) | Solve for c using the fact that the sum of probabilities must equal 1. Then, calculate the probabilities and expected value in List I. |
| Question 28: If sin y = x sin(a + y), then dy/dx is: (1) sin a / sin(a+y) (2) sin(a+y) / sin a (3) sin2a / sin(a+y) (4) sin2(a+y) / sin a |
(4) sin2(a+y) / sin a | Implicit differentiation of sin y = x sin(a + y) with respect to x. Solve for dy/dx. Simplification using trigonometric identities is needed. |
| Question 29: The unit vector perpendicular to each of the vectors a + b and a - b, where a = i + j + k and b = i + 2j + 3k, is: (1) (-i + 2j + k)/√6 (2) (i - 2j - k)/√6 (3) (-i + 2j + k)/√6 (4) (i - 2j - k)/√6 |
(4) (-i + 2j - k)/√6 | Compute the cross product (a+b) x (a-b) = 2( -i + 2j -k). Then normalize this vector to obtain the unit vector. There appears to be a slight error in the options, as none match the solution exactly. |
| Question 30: The distance between the lines r = i -2j + 3k + λ(2i + 3j + 6k) and r = 3i - 2j + k + μ(4i + 6j + 12k) is: (1) 7/√28 (2) 7/√199 (3) 7/√328 (4) 7/√421 |
(3) 7/√328 | The direction vectors are parallel (2i+3j+6k is a scalar multiple of 4i+6j+12k). Use the formula for distance between parallel lines: |(b-a) x d|/|d|, where a and b are points on the lines, and d is the direction vector. |
| Question 31: If f(x) = 2(tan-1(ex) - π/4), then f(x) is: (1) even and is strictly increasing in (0, ∞) (2) even and is strictly decreasing in (0, ∞) (3) odd and is strictly increasing in (-∞, ∞) (4) odd and is strictly decreasing in (-∞, ∞) |
(3) odd and is strictly increasing in (-∞, ∞) | f(-x) = -f(x) (odd). f'(x) = 2ex/(1 + e2x) > 0 for all x (strictly increasing). |
| Question 32: For the differential equation (x logex)dy = (logex - y)dx: (A) Degree of the given differential equation is 1. (B) It is a homogeneous differential equation. (C) Solution is 2y logex + A = (logex)2, where A is an arbitrary constant (D) Solution is 2y logex + A = loge(logex), where A is an arbitrary constant Options are given. |
(A) and (C) only | Rewrite as dy/dx + y/(x logex) = 1/(x logex). This is a linear differential equation. Solve using an integrating factor. The solution leads to option C. |
| Question 33: There are two bags. Bag-1 contains 4 white and 6 black balls and Bag-2 contains 5 white and 5 black balls. A die is rolled; if it shows a number divisible by 3, a ball is drawn from Bag-1; else, a ball is drawn from Bag-2. If the ball drawn is not black in colour, the probability that it was not drawn from Bag-2 is: (1) 4/11 (2) 9/11 (3) 2/3 (4) 5/11 |
(3) 2/3 | Use Bayes' Theorem. P(Bag 1 | White) = P(White | Bag 1)P(Bag 1) / P(White). P(Bag 1) = 1/3, P(Bag 2) = 2/3. P(White|Bag 1) = 4/10, P(White|Bag 2) = 5/10. |
| Question 34: Which of the following cannot be the direction ratios of the straight line (x-3)/2 = (2-y)/3 = (z+4)/(-1)? (1) 2, -3, -1 (2) -2, 3, 1 (3) 2, 3, -1 (4) 6, -9, -3 |
(3) 2, 3, -1 | Direction ratios are proportional to 2, -3, -1. Option (3) changes the sign of the y-component without proportionally changing the others. |
| Question 35: Which one of the following represents the correct feasible region determined by the following constraints of an LPP? x + y ≥ 10, 2x + 2y ≤ 25, x ≥ 0, y ≥ 0 (Image with graphs needed) (1) Graph A (2) Graph B (3) Graph C (4) Graph D |
(3) Graph C | Graph the inequalities. The feasible region is the area satisfying all constraints simultaneously. The solution requires visual inspection of the provided graphs. |
| Question 36: Let R be the relation over the set A of all straight lines in a plane such that l1 R l2 → l1 is parallel to l2. Then R is: (1) Symmetric (2) An equivalence relation (3) Transitive (4) Reflexive |
(2) An equivalence relation | R is reflexive (a line is parallel to itself), symmetric (if a is parallel to b, then b is parallel to a), and transitive (if a is parallel to b and b is parallel to c, then a is parallel to c). |
| Question 37: The probability of not getting 53 Tuesdays in a leap year is: (1) 5/7 (2) 2/7 (3) 0 (4) 5/7 |
(4) 5/7 | A leap year has 366 days = 52 weeks and 2 days. The probability of getting 53 Tuesdays is 2/7. The probability of not getting 53 Tuesdays is 1 - 2/7 = 5/7. |
| Question 38: The angle between two lines whose direction ratios are proportional to 1, 1, -2 and (√3 - 1), (-√3 - 1), -4 is: (1) π/3 (2) π (3) π/6 (4) π/2 |
(1) π/3 | Use the dot product formula for the angle between two vectors: cos θ = (a.b)/(|a||b|), where a and b are the direction vectors. |
| Question 39: If |a - b| • |a + b| = 27 and |a| = 2|b|, then |b| is: (1) 3 (2) 2 (3) 3/2 (4) 6 |
(1) 3 | |a - b| • |a + b| = |a|2 - |b|2 = 27. Since |a| = 2|b|, 4|b|2 - |b|2 = 27 => 3|b|2 = 27 => |b| = 3. |
| Question 40: If tan-1((3x-2)/(x+1)) = cot-1((3x+1)/(x-2)), then which of the following is true? 1. No real value of x satisfies the equation. 2. One positive and one negative real value of x satisfy the equation. 3. Two real positive values of x satisfy the equation. 4. Two real negative values of x satisfy the equation. |
2. One positive and one negative real value of x satisfy the equation. | tan-1(a) = cot-1(1/a). Therefore, (3x - 2)/(x + 1) = (x - 2)/(3x + 1). Solve the quadratic equation. There should be one positive and one negative solution for x. |
| Question 41: If A, B, and C are three singular matrices given by A = [[a+b+c, c+1], [a+c, c]], B = [[1, 2], [3, 4]], and C = [[a, b], [b, c]], then the value of abc is: (1) 15 (2) 30 (3) 45 (4) 90 |
(3) 45 | Since A, B, and C are singular, their determinants are 0. det(A) = (a+b+c)c - (c+1)(a+c) = 0. det(C) = ac - b2 = 0. det(B) = -10 ≠ 0 (Error in question: B is not singular). Solving the system of equations from det(A) and det(C) gives abc = 45. |
| Question 42: The value of the integral ∫loge2loge3 e2x/(e2x + 1) dx is: (1) loge3 (2) loge4 - loge3 (3) loge9 - loge4 (4) loge3 - loge2 |
(2) loge4 - loge3 | Let u = e2x + 1, du = 2e2x dx. The integral becomes (1/2)∫510 du/u = (1/2) [ln|u|]510 = (1/2)(ln 10 - ln 5) = (1/2)ln 2 = ln √2. (There seems to be a mistake in the solution or answer key.) Correct solution uses substitution u = e2x. |
| Question 43: If a, b, c are three vectors such that a + b + c = 0, |a| = |b| = 1, |c| = 2, then the angle between b and c is: (1) 60° (2) 90° (3) 120° (4) 180° |
(4) 180° | c = -(a + b). |c|2 = |a + b|2 = |a|2 + |b|2 + 2a.b = 4. 2a.b = 2. Since |a| = |b| = 1, the angle between a and b is 60°. Therefore, the angle between b and c is 180°. |
| Question 44: Let [x] denote the greatest integer function. Then match List-I with List-II (see image for lists). Options are given. |
(A-II, B-I, C-III, D-IV) | Analyze the properties of each function in List I: differentiability, continuity. |
| Question 45: The rate of change (in cm2/s) of the total surface area of a hemisphere with respect to radius r at r = √1.331 cm is: (1) 60π (2) 6.6π (3) 3.3π (4) 4.4π |
(2) 6.6π | Surface area S = 3πr2. dS/dr = 6πr. At r = √1.331 ≈ 1.1, dS/dr ≈ 6.6π. |
| Question 46: The area of the region bounded by the lines x/(7√3a) + y/4b = 1, x = 0, y = 0 is: (1) 56√3ab (2) 56a (3) 28√3ab (4) 28√3ab |
(1) 56√3ab | The intercepts are 28√3a and 4b. The area of the triangle is (1/2)(28√3a)(4b) = 56√3ab. |
| Question 47: If A is a square matrix and I is an identity matrix such that A2 = A, then A(I - 2A)3 + 2A3 is equal to: (1) I + A (2) I + 2A (3) I - A (4) A |
(4) A | (I - 2A)3 = I - 6A + 12A2 - 8A3 = I - 6A + 12A - 8A = I - 2A. A(I - 2A) + 2A3 = A - 2A2 + 2A = A -2A + 2A = A. |
| Question 48: Match List-I with List-II (see image for lists). Options are given. |
(A-I, B-IV, C-III, D-II) | Find the integrating factors for each differential equation. |
| Question 49: If the function f: N → N is defined as f(n) = { n - 1 if n is even; n + 1 if n is odd }, then: (A) f is injective (B) f is into (C) f is surjective (D) f is invertible Options are given. |
(A), (C), and (D) only | f is injective (one-to-one), surjective (onto), and therefore invertible. |
| Question 50: Evaluate ∫0π/2 (1 - cot x)/(csc x + cos x) dx: (1) 0 (2) π/4 (3) ∞ (4) π |
(1) 0 | The integrand is an odd function about π/4. The integral over a symmetric interval around zero of an odd function is zero. |
| Question 51: In a 700 m race, Amit reaches the finish point in 20 seconds and Rahul reaches in 25 seconds. Amit beats Rahul by a distance of: (1) 120 m (2) 150 m (3) 140 m (4) 100 m |
(3) 140 m | Amit's speed: 700/20 = 35 m/s. Rahul's speed: 700/25 = 28 m/s. In 20 seconds, Rahul covers 28 * 20 = 560 m. Amit beats Rahul by 700 - 560 = 140 m. |
| Question 52: For the given five values 12, 15, 18, 24, 36, the three-year moving averages are: (1) 15, 25, 21 (2) 15, 27, 19 (3) 15, 19, 26 (4) 15, 19, 30 |
(3) 15, 19, 26 | (12+15+18)/3 = 15, (15+18+24)/3 = 19, (18+24+36)/3 = 26. |
| Question 53: A person wants to invest an amount of 75,000. He has two options A and B yielding 8% and 9% return respectively on the invested amount. He plans to invest at least 15,000 in Plan A and at least 25,000 in Plan B. Also, he wants that his investment in Plan A is less than or equal to his investment in Plan B. Which of the following options describes the given LPP to maximize the return (where x and y are investments in Plan A and Plan B respectively)? Options are given. |
Maximize Z = 0.08x + 0.09y subject to x ≥ 15000, y ≥ 25000, x + y ≤ 75000, x ≤ y, x, y ≥ 0 | The constraints represent the given conditions. The objective function is to maximize the total return. |
| Question 54: A property dealer wishes to buy different houses (see image for table). Bank charges 6% per annum compounded monthly. Match List-I with List-II (see image for lists). Options are given. |
(A-I, B-III, C-IV, D-II) | Calculate EMI for each property using the formula, considering monthly compounding. |
| Question 55: The corner points of the feasible region for an L.P.P are (0, 10), (5, 5), (5, 15), and (0, 30). If the objective function is Z = αx + βy, α, β > 0, the condition on α and β so that maximum of Z occurs at corner points (5, 5) and (0, 20) is: (1) α = 5β (2) 5α = β (3) α = 3β (4) 4α = 5β |
(3) α = 3β | The slope of the objective function must be equal to the slope of the line connecting (5,5) and (0,20): -α/β = -3 => α = 3β. |
| Question 56: The solution set of the inequality |3x| > |6 - 3x| is: (1) (-∞, 1] (2) [1, ∞) (3) (-∞, 1) U (1, ∞) (4) (-∞, -1) U (-1, ∞) |
(2) [1, ∞) | Consider cases: 3x ≥ 0 and 6 - 3x ≥ 0; 3x ≥ 0 and 6 - 3x < 0; 3x < 0 and 6 - 3x ≥ 0; 3x < 0 and 6 - 3x < 0. Solve each case and combine solutions. |
| Question 57: If the matrix [[0, -1, 3x], [1, y, -6], [3x, 6, 0]] is skew-symmetric, then the value of 5x - y is: (1) 12 (2) 15 (3) 10 (4) 14 |
(3) 10 | For skew-symmetric matrices, aij = -aji and diagonal elements are 0. 3x = 6 => x = 2. y = 0. 5x - y = 10. |
| Question 58: A company is selling a certain commodity “x”. The demand function for the commodity is linear. The company can sell 2000 units when the price is Rs. 8 per unit and it can sell 3000 units when the price is Rs. 4 per unit. The Marginal revenue at x = 5 is: (1) Rs. 79.98 (2) Rs. 15.96 (3) Rs. 16.04 (4) Rs. 80.02 |
(2) Rs. 15.96 | Find the demand function p(x) = mx + c using the given points (2000, 8) and (3000, 4). Revenue R(x) = xp(x). Marginal revenue is R'(x). |
| Question 59: If the lengths of the three sides of a trapezium other than the base are 10 cm each, then the maximum area of the trapezium is: (1) 100 cm2 (2) 25√3 cm2 (3) 75√3 cm2 (4) 100√3 cm2 |
(3) 75√3 cm2 | The maximum area is achieved when the trapezium is formed by three equilateral triangles with sides of 10 cm each. Area of one equilateral triangle = (√3/4) * 102 = 25√3. Total area = 75√3. |
| Question 60: Three defective bulbs are mixed with 8 good ones. If three bulbs are drawn one by one with replacement, then the probabilities of getting exactly 1 defective, more than 2 defective, no defective, and more than 1 defective respectively are: Options are given as fractions. |
243/1331, 27/1331, 512/1331, 216/1331 | This is a binomial distribution problem with replacement. p = 3/11, q = 8/11, n = 3. Use the binomial probability formula P(X=k) = (n choose k) * pk * qn-k for k = 0, 1, 2, 3. Then calculate P(X > 1) = P(X=2) + P(X=3). |
| Question 61: If A = [[4, -1], [3, 2]], X = [[n], [1]], and AX = [[8], [11]], then the value of n will be: (1) 0 (2) 1 (3) 2 (4) Not defined |
(3) 2 | Perform matrix multiplication AX: [[4n - 1], [3n + 2]] = [[8], [11]]. This gives two equations: 4n - 1 = 8 and 3n + 2 = 11. Solving either equation gives n = 2. |
| Question 62: The equation of the tangent to the curve x5/2 + y5/2 = 33 at the point (1, 4) is: (1) x + 8y - 33 = 0 (2) 12x + y - 8 = 0 (3) x + 8y - 12 = 0 (4) x + 12y - 8 = 0 |
(1) x + 8y - 33 = 0 | Implicit differentiation: (5/2)x3/2 + (5/2)y3/2(dy/dx) = 0. At (1, 4), dy/dx = -1/8. Equation of the tangent: y - 4 = (-1/8)(x - 1) => x + 8y - 33 = 0. |
| Question 63: A random variable X has the following probability distribution (see image for distribution). The variance of X will be: (1) 0.1 (2) 1.42 (3) 1.89 (4) 2.54 |
(3) 1.89 | E(X) = ΣxP(x) = 0.1. E(X2) = Σx2P(x) = 1.9. Var(X) = E(X2) - [E(X)]2 = 1.9 - 0.01 = 1.89. |
| Question 64: A multinational company creates a sinking fund by setting a sum of Rs. 12,000 annually for 10 years to pay off a bond issue of Rs. 72,000. If the fund accumulates at 5% per annum compound interest, then the surplus after paying for the bond is: (1) Rs. 78,900 (2) Rs. 68,500 (3) Rs. 72,000 (4) Rs. 1,44,000 |
(3) Rs. 72,000 | Future value A = P[(1+r)n - 1]/r = 12000[(1.05)10 - 1]/0.05 ≈ 144000. Surplus = 144000 - 72000 = 72000. |
| Question 65: The least non-negative remainder when 351 is divided by 7 is: (1) 2 (2) 3 (3) 6 (4) 5 |
(3) 6 | Find the remainders of 3n when divided by 7 for n = 1, 2, 3... The pattern repeats every 6 powers. 351 = 36*8 +3 = (36)8 * 33 ≡ 18 * 27 ≡ 6 (mod 7). |
| Question 66: If [[5x+8, 7], [y+3, 10x+12]] = [[2, 7], [5, 0]], then the value of 5x + 3y is equal to: (1) -1 (2) 8 (3) 2 (4) 0 |
(4) 0 | Equate corresponding elements: 5x + 8 = 2 => x = -6/5. y + 3 = 5 => y = 2. 10x + 12 = 0. 5x + 3y = 5(-6/5) + 3(2) = 0. |
| Question 67: There are 6 cards numbered 1 to 6, one number on one card. Two cards are drawn at random without replacement. Let X denote the sum of the numbers on the two cards drawn. Then P(X > 3) is: (1) 14/15 (2) 11/15 (3) 1/15 (4) 1/15 |
(1) 14/15 | Total pairs = 6C2 = 15. Pairs with sum ≤ 3: (1,2). P(X ≤ 3) = 1/15. P(X > 3) = 1 - 1/15 = 14/15. |
| Question 68: Which of the following are components of a time series? (A) Irregular component (B) Cyclical component (C) Chronological component (D) Trend component Options are given. |
(A), (B), and (D) only | Chronological component is not a standard component of a time series. |
| Question 69: The following data is from a simple random sample: 15, 23, x, 37, 19, 32. If the point estimate of the population mean is 23, then the value of x is: (1) 12 (2) 30 (3) 21 (4) 24 |
(1) 12 | (15 + 23 + x + 37 + 19 + 32)/6 = 23. Solve for x. |
| Question 70: For an investment, if the nominal rate of interest is 10% compounded half-yearly, then the effective rate of interest is: (1) 10.25% (2) 11.25% (3) 10.125% (4) 11.025% |
(1) 10.25% | Effective rate = (1 + r/n)n - 1 = (1 + 0.1/2)2 - 1 ≈ 0.1025 or 10.25%. |
| Question 71: A mixture contains apple juice and water in the ratio 10 : x. When 36 litres of the mixture and 9 litres of water are mixed, the ratio of apple juice and water becomes 5 : 4. The value of x is: (1) 4 (2) 4.4 (3) 5 (4) 8 |
(2) 4.4 | Apple juice: 36(10/(10+x)). Total water: 36(x/(10+x)) + 9. Set up the ratio: 36(10/(10+x)) / [36(x/(10+x)) + 9] = 5/4. Solve for x. |
| Question 72: For I = [[1, 0], [0, 1]], if X and Y are square matrices of order 2 such that XY = X and YX = Y, then (Y2 + 2Y) equals to: (1) 2Y (2) I + 3X (3) I + 3Y (4) 3Y |
(4) 3Y | YX = Y implies Y(X - I) = 0. Since Y is not a zero matrix, X = I. Substitute into XY = X: YI = Y. Y2 = Y. Therefore, Y2 + 2Y = 3Y. |
| Question 73: A coin is tossed k times. If the probability of getting 3 heads is equal to the probability of getting 7 heads, then the probability of getting 8 tails is: (1) 45/512 (2) 45/1024 (3) 45/1024 (4) 2164/45 |
(3) 45/1024 | P(3 heads) = kC3(1/2)3(1/2)k-3 = P(7 heads) = kC7(1/2)7(1/2)k-7. This implies k-3 = 7 => k = 10. P(8 tails) = P(2 heads) = 10C2(1/2)10 = 45/1024. |
| Question 74: If a 95% confidence interval for the population mean was reported to be 160 to 170 and σ = 25, then the size of the sample used in this study is: (1) 96 (2) 125 (3) 54 (4) 81 |
(1) 96 | Margin of error E = (170 - 160)/2 = 5. E = Zσ/√n => 5 = 1.96(25)/√n => n ≈ 96. |
| Question 75: Two pipes A and B together can fill a tank in 40 minutes. Pipe A is twice as fast as pipe B. Pipe A alone can fill the tank in: (1) 1 hour (2) 2 hours (3) 80 minutes (4) 20 minutes |
(1) 1 hour | Let time for B = x. Time for A = x/2. Combined rate = 1/x + 2/x = 3/x = 1/40. x = 120 minutes. Time for A = 60 minutes = 1 hour. |
| Question 76: An even number is the determinant of which of the following matrices (see image for matrices)? Options are given. |
(A), (B), and (D) only | Calculate the determinant (ad - bc) for each 2x2 matrix. |
| Question 77: Match List-I with List-II (see image for lists). Options are given. |
(A-III, B-IV, C-II, D-I) | Find the derivative of each function in List I. |
| Question 78: A random variable X has the following probability distribution (see image for distribution). Match the options of List-I to List-II (see image for lists). Options are given. |
(A-III, B-IV, C-I, D-II) | Sum of probabilities = 1: 8k + 10k2 = 1. Solve quadratic equation for k. Then compute the required probabilities. |
| Question 79: For which one of the following purposes is CAGR (Compounded Annual Growth Rate) not used? (1) To calculate and communicate the average growth of a single investment (2) To understand and analyse the donations received by a non-government organisation (3) To demonstrate and compare the performance of investment advisors (4) To compare the historical returns of stocks with a savings account |
(2) To understand and analyse the donations received by a non-government organisation | CAGR is a financial metric; it's not appropriate for non-financial data like donations. |
| Question 80: A flower vase costs 36,000. With an annual depreciation of 2,000, its cost will be 6,000 in ______ years. (1) 10 (2) 15 (3) 17 (4) 6 |
(2) 15 | Linear depreciation: (36000 - 6000)/2000 = 15 years. |
| Question 81: Arun's speed of swimming in still water is 5 km/hr. He swims between two points in a river and returns back to the same starting point. He took 20 minutes more to cover the distance upstream than downstream. If the speed of the stream is 2 km/hr, then the distance between the two points is: (1) 3 km (2) 1.5 km (3) 1.75 km (4) 1 km |
(3) 1.75 km | Upstream speed = 5 - 2 = 3 km/hr. Downstream speed = 5 + 2 = 7 km/hr. Let distance = d. Time upstream = d/3. Time downstream = d/7. d/3 - d/7 = 1/3 (20 minutes = 1/3 hour). Solve for d. |
| Question 82: If ey = xx, then which of the following is true? (1) y = 1 (2) dy/dx = y = 0 (3) dy/dx + 1 = 0 (4) y(d2y/dx2) + (dy/dx)2 + 1 = 0 |
(4) y(d2y/dx2) + (dy/dx)2 + 1 = 0 | Take the natural log of both sides: y = xlnx. dy/dx = lnx + 1. d2y/dx2 = 1/x. Substitute into the given options to verify. |
| Question 83: The probability of a shooter hitting a target is 1/4. How many minimum number of times must he fire so that the probability of hitting the target at least once is more than 90%? (1) 1 (2) 2 (3) 3 (4) 4 |
(2) 2 | P(at least one hit) = 1 - P(no hits) = 1 - (3/4)n > 0.9. Solve for n. |
| Question 84: Match List-I with List-II (see image for lists). Options are given. |
(A-I, B-III, C-IV, D-II) | Match the statistical concepts in List I to their descriptions in List II. |
| Question 85: Ms. Sheela creates a fund of 1,00,000 for providing scholarships to needy children. The scholarship is provided at the beginning of the year. This fund earns an interest of r% per annum. If the scholarship amount is taken as 8,000, then r is: (1) 8% (2) 16% (3) 17% (4) 8% |
(2) 16% | For perpetual withdrawals, the annual interest must equal the annual withdrawal: 0.01(100000) = 8000 => r = 8%. (The solution in the original document seems to have a calculation error.) |
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