| Updated On - Nov 21, 2024
CUET Mathematics Question Paper 2024 (Set B) is available for download here. NTA conducted CUET 2024 Mathematics paper on 16 May in Shift 2B from 5:15 PM to 6:15 PM. CUET Mathematics Question Paper 2024 is based on objective-type questions (MCQs). Candidates get 60 minutes to solve 40 MCQs out of 50 in CUET 2024 question paper for Mathematics.
CUET Mathematics Question Paper 2024 (Set B) PDF Download
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CUET 2024 Mathematics Question Paper with Solution
| Serial no. | Questions | Solution | Detailed Solution |
|---|---|---|---|
| 1 | 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 | For symmetric matrices A and B, the matrix AB - BA is skew-symmetric. A matrix M is said to be skew-symmetric if Mᵀ = -M. In this case, (AB - BA)ᵀ = (BᵀAᵀ - AᵀBᵀ) = BA - AB = -(AB - BA). Hence, AB - BA is skew-symmetric. |
| 2 | If [A]₃×₂, [B]ₓ×ᵧ = [C]₃×₁, 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 AB = C, the number of columns of A must be equal to the number of rows of B. A is 3×2, and B must be 2×y. Since the result is 3×1, B must have dimensions 2×1. Therefore, x = 3 and y = 1. |
| 3 | 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. At x = 1, f'(1) = 2 + b, and at x = 2, f'(2) = 4 + b. For the function to be increasing, we need f'(1) ≥ 0 and f'(2) ≥ 0. This gives b ≥ -2. Hence, the least value of b is -2. |
| 4 |
Two dice are thrown simultaneously. If X denotes the number of fours, then the expectation of X will be: (1) 5/9 |
(2) 1/3 | The probability of getting a four on a die is 1/6, and the expectation of a binomial random variable is given by E(X) = np. Here, n = 2 (two dice), p = 1/6, so E(X) = 2 × (1/6) = 1/3. |
| 5 | For the function f(x) = 2x³ - 9x² + 12x - 5, x ∈ [0, 3], match List-I with List-II: (A) Absolute maximum value (B) Absolute minimum value (C) Point of maxima (D) Point of minima |
(1) (A) - (IV), (B) - (II), (C) - (I), (D) - (III) | First, find the critical points by setting f'(x) = 0: f'(x) = 6x² - 18x + 12 = 0. Solve to get x = 1 and x = 2. Next, compute the values of f(x) at these points and the boundaries (x = 0 and x = 3). The absolute maximum occurs at x = 2, and the absolute minimum occurs at x = 1. Hence, match (A) with IV, (B) with II, (C) with I, and (D) with III. |
| 6 | An objective function Z = ax + by is maximum at points (8, 2) and (4, 6). If a ≥ 0 and b ≥ 0 and ab = 25, then the maximum value of the function is equal to: (1) 60 (2) 50 (3) 40 (4) 80 |
(1) 60 | The function Z = ax + by is maximum at both (8, 2) and (4, 6), so Z = a × 8 + b × 2 = a × 4 + b × 6. Solving the system of equations gives a = 5 and b = 5. Therefore, the maximum value of Z is Z = 5 × 8 + 5 × 2 = 60. |
| 7 | 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 |
(3) 24 sq units | The area of the triangle formed by the lines can be found by integrating or using geometric formulas. The vertices are (2, 0), (6, 0), and (6, 3). The area is ½ × base × height = ½ × 4 × 6 = 24 square units. |
| 8 | 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 |
(3) 24 sq units | The area of the triangle formed by the lines can be found by integrating or using geometric formulas. The vertices are (2, 0), (6, 0), and (6, 3). The area is ½ × base × height = ½ × 4 × 6 = 24 square units. |
| 9 | A die is rolled thrice. What is the probability of getting a number greater than 4 in the first and second throw and a number less than 4 in the third throw? (1) 1/3 (2) 1/6 (3) 1/9 (4) 1/18 |
(3) 1/9 | The probability of getting a number greater than 4 (i.e., 5 or 6) is 2/6 = 1/3. The probability of getting a number less than 4 (i.e., 1, 2, or 3) is 3/6 = 1/2. So, the probability for the first two throws is (1/3) × (1/3), and for the third throw, it is 1/2. The overall probability is (1/3) × (1/3) × (1/2) = 1/9. |
| 10 | 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 maximum value of the objective function occurs on the line segment AB. The slope of the line segment AB is (8 - 0)/(0 - 4) = -2. For the objective function to have its maximum on this line, the ratio of a to b must match the slope of the line, i.e., a/b = 2, which implies a = 2b. |
| 11 | If t = e²ˣ and y = logₑ(t²), then d²y/dx² is: (1) 0 (2) 4t (3) 2t⁴e/t (4) 2t²e (4t - 1)/t |
(4) 2t²e (4t - 1)/t | First, we find dy/dx = (d/dx)[logₑ(t²)] = (2/t) × dt/dx = (2/t) × (2e²ˣ) = 4e²ˣ/t. Now, differentiating this with respect to x again gives d²y/dx² = 2t²e (4t - 1)/t. |
| 12 | ∫₀^π (n/x + π) dx = ? | n/eⁿ × (x - 1) log(Cπ) | This is an integral involving logarithms, so we need to use standard integral results involving logarithmic functions. After applying integration techniques, the result is option (1). |
| 13 | The value of ∫₀^∞ (a² - bx²) dx/(a² + bx²)² is: | (a - b)/(a + b) | To evaluate this integral, we use the standard results of definite integrals involving rational functions. After simplification, the answer is (a - b)/(a + b). |
| 14 | The second order derivative of which of the following functions is 5ˣ? (1) 5ˣ logₑ 5 (2) 5ˣ(logₑ 5)² (3) e⁵ˣ log 5 (4) e²ˣ log 5 |
(2) 5ˣ(logₑ 5)² | The second-order derivative of 5ˣ is obtained by differentiating twice. Since the derivative of 5ˣ is 5ˣ logₑ 5, and the second derivative involves another multiplication by logₑ 5, the answer is (logₑ 5)² × 5ˣ. |
| 15 | The degree of the differential equation (d²y/dx²)^(3/2) + k = 1 is: (1) 1 (2) 2 (3) 3 (4) 3/2 |
(2) 2 | The degree of a differential equation is the highest power of the highest-order derivative after making it polynomial in derivatives. Here, the highest order derivative is (d²y/dx²), and it appears raised to the power 3/2. Hence, the degree is 2 after making the equation polynomial. |
| 16 | Let R be the relation over the set A of all straight lines in a plane such that l₁ R l₂ if and only if l₁ is parallel to l₂. Then R is: (1) Symmetric (2) An Equivalence relation (3) Transitive (4) Reflexive |
(2) An Equivalence relation | A relation is an equivalence relation if it is reflexive, symmetric, and transitive. The relation "is parallel to" satisfies all these properties, hence R is an equivalence relation. |
| 17 | The probability of not getting 53 Tuesdays in a leap year is: (1) 2/7 (2) 1/7 (3) 0 (4) 5/7 |
(4) 5/7 | A leap year has 366 days, which gives 52 full weeks and 2 extra days. These extra days can be any pair from {Sunday, Monday, ..., Saturday}. The probability of not getting 53 Tuesdays is 5 favorable cases out of 7, giving a probability of 5/7. |
| 18 | The angle between two lines whose direction ratios are proportional to (1, 1, -2) and (3, -1, -4) is: (1) π/3 (2) π (3) π/6 (4) π/2 | (1) π/3 | The cosine of the angle between two vectors can be found using the dot product formula. After calculating the dot product and magnitudes, the angle comes out to be π/3. |
| 19 | The angle between two lines whose direction ratios are proportional to 1, 1, −2 and ( √ 3 − 1),(− √ 3 − 1), −4 is: | π/3 |
To find the angle θ between two lines with direction ratios ⃗d1 = (1, 1, −2) and ⃗d2 = (√3 − 1, −√3 − 1, −4), use the formula: cos θ = (⃗d1 · ⃗d2) / (|⃗d1| |⃗d2|) Step 1: Compute the dot product: Step 2: Compute the magnitudes: Step 3: Apply the formula: Thus, θ = cos⁻¹(1/2) = π/3. |
| 20 | If tan⁻¹(-x/√(2 + x²)) = cot⁻¹(x/√(3 + x²)), then which one of the following is true? (1) There is no real value of x satisfying the equation. (2) One positive and one negative real value of x (3) Two real positive values (4) Two real negative values | (1) No real value of x | After simplifying and solving the given inverse trigonometric equation, we find that there is no real solution for x. Hence, the correct option is (1). |
| 21 | If A, B, and C are three singular matrices given by A = \[1,4, 3,2a], B = \[3b,5, a,2], and C = \[a+b+c,c,1, a,c,c], then the value of abc is: (1) 15 (2) 30 (3) 45 (4) 90 | (3) 45 | A matrix is singular if its determinant is zero. We calculate the determinant of each matrix and equate them to zero. By solving the resulting equations for the values of a, b, and c, we find that the value of abc is 45. |
| 22 | The value of the integral ∫ee⋅log32xlog2+e−12xdx is: (1) loge3 (2) loge4−loge3 (3) loge9−loge4 (4) loge3−loge2 | (4) loge3−loge2 | To solve this integral, we can use substitution and properties of logarithmic integrals. The result simplifies to loge3−loge2. |
| 25 | The rate of change (in cm²/s) of the total surface area of a hemisphere with respect to the radius r at r=323 cm is: (1) 66π (2) 6.6π (3) 3.3π (4) 4.4π | (2) 6.6π | The total surface area of a hemisphere is given by A=3πr2. Differentiating with respect to r, we get dA/dr=6πr. Substituting r=323, we find the rate of change to be 6.6π. |
| 26 | The area of the region bounded by the lines 7x+3ay=4, x=0, and y=0 is: (1) 356ab (2) 56a (3) 2ab (4) 3ab | (1) 356ab | The given equation is that of a line in intercept form. The region described forms a triangle with the x-axis and y-axis as boundaries. The area of the triangle is calculated using the formula for the area of a triangle, 21×base×height, yielding 356ab. |
| 27 | 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 | (1) I+A | Using the condition A2=A, we simplify A(I−2A)3+2A3. Expanding the expression and substituting A2=A and A3=A, the result simplifies to I+A. |
| 28 | Match List-I with List-II: (A) Integrating factor of xdy−(y+2x2)dx=0 (I) x1, (B) Integrating factor of (2x2−3y)dx=xdy (II) x, (C) Integrating factor of (2y+3x2)dx+xdy=0 (III) x2, (D) Integrating factor of 2xdy+(3x3+2y)dx=0 (IV) x3 | (1) (A)-(I), (B)-(III), (C)-(IV), (D)-(II) | We solve each differential equation and find the integrating factor by converting them into an exact form. After solving, we match the corresponding integrating factors with the correct options. |
| 29 | If the function f:N→N is defined as f(n)={n+1n−1if n is odd,if n is even,, then: (A) f is injective, (B) f is into, (C) f is surjective, (D) f is invertible. Choose the correct answer from the options: (1) (B) only (2) (A), (B), and (D) only (3) (A) and (C) only (4) (A), (C), and (D) only | (4) (A), (C), and (D) only | The function is bijective (both injective and surjective), making it invertible. It maps each odd number to the next even number and each even number to the previous odd number, ensuring it is both one-to-one and onto. Thus, the function satisfies (A), (C), and (D). |
| 30 | ∫0π2+1−cotxcscx⋅cosxdx=? (1) 0 (2) 4π (3) ∞ (4) 12π | (2) 4π | This integral can be solved by trigonometric identities and substitution. After simplifying the expression, the integral evaluates to 4π. |
| 31 | If the random variable X has the following distribution:X: 0, 1, 2, OtherwiseP(X): k, 2k, 3k, 0Match List-I with List-II:(A) k(B) P(X < 2)(C) E(X)(D) P(1 ≤ X ≤ 2)(I) 6/5(II) 3/4(III) 2/1Choose correct answer. | (A) - (IV), (B) - (III), (C) - (II), (D) - (I) | Step 1: Solve for k by using the probability sum rule, k+2k+3k=1. This gives k=61.Step 2: Use the value of k to calculate P(X < 2), E(X), and P(1 ≤ X ≤ 2). Compare them with List-II options to match. |
| 33 | The matrix 001010100 is:(A) scalar matrix(B) diagonal matrix(C) skew-symmetric matrix(D) symmetric matrixChoose correct answer. | (B), (C), and (D) only | Step 1: A scalar matrix has all diagonal elements equal, which is not the case here.Step 2: Check if the matrix satisfies properties of diagonal, skew-symmetric, and symmetric matrices. |
| 34 | Feasible region of LPP with constraints 4x+y≥80, x+5y≥115, 3x+2y≤150, x,y≥0. | Region C | Step 1: Plot the inequalities on a graph.Step 2: The feasible region is the intersection of the areas. Compare with given options. |
| 35 | Area of the region enclosed between curves 4x2=y and y=4. | 16 sq. units | Step 1: Find the intersection points of the curves.Step 2: Integrate to calculate the area between the curves. |
| 36 | ∫2xex(2x+1)dx | exx+C | Step 1: Use substitution and simplification techniques for integration.Step 2: Apply the result to arrive at the final answer. |
| 37 | If f(x)={kx+1cosxif x≤πif x>π is continuous at x=π, find k. | k=−2π | Step 1: Apply continuity condition limx→π−f(x)=limx→π+f(x).Step 2: Solve for k using the equation. |
| 38 | If P=[12−1] and Q=[2−41], find (PQ)′. | −87−2−254678 | Step 1: Multiply matrices P and Q.Step 2: Take the transpose of the resulting matrix. |
| 39 | Δ=−1−cosx1−cosx1cosx1cosx1 and find the minimum and maximum values. | (A), (C), and (D) only | Step 1: Compute the determinant Δ and simplify.Step 2: Find the minimum and maximum values of Δ. |
| 40 | For f(x)=sinx+21cos2x, find critical points and extreme values. | (A), (B), and (C) only | Step 1: Find f′(x)=cosx−sin2x and solve for critical points.Step 2: Evaluate the function at the critical points for minimum and maximum values. |
| 41 | The direction cosines of the line which is perpendicular to the lines with direction ratios 1,−2,−2 and 0,2,1 are: | 32,−31,32 | Step 1: The direction ratios of the perpendicular line are the cross product of the given direction ratios.Step 2: Normalize the direction ratios to obtain direction cosines. |
| 42 | 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, where c is some constant:P(X=x)=⎩⎨⎧c(5−x),cx,0.1,0,if x=3 or 4if x=1 or 2if x=0otherwiseMatch List-I with List-II. | (A) - (III), (B) - (IV), (C) - (II), (D) - (I) | Step 1: Solve for c using the probability sum rule: P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)=1.Step 2: Calculate P(X≤2),P(X=2),P(X≥2) using the value of c. Match results with the list. |
| 43 | If siny=xsin(a+y), then dxdy is: | sin(a+y)sina | Step 1: Differentiate both sides of the given equation implicitly with respect to x.Step 2: Solve for dxdy to obtain the required result. |
| 44 | 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: | −61i^+62j^+61k^ | Step 1: Find the cross product of the vectors a+b and a−b.Step 2: Normalize the resulting vector to get the unit vector. |
| 45 | The distance between the lines r=i^−2j^+3k^+λ(2i^+3j^+6k^) and r=3i^−2j^+k^+μ(4i^+6j^+12k^) is: | 3287 | Step 1: Use the formula for the distance between two skew lines: ( d = \frac{ |
| 46 | If f(x)=2tan−1(e4π−x), then f(x) is: | Even and strictly decreasing in (0,∞) | Step 1: Check the function for evenness by evaluating f(−x)=f(x).Step 2: Differentiate to determine if it is increasing or decreasing. |
| 47 | For the differential equation (xlogx)dxdy=(logx−y)dx:(A) Degree is 1.(B) It is homogeneous.(C) Solution is 2ylogx+A=(logx)2.(D) Solution is 2ylogx+A=log(logx). Choose correct answer. | (A), (B), and (C) only | Step 1: Confirm that the degree of the differential equation is 1.Step 2: Verify that the differential equation is homogeneous.Step 3: Solve the differential equation to verify the solution form. |
| 48 | Two bags: Bag 1 contains 4 white and 6 black balls; 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 from Bag 2. If the ball drawn is not black, find the probability it was not drawn from Bag 2. | 94 | Step 1: Use conditional probability: ( P(\text{Not Bag 2} |
| 49 | Which of the following cannot be the direction ratios of the straight line x−32=3−y2=z+4−1? | 6, -9, -3 | Step 1: The direction ratios of the line are proportional to 2,−3,−1.Step 2: Check each option for proportionality to 2,−3,−1. |
| 50 | 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 | (Graph-dependent answer) | Step 1: Plot the lines x+y=10 and 2x+2y=25.Step 2: Determine the feasible region based on the inequalities and constraints x≥0, y≥0. The correct graph will show the region satisfying all constraints. |
| 51 | The least non-negative remainder when 351 is divided by 7 is: | 2 | Step 1: Perform the division: 351÷7=50 with a remainder of 2.Step 2: Hence, the least non-negative remainder is 2. |
| 52 | If [5x+810x+125xy+33y+1]=[5203y+1], then the value of 5x+3y is: | 8 | Step 1: Compare the elements of the two matrices.Step 2: Solve the equation 5x+810x+12=5 and find x=1.Step 3: Solve y+3=0 to get y=−3.Step 4: Substitute x=1 and y=−3 into 5x+3y to get 5(1)+3(−3)=8. |
| 53 | There are 6 cards numbered 1 to 6. 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: | 1515 | Step 1: Compute the total possible outcomes: (26)=15.Step 2: Identify favorable outcomes where the sum is greater than 3 (all outcomes).Step 3: Probability P(X>3) is 1515=1. |
| 54 | Which of the following are components of a time series?(A) Irregular component(B) Cyclical component(C) Chronological Component(D) Trend Component | (A), (B), and (D) only | Step 1: Time series analysis includes trend, seasonal, cyclical, and irregular components.Step 2: Chronological component is not part of time series analysis. |
| 55 | 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: | 12 | Step 1: The sample mean is given as 23. The sample has 6 numbers.Step 2: Set up the equation for the mean: 615+23+x+37+19+32=23.Step 3: Solve for x to get x=12. |
| 56 | For an investment, if the nominal rate of interest is 10% compounded half-yearly, then the effective rate of interest is: | 10.25% | Step 1: Use the formula for effective rate: (1+nr)n−1, where r=0.1 and n=2.Step 2: Calculate: (1+0.05)2−1=0.1025=10.25%. |
| 57 | A mixture contains apple juice and water in the ratio 10:x. When 36 liters of the mixture and 9 liters of water are mixed, the ratio becomes 5:4. The value of x is: | 5 | Step 1: Let the amount of water be w=x×apple juice.Step 2: After adding 9 liters, the new ratio is 5:4. Solve to get x=5. |
| 58 | For I=[0110], if X and Y are square matrices of order 2 such that XY=X and YX=Y, then Y2+2Y equals: | I+3Y | Step 1: Given XY=X and YX=Y, apply the matrix operations.Step 2: Solve for Y2+2Y and find it equal to I+3Y. |
| 59 | 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: | 102445 | Step 1: Use binomial probability: P(X=k)=(kK)pk(1−p)K−k.Step 2: Solve for K=10, find the probability of getting 8 tails. |
| 60 | 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 is:(Given Z0.025=1.96) | 96 | Step 1: Use the formula for confidence interval: xˉ±Znσ.Step 2: Solve for n using the given interval and standard deviation. |
| 61 | 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 hour | Step 1: Let tB be the time taken by Pipe B, then tA=2tB.Step 2: Set up the equation for combined flow rate and solve for tA. |
| 62 | Which matrix determinant gives an even number?(A) [1−1−15](B) [13−1−115](C) [16−11−115](D) [611−1215] | (A), (B), (C) only | Step 1: Calculate the determinants of all four matrices.Step 2: Check which determinants are even numbers. |
| 63 | Match List-I with List-II:(A) xe5log5(B) loge5(C) 5xloge5(D) 5x | (A)-(I), (B)-(III), (C)-(II), (D)-(IV) | Step 1: Differentiate each function with respect to x.Step 2: Match the derivatives with the corresponding options in List-II. |
| 64 | A random variable X has the following probability distribution:X:1,2,3,4,5,6,7P(X):k,2k,2k,3k,k,2k2,7k2+k | k=101, P(X<3)=105, P(X>2)=53, P(2| Step 1: Normalize the probabilities by setting k+2k+2k+3k+k+2k2+(7k2+k)=1. Solve for k.Step 2: Compute each probability from the distribution for matching. |
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| 65 | For which one of the following purposes is CAGR (Compounded Annual Growth Rate) not used? | (2) To understand and analyze the donations received by a non-government organization | Step 1: CAGR is typically used for investments and returns analysis.Step 2: It’s not suitable for analyzing donations, which can be irregular. |
| 66 | A flower vase costs ₹ 36,000. With an annual depreciation of ₹ 2,000, its cost will be ₹ 6,000 in ______ years. | 15 | Step 1: The depreciated value can be calculated as 36,000−2,000n=6,000.Step 2: Solve for n to find n=15. |
| 67 | 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: | 1.5 km | Step 1: Let the distance be d. Speed downstream = 5+2=7 km/hr, and upstream = 5−2=3 km/hr.Step 2: Set up the equation based on the time difference: 3d−7d=6020. Solve for d. |
| 68 | If ey=xx, then which of the following is true? | (2) dx2d2y−y=0 | Step 1: Take logarithm on both sides: y=xlogx.Step 2: Differentiate twice to get dx2d2y−y=0. |
| 69 | The probability of a shooter hitting a target is 43. How many minimum times must he fire so that the probability of hitting the target at least once is more than 90%? | 4 | Step 1: The probability of missing is 41.Step 2: Set up 1−(41)n>0.9 and solve for n. The smallest integer satisfying this is n=4. |
| 70 | Match List-I with List-II:(A) Distribution of a sample leads to becoming a normal distribution(B) Some subset of the entire population(C) Population mean(D) Some assumptions about the population | (2) (A) - (I), (B) - (III), (C) - (IV), (D) - (II) | Step 1: Understand concepts: A sample’s distribution becomes normal as sample size increases due to the Central Limit Theorem.Step 2: Match based on definitions and concepts of statistics |
CUET Questions
1.
A furniture trader deals in tables and chairs. He has Rs. 75,000 to invest and a space to store at most 60 items. A table costs him Rs. 1,500 and a chair costs him Rs. 1,000. The trader earns a profit of Rs. 400 and Rs. 250 on a table and chair, respectively. Assuming that he can sell all the items that he can buy, which of the following is/are true for the above problem:
(A) Let the trader buy \( x \) tables and \( y \) chairs. Let \( Z \) denote the total profit. Thus, the mathematical formulation of the given problem is:
\[
Z = 400x + 250y,
\]
subject to constraints:
\[
x + y \leq 60, \quad 3x + 2y \leq 150, \quad x \geq 0, \quad y \geq 0.
\]
(B) The corner points of the feasible region are (0, 0), (50, 0), (30, 30), and (0, 60).
(C) Maximum profit is Rs. 19,500 when trader purchases 60 chairs only.
(D) Maximum profit is Rs. 20,000 when trader purchases 50 tables only.
Choose the correct answer from the options given below:
A furniture trader deals in tables and chairs. He has Rs. 75,000 to invest and a space to store at most 60 items. A table costs him Rs. 1,500 and a chair costs him Rs. 1,000. The trader earns a profit of Rs. 400 and Rs. 250 on a table and chair, respectively. Assuming that he can sell all the items that he can buy, which of the following is/are true for the above problem:
(A) Let the trader buy \( x \) tables and \( y \) chairs. Let \( Z \) denote the total profit. Thus, the mathematical formulation of the given problem is:
\[ Z = 400x + 250y, \]
subject to constraints:
\[ x + y \leq 60, \quad 3x + 2y \leq 150, \quad x \geq 0, \quad y \geq 0. \]
(B) The corner points of the feasible region are (0, 0), (50, 0), (30, 30), and (0, 60).
(C) Maximum profit is Rs. 19,500 when trader purchases 60 chairs only.
(D) Maximum profit is Rs. 20,000 when trader purchases 50 tables only.
Choose the correct answer from the options given below:
- (A), (B), and (C) only
- (A), (B), and (D) only
- (B) and (C) only
- (B), (C), and (D) only
2. The correct solution of \(-22 < 8x - 6 \leq 26\) is the interval:
The correct solution of \(-22 < 8x - 6 \leq 26\) is the interval:
- \([-2, 4]\)
- \((-2, 4]\)
- \((-2, 4)\)
- \([-2, 4)\)
3. In a series of 4 trials, the probability of getting two successes is equal to the probability of getting three successes. The probability of getting at least one success is:
In a series of 4 trials, the probability of getting two successes is equal to the probability of getting three successes. The probability of getting at least one success is:
- \(\frac{609}{625}\)
- \(\frac{16}{625}\)
- \(\frac{513}{625}\)
- \(\frac{112}{625}\)
4. If \( y = e^{{2}\log_e t} \) and \( x = \log_3(e^{t^2}) \), then \( \frac{dy}{dx} \) is equal to:
If \( y = e^{{2}\log_e t} \) and \( x = \log_3(e^{t^2}) \), then \( \frac{dy}{dx} \) is equal to:
- \( \frac{1}{4t\sqrt{t}} \)
- \({2t^2} \)
- \( \frac{\log_e 3}{4t\sqrt{t}} \)
- \( \frac{2t^2}{e^{\frac{1}{2}\log_e t}} \)
5. Match List-I with List-II.List-I List-II (A) Confidence level (I) Percentage of all possible samples that can be expected to include the true population parameter (B) Significance level (III) The probability of making a wrong decision when the null hypothesis is true (C) Confidence interval (II) Range that could be expected to contain the population parameter of interest (D) Standard error (IV) The standard deviation of the sampling distribution of a statistic
Choose the correct answer from the options given below:
Match List-I with List-II.
| List-I | List-II |
|---|---|
| (A) Confidence level | (I) Percentage of all possible samples that can be expected to include the true population parameter |
| (B) Significance level | (III) The probability of making a wrong decision when the null hypothesis is true |
| (C) Confidence interval | (II) Range that could be expected to contain the population parameter of interest |
| (D) Standard error | (IV) The standard deviation of the sampling distribution of a statistic |
Choose the correct answer from the options given below:
- (A)-(I), (B)-(III), (C)-(II), (D)-(IV)
- (A)-(I), (B)-(III), (C)-(IV), (D)-(II)
- (A)-(I), (B)-(II), (C)-(IV), (D)-(III)
- (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
6. The cost of a machinery is ₹8,00,000. Its scrap value will be one-tenth of its original cost in 15 years. Using the linear method of depreciation, the book value of the machine at the end of the 10th year will be:
The cost of a machinery is ₹8,00,000. Its scrap value will be one-tenth of its original cost in 15 years. Using the linear method of depreciation, the book value of the machine at the end of the 10th year will be:
- ₹4,80,000
- ₹3,20,000
- ₹3,68,000
- ₹4,32,000
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