Mathematics

The total runs after 10 innings is 10 × 32 = 320. To increase the average by 4, the new average should be 36 after 11 innings, so total runs should be 11 × 36 = 396. Therefore, runs needed in the next innings = 396 - 320 = 76.

First, calculate the perimeter of the rectangle: 2 × (16 + 14) = 60 cm. Since the square's perimeter equals 60 cm, each side is 60 ÷ 4 = 15 cm. The diameter of the semicircle is 15 cm, so its circumference is half the circumference of a full circle plus the diameter: (π × 15) ÷ 2 + 15 ≈ 23.56 cm, rounded to 23.57 cm (option A).

Let the weights of A, B, and C be a, b, and c respectively. Given: (a + b + c)/3 = 45, so a + b + c = 135. Also, (a + b)/2 = 40, so a + b = 80, and (b + c)/2 = 43, so b + c = 86. Adding a + b = 80 and b + c = 86 gives a + 2b + c = 166. Subtracting a + b + c = 135 from this, we get b = 31. However, this contradicts the initial calculation. Rechecking: From a + b = 80 and b + c = 86, adding gives a + 2b + c = 166. Since a + b + c = 135, subtracting gives b = 31. So the weight of B is 31 kg, which corresponds to option D. Therefore, the original correct answer D is correct.

The quadratic equation x² + 2x - 15 = 0 can be factored as (x + 5)(x - 3) = 0. Setting each factor equal to zero gives the roots x = -5 and x = 3. Therefore, the correct roots are -5 and 3, which corresponds to option A.

Solving the quadratic equation a² + 11a + 30 = 0 by factoring gives (a + 5)(a + 6) = 0, so the roots are a = -5 and a = -6. Solving the quadratic equation b² + 6b + 5 = 0 by factoring gives (b + 1)(b + 5) = 0, so the roots are b = -1 and b = -5. Comparing the roots, the largest root of a is -5 and the largest root of b is -1, so a < b. Therefore, the correct relationship is a < b, which corresponds to option A.

The discriminant of the quadratic equation 3x² – 12x + 10 = 0 is Δ = (-12)² - 4*3*10 = 144 - 120 = 24, which is positive but not a perfect square. This means the roots are real, unequal, and irrational.

Given the quadratic equation x² + 8x + 4 = 0, the sum of roots a + b = -8 and the product ab = 4. The expression a/b + b/a can be rewritten as (a² + b²)/ab. Using the identity a² + b² = (a + b)² - 2ab, we get ( (-8)² - 2*4 ) / 4 = (64 - 8) / 4 = 56 / 4 = 14.

The total marks for each batch are calculated by multiplying the number of students by their average marks: (55 × 50) = 2750, (60 × 55) = 3300, and (45 × 60) = 2700. Adding these gives the combined total marks: 2750 + 3300 + 2700 = 8750. The total number of students is 55 + 60 + 45 = 160. The average marks of all students combined is the total marks divided by the total number of students: 8750 ÷ 160 = 54.6875, which rounds to 54.68. Therefore, option B (54.68) is the correct answer, not option A (53.33).

Let the number of papers be n. The total marks initially are 63n. After adding 20 marks to the Geography paper and 2 marks to the History paper, the total marks become 63n + 22. The new average is 65, so (63n + 22)/n = 65. Solving this gives 63n + 22 = 65n, or 22 = 2n, so n = 11. Therefore, the number of papers in the examination is 11, which corresponds to option D.

The ratio of volumes of two cubes is 729 : 1331, which corresponds to the cubes of their side lengths. Taking cube roots, the ratio of their sides is 9 : 11. Since surface area scales with the square of the side length, the ratio of their total surface areas is (9^2) : (11^2) = 81 : 121.