Mathematics

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.

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).

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.

Let the marks obtained in Physics be P. The total marks in Physics, Chemistry, and Mathematics is P + 150. Therefore, the sum of marks in Chemistry and Mathematics is (P + 150) - P = 150. The average mark in Chemistry and Mathematics is 150 ÷ 2 = 75.

The total age of 15 students is 15 × 15 = 225 years. The total age of 5 students is 5 × 14 = 70 years, and the total age of 9 students is 9 × 16 = 144 years. The 15th student's age = 225 - (70 + 144) = 11 years. However, since the sum of 5 and 9 is 14, not 15, the question likely meant 5 students, 9 students, and the 15th student separately. The 15th student's age is 225 - (70 + 144) = 11 years, which corresponds to option A. But option A is 11 years, so correct answer is A, not D. Correction: The sum of 5 and 9 is 14, so the 15th student is the one left. So the age is 225 - (70 + 144) = 11 years. Therefore, correct answer is A.

The area of a trapezium is calculated using the formula: Area = 1/2 × (sum of parallel sides) × height. Here, (20 + 18) = 38 cm, and height = 15 cm, so area = 1/2 × 38 × 15 = 285 cm². However, the calculation shows 285 cm², which corresponds to option C, so the original correct answer C is correct. The explanation was missing and has been added. Difficulty is easy given the straightforward calculation.

Given the quadratic equation x² – 9x + 20 = 0, the sum of roots a + b = 9 and the product ab = 20. Using the identity a² + b² = (a + b)² - 2ab, we get a² + b² = 9² - 2×20 = 81 - 40 = 41. Therefore, a² + b² + ab = 41 + 20 = 61. Hence, the correct answer is option C.

The area of a triangle can be found using the formula Area = inradius × semiperimeter. Given the perimeter is 28 cm, the semiperimeter is 14 cm. Multiplying by the inradius 2.5 cm gives an area of 35 cm². However, since 35 cm² is not listed, the closest correct calculation is 2.5 × 14 = 35 cm², so the correct answer should be 35 cm². Since 35 cm² is not an option, the options are incorrect. The best approach is to correct the options to include 35 cm² and select it as correct. Alternatively, if options cannot be changed, then 'None of these' would be correct. But here, options do not include 35 cm², so the correct answer is 'None of these'.

First, find the radii from the circumferences: r = C/(2π). For the smaller circle, r = 264/(2π) = 42 meters; for the larger, r = 352/(2π) = 56 meters. Then, calculate areas: A = πr². Smaller area = π×42² = 1764π; larger area = π×56² = 3136π. The difference is 3136π - 1764π = 1372π ≈ 4304 sq m.

The production forms an arithmetic progression starting at 5000 tonnes in January, increasing by 100 tonnes each month. The total production for the year is the sum of 12 terms: S = 12/2 * (2*5000 + (12-1)*100) = 6 * (10000 + 1100) = 6 * 11100 = 66600 tonnes. The average monthly production is 66600 / 12 = 5550 tonnes. However, since the options do not include 5550 as the average but as an option, rechecking the calculation shows the sum is 66600, average 5550, which matches option C. Therefore, the correct answer is C.

Let the original number of boys be n. The total weight is 30n. After adding a boy weighing 35 kg, the average becomes 31 kg with n+1 boys. So, (30n + 35) / (n + 1) = 31. Solving gives n = 6.

The total runs scored in the first 20 matches is 20 × 40 = 800. In the next 10 matches, the batsman scored an average of 13 runs, so total runs = 10 × 13 = 130. Total runs in all 30 matches = 800 + 130 = 930. Therefore, the average runs in 30 matches = 930 ÷ 30 = 31. Hence, the correct answer is A (31).

The original total sum of observations is 50 × 36 = 1800. Since the observation 48 was wrongly recorded as 23, the sum was understated by 48 - 23 = 25. Adding this difference to the original sum gives 1800 + 25 = 1825. The corrected mean is then 1825 ÷ 50 = 36.5. Therefore, the correct answer is option C (36.5).

The original total sum of the 18 observations is 18 × 124 = 2232. The incorrect observations 46 and 82 sum to 128, while the correct observations 64 and 28 sum to 92. Adjusting the total sum: 2232 - 128 + 92 = 2196. The corrected average is 2196 ÷ 18 = 122. However, since 2196 ÷ 18 = 122, the correct average is 122, which corresponds to option D. Therefore, the correct answer is D, not C.

Let the breadth be x meters. Then the length is 3x meters. Area = length × breadth = 3x × x = 3x² = 867. Solving for x, x² = 289, so x = 17 meters. Therefore, the breadth is 17 meters.

The side length of the square plot is the square root of 289, which is 17 ft. The perimeter is 4 times the side, so 4 × 17 = 68 ft. The cost of building the fence is 68 ft × Rs. 58 = Rs. 3944, which corresponds to option A.

The error in marks is 83 - 63 = 20. The average marks for the class increased by half (0.5) due to this error. Since the increase in total marks divided by the number of pupils equals the increase in average, we have 20 / number_of_pupils = 0.5. Solving for the number of pupils gives number_of_pupils = 20 / 0.5 = 40. Therefore, the number of pupils in the class is 40, which corresponds to option C.

The volume of the metallic sphere is conserved when it is melted and drawn into a wire. The volume of the sphere is given by $\frac{4}{3}\pi r^3=\frac{4}{3}\pi (12{)}^3=7238.23{\text{ cm}}^3$. The wire is a cylinder with volume $\pi r^{2}h$, where the radius of the wire's cross section is 1.6 cm and $h$ is the length. Equating the volumes: $7238.23=\pi (1.6{)}^{2}h$, solving for $h$ gives $h=\frac{7238.23}{\pi \times 2.56}\approx 900\text{ cm}$ or 9 meters. Since none of the given options (45 cm, 18 cm, 90 cm) are close to 900 cm, the correct choice is "None of these" (option D).

The area of a rhombus is given by the formula: Area = (1/2) × (product of diagonals). Given the side length is 25 cm and one diagonal is 30 cm, we use the Pythagorean theorem to find the other diagonal. Half of the known diagonal is 15 cm. Using the right triangle formed by half the diagonals and the side: (other diagonal/2)^2 + 15^2 = 25^2, so (other diagonal/2)^2 = 625 - 225 = 400, thus other diagonal = 40 cm. The area is then 0.5 × 30 × 40 = 600 cm². Therefore, the correct answer is option C (600 sq.cm). The units in the options should be cm², not sq.m as originally stated.

Let the three consecutive even natural numbers be x, x+2, and x+4. Their squares sum to x^2 + (x+2)^2 + (x+4)^2 = 1460. Simplifying leads to 3x^2 + 12x + 20 = 1460, or 3x^2 + 12x - 1440 = 0. Dividing by 3 gives x^2 + 4x - 480 = 0. Solving this quadratic yields x = 20 (positive root). Thus, the numbers are 20, 22, and 24, which matches option B.

Let the sides of the rectangle be x and y. The perimeter is 2(x + y) = 28, so x + y = 14. The area of the rectangle is xy. The square formed on the diagonal has side equal to the diagonal length, which is $\sqrt{x^2+y^2}$, so its area is x^2 + y^2. The problem states this area is 108 1/3 % (which is 13/12 times) more than the area of the rectangle, so: x^2 + y^2 = xy + (13/12)xy = (25/12)xy. Using (x + y)^2 = x^2 + y^2 + 2xy, we get 14^2 = (25/12)xy + 2xy = (49/12)xy, so xy = (14^2 * 12)/49 = 48. Then, (x - y)^2 = (x + y)^2 - 4xy = 196 - 192 = 4, so x - y = 2. Hence, the difference between the sides is 2 units.

The total present age of the family is 26 × 5 = 130 years. Since the youngest member is 10 years old, 10 years ago the total age was 130 - (5 × 10) = 130 - 50 = 80 years. Therefore, the average age at that time was 80 ÷ 5 = 16 years.

Let the original price per notebook be x. Then, the number of notebooks bought is 300/x. When the price increases by Rs.5, the price per notebook is x+5, and the number of notebooks bought is 300/(x+5), which is 10 less than before. So, 300/x - 300/(x+5) = 10. Solving this equation gives x = 15. Hence, the original price per notebook is Rs.15.

The quadratic equation is 2x² + 3x - 9 = 0. Using the quadratic formula x = [-b ± sqrt(b² - 4ac)] / 2a with a=2, b=3, and c=-9, the discriminant is 81. The roots are x = ( -3 + 9 ) / 4 = 3/2 and x = ( -3 - 9 ) / 4 = -3. Therefore, the correct roots are 3/2 and -3, which corresponds to option B.

Given the average monthly incomes: (P + Q)/2 = 5050, so P + Q = 10100; (Q + R)/2 = 6250, so Q + R = 12500; (P + R)/2 = 5200, so P + R = 10400. Adding all three equations gives 2(P + Q + R) = 10100 + 12500 + 10400 = 33000, so P + Q + R = 16500. Subtracting Q + R = 12500 from this total gives P = 16500 - 12500 = 4000. Therefore, the monthly income of P is Rs. 4000, which corresponds to option B.