Averages

The total initial sum of heights is 35 × 180 = 6300 cm. The incorrect height recorded was 166 cm instead of 106 cm, so the total sum should be corrected by subtracting 166 and adding 106, resulting in 6300 - 166 + 106 = 6240 cm. The actual average height is 6240 ÷ 35 = 178.2857, which rounds to 178.29 cm. However, since the calculation shows 178.29 cm, option B matches this value exactly, but the original correct_answer was B. Rechecking the calculation: 6300 - 166 + 106 = 6240, 6240/35 = 178.2857 ≈ 178.29 cm, so the correct answer is B, not D. The initial verification incorrectly marked D. Therefore, the correct answer is B.

The total salary of 20 employees is 20 × 1500 = Rs. 30,000. When the manager's salary is added, the number of employees becomes 21 and the average salary increases by Rs. 100, making the new average 1600. Therefore, the total salary for 21 employees is 21 × 1600 = Rs. 33,600. The manager's salary is the difference between the new total and the original total, which is 33,600 - 30,000 = Rs. 3,600. Hence, the manager's monthly salary is Rs. 3,600, which corresponds to option C.

Let the weights of A, B, and C be a, b, and c respectively. From the averages: (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 yields b = 31. However, this contradicts the options. Rechecking calculations: a + b = 80, b + c = 86, a + b + c = 135. Subtracting a + b = 80 from a + b + c = 135 gives c = 55. Subtracting b + c = 86 from a + b + c = 135 gives a = 49. Using a + b = 80, b = 80 - a = 80 - 49 = 31. So b = 31 kg, which matches option D. Therefore, the original correct answer D is correct. The initial confusion was in the explanation. The correct weight of B is 31 kg.

Let the original average be x. After 10 innings, total runs = 10x. After scoring 108 in the 11th inning, the new average is x + 6. So, (10x + 108)/11 = x + 6. Solving gives x = 46, so new average = 52 runs.

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 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 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 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.

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 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.

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 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.

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.

Let the present ages of husband, wife, and child be H, W, and C respectively. Three years ago, their average age was 27, so (H-3 + W-3 + C-3)/3 = 27, which simplifies to H + W + C = 90. Five years ago, the average age of wife and child was 20, so (W-5 + C-5)/2 = 20, which simplifies to W + C = 50. Substituting W + C = 50 into H + W + C = 90 gives H = 40 years.

The total sum of the five numbers is 5 × 27 = 135. When one number is excluded, the average of the remaining four numbers is 25, so their total sum is 4 × 25 = 100. The excluded number is the difference: 135 - 100 = 35.

Let the total number of workers be x. The total salary is 8000x. The salary of 7 technicians is 7 × 12000 = 84000, and the salary of the rest (x - 7) workers is (x - 7) × 6000. So, 8000x = 84000 + 6000(x - 7). Solving gives x = 21.

The total age of all 35 students is 35 × 16 = 560 years. The total age of 21 students is 21 × 14 = 294 years. The total age of the remaining 14 students is 560 - 294 = 266 years. Therefore, the average age of the remaining 14 students is 266 ÷ 14 = 19 years.

The total weight of 16 boys is 16 × 50.25 = 804 kg, and the total weight of 8 boys is 8 × 45.15 = 361.2 kg. The combined total weight is 804 + 361.2 = 1165.2 kg. The total number of boys is 16 + 8 = 24. Therefore, the average weight is 1165.2 ÷ 24 = 48.55 kg, which matches option C. However, rechecking the calculation shows 1165.2 ÷ 24 = 48.55, so option C is correct. The initial answer was incorrect; the correct average is 48.55 kg (option C).

The total age of all seven persons is 7 × 28 = 196 years. The total age of the first three persons is 3 × 21 = 63 years, and the total age of the last three persons is 3 × 34 = 102 years. Adding these gives 63 + 102 = 165 years for six persons. Subtracting from the total, the middle person's age is 196 - 165 = 31 years. Therefore, the age of the person sitting in the middle of the row is 31 years, which corresponds to option B.

To find the average temperature, sum all daily temperatures: 6 + 4 + (-2) + (-3) + 0 + 3 + 6 = 14. Then divide by 7 days: 14 ÷ 7 = 2 degrees. Thus, the average temperature is 2 degrees.

The average of two numbers is given as XY, so the sum of the two numbers is 2 * XY. If one number is X, then the other number must be the total sum minus X, which is 2XY - X. This matches option D exactly. The notation XY is treated as a single average value, not a product of X and Y. Therefore, the correct answer is option D: 2XY - X.

The average is calculated by summing the numbers (1+2+3+4+5 = 15) and dividing by the count (5), resulting in 15/5 = 3. Therefore, option C is correct.