Practice Basic Average Aptitude Questions with answers and step-by-step solutions to strengthen your quantitative aptitude skills. Average is an important topic commonly asked in SSC, Banking, Railways, CAT, UPSC, and other competitive exams. This practice set will help you understand key concepts, improve calculation speed, and boost accuracy for exams.
A shopkeeper records the daily sales (in ₹) for 7 days: 354, 281, 623, 518, 447, 702, and 876. What is the average daily sale?
A. ₹528
B. ₹543
C. ₹557
D. ₹512
₹543
Sum = 354 + 281 + 623 + 518 + 447 + 702 + 876 = 3801. Average = 3801 ÷ 7 = 543.
The weights of six boys are 54 kg, 64 kg, 75 kg, 67 kg, 45 kg and 91 kg. What is their average weight?
A. 62 kg
B. 64 kg
C. 66 kg
D. 68 kg
66 kg
Total weight = 54 + 64 + 75 + 67 + 45 + 91 = 396 kg. Average = 396 ÷ 6 = 66 kg.
There are six numbers: 30, 72, 53, 68, x and 87. Their average is 60. What is the value of x?
A. 40
B. 45
C. 50
D. 55
50
Total sum = 60 × 6 = 360. Known sum = 30 + 72 + 53 + 68 + 87 = 310. So x = 360 − 310 = 50.
A teacher asks students to find the average of all prime numbers between 30 and 50. What is the correct answer?
A. 38.8
B. 39.8
C. 40.2
D. 41.0
39.8
Primes between 30 and 50: 31, 37, 41, 43, 47. Sum = 199. Average = 199 ÷ 5 = 39.8.
What is the average of the first 40 natural numbers?
A. 19.5
B. 20.0
C. 20.5
D. 21.0
20.5
Sum of first n natural numbers = n(n+1)/2 = 40×41/2 = 820. Average = 820 ÷ 40 = 20.5.
What is the average of the first 20 multiples of 7?
A. 70.5
B. 73.5
C. 77.0
D. 78.5
73.5
First 20 multiples of 7: 7, 14, …, 140. Average = 7 × (1+2+…+20)/20 = 7 × (20×21/2)/20 = 7 × 10.5 = 73.5.
A man bought 5 shirts at ₹450 each, 4 trousers at ₹750 each and 12 pairs of shoes at ₹750 each. What is the average expenditure per article?
A. ₹642.86
B. ₹657.14
C. ₹671.43
D. ₹678.57
₹678.57
Shirts: 5×450 = ₹2250. Trousers: 4×750 = ₹3000. Shoes: 12×750 = ₹9000. Total = ₹14,250. Total articles = 21. Average = 14250 ÷ 21 = ₹678.57.
13 chairs and 5 tables were bought for ₹8,280. If the average cost of a table is ₹1,227, what is the average cost of a chair?
A. ₹145
B. ₹155
C. ₹165
D. ₹175
₹165
Cost of 5 tables = 5 × 1227 = ₹6135. Cost of 13 chairs = 8280 − 6135 = ₹2145. Average chair cost = 2145 ÷ 13 = ₹165.
The average of five consecutive numbers A, B, C, D and E is 48. What is the product of A and E?
A. 2200
B. 2250
C. 2300
D. 2350
2300
Middle number C = 48. So numbers are 46, 47, 48, 49, 50. A = 46, E = 50. Product = 46 × 50 = 2300.
The average of five numbers is 58. The average of the first two is 48.5 and the average of the last two is 53.5. What is the third number?
A. 78
B. 82
C. 86
D. 90
86
Total sum = 5×58 = 290. Sum of first two = 2×48.5 = 97. Sum of last two = 2×53.5 = 107. Third number = 290 − 97 − 107 = 86.
The average marks of 13 papers is 40. The average of the first 7 is 42 and the last 7 is 35. What are the marks of the 7th paper?
A. 15
B. 17
C. 19
D. 21
19
Total = 13×40 = 520. First 7 sum = 7×42 = 294. Last 7 sum = 7×35 = 245. Paper 7 = 294 + 245 − 520 = 19.
The monthly incomes of five persons are ₹1132, ₹1140, ₹1144, ₹1136 and ₹1148. What is their arithmetic mean?
A. ₹1136
B. ₹1138
C. ₹1140
D. ₹1142
₹1140
Total = 1132+1140+1144+1136+1148 = 5700. Mean = 5700 ÷ 5 = ₹1140.
The arithmetic mean of 15 numbers is 41.4. What is the sum of these numbers?
A. 606
B. 612
C. 618
D. 621
621
Sum = Mean × Count = 41.4 × 15 = 621.
If 25a + 25b = 115, what is the average of a and b?
A. 2.1
B. 2.3
C. 2.5
D. 2.7
2.3
25(a+b) = 115 → a+b = 4.6. Average = (a+b) ÷ 2 = 4.6 ÷ 2 = 2.3.
The average of the reciprocals of x and y is:
A. (x+y) / xy
B. (x+y) / 2xy
C. 2xy / (x+y)
D. xy / (x+y)
(x+y) / 2xy
Average of 1/x and 1/y = (1/x + 1/y) ÷ 2 = ((x+y)/xy) ÷ 2 = (x+y) / 2xy.
The average of two numbers is XY. If one number is X, what is the other number?
A. X(2Y+1)
B. 2XY + X
C. X(2Y−1)
D. 2Y − X
X(2Y−1)
Sum of two numbers = 2 × XY = 2XY. Other number = 2XY − X = X(2Y − 1).
What is the mean of the first ten even natural numbers?
A. 9
B. 10
C. 11
D. 12
11
First 10 even natural numbers: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20. Sum = 110. Mean = 110 ÷ 10 = 11.
What is the average of all numbers between 6 and 34 that are divisible by 5?
A. 18
B. 20
C. 22
D. 25
20
Numbers between 6 and 34 divisible by 5: 10, 15, 20, 25, 30. Sum = 100. Average = 100 ÷ 5 = 20.
What is the average of the first nine prime numbers?
A. 10.11
B. 11.11
C. 12.11
D. 13.11
11.11
First 9 primes: 2, 3, 5, 7, 11, 13, 17, 19, 23. Sum = 100. Average = 100 ÷ 9 ≈ 11.11.
What is the average of the first 100 positive integers?
A. 49.5
B. 50.0
C. 50.5
D. 51.0
50.5
Sum of first 100 positive integers = 100×101/2 = 5050. Average = 5050 ÷ 100 = 50.5.
If a, b, c, d, e are five consecutive odd numbers, their average is:
A. a + 4
B. (abcde)^(1/5)
C. 5(a+4)
D. a + 2
a + 4
Five consecutive odd numbers: a, a+2, a+4, a+6, a+8. Sum = 5a+20. Average = (5a+20) ÷ 5 = a+4 (the middle number).
The average of a non-zero number and its square is 5 times the number. What is the number?
A. 7
B. 8
C. 9
D. 10
9
(n + n²) ÷ 2 = 5n → n + n² = 10n → n² − 9n = 0 → n(n−9) = 0. Since n ≠ 0, n = 9.
The average of two numbers is 6.5 and the square root of their product is 6. What are the two numbers?
A. 3 and 10
B. 4 and 9
C. 5 and 8
D. 6 and 7
4 and 9
a+b = 13 (since average is 6.5). √(ab) = 6 → ab = 36. Solving x²−13x+36 = 0: (x−4)(x−9) = 0 → numbers are 4 and 9.
The average of 7 consecutive numbers is 20. What is the largest of these numbers?
A. 21
B. 22
C. 23
D. 24
23
The middle (4th) number = 20. So the numbers are 17, 18, 19, 20, 21, 22, 23. Largest = 23.
The average of five consecutive odd numbers is 95. What is the fourth number in descending order?
A. 91
B. 93
C. 95
D. 97
93
Middle number = 95. Numbers: 91, 93, 95, 97, 99. Descending order: 99, 97, 95, 93, 91. Fourth number = 93.
