SSC CGL (Prelims) Maths: Average questions with solutions

Q1. 12 kg of rice costing ₹30 per kg is mixed with 8 kg of rice costing ₹40 per kg. The average price per kg of the mixed rice is:

1. ₹38
2. ₹37
3. ₹35
4. ₹34

Answer: ₹34

The total cost is 12 × 30 + 8 × 40 = 360 + 320 = 680. The total quantity is 12 + 8 = 20 kg, so the average price per kg is 680/20 = 34.

Q2. The average of 30 results is 20 and the average of the other 20 results is 30. What is the average of all the results?

1. 24
2. 48
3. 25
4. 50

Answer: 24

The sum of the first 30 results is 30 × 20 = 600, and the sum of the other 20 results is 20 × 30 = 600. So the total sum is 1200 for 50 results, giving an average of 1200/50 = 24.

Q3. The average marks obtained in an examination by 8 students was 51 and by 9 other students was 68. The average marks of all 17 students were:

1. 59
2. 59.5
3. 60
4. 60.5

Answer: 60

The total marks of 8 students is 8 × 51 = 408 and of 9 students is 9 × 68 = 612. Their combined total is 1020, and 1020/17 = 60.

Q4. Out of four numbers, the average of the first three is 16 and that of the last three is 15. If the last number is 20, the first number is:

1. 23
2. 25
3. 28
4. 21

Answer: 23

Let the numbers be a, b, c, d. Then a+b+c = 48 and b+c+d = 45. Since d = 20, we get b+c = 25, so a = 48 - 25 = 23.

Q5. The average of 13 results is 70. The average of the first seven is 65 and that of the last seven is 75. The seventh result is:

1. 67
2. 70
3. 68
4. 70.5

Answer: 70

Total of 13 results = 13 × 70 = 910. Sum of first seven = 7 × 65 = 455 and sum of last seven = 7 × 75 = 525. Since the 7th result is counted twice in these two sums, it equals 455 + 525 - 910 = 70.

Q6. The average of 12 numbers is 15 and the average of the first two is 14. What is the average of the remaining numbers?

1. 15
2. 15.2
3. 14
4. 14.5

Answer: 15.2

The total of 12 numbers is 12 × 15 = 180. The first two numbers total 2 × 14 = 28, so the remaining 10 numbers total 180 - 28 = 152. Their average is 152/10 = 15.2.

Q7. Out of four numbers, the average of the first three is 18 and that of the last three is 16. If the last number is 19, the first number is:

1. 19
2. 18
3. 20
4. 25

Answer: 25

Let the numbers be a, b, c, d. Then a+b+c = 54 and b+c+d = 48. Since d = 19, b+c = 29, so a = 54 - 29 = 25.

Q8. The average of three numbers is 135. The largest number is 195 and the difference between the other two is 20. The smallest number is:

1. 65
2. 105
3. 95
4. 115

Answer: 95

The sum of the three numbers is 3 × 135 = 405. Removing the largest number 195 leaves 210 for the other two numbers. If their difference is 20, the smaller one is (210 - 20)/2 = 95.

Q9. The average marks of a class of 35 children is 35. The marks of one student, who scored 35, were incorrectly entered as 65. What is the correct average of the class?

1. 34.14
2. 28.20
3. 42.21
4. 38.14

Answer: 34.14

The entered mark 65 should have been 35, so the total was overstated by 30. The incorrect total is 35 × 35 = 1225, so the correct total is 1225 − 30 = 1195. Dividing by 35 gives the correct average.

Q10. The average marks secured by 36 students were 52. But it was discovered that 64 was misread as 46. What is the correct mean of marks?

1. 52.0
2. 52.5
3. 53.0
4. 53.5

Answer: 52.5

The value 64 was misread as 46, so the total marks were understated by 18. The incorrect total is 36 × 52 = 1872, so the correct total is 1872 + 18 = 1890. The correct mean is 1890 ÷ 36 = 52.5.

Q11. A batsman scores 87 runs in the 17th innings and thus increases his average by 3. Find his average after the 17th innings.

1. 33
2. 39
3. 84
4. 90

Answer: 39

If the average before the 17th innings is x, then runs in 16 innings = 16x. After scoring 87, the new average becomes x + 3, so total runs = 17(x + 3). Equating totals gives 16x + 87 = 17x + 51, hence x = 36 and the new average is 39.

Q12. The average runs of a player is 32 after 10 innings. How many runs must he score in the next innings to increase his average by 6?

1. 6
2. 38
3. 40
4. 98

Answer: 98

Current total runs = 10 × 32 = 320. To increase the average by 6, the new average must be 38 after 11 innings, so required total = 11 × 38 = 418. Therefore, runs needed in the next innings = 418 − 320 = 98.

Q13. The batting average for 30 innings of a cricket player is 40 runs. His highest score exceeds his lowest score by 100 runs. If these two innings are not included, the average of the remaining 28 innings is 38 runs. The lowest score of the player is:

1. 12
2. 15
3. 18
4. 20

Answer: 18

Total runs in 30 innings = 30 × 40 = 1200. Total of remaining 28 innings = 28 × 38 = 1064, so highest + lowest = 1200 − 1064 = 136. Also, highest − lowest = 100. Solving these two equations gives the lowest score as 18.

Q14. The bowling average of a cricketer was 12.4. He improves his bowling average by 0.2 points when he takes 5 wickets for 26 runs in his last match. The number of wickets taken by him before the last match was:

1. 125
2. 150
3. 175
4. 200

Answer: 175

Let the number of wickets before the last match be x. Then runs conceded before the match = 12.4x, and after taking 5 wickets for 26 runs, the new average becomes 12.2, so (12.4x + 26)/(x + 5) = 12.2. Solving gives x = 175.

Q15. The average weight of 25 students of a class is 50 kg. If the weight of the class teacher is included, the average increases by 1 kg. The weight of the teacher is:

1. 74 kg
2. 75 kg
3. 76 kg
4. 77 kg

Answer: 76 kg

Total weight of 25 students = 25 × 50 = 1250 kg. With the teacher included, average becomes 51 kg for 26 people, so new total = 26 × 51 = 1326 kg. Therefore, teacher’s weight = 1326 − 1250 = 76 kg.

Q16. The average of five numbers is 7. When three new numbers are included, the average of the eight numbers becomes 8.5. The average of the three new numbers is:

1. 9
2. 10.5
3. 11
4. 11.5

Answer: 11

Sum of the first five numbers = 5 × 7 = 35. Sum of all eight numbers = 8 × 8.5 = 68. So, sum of the three new numbers = 68 − 35 = 33, and their average = 33 ÷ 3 = 11.

Q17. The average age of a mother and her six children is 12 years, which is reduced by 5 years if the age of the mother is excluded. The age of the mother (in years) is:

1. 40
2. 42
3. 48
4. 50

Answer: 42

The average age of 7 people is 12, so their total age is 7 × 12 = 84. Excluding the mother, the average of 6 children becomes 7, so their total age is 6 × 7 = 42. Therefore, the mother’s age is 84 - 42 = 42 years.

Q18. The average age of a husband and his wife was 23 years at the time of their marriage. After five years, they have a one-year-old child. The average age of the family now is:

1. 19 years
2. 23 years
3. 29.3 years
4. 28.5 years

Answer: 19 years

At the time of marriage, the combined age of husband and wife was 23 × 2 = 46. After 5 years, their combined age becomes 56, and the child is 1 year old, so total family age is 57. The average age of the 3 family members is 57/3 = 19 years.

Q19. Three years ago, the average age of a family of 5 members was 17 years. A baby having been born, the average age of the family remains the same today. The age of the baby today is:

1. 1 year
2. 2 years
3. 3 years
4. 1.5 years

Answer: 2 years

Three years ago, the total age of 5 members was 5 × 17 = 85. Today, those 5 members together are 85 + 15 = 100 years old. If the baby’s age is x and the average of 6 members is still 17, then total age = 6 × 17 = 102, so x = 2 years.