📑 Topics on this page (Jump to Topic):
📌 1. Basic Formulas
📘 Formulas:
- Average = sum of observations / number of observations
- Average of first n natural numbers = (n + 1)/2
- Average of first n even numbers = n + 1; of first n odd numbers = n
- Average of an AP = (first term + last term)/2
- New member joins: newcomer = old average + (n + 1) × rise in average
✏️ Example: Average of 5 numbers is 20; removing one makes it 18 → removed = 100 − 72 = 28.
📌 2. Special Problems
- Wrong entry: correct average = (wrong sum − wrong value + correct value)/n.
- Average speed: two speeds over equal distances → 2xy/(x + y) (not the simple average).
- Cricket: run rate = runs/overs; runs required = target − runs scored.
⚠️ Exam Traps:
- Adding k to each number raises the average by k; multiplying each by k multiplies the average by k.
📌 3. Practice MCQs
📝 Practice MCQs (prepared by SF Prep — not PYQs)
Q1. The average of 9 consecutive numbers is 55. The largest number is: Practice
Correct answer: (B)
Solution: Average of consecutive numbers = middle number = 55; numbers run from 51 to 59 → answer 59.
Solution: Average of consecutive numbers = middle number = 55; numbers run from 51 to 59 → answer 59.
Q2. The average age of 10 students is 30 years. Including the teacher, the average rises by 2 year(s). The teacher's age is: Practice
Correct answer: (C)
Solution: Teacher = old average + (n + 1) × rise = 30 + 11 × 2 = 52 years.
Solution: Teacher = old average + (n + 1) × rise = 30 + 11 × 2 = 52 years.
Q3. The mean of 25 observations was 60. Later it was found that 84 had been wrongly read as 36. The correct mean is: Practice
Correct answer: (D)
Solution: Correct sum = 1500 − 36 + 84 = 1548; mean = 1548/25 = 61.92.
Solution: Correct sum = 1500 − 36 + 84 = 1548; mean = 1548/25 = 61.92.
Q4. The average of 10 numbers is 30. When one number is removed, the average of the rest becomes 33. The removed number is: Practice
Correct answer: (A)
Solution: 10 × 30 − 9 × 33 = 300 − 297 = 3.
Solution: 10 × 30 − 9 × 33 = 300 − 297 = 3.
Q5. A man goes from A to B at 20 km/h and returns at 30 km/h. His average speed for the whole journey is: Practice
Correct answer: (B)
Solution: Average speed = 2xy/(x + y) = 2×20×30/50 = 24 km/h.
Solution: Average speed = 2xy/(x + y) = 2×20×30/50 = 24 km/h.
Q6. In a 50-over match the run rate in the first 20 overs was 3.2. To reach a target of 250, the required run rate in the remaining 30 overs is: Practice
Correct answer: (C)
Solution: Runs made = 20 × 3.2 = 64; needed = 186; rate = 186/30 = 6.2.
Solution: Runs made = 20 × 3.2 = 64; needed = 186; rate = 186/30 = 6.2.
Q7. The average of the first 25 odd numbers is: Practice
Correct answer: (D)
Solution: n = 25.
Solution: n = 25.
Q8. The average weight of A, B and C is 34 kg. The average of A and B is 36 kg and of B and C is 33 kg. B's weight is: Practice
Correct answer: (A)
Solution: B = 2×36 + 2×33 − 3×34 = 36 kg.
Solution: B = 2×36 + 2×33 − 3×34 = 36 kg.
Q9. The average weight of 15 persons increases by 3 kg when a person weighing 55 kg is replaced by a new person. The new person's weight is: Practice
Correct answer: (B)
Solution: New = old + n × increase = 55 + 15 × 3 = 100 kg.
Solution: New = old + n × increase = 55 + 15 × 3 = 100 kg.
Q10. The average of 9 consecutive numbers is 44. The largest number is: Practice
Correct answer: (C)
Solution: Average of consecutive numbers = middle number = 44; numbers run from 40 to 48 → answer 48.
Solution: Average of consecutive numbers = middle number = 44; numbers run from 40 to 48 → answer 48.
Q11. The average age of 24 students is 40 years. Including the teacher, the average rises by 1 year(s). The teacher's age is: Practice
Correct answer: (D)
Solution: Teacher = old average + (n + 1) × rise = 40 + 25 × 1 = 65 years.
Solution: Teacher = old average + (n + 1) × rise = 40 + 25 × 1 = 65 years.
Q12. The mean of 25 observations was 60. Later it was found that 84 had been wrongly read as 45. The correct mean is: Practice
Correct answer: (A)
Solution: Correct sum = 1500 − 45 + 84 = 1539; mean = 1539/25 = 61.56.
Solution: Correct sum = 1500 − 45 + 84 = 1539; mean = 1539/25 = 61.56.
Q13. The average of 10 numbers is 20. When one number is removed, the average of the rest becomes 18. The removed number is: Practice
Correct answer: (B)
Solution: 10 × 20 − 9 × 18 = 200 − 162 = 38.
Solution: 10 × 20 − 9 × 18 = 200 − 162 = 38.
Q14. The average of the first 40 odd numbers is: Practice
Correct answer: (C)
Solution: n = 40.
Solution: n = 40.
Q15. The average weight of 10 persons increases by 2 kg when a person weighing 55 kg is replaced by a new person. The new person's weight is: Practice
Correct answer: (D)
Solution: New = old + n × increase = 55 + 10 × 2 = 75 kg.
Solution: New = old + n × increase = 55 + 10 × 2 = 75 kg.
Q16. The average of 11 consecutive numbers is 36. The largest number is: Practice
Correct answer: (A)
Solution: Average of consecutive numbers = middle number = 36; numbers run from 31 to 41 → answer 41.
Solution: Average of consecutive numbers = middle number = 36; numbers run from 31 to 41 → answer 41.
Q17. The average age of 15 students is 30 years. Including the teacher, the average rises by 1 year(s). The teacher's age is: Practice
Correct answer: (B)
Solution: Teacher = old average + (n + 1) × rise = 30 + 16 × 1 = 46 years.
Solution: Teacher = old average + (n + 1) × rise = 30 + 16 × 1 = 46 years.
Q18. The mean of 20 observations was 60. Later it was found that 27 had been wrongly read as 72. The correct mean is: Practice
Correct answer: (C)
Solution: Correct sum = 1200 − 72 + 27 = 1155; mean = 1155/20 = 57.75.
Solution: Correct sum = 1200 − 72 + 27 = 1155; mean = 1155/20 = 57.75.
Q19. A man goes from A to B at 60 km/h and returns at 90 km/h. His average speed for the whole journey is: Practice
Correct answer: (D)
Solution: Average speed = 2xy/(x + y) = 2×60×90/150 = 72 km/h.
Solution: Average speed = 2xy/(x + y) = 2×60×90/150 = 72 km/h.
Q20. In a 50-over match the run rate in the first 20 overs was 4.5. To reach a target of 300, the required run rate in the remaining 30 overs is: Practice
Correct answer: (A)
Solution: Runs made = 20 × 4.5 = 90; needed = 210; rate = 210/30 = 7.
Solution: Runs made = 20 × 4.5 = 90; needed = 210; rate = 210/30 = 7.