Averages
CSAT · Quantitative Aptitude
Introduction
Average questions are among the fastest in CSAT if you work with totals and deviations instead of long sums. They also feed directly into data interpretation.
Key concepts
The basic idea
Average = sum ÷ count, so sum = average × count. Almost every average problem is solved by comparing two totals.
New member, removed member, replacement
- Someone joins and the average of n rises by d: newcomer = old average + (n + 1) × d.
- One person is replaced and the average of n rises by d: new person = old person + n × d.
- Someone leaves: their value = old total − new total.
Equally spaced numbers
For an arithmetic sequence the average is the middle term (or the mean of the two middle terms), i.e. (first + last)/2.
Weighted average
When groups of different sizes are combined, average = (n₁a₁ + n₂a₂)/(n₁ + n₂). The combined average always lies between the two group averages, closer to the larger group.
Average speed
For equal distances at speeds x and y, average speed = 2xy/(x + y), not (x + y)/2.
Cricket and run rates
Run rate = runs ÷ overs. Required rate = (target − runs scored) ÷ overs left.
- Sum = average × number
- Joiner = A + (n + 1)d; replacement = old + nd
- AP average = (first + last)/2
- Weighted: (n₁a₁ + n₂a₂)/(n₁ + n₂)
Solved examples
Example 1
The average of 5 consecutive odd numbers is 27. What is the largest of them?
(a) 32 (b) 33 (c) 31 (d) 29
- For an odd count of equally spaced numbers, the average is the middle number: 27.
- There are 2 numbers above it, each 2 apart.
- Largest = 27 + 2 × 2 = 31.
Answer: (c) 31
Example 2
The average age of 8 students is 41 years. When the teacher's age is included, the average rises by 2 years. What is the teacher's age?
(a) 59 years (b) 57 years (c) 43 years (d) 60 years
- Total of 8 = 8 × 41 = 328.
- New total of 9 = 9 × 43 = 387.
- Teacher = 387 − 328 = 59.
- Shortcut: A + (n + 1) × increase = 41 + 9 × 2 = 59.
Answer: (a) 59 years
Example 3
The average weight of 6 people increases by 2 kg when a person weighing 79 kg is replaced by a new person. What is the weight of the new person?
(a) 91 kg (b) 67 kg (c) 93 kg (d) 81 kg
- Total weight rises by 6 × 2 = 12 kg.
- New person = 79 + 12 = 91 kg.
Answer: (a) 91 kg
Example 4
In the first 20 overs of a 50-over match the run rate was 4.8. What run rate is needed in the remaining 30 overs to reach 260 runs?
(a) 5.2 (b) 5.47 (c) 5.97 (d) 3.28
- Runs so far = 4.8 × 20 = 96.
- Runs needed = 260 − 96 = 164.
- Required rate = 164 ÷ 30 = 5.47.
Answer: (b) 5.47
Example 5
In the first 10 overs of a 50-over match the run rate was 3.2. What run rate is needed in the remaining 40 overs to reach 282 runs?
(a) 5 (b) 6.75 (c) 5.64 (d) 6.25
- Runs so far = 3.2 × 10 = 32.
- Runs needed = 282 − 32 = 250.
- Required rate = 250 ÷ 40 = 6.25.
Answer: (d) 6.25
Example 6
In the first 10 overs of a 50-over match the run rate was 4.5. What run rate is needed in the remaining 40 overs to reach 250 runs?
(a) 5 (b) 5.62 (c) 4.1 (d) 5.12
- Runs so far = 4.5 × 10 = 45.
- Runs needed = 250 − 45 = 205.
- Required rate = 205 ÷ 40 = 5.12.
Answer: (d) 5.12
Example 7
The average weight of 12 people increases by 2.5 kg when a person weighing 54 kg is replaced by a new person. What is the weight of the new person?
(a) 84 kg (b) 24 kg (c) 56.5 kg (d) 86.5 kg
- Total weight rises by 12 × 2.5 = 30 kg.
- New person = 54 + 30 = 84 kg.
Answer: (a) 84 kg
Example 8
The average age of 9 students is 44 years. When the teacher's age is included, the average rises by 1 year. What is the teacher's age?
(a) 45 years (b) 54 years (c) 55 years (d) 53 years
- Total of 9 = 9 × 44 = 396.
- New total of 10 = 10 × 45 = 450.
- Teacher = 450 − 396 = 54.
- Shortcut: A + (n + 1) × increase = 44 + 10 × 1 = 54.
Answer: (b) 54 years
- Assume a base value and add deviations: average of 47, 52, 55, 46 = 50 + (−3 + 2 + 5 − 4)/4 = 50.
- Ask how the total changed – that is usually the answer.
- Equal distances → harmonic mean for speed.
- Averaging averages of unequal groups.
- Using n instead of n + 1 when someone joins.
- Taking (x + y)/2 for average speed.
Practice set
Attempt all 12 questions (about 90 seconds each), then check the worked solutions below.
Q1. In the first 10 overs of a 50-over match the run rate was 3.5. What run rate is needed in the remaining 40 overs to reach 282 runs?
(a) 5.64 (b) 4.94 (c) 6.17 (d) 6.67
Q2. In the first 20 overs of a 50-over match the run rate was 4.2. What run rate is needed in the remaining 30 overs to reach 250 runs?
(a) 5 (b) 5.53 (c) 6.03 (d) 3.32
Q3. In the first 20 overs of a 50-over match the run rate was 4.2. What run rate is needed in the remaining 30 overs to reach 300 runs?
(a) 4.32 (b) 6 (c) 7.7 (d) 7.2
Q4. The average weight of 6 people increases by 4 kg when a person weighing 66 kg is replaced by a new person. What is the weight of the new person?
(a) 42 kg (b) 90 kg (c) 70 kg (d) 94 kg
Q5. The average weight of 9 people increases by 3 kg when a person weighing 42 kg is replaced by a new person. What is the weight of the new person?
(a) 15 kg (b) 69 kg (c) 72 kg (d) 45 kg
Q6. The average of 7 consecutive odd numbers is 55. What is the largest of them?
(a) 61 (b) 62 (c) 59 (d) 63
Q7. The average weight of 11 people increases by 3 kg when a person weighing 78 kg is replaced by a new person. What is the weight of the new person?
(a) 114 kg (b) 81 kg (c) 111 kg (d) 45 kg
Q8. In the first 10 overs of a 50-over match the run rate was 4.5. What run rate is needed in the remaining 40 overs to reach 300 runs?
(a) 6.88 (b) 5.1 (c) 6 (d) 6.38
Q9. The average of 5 consecutive odd numbers is 47. What is the largest of them?
(a) 49 (b) 52 (c) 51 (d) 53
Q10. The average weight of 12 people increases by 4 kg when a person weighing 47 kg is replaced by a new person. What is the weight of the new person?
(a) 95 kg (b) -1 kg (c) 51 kg (d) 99 kg
Q11. The average weight of 9 people increases by 2.5 kg when a person weighing 59 kg is replaced by a new person. What is the weight of the new person?
(a) 84 kg (b) 36.5 kg (c) 61.5 kg (d) 81.5 kg
Q12. The average of 9 consecutive odd numbers is 37. What is the largest of them?
(a) 45 (b) 47 (c) 43 (d) 46
Q1 – (c) 6.17. Runs so far = 3.5 × 10 = 35. Runs needed = 282 − 35 = 247. Required rate = 247 ÷ 40 = 6.17.
Q2 – (b) 5.53. Runs so far = 4.2 × 20 = 84. Runs needed = 250 − 84 = 166. Required rate = 166 ÷ 30 = 5.53.
Q3 – (d) 7.2. Runs so far = 4.2 × 20 = 84. Runs needed = 300 − 84 = 216. Required rate = 216 ÷ 30 = 7.2.
Q4 – (b) 90 kg. Total weight rises by 6 × 4 = 24 kg. New person = 66 + 24 = 90 kg.
Q5 – (b) 69 kg. Total weight rises by 9 × 3 = 27 kg. New person = 42 + 27 = 69 kg.
Q6 – (a) 61. For an odd count of equally spaced numbers, the average is the middle number: 55. There are 3 numbers above it, each 2 apart. Largest = 55 + 3 × 2 = 61.
Q7 – (c) 111 kg. Total weight rises by 11 × 3 = 33 kg. New person = 78 + 33 = 111 kg.
Q8 – (d) 6.38. Runs so far = 4.5 × 10 = 45. Runs needed = 300 − 45 = 255. Required rate = 255 ÷ 40 = 6.38.
Q9 – (c) 51. For an odd count of equally spaced numbers, the average is the middle number: 47. There are 2 numbers above it, each 2 apart. Largest = 47 + 2 × 2 = 51.
Q10 – (a) 95 kg. Total weight rises by 12 × 4 = 48 kg. New person = 47 + 48 = 95 kg.
Q11 – (d) 81.5 kg. Total weight rises by 9 × 2.5 = 22.5 kg. New person = 59 + 22.5 = 81.5 kg.
Q12 – (a) 45. For an odd count of equally spaced numbers, the average is the middle number: 37. There are 4 numbers above it, each 2 apart. Largest = 37 + 4 × 2 = 45.
- Work with totals
- Join: A + (n+1)d; replace: old + nd
- AP average = middle term
- Weighted average lies nearer the bigger group
