CSAT· Prelims
Average for CSAT
Average for CSAT: the total trick, weighted averages, replacement shortcuts and sequence averages, with worked examples, traps and practice MCQs.
Average is the value you get by dividing the sum of a set of quantities by the number of quantities: average = total sum / number of items. It is also called the arithmetic mean. It matters for CSAT because Paper II is a qualifying paper needing a 33 percent score, and average questions are quick marks: once you read 'average', you know the total is hidden one multiplication away, and almost every question reduces to tracking that total.
The core idea: total = average x count
Every average question is a question about the total. If the average of 8 numbers is 15, their total is 8 x 15 = 120, and any new average after adding or removing numbers comes from adjusting this total. This single rearrangement, total = average x number of items, solves the great majority of CSAT average questions.
Example 1. The average of 7 numbers is 12. What is their sum? Sum = 7 x 12 = 84. If an eighth number 20 is added, the new average is (84 + 20) / 8 = 104 / 8 = 13.
Weighted average
A weighted average is the average of groups of different sizes, where each group's average counts in proportion to its size. If one group of m items averages A and another group of n items averages B, the combined average is (m x A + n x B) / (m + n), not simply (A + B)/2. The plain mean of two averages is correct only when the groups are equal in size.
Example 2. In a class, 30 boys average 70 marks and 20 girls average 80 marks. What is the class average? Total = 30 x 70 + 20 x 80 = 2,100 + 1,600 = 3,700. Divide by 50 students: 3,700 / 50 = 74. The class average is 74, pulled toward the larger group.
Replacing a person: the standard shortcut
When one member of a group is replaced and the average changes, the difference in their values equals the change in average times the group size: (new value - old value) = change in average x n. This is because the total changed by exactly the difference between the two members. The same logic covers a newcomer joining: newcomer's value = old total + change in average x new group size.
Example 3. The average age of 5 people is 26. When one person is replaced by a new person, the average becomes 28. What is the age difference between the two? The total rose by 2 x 5 = 10, so the new person is 10 years older than the one replaced.
Example 4. The average of A, B, C and D is 50. When E joins, the average of all five becomes 49. How old is E? Old total = 4 x 50 = 200. New total = 5 x 49 = 245. So E = 245 - 200 = 45 years.
Averages of number sequences
For an arithmetic sequence (numbers with a constant gap, like 3, 7, 11, 15), the average is simply (first term + last term) / 2, which is also the middle term when the count is odd. Useful instant results: the average of the first n natural numbers is (n+1)/2; of the first n even numbers is n+1; of the first n odd numbers is n.
Example 5. Find the average of all even numbers from 2 to 40. First term 2, last term 40, so the average is (2 + 40) / 2 = 21. Check: there are 20 even numbers, total = 20 x 21 = 420.
Common traps and speed tips
- Trap 1: averaging the averages. If two groups differ in size, you cannot average their averages. Weight each by its group size.
- Trap 2: forgetting the count changed. After a member joins or leaves, divide the new total by the NEW count, not the old one.
- Trap 3: average of speeds. The average speed of a round trip is not the mean of the two speeds; it is the harmonic mean 2ab/(a+b).
- Tip 1: assume a convenient total. When only the average is given, pick the total as average x count and work with it directly.
- Tip 2: track differences from the average. Numbers above and below the average must balance: if the average of five numbers is 20 and four are 18, 22, 19, 21 (differences -2, +2, -1, +1 sum to 0), the fifth is 20.
Situation | Formula |
|---|---|
Basic average | sum / n |
Find the total | average x n |
Weighted average | (mA + nB) / (m + n) |
Replacement | new - old = change x n |
First n natural numbers | (n + 1) / 2 |
First n even numbers | n + 1 |
First n odd numbers | n |
The average of 9 numbers is 27. If one number is removed, the average becomes 26. The removed number is:
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Answer: (A) Old total = 9 x 27 = 243. New total = 8 x 26 = 208. Removed number = 243 - 208 = 35.
The average weight of 40 students is 45 kg. If the teacher's weight is included, the average becomes 46 kg. The teacher's weight is:
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Answer: (A) Old total = 40 x 45 = 1,800. New total = 41 x 46 = 1,886. Teacher's weight = 1,886 - 1,800 = 86 kg.
The average of the first 50 natural numbers is:
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Answer: (B) Average of first n natural numbers = (n+1)/2 = 51/2 = 25.5.
A batsman scores 87 runs in his 17th innings and thereby increases his average by 3. His average after the 17th innings is:
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Answer: (A) Let the average after 16 innings be x. Then 16x + 87 = 17(x + 3), so 16x + 87 = 17x + 51, giving x = 36. The new average is 36 + 3 = 39.
In a class of 60 students, 35 boys have an average score of 72 and the girls have an average score of 84. The class average is:
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Answer: (A) Total = 35 x 72 + 25 x 84 = 2,520 + 2,100 = 4,620. Class average = 4,620 / 60 = 77.
The average of 5 consecutive odd numbers is 41. The largest of them is:
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Answer: (B) Five consecutive odd numbers with average 41: the middle one is 41, so the numbers are 37, 39, 41, 43, 45. The largest is 45.
Answer key
- Q1 - (a). Old total = 9 x 27 = 243. New total = 8 x 26 = 208. Removed number = 243 - 208 = 35.
- Q2 - (a). Old total = 40 x 45 = 1,800. New total = 41 x 46 = 1,886. Teacher's weight = 1,886 - 1,800 = 86 kg.
- Q3 - (b). Average of first n natural numbers = (n+1)/2 = 51/2 = 25.5.
- Q4 - (a). Let the average after 16 innings be x. Then 16x + 87 = 17(x + 3), so 16x + 87 = 17x + 51, giving x = 36. The new average is 36 + 3 = 39.
- Q5 - (a). Total = 35 x 72 + 25 x 84 = 2,520 + 2,100 = 4,620. Class average = 4,620 / 60 = 77.
- Q6 - (b). Five consecutive odd numbers with average 41: the middle one is 41, so the numbers are 37, 39, 41, 43, 45. The largest is 45.
Why can't I just average two group averages?
Because each group's average represents a different number of people. Thirty students averaging 70 contribute 2,100 marks while twenty averaging 80 contribute 1,600; the plain mean 75 ignores that the boys' total is larger. The weighted average 74 is the true class mean.
What happens to the average when every number is increased by 5?
The average also increases by exactly 5, because the total increases by 5 times the count and dividing by the count leaves 5. Similarly, multiplying every number by 3 multiplies the average by 3.
How do I handle 'average of the remaining' questions?
Compute the old total (average x old count), subtract the removed item, and divide by the new count. Example: average of 10 numbers is 30 (total 300); remove 45; the average of the remaining 9 is (300 - 45)/9 = 255/9 = 28.33.
Is the median the same as the average?
No. The average is the total divided by the count; the median is the middle value when the numbers are sorted. For the symmetric sequences CSAT asks about (like consecutive odd numbers), the two coincide at the middle term, but in general they differ.