Averages — SSC CGL Quant Notes

Concepts and rules

Average (arithmetic mean) = Sum of observations ÷ Number of observations. Reversed, Sum = Average × Number. Almost every SSC average question is really a question about sums.

The average always lies between the smallest and the largest observation. It need not be one of the observations.

For numbers in arithmetic progression (consecutive numbers, consecutive even or odd numbers), Average = (First term + Last term)/2. With an odd count it is the middle term; with an even count it is the mean of the two middle terms.

Standard results: average of the first n natural numbers = (n + 1)/2; of the first n even numbers = n + 1; of the first n odd numbers = n.

Average of the squares of the first n natural numbers = (n + 1)(2n + 1)/6; average of their cubes = n(n + 1)²/4.

If every observation is increased (or decreased) by k, the average increases (or decreases) by k. If every observation is multiplied by k, the average is multiplied by k; dividing every observation by k (k ≠ 0) divides the average by k.

Wrong entry: if a value was recorded as W instead of the correct C, the count does not change, so Correct average = Wrong average + (C − W)/n.

Ages: the average age of a fixed group rises by exactly 1 year every year, so after t years it rises by t years.

Method and shortcuts

Step 1: write every given average as a sum (Average × count). Step 2: add or subtract the sums as the story requires. Step 3: divide by the final count.

Deviation method: choose a convenient assumed average, write each value's deviation from it, then Average = Assumed value + (sum of deviations)/n. It works because the deviations from the true average add up to 0.

One value joins: if x joins n values and the average rises by d, then x = New average + n × d (n is the OLD count). The newcomer brings the new average for himself plus d for each of the n old members.

One value leaves: if x leaves n values and the average of the rest rises by d, then x = New average − n × d (again n is the old count). If the average falls, take d as negative.

Replacement: if one member of a group of n is replaced and the average rises by d, then New member − Old member = n × d. The count does not change.

Overlapping groups: if the first p and the last q of n values share exactly one middle value (p + q = n + 1), that value = p × (first average) + q × (last average) − n × (overall average), because it was counted twice.

Example 1: Direct average

Question: Find the average of 12, 18, 25, 31 and 44.

Solution: Sum = 12 + 18 + 25 + 31 + 44 = 130 and count = 5, so average = 130 ÷ 5 = 26. Check by deviations from 26: −14, −8, −1, +5, +18 add up to 0.

Answer: 26

Example 2: Odd numbers in a range

Question: Find the average of all odd numbers between 20 and 60.

Solution: The odd numbers are 21, 23, …, 59, an arithmetic progression. Average = (First + Last)/2 = (21 + 59)/2 = 80/2 = 40. You do not need the count (it is 20) at all.

Answer: 40

Example 3: A new member joins

Question: The average age of 24 students is 15 years. When the teacher's age is included, the average rises by 1 year. Find the teacher's age.

Solution: New average = 16. Shortcut: teacher = New average + n × d = 16 + 24 × 1 = 40 years. Check with sums: 25 × 16 − 24 × 15 = 400 − 360 = 40.

Answer: 40 years

Example 4: Overlapping groups

Question: The average of 11 results is 60. The average of the first 6 is 58 and that of the last 6 is 63. Find the 6th result. (a) 64 (b) 62 (c) 68 (d) 66

Solution: First 6 sum = 6 × 58 = 348; last 6 sum = 6 × 63 = 378; all 11 sum = 11 × 60 = 660. The 6th result is in both groups, so 6th = 348 + 378 − 660 = 66.

Answer: (d) 66

Example 5: Corrections and a removal

Question: The average of 40 numbers is 52. Later it was found that 63 and 37 had been copied as 36 and 73, and that a number 82 had been included by mistake. Find the correct average of the remaining 39 numbers. (a) 52 (b) 51.5 (c) 51 (d) 50.5

Solution: Recorded sum = 40 × 52 = 2080. Remove the wrong copies and add the true values: 2080 − (36 + 73) + (63 + 37) = 2080 − 109 + 100 = 2071.

Now drop 82: 2071 − 82 = 1989, and the count is 39. Correct average = 1989 ÷ 39 = 51.

Answer: (c) 51

Example 6: Batsman across two innings

Question: A batsman's average after some innings is 36. He scores 90 in the next innings and 22 in the one after that, and his average becomes 37. How many innings has he played in all now? (a) 38 (b) 40 (c) 42 (d) 36

Solution: Let the first n innings total 36n. After 2 more innings the total is 36n + 90 + 22 = 36n + 112 over n + 2 innings, so 36n + 112 = 37(n + 2) = 37n + 74.

Hence n = 112 − 74 = 38 and the innings played now = 38 + 2 = 40. Shortcut view: the 2 new scores bring 112, which must cover 37 each for themselves (74) and +1 for each old innings, so n = 38.

Answer: (b) 40

Common mistakes

Dividing by the old count after someone joins. When the teacher joins 24 students, divide the new total by 25, not 24.

In replacement questions using n − 1 instead of n. The whole group of n shares the change, so the difference is n × d.

Forgetting that the middle value is counted twice in overlapping groups. Subtract the overall sum once, as in Example 4.

Taking the plain mean of two group averages when the group sizes differ. That is valid only for equal groups; otherwise use the weighted average.

For a wrong entry, changing the count. A miscopied number changes only the sum; the count stays n.

Practice questions

Q1. The average of 7 consecutive even numbers is 36. What is the largest of them? (a) 40 (b) 42 (c) 44 (d) 38

Q2. The average weight of 8 persons increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg. What is the weight of the new person? (a) 80 kg (b) 75 kg (c) 85 kg (d) 90 kg

Q3. What is the average of the squares of the first 25 natural numbers? (a) 221 (b) 225 (c) 210 (d) 231

Answers

Q1. (b) The middle (4th) number is 36; the largest is 3 steps of 2 above it: 36 + 6 = 42.

Q2. (c) New − Old = n × d = 8 × 2.5 = 20, so new person = 65 + 20 = 85 kg.

Q3. (a) (n + 1)(2n + 1)/6 = 26 × 51 ÷ 6 = 1326 ÷ 6 = 221.

Quick revision

Sum = Average × Number; work in sums, divide at the end.

Arithmetic progression: average = (first + last)/2 = middle term.

Joining: x = new average + n × d; leaving: x = new average − n × d; replacement: new − old = n × d (n = old count).

Wrong entry: correct average = wrong average + (C − W)/n.

Overlap: middle value = p × A1 + q × A2 − n × A.

Concepts and rules

Weighted average = (w1x1 + w2x2 + …)/(w1 + w2 + …), where each value x is multiplied by its weight w (number of people, kilograms, hours). A simple average is the special case where all weights are equal.

Combined average of two groups: A = (n1A1 + n2A2)/(n1 + n2). It always lies between A1 and A2 and is closer to the average of the larger group.

Size ratio rule: n1 : n2 = (A2 − A) : (A − A1). Each group's size is proportional to the OTHER group's distance from the combined average. This is the alligation rule in disguise.

The plain mean (A1 + A2)/2 is correct only when n1 = n2. Otherwise it gives a wrong answer.

Removing a sub-group: if a group of N has average A and a part of n with average a leaves, the rest have average (NA − na)/(N − n).

Average speed = Total distance ÷ Total time. It is a weighted average of the speeds with TIME as the weight, so for equal times it is (x + y)/2.

For two equal distances at speeds x and y, average speed = 2xy/(x + y); for three equal distances at x, y, z it is 3xyz/(xy + yz + zx). These are always less than the plain mean unless the speeds are equal.

Method and shortcuts

Step 1: identify the weight of each value (how many students, how many hours, how many kilograms). Step 2: multiply each average by its weight to get sums. Step 3: total sum ÷ total weight.

Distance shortcut for two groups: write A1, A, A2 on a line; the gaps (A − A1) and (A2 − A) give the size ratio crosswise. Use it whenever the combined average is known and a count is asked.

Deviation shortcut: combined average = A1 + n2(A2 − A1)/(n1 + n2). Only the difference A2 − A1 has to be split, which keeps numbers small.

For journeys, find the time of each leg (distance ÷ speed) before averaging. If times are given, find the distances (speed × time) first.

When it fails: the size-ratio rule needs exactly two groups. With three groups, convert them into sums or combine two of them first.

Example 1: Two sections combined

Question: Section A has 30 students with average marks 60 and section B has 20 students with average marks 75. Find the average of all the students.

Solution: Total marks = 30 × 60 + 20 × 75 = 1800 + 1500 = 3300 for 50 students, so average = 3300 ÷ 50 = 66. Shortcut: 60 + 20 × 15 ÷ 50 = 60 + 6 = 66.

Answer: 66

Example 2: Average speed over equal distances

Question: A car goes 120 km from Delhi to Agra at 40 km/h and returns at 60 km/h. Find its average speed for the round trip.

Solution: Time = 120/40 + 120/60 = 3 + 2 = 5 hours for 240 km, so average speed = 240 ÷ 5 = 48 km/h. Formula check: 2 × 40 × 60/(40 + 60) = 4800/100 = 48.

Answer: 48 km/h

Example 3: Finding group sizes

Question: In a factory the average monthly salary of all employees is ₹12,000. Officers average ₹30,000 and workers average ₹10,000. If there are 15 officers, how many employees are there in all?

Solution: Officers : Workers = (12,000 − 10,000) : (30,000 − 12,000) = 2,000 : 18,000 = 1 : 9. With 15 officers, workers = 135 and total = 150. Check: (15 × 30,000 + 135 × 10,000)/150 = 18,00,000/150 = 12,000.

Answer: 150

Example 4: Boys and girls in a class

Question: The average score of a class of 40 students is 54. Boys average 60 and girls average 50. How many girls are in the class? (a) 24 (b) 16 (c) 20 (d) 18

Solution: Boys : Girls = (54 − 50) : (60 − 54) = 4 : 6 = 2 : 3. Girls = 3/5 × 40 = 24. Check: (16 × 60 + 24 × 50)/40 = (960 + 1200)/40 = 54.

Answer: (a) 24

Example 5: Three sections and a transfer

Question: Sections A, B and C have 30, 40 and 50 students with average marks 62, x and 70, and the overall average is 66. Then 10 students of C whose average is 82 move to B. What is the new average of B? (a) 67.6 (b) 66.4 (c) 68.2 (d) 65.8

Solution: Total = 120 × 66 = 7920. A's sum = 30 × 62 = 1860 and C's sum = 50 × 70 = 3500, so B's sum = 7920 − 1860 − 3500 = 2560 and x = 2560 ÷ 40 = 64.

After the move B has 50 students with sum 2560 + 10 × 82 = 2560 + 820 = 3380, so the new average = 3380 ÷ 50 = 67.6.

Answer: (a) 67.6

Example 6: Mixed journey

Question: Ravi drives the first 2 hours at 45 km/h, the next 180 km at 60 km/h and the remaining 150 km at 50 km/h. What is his average speed for the whole journey? (a) 51 km/h (b) 55 km/h (c) 52.5 km/h (d) 53.75 km/h

Solution: Leg 1: distance = 45 × 2 = 90 km, time 2 h. Leg 2: time = 180 ÷ 60 = 3 h. Leg 3: time = 150 ÷ 50 = 3 h.

Total distance = 90 + 180 + 150 = 420 km; total time = 2 + 3 + 3 = 8 h; average speed = 420 ÷ 8 = 52.5 km/h. The plain mean of 45, 60 and 50 (51.67) is wrong because the times differ.

Answer: (c) 52.5 km/h

Common mistakes

Averaging two group averages directly. (60 + 75)/2 = 67.5 in Example 1 is wrong; the larger section pulls the answer to 66.

Reversing the size ratio. The group whose average is FARTHER from the combined average is the SMALLER group.

Taking average speed as the mean of speeds. Over equal distances use 2xy/(x + y); in general always use total distance ÷ total time.

After moving members between groups, forgetting to change both the sum and the count of each group.

Using percentages of groups as if they were counts. Pass percentages of two schools combine with their numbers of students as weights.

Practice questions

Q1. The average marks of 60 students is 48. Those who passed averaged 55 and those who failed averaged 34. How many students passed? (a) 36 (b) 40 (c) 42 (d) 45

Q2. The average weight of 25 boys is 48 kg and that of 15 girls is 40 kg. What is the average weight of all 40 children? (a) 44 kg (b) 46 kg (c) 44.5 kg (d) 45 kg

Q3. A cyclist climbs a hill at 24 km/h and comes down the same road at 36 km/h. What is his average speed? (a) 28.8 km/h (b) 30 km/h (c) 29.6 km/h (d) 28 km/h

Answers

Q1. (b) Passed : Failed = (48 − 34) : (55 − 48) = 14 : 7 = 2 : 1, so passed = 2/3 × 60 = 40.

Q2. (d) (25 × 48 + 15 × 40)/40 = (1200 + 600)/40 = 1800/40 = 45 kg.

Q3. (a) Equal distances: 2 × 24 × 36/(24 + 36) = 1728/60 = 28.8 km/h.

Quick revision

Combined average = (n1A1 + n2A2)/(n1 + n2); it lies between A1 and A2, nearer the bigger group.

n1 : n2 = (A2 − A) : (A − A1); sizes cross the distances.

Average speed = total distance ÷ total time; equal distances 2xy/(x + y); equal times (x + y)/2.

The plain mean of averages works only for equal weights.